Solucionario Procesos de Transporte y Principios de Procesos de Separación 4ta Edicion Christie Geankopolis

SIGUENOS EN:

LIBROS UNIVERISTARIOS Y SOLUCIONARIOS DE MUCHOS DE ESTOS LIBROS GRATIS EN DESCARGA DIRECTA VISITANOS PARA DESARGALOS GRATIS.

Supplemental Material for Transport Process and Separation Process Principles

Chapter 1 Introduction to Engineering Principles and Units This chapter presents a series of problems that introduce the student to the use of units and methods for expressing different variables such as temperature and composition. Likewise, implementation of material balance and energy in fuel cells are illustrated in the following set of problem modules. 1.3-1 Determination of a Solution Density 1.4-1 Gas-Law Constant 1.4-2 Composition of a Gas Mixture 1.5-3 Combustion of Fuel Gas 1.6-1 Pre-heating of Methane and Steam 1.6-2 Heating of an Ethanol Solution 1.6-3 Calculation of Heat Transfer Rate using Steam Tables 1.6-4 Incomplete Combustion of Methane 1.6-5 Standard Enthalpy of Reaction 1.7-1 Cooling of a Fuel Cell 1.7-2 Simultaneous Material and Energy Balances 1.7-3 Oxidation of Woody Biomass

Daniel López Gaxiola Jason M. Keith

1

http://librosysolucionarios.net

Student View

Introduction to Engineering Principles and Units Example 1.3-1: Composition of a Methanol Solution A Direct-Methanol Fuel Cell is being fed with a liquid mixture of 80 mol % methanol and 20 mol % water at a temperature of 20 °C. Calculate the mass fractions and the mass of each component of this mixture in g and lbm. Strategy We will select a basis of 100 moles of mixture and use the molecular weights of methanol and water to determine the mass. Solution In the basis of 100 moles of mixture we selected, there will be 80 moles of methanol and 20 moles of water. We need to use the molecular weight of each component to determine the mass of methanol and water in the mixture, as shown in the following equations: mH

O

2

= n H OMH 2

O

2

m CH OH = n CH OH M CH OH 3

3

3

Substituting the corresponding quantities into these equations yields:

mH

O

2

mH

O

2

 18 g H 2 O  = ( 20 moles H 2 O )    1 mol H 2 O 

= __________ g H 2 O

 32 g CH 3OH  m CH OH = ( _____ moles CH 3OH )   1 mol CH OH  3  3  m CH OH = ____________ g CH 3OH 3

These values can be converted into lbm by multiplying the results by the conversion factor between lbm and g: mH

O

2

mH

O

2

  1 lb m = 360 g H 2 O    ___________ g 

= _________ lb m


2


Student View


 1 lb m  m CH OH = ____________ g CH 3OH   3  453.59 g  m CH OH = ___________ lb m 3

To determine the mass fractions of methanol and water, first we need to calculate the mass of mixture by adding the individual weights of its components: m mixture = m H O + m CH OH 2

3

m mixture = __________ lb m + __________ lb m m mixture = 6.438 lb m

Now we can calculate the mass fractions by dividing the mass of each component by the mass of the mixture: xH

O

2

=

mH

O

2

m mixture

=

________ lb m 6.438 lb m

x H O = 0.123 2

x CH OH = 3

m CH OH 3

m mixture

=

_________ lb m 6.438 lb m

x CH OH = ____________ 3


3


Student View

Introduction to Engineering Principles and Units Example 1.4-1: Gas-Law Constant Calculate the value of the gas-constant R if the pressure is in mm Hg, the moles are measured in cal g mol units, the volume is in liters and the temperature is in K. Convert this value to . mol ⋅ K

Strategy We can calculate the constant R by solving it from the ideal-gas law at standard conditions.

Solution The ideal-gas equation of state can be solved for R to yield: R=

pV nT

At standard conditions, P = 760 mm Hg, V = 22.414 L, n = 1 mol and T = 273.15 K. Substituting these values into the ideal gas equation of state, we have:

R=

( 760 mm Hg )( 22.414 L ) (1 mol )( 273.15 K )

R = _____________

To obtain the value of R in

R = 62.364

L ⋅ mm Hg mol ⋅ K

cal , we can start by converting it to SI units as shown below: mol ⋅ K

L ⋅ mm Hg  ________________ Pa   1 m3     mol ⋅ K  760 mm Hg   1000 L 

Pa ⋅ m3 R = _____________ mol ⋅ K

A Pascal is defined as: Pa =

N m2

If we multiply the Pa by m3, we get: Pa ⋅ m3 =

N 3 m = N ⋅ m = Joule(J) m2


4


Student View


Now we can use a conversion factor between Joules and calories to determine the value of R in the desired units: R = ____________

R = 1.987

 J  1 cal   mol ⋅ K  ____________ J 

cal mol ⋅ K


5


Student View

Introduction to Engineering Principles and Units Example 1.4-2: Composition of a Gas Mixture Hydrogen can be produced by an ethanol reforming process. The gas produced by the reformer is exiting at a pressure of 2144 kPa and has the following composition. Component

Molar Fraction

H2

0.392

H2 O

0.438

CO

0.081

CO2

0.080

CH4

0.009

Determine the partial pressure of each component.

Strategy The definition of partial pressure will allow us to solve this problem.

Solution To calculate the partial pressure of each gas in the stream exiting the reformer, we can use the following equation: Pi = y i P

where: yi = molar fraction of the species i present in the gas mixture. Pi = Partial pressure of the species i present in the gas mixture. P = absolute pressure of the system. Substituting the corresponding molar fraction and the absolute pressure of the system into the definition of partial pressure, we have:

PH = y H P = 0.392 ( _____________ kPa ) 2

2

PH = 840.45 kPa 2

PH

O

2

= y H O P = 0.438 ( _____________ kPa ) 2

PH O = ____________ kPa 2


6


Student View


PCO = yCO P = _________ ( _____________ kPa ) PCO = 173.66 kPa

PCO = y CO P = _________ ( _____________ kPa ) 2

2

PCO = __________ kPa 2

PCH = y CH P = __________ ( _____________ kPa ) 4

4

PCH = __________ kPa 4


7


Student View

Introduction to Engineering Principles and Units Example 1.5-3: Combustion of Fuel Gas In the steam-methane reforming process for producing hydrogen, part of the reformer exit gas is being burned with 25 % excess oxygen from air in order to supply heat for the reforming reaction to occur. The fuel being burned has the following composition: Component

Mol %

H2

41.9

H2 O

5.1

CO

1.7

CO2

41.9

CH4

9.4

The combustion reactions occurring inside the firebox are shown below. The combustion of methane is only 87 % complete. 1) CH4 + 2O2 2) CO +

3) H2 +

1 O2 2

1 O2 2

CO2 + 2H2O

CO2 H2 O

Determine the composition of the gas produced by the combustion reaction assuming an air composition of 21 mol % oxygen and 79 mol % nitrogen.

Strategy We can perform molecular material balances around the combustion chamber to determine the amounts of each species in the exhaust gases. A basis of 100 moles of fuel will be selected for simplicity.

Solution We can start by performing a methane balance around the combustion chamber:

CH4 balance Input = Output + Consumption Input = 9.4 moles CH 4


8


Student View


The amount of methane exiting in the flue gas can be determined using the definition of fractional conversion:

x CH =

n CH

4

,in

n CH

4

Solving for the n CH n CH

n CH n CH

− n CH

4

4

4

,out

,in

and substituting the corresponding quantities yields:

,out

4

,out

= n CH

4

,out

= 9.4 moles − ______ ( 9.4 moles )

,out

= ______ moles

4

4

,in

− x CH n CH 4

4

,in

Since no information is given about the fractional conversion for reactions 2) and 3), complete combustion will be assumed. Thus, n CO,out = n H

2

,out

=0

We can proceed to perform a material balance on carbon dioxide as follows:

CO2 balance Input + Generation = Output Input = __________ moles By looking at the chemical reactions, it can be seen that both reactions 1) and 2) are generating carbon dioxide. From the stoichiometric coefficients of these reactions, we can see that one mole of fuel is producing one mole of CO2. Thus, n CO

2

,generated

= n CH

4

,reacted

+ n CO,reacted

Substituting numeric values into this equation, we get:

n CO n CO

2

,generated

= ________ ( 9.4 moles ) + ________ moles

2

,generated

= 9.88 moles CO 2

We can substitute this result into the material balance equation for CO2 to yield: n CO

2

,out

= ________ moles + 9.88 moles


9


Student View

Introduction to Engineering Principles and Units n CO

2

,out

= __________ moles

In a similar way, we can perform material balances for water, considering the generation of water by reactions 1) and 3). In these reactions, it can be seen that 1 mole of fuel is producing 1 mole of water:

H2O balance Input + Generation = Output Input = 5.1 moles The amount of water generated by the chemical reactions will be given by: nH

O,generated

2

= nH

2

,reacted

+ 2n CH

4

,reacted

Substituting numeric values into this equation, we have:

nH nH

O,generated

= _________ moles + 2 ( ________ )( 9.4 moles )

O,generated

= ____________ moles

2

2

We can substitute this result into the material balance equation for H2O to get: nH nH

O,out

= 5.1 moles + ___________ moles

O,out

= ___________ moles

2

2

To determine the oxygen exiting in the product stream, we will perform a material balance for molecular oxygen:

O2 balance Input = Output + Consumption To determine the amount of oxygen that must be fed into the reactor, first we need to determine the theoretical amount of oxygen required by each reaction. This value can be calculated by multiplying the amount of fuel in the feed (carbon monoxide, methane or hydrogen) by the stoichiometric ratio of fuel to oxygen. Thus, nO

2

,r1

 2 moles O 2  = 41.9 moles CH 4   = ___________ moles O 2  mol CH 4 


10


Student View


nO

nO

2

,r 2

2

,r3

 ________ moles O 2  = _______ moles CO   = 0.85 moles O 2 mol CO    ________ moles O 2  = 41.9 moles H 2   = ___________ moles O 2 mol H  2 

Knowing the values of oxygen consumed by each reaction, we can calculate the total amount of oxygen required by this process: Consumption = n O

2

,r1

+ nO

2

,r 2

+ nO

2

,r3

Consumption = _________ moles O 2 + 0.85 moles O 2 + _________ moles O 2 Consumption = 105.6 moles O 2

Since there is 25 % excess oxygen, the number of moles of oxygen entering the reactor will be given by:

Input = ________ ( Consumption ) = ________ (105.6 moles O2 ) Input = _________ moles O 2

Solving for the amount of oxygen exiting the reactor in the balance equation for molecular oxygen and substituting the amount of oxygen fed and reacted, we get: Output = Input – Consumption nO nO

,out

= ________ moles − 105.6 moles

2

,out

= ________ moles

2

Finally, since the oxygen is entering the process as air, we need to take into account that nitrogen is entering into the firebox in this process. However, all the nitrogen will exit in the product stream since it is not being consumed nor generated by the chemical reactions. Thus,

N2 balance nN

2

,in

= nN

2

,out

The amount of nitrogen fed into the system can be determined by multiplying the molar fraction of nitrogen in the air by the amount of air fed:


11


Student View

Introduction to Engineering Principles and Units nN nN

2

,in

= nN

2

,in

= nN

2

,out

2

,out

 _______ moles N 2  =   ( ________ moles O 2 )  0.21 moles O 2  = 496.57 moles N 2

Now that we know the amount of moles of all the species, we can calculate the total number of moles exiting the combustion chamber. n out = n CH

4

,out

+ n CO

2

,out

+ nH

O,out

2

+ nN

2

,out

+ nO

2

,out

n out = __________ moles + __________ moles + 63.36 moles + 496.57 moles + __________ moles n out = __________ moles

The molar fraction of each component of the exhaust gases can be obtained by dividing the number of moles of each component by the total number of moles. Thus, y CH = 4

n CH

4

,out

n out

=

n H O,out n O ,out __________ moles 63.36 moles __________ moles = yO = 2 = yH O = 2 2 n out __________ moles n out __________ moles __________ moles 2

yH

y CH = 0.002 4

y CO = 2

n CO

2

,out

n out

=

O

= __________

2

y N = __________ 2

n __________ moles y = N2 ,out = 496.57 moles N 2 n out __________ moles __________ moles

y CO = 0.081 2


y N = __________ 2

12


Student View


Example 1.6-1: Pre-heating of Methane and Steam a) The steam used for producing hydrogen by steam-methane reforming process is leaving the boiler kmol at a temperature of 400 °C, a pressure of 1 atm, and a flow rate of 38.31 . However, before hr entering the reformer, it must be heated to a temperature of 450 °C at constant pressure. Determine the power required in kW to heat the steam to this temperature. Strategy Since there is no phase change involved in this heating process, the definition of sensible heat can be used to determine the power required.

Solution The following equation is used to determine the heat required for increasing the temperature of a substance:

= nC ˆ p,i (T2 − T1 ) Q where:

= Power required for changing the temperature of a mass of substance from T1 to T2. Q n = Molar flow rate of substance i.

Cp,i = Heat capacity of substance i. T1 = Initial temperature T2 = Final temperature The value of the heat capacity of steam at the final temperature of 450°C can be obtained using linear interpolation between 400°C and 500°C with the data found in Table 1.6-1 of Geankoplis as shown in the following steps: Tmid − Tlow Thigh − Tlow

=

Cp,H

O 2 @ Tmid

Cp,H O @ T 2

high

− C p,H O @ T 2

− Cp,H

low

O 2 @ Tlow

Inserting the numbers from Table 1.6-1, we have: 450 °C − 400 °C C p,H2O @ 450 ° C − 35.21 = 500 °C − 400 °C __________ − 35.21


13


Student View

Introduction to Engineering Principles and Units Solving for the heat capacity at the process temperature of 450 °C, we get:

Cp,H

O

= __________

2

J mol ⋅ K

Substituting all known values into the equation for the power yields: =  __________ kmol  1 hr   1000 moles    __________ J   1 kJ  (450°C − 400°C) Q        hr  3600 s   1 kmol    mol ⋅ K   1000 J   = __________ kW Q

b) Determine the heat transfer rate required to bring methane from the gas lines at room temperature to the operating temperature of the steam-methane reforming process from part a). Methane is being kmol consumed at a rate of 12.8 hr Strategy In a similar way to part a) of this problem, we can use the definition of sensible heat to calculate the heat required.

Solution We can substitute the corresponding values for the initial and final conditions of methane into the equation used to calculate sensible heat in part a). = 12.8 kmol  1 hr   1000 moles    __________ J   1 kJ  (450°C − 25°C) Q        hr  3600 s   1 kmol    mol ⋅ K   1000 J   = __________ kW Q

The heat capacity value for methane was obtained through linear interpolation using data from Table 1.6-1 of Geankoplis, at a temperature of 450°C.


14


Student View


Example 1.6-2: Heating of an Ethanol Solution A gas mixture containing 87.5 mole % water and 12.5 mole % ethanol is being heated from 210.4 °C to 350 °C before entering a pre-reformer unit to produce hydrogen for use in ProtonExchange Membrane Fuel Cells. This mixture will enter the pre-reformer at a flow rate of kmol 61.4 hr Calculate the heat required to bring the ethanol/water mixture to the operating conditions of the ethanol-reforming process. The heat capacity of ethanol vapor at the average temperature of J 280.2 °C is 98.9 . This value was calculated using the equation for heat capacity of ethanol mol ⋅ K vapor as a function of temperature, with parameters obtained from Table B.2 of Elementary Principles of Chemical Processes by Felder & Rousseau.

Strategy This problem can be solved by calculating and adding the sensible heat of each component of the mixture.

Solution The sensible heat of a mixture can be obtained from the following equation: = Q mixture

( ∑ n Cˆ ) (T − T ) i

2

p,i

1

Applying this equation to the number of components of the mixture in this problem, we get:

(

)

ˆ ˆ = n H O C + n C H OH C Q (T2 − T1 ) mixture p,H O p,C H OH 2

2

2 5

2 5

The flow rate of water and ethanol can be calculated by multiplying their corresponding molar fractions by the overall flow rate of the mixture: n H

n H

n H

O

2

O

2

O

= y H O n

n C

kmol   = 0.875  61.4  hr  

n C

kmol hr

n C

2

= __________

2

H OH 2 5

H OH

2 5

H OH

2 5

= y C H OH n 2 5

kmol   = __________  61.4  hr  

= __________

kmol hr

The specific heat of water can be obtained using linear interpolation between the temperatures of 300 °C and 400 °C with data from Table 1.6-1 of Geankoplis. Thus,


15


Student View

Introduction to Engineering Principles and Units Cp,H

O @ 350 ° C

= __________

2

J mol ⋅ K

Now we can substitute all the values we found into the equation for the sensible heat of the mixture to yield:

 kmol   1 hr  1000 moles  J  1 kJ  Q = (350°C − ________ °C)  ________    ________   mixture hr   3600 s  1 kmol  mol ⋅°C  1000 J   kmol   1 hr  1000 moles   J    1 kJ   +  _________     98.9   hr   3600 s  1 kmol   mol ⋅°C   1000 J   

= 102.0 kW Q mixture


16


Student View


Example 1.6-3: Calculation of Heat Transfer Rate using Steam Tables Liquid water at 30 °C is fed into a steam-methane reforming plant for producing syngas which can be used as fuel for solid oxide fuel cells. Before entering the reformer, the water must be boiled and heated to a temperature of 450 °C and at a pressure of 2180 kPa. Use steam tables to determine the following:

a) The amount of heat required for heating 1 mol of water from 30 °C to the boiling point at 2180 kPa. b) The amount of heat required to vaporize 1 mol of water. c) The amount of heat needed for heating 1 mol of saturated steam to a temperature of 450 °C. Strategy To determine the amount of heat required for parts a) to c) in this problem we will make use of thermodynamic properties of water and steam.

Solution To solve parts a), b) and c) of this problem, we need to look for 4 different enthalpy values of water and steam, which are described below: Ha,l = Enthalpy of liquid water at 30 °C Hb,l = Enthalpy of saturated liquid water at the boiling temperature at a pressure of 2180 kPa Hb,v = Enthalpy of saturated steam at the boiling temperature at a pressure of 2180 kPa Hc,v = Enthalpy of superheated steam at the process temperature of 450 °C After finding these 4 enthalpies in the steam tables, the enthalpies we need to calculate to solve this problem will be given by the following equations: Ha = Hb,l – Ha,l Hb = Hb,v – Hb,l Hc = Hc,v – Hb,v


17


Student View

Introduction to Engineering Principles and Units The enthalpies of liquid water can be found in Table 2-305 of Perry’s Chemical Engineers’ Handbook 8th Edition. However, we will have to use linear interpolation for determining the enthalpies at the temperature and pressure given in the problem statement. The data given in this table is shown below:

Pressure (kPa)

Temperature (K)

1000 kJ mol

300

2.0444

303.15

ˆ H @ 1000 kPa ,303.15K

400

9.6106

kJ mol

2180

5000

ˆ H @ 2180 kPa,300K

____________

ˆ H @ 2180 kPa,303.15K

ˆ H @ 5000 kPa,303.15K

ˆ H @ 2180 kPa,400K

9.6601

kJ mol

kJ mol

Where the caret (^) indicates the enthalpy is per mole. From the table shown above, we need to calculate first the values of the enthalpy at a constant temperature of 303.5 K at a pressure of 1000 kPa. We setup the linear interpolation as:

Tmid − Tlow Thigh − Tlow

=

ˆ H @T

ˆ −H @T

ˆ H @T

ˆ −H @T

mid

high

low low

Inserting the numbers from the table, we have: ˆ − 2.0444 _________ − 300 K H @ 1000 kPa ,303.15K = 400 K − 300 K _________ − 2.0444

Solving for H @ 1000 kPa, 303.15, K: kJ  _________ K − 300 K ˆ H = ( _________ − 2.0444 )  + 2.0444 = 2.2827 @ 1000 kPa ,303.15K 400 K − 300 K mol  

Repeating this procedure for the enthalpies at a pressure of 5000 kPa we have: kJ  _________ K − 300 K ˆ H = ( _________ − 2.1106 )  + 2.1106 = _________ @ 5000 kPa,303.15K 400 K − 300 K mol  


18


Student View


After substituting the calculated values for the enthalpies at T = 303.15 K in the table, we get:

Pressure (kPa)

Temperature (K)

300

2180

5000

kJ mol kJ 2.2827 mol kJ 9.6106 mol

ˆ H @ 2180 kPa,300K

kJ mol kJ _________ mol kJ 9.6601 mol

2.0444

303.15 400

1000

ˆ H @ 2180 kPa,303.15K ˆ H @ 2180 kPa,400K

_________

At this point, we can solve for the enthalpy at the operation conditions in the process described in this problem.

ˆ yields: Interpolating across at a constant temperature and solving for H @ 2180 kPa,303.15K  _________ kPa − 1000 kPa  kJ ˆ H = ( _________ − 2.2827 ) + 2.2827 @ 2180 kPa ,303.15 K mol  _________ kPa − 1000 kPa  Thus, the enthalpy of liquid water at 30°C is estimated to be:

kJ ˆ =H ˆ H ≈ __________ a,l @ 2180 kPa,303.15 K mol

Pressure (kPa)

Now we can proceed to look in Table A.2-9 of Geankoplis for the enthalpy values of saturated water and saturated steam. Again, we will have to use linear interpolation to calculate the enthalpies of saturated water and steam at a pressure of 2180 kPa using the values from the following table.


 kJ  Hl    kg 

 kJ  Hv    kg 

1553.8

____________

2793.2

2180

= 924.46 H b,l

= ____________ H b,v

2548

966.78

____________

19


Student View

Introduction to Engineering Principles and Units The unknown enthalpies were calculated by interpolation as shown in the following equations: =  2180 kPa − 1553.8 kPa ( 966.78 − ___________ )  + ___________ = 924.46 kJ H b,l  2548 kPa − 1553.8 kPa  kg =  2180 kPa − 1553.8 kPa ( __________ − 2793.2 )  + 2793.2 = __________ kJ H b,v  2548 kPa − 1553.8 kPa  kg

To calculate the remaining unknown enthalpy, we need to look for data for superheated steam as we did for calculating Ha,l. With the values obtained from Table A.2-10 of Geankoplis, the table for the process conditions will be given by:

Pressure (kPa) 2180

2000

Temperature (°C)

420

450

500

kJ kg kJ 3357.6 kg kJ 3467.6 kg

2500 3284.0

3291.6

= ______________ kJ H c,v kg

kJ kg

_________________ 3462.1

kJ kg

kJ kg

where the values in bold were calculated by interpolation as shown in the steps below: kJ  450°C − 420°C = H ( 3467.6 − 3291.6 ) + 3291.6 = 3357.6 @ 2000 kPa,450° C kg  500°C − 420°C 

 ________ °C − ________ °C  kJ H = ( 3462.1 − 3284.0 )  + 3284.0 = _____________ @ 2500 kPa,450° C kg  500°C − ________ °C   _________ kPa − 2000 kPa  kJ =H H = ( ___________ − 3357.6 )  + 3357.6 = ___________ c,v @ 2180 kPa,450° C kg  __________ − 2000 kPa 

Finally, we can apply the equations shown above for the enthalpies Ha, Hb and Hc to yield:

ˆ = H ˆ – H ˆ = 1 mol 924.46 kJ  18 kg  1 kmol  − ______________ kJ  H a     b,l a ,l kg  1 kmol  1000 mol  mol   ˆ = 14.34 kJ H a Daniel López Gaxiola Jason M. Keith

20


Student View


ˆ =H ˆ –H ˆ = 1 mol  ______________ kJ − 924.46 kJ   18 kg   1 kmol  H     b b,v b,l kg kg   1 kmol   1000 mol  

ˆ = ______________ kJ H b ˆ =H ˆ –H ˆ = 1 mol  ______________ kJ − ______________ kJ   18 kg   1 kmol  H c c,v     b,v kg kg   1 kmol   1000 mol   H c = ______________ kJ


21


Student View

Introduction to Engineering Principles and Units Example 1.6-4: Incomplete Combustion of Methane A mixture of air and methane is being fed to the firebox in a steam-methane reforming process. The methane in this mixture is being burned in order to provide heat for the steam-methane reforming reaction to occur. kmol and has a molar fraction of 0.0111 of methane. hr 90 % of the methane is burning to produce CO2, and the rest is undergoing incomplete combustion to produce CO. These combustion reactions are shown below:

The fuel mixture is entering at a rate of 213.5

1) CH4 + 2O2 2) CH4 +

CO2 + 2H2O(g)

ˆ o = −802600 ∆H r,1

kJ kmol

CO + 2H2O(g)

ˆ o = −519670 ∆H r,2

kJ kmol

3 O2 2

Calculate the amount of heat transferred by these two reactions in

kcal . hr

Note: The standard heats of reaction may be obtained from Table A.3-2 of Geankoplis or using standard heats of formation of the molecules involved in the chemical reaction, shown in Table A.31. Strategy This problem can be solved by using tabulated data for standard heats of reaction. We need to take into account the selectivity of CO2 to CO.

Solution The amount of heat transferred by the two reactions will be given by the following equation:

(

)

(

= ∆H = n ˆ o + n ô Q ∆H ∆H CO r,1 CO r,2 2

)

where: n CO = Molar flow rate of carbon monoxide being produced by reaction 2 2

( ∆Hˆ ) = Standard enthalpy of reaction 1 o r,1

n CO = Molar flow rate of carbon monoxide being produced by reaction 2

( ∆Hˆ ) = Standard enthalpy of reaction 2 o r,2


22


Student View


The flow rates of carbon dioxide and carbon monoxide can be calculated by multiplying the overall flow rate of fuel/air mixture by the molar fraction of methane. Then, we multiply this amount by the amount of methane being converted to carbon dioxide and carbon monoxide respectively. If we look at the stoichiometric coefficient of both chemical reactions, it can be seen that the ratio of carbon dioxide to methane in reaction 1 and the ratio carbon monoxide to methane in reaction 2 are both equal to one. Thus, the amount of methane consumed by reaction 1 will be equal to the amount of carbon dioxide produced. Similarly, the amount of carbon monoxide produced by reaction 2 will be equal to the amount of methane consumed by reaction 2. n CO = n CH 2

4

, r1

n CO = 2.13 2

kmol CH 4    kmol  = ______  ____________    213.5 kmol   hr  

kmol CO 2 hr

For carbon monoxide: n CO = n CH

4

, r2

kmol CH 4    kmol  = ______  ___________    213.5 kmol   hr  

n CO = __________

kmol CO hr

Now we can substitute the molar flow rates and the enthalpies of reaction into the equation for the to get: heat transfer rate Q

= ∆H =  2.13 kmol CO 2  −802600 kJ  + ________ kmol CO  ___________ kJ    1 kcal  Q        hr kmol  hr kmol    _______ kJ    

= ______________ kcal Q hr


23


Student View

Introduction to Engineering Principles and Units Example 1.6-5: Standard Enthalpy of Reaction The water-gas shift reaction produces hydrogen from steam and carbon monoxide and is described by the following equation: CO + H2O(g)

CO2 + H2

Determine the standard heat of this reaction.

Strategy The heat of a given reaction can be obtained based on the stoichiometric coefficient of the species involved in the chemical reaction and their individual heats of formation.

Solution The heat of reaction can be calculated using the following equation:

∑

ô = ∆H r

products

ô − | ν i |∆H f ,i

∑ reactan ts

ô | ν i |∆H f ,i

Applying this equation to the number of products and reactants for the reaction of methanol in the fuel cell yields:

ˆ o =| ν | ∆H ˆ o + | ν | ∆H ˆ o − | ν | ∆H ˆ o − | ν | ∆H ô ∆H r H CO H O CO f ,CO f ,H f ,H O f ,CO 2

2

2

2

2

2

Since an element is a pure chemical substance, there is no energy transfer involved in its formation. Thus, the heat of reaction equation will be reduced to:

ˆ o =| ν | ∆H ˆ o − | ν | ∆H ˆ o − | ν | ∆H ô ∆H r CO H O CO f ,CO f ,H O f ,CO 2

2

2

2

The individual heats of formation for each one of these molecules can be found in Table A.3-1 of Geankoplis: ô ∆H = −393.51 f ,CO 2

ô ∆H f ,H

O

kJ mol

= ______________

2


kJ mol

24


Student View


ˆ o = ______________ ∆H f ,CO

kJ mol

Substituting these values and the stoichiometric coefficients into the heat of reaction equation, we have: ˆ o = (1)  −393.51 kJ  − (1)  − ______________ kJ  − (____)  − ______________ kJ  ∆H r       mol  mol  mol    

ˆ o = −41.16 ∆H r


kJ mol

25


Student View

Introduction to Engineering Principles and Units Example 1.7-1: Cooling of a Fuel Cell The bipolar plates are the component of a fuel cell responsible of transferring the heat out of the fuel cell. A modified design of the bipolar plate is shown in the following figure. This design has additional channels which allow cooling of the fuel cell by circulating a fluid such as air or water. Cooling air/water channels

Hydrogen channels

Reactant air channels

A fuel cell stack in a vehicle initially at a temperature of 80 °C is being cooled by water at room temperature. The fuel cell is being cooled down to a temperature of 45 °C. Determine the water flow L rate required in if the fuel cell reaches the temperature of 45 °C in 10 seconds. min The following data is available for the bipolar plate. V = 168.75 cm3

ρ = 1.3413

g cm3

= 1.62 C p

kJ kg ⋅ °C

The cooling water is exiting the cooling channels at a temperature of 33 °C.

Strategy We can set the energy balance by equaling the amount of heat lost by the fuel cell to the amount of heat gained by the water. Since only liquid and solid phases are present in this problem, the heat capacities of water and the bipolar plates may be assumed constant.

Solution The energy balance for this problem may be written as follows: Daniel López Gaxiola Jason M. Keith

26


Student View


Heat gained by water = Heat lost by fuel cell ∆H H

O

2

= ∆H fuel cell

Both amounts of heat in the energy balance can be obtained using the equation for sensible heat. ∆H H

O

2

(

H OC = m ∆TH p,H O 2

2

O

2

)t

∆H fuel cell = _______ ( _________ ) ∆Tfuel cell where:

H m

= mass flow rate of water

O

2

C p,H

O

= mass specific heat of liquid water

2

t = time required for cooling the bipolar plate to the final temperature of 45 °C

m bp = mass of bipolar plate C p,H

O

= mass specific heat of bipolar plates

2

Note that the equation for the change in enthalpy of water is multiplied by a time period t. This is done so we can obtain the total amount of energy transferred as heat to the water instead of the heat transfer rate. We can start by calculating the change in enthalpy of water as shown in the following steps: ∆H H

O

2

(

)

H OC = m ∆T t p,H O 2

2

The specific heat of water can be obtained from Table A.2-5 of Geankoplis to be:

C p,H

O

= ______________

2

kJ kg ⋅°C

Substituting this value, the time period t and the corresponding temperatures into the equation for the change in enthalpy of water yields: ∆H H

O

2

 kJ  H O  ______________ =m  ( 33°C − 25°C )( ______ s ) 2 kg ⋅°C  


27


Student View

Introduction to Engineering Principles and Units Now we proceed to calculate the change in the enthalpy of the fuel cell:

∆T ∆H fuel cell = m bp C p,bp The mass of the bipolar plate can be calculated using the volume and density data given in the problem statement:

 _________  g   m bp = ρbp Vbp = 1.3413 _________ cm3 )   = 0.226kg 3 ( cm    _________  Substituting this result, the heat capacity value of the bipolar plates and the initial and final temperatures of the fuel cell into the equation for ∆H fuel cell we get:

 kJ  ∆H fuel cell = ( _______ kg )  1.62  ( _____ °C − _____ °C ) kg ⋅°C   ∆H fuel cell = __________ kJ Now we can substitute the equations for ∆H fuel cell and ∆H H O into the energy balance equation: 2

 kJ  H O  ___________ m  ( 33°C − 25°C )(10 s ) = _______ kJ 2 kg ⋅°C   Solving for the mass flow rate of water we have:

H m

H m

_______ kJ  kJ   ________ kg ⋅°C  ( 33°C − 25°C )(10 s )  

O

=

O

= 0.038

2

2

kg s

We need to convert the mass flow rate to volumetric flow rate as follows:

V H

O

2

=

HO m 2

ρH O

=

2

V H

O

kg  ________  s  ________  kg _______ L

0.038

= _______

2


L min

28


Student View


Example 1.7-2: Simultaneous Material and Energy Balances The steam-methane reforming process is carried out in a plant operating on the scale of a large kg H 2 central station, which is capable of producing 1× 106 (equivalent to 1 million gallons of day gasoline). Since the reaction for producing hydrogen from methane and steam is endothermic, heat is being added to the reactor. The following diagram is describing this process.

CH4 + H2O(g)

∆H or,298 K = 206.16

CO (g) + 3H2(g)

kJ mol

Q

mol 439.5 min mol CH 4 0.237 mol mol H 2 O 0.763 mol T = 431 °C

n H

2

,out

n CO,out T = 810°C

Determine the heat added to the reactor if all of the methane is converted into products.

Strategy by performing material and energy balances around the We can determine the heat transfer rate Q reaction chamber.

Solution The flow rates of hydrogen and carbon monoxide out of the reactor can be determined by the ratio of their stoichiometric coefficients to the stoichiometric coefficient of methane (limiting reactant). Thus,

n H

n CO =1 n CH

2

n CH

4

= ______

4

Solving these equations for the unknown molar flow rates n H , n CO , we get: 2

n CO = n CH

4

n H = ____ n CH 2

4

The molar flow rate of methane entering the reactor can be obtained by multiplying the overall flow rate by the molar fraction of methane to yield:


29


Student View

Introduction to Engineering Principles and Units

n CH = yCH n = 0.237 4

4

n CH = ________

mol CH 4  mol   ________  mol  min 

mol CH 4

4

min

Substituting this result into the equations for the molar flow rates n H , n CO we can obtain the 2

individual flow rates of each gas in the product stream: n CO =

mol CH 4  1 mol CO   __________  mol CH 4  min 

n CO = __________

mol CO min

n H = 2

____ moles H 2  mol CH 4   __________  mol CH 4 min  

n H = __________ 2

mol H 2 min

Now we can proceed to perform energy balances to determine the heat transfer rate. For this we will select a reference state of 25 °C (298 K). The general balance equation is given by:

∑ H + ( −∆H ) + q = ∑ H R

o 298

P

Input items The heat entering the reactor is equal to the sum of the changes in the enthalpies of the reactants between the reference state and the process conditions. Thus, mol CH 4    kJ   1 min  −2 ˆ p,CH ∆T =  __________ H = nC   ( ______ °C − 25°C )    4.680 × 10 CH 4 4 min   mol ⋅°C   60 s  

= 32.99 kW H CH 4

H H

H H

O

2

O

mol H 2 O  mol   kJ   1 min  ˆ p,H O ∆T = 0.763 = nC   439.5   ______________  ( ______ °C − 25°C )  2 mol  min   mol ⋅ °C   60 s  = __________ kW

2

The heat capacities were obtained using linear interpolation with the data from Table 1.6-1 of Geankoplis at the temperature of 431°C (704 K). The standard enthalpy of reaction must be included in the input items as shown in the general energy balance equation. Daniel López Gaxiola Jason M. Keith

30


Student View


kJ   mol  1 min   o −∆H =  __________   __________   298 K mol   min  60 s  

o −∆H = ____________ kW 298 K Output items In a similar way we can obtain the enthalpies of the gases leaving in the product stream as shown in the following steps. The heat capacities of carbon monoxide and hydrogen were obtained by linear interpolation of the data in Table 1.6-1 of Geankoplis at a temperature of 810 °C (1083 K). mol CO  kJ   1 min  = nC ˆ p,CO ∆T =  __________ H   _____________  ( 810°C − 25°C )  CO min  mol ⋅°C   60 s  

= 42.50 kW H CO mol H 2    kJ   1 min  = nC ˆ p,H ∆T =  __________ H _______________ 810°C − 25°C )  (     H 2 2 min   mol ⋅°C   60 s  

= __________ kW H H 2

Now we can substitute the enthalpies we calculated into the general balance equation and solve it for the heat added to the system q to get:

∑ H + ( −∆H ) + q = ∑ H o 298

R

P

q = ∑ H P − ∑ H R − ( −∆H o298 )

+H −H q = H −H − ( −∆H o298 ) H CO CH H O 2

4

2

q = 42.50 kW + __________ kW − 32.99 kW − __________ kW + __________ kW q = __________ kW

The positive sign of q indicates that heat is being added to the system.


31


Student View

Introduction to Engineering Principles and Units Example 1.7-3: Oxidation of Woody Biomass Biomass gasification is a process used for producing hydrogen in large scale for use in fuel cells. The elemental analysis of dry woody biomass used in the gasification process yielded the following results:

Element C H O

Mol % 33.3 46.7 20.0

The following equation represents the combustion reaction of woody biomass.

C0.333H0.467O0.200 + 0.3498O2

0.333CO2 + 0.2335H2O(g)

∆H or,25° C = −116.62

kJ mol

Calculate the molar heat of combustion at the temperature of 600 °C.

The following equation may be used for estimating the heat capacity of dry wood.

 J  = 0.1031 + ( 3.867 × 10−3 ) T C  p  g ⋅ °C  where: T = Temperature in degrees Kelvin.

This equation was obtained from the book Wood as an Engineering Material published by the U.S. Department of Agriculture. A diagram of the biomass oxidation process is shown in the figure below: o q = −∆H r,600 °C

n1 , mol biomass T = 40 °C

n 3 , mol CO 2 Combustion

n 2 , mol O 2 T = 40 °C


n 4 , mol H 2 O T = 600 °C

32


Student View


Strategy This problem can be solved by using energy balances around the combustion chamber. The reference temperature will be selected to be 25 °C and a basis of 1 mol of biomass will be selected.

Solution First we need to determine the amount of steam and carbon dioxide in the exhaust gases as well as the amount of oxygen required by this process. These amounts can be calculated based on the stoichiometric coefficients of each molecule involved in the chemical reaction. Thus, n2

n1

n3

= 0.3498

n1

= __________

n4

n1

= __________

Since we selected a basis of n1 = 1 mol, the amount of moles of oxygen n 2 , carbon dioxide n 3 and steam n 4 are given by: n 2 = 0.3498 moles O 2

n 3 = __________ moles CO 2

n 4 = __________ moles H 2 O

Now we can proceed to calculate the amount of heat entering through the reactants. Since there are no phase changes occurring for any of the species, the amount of heat can be calculated using the definition of sensible heat. We can start from the general energy balance equation for a reactive process shown below:

∑ H + ( −∆H R

o r,25° C

)+ q = ∑H

P

Input items H ( biomass ) = m biomass C ∆T p,biomass The mass heat capacity of biomass can be obtained by substituting the temperature value of 40 °C (313.15 K) into the expression given in the problem statement:

 J  = __________ + ( _______________ )( 313.15 K ) C  p  g ⋅ °C  = __________ C p

J g ⋅ °C

Substituting this value into the equation for the enthalpy of biomass entering in the reactants stream we get: Daniel López Gaxiola Jason M. Keith

33


Student View

Introduction to Engineering Principles and Units J   __________ g   H ( biomass ) = 1 mol biomass   ( _____ °C − 25°C )   __________ g ⋅ °C   1 mol biomass  

H ( biomass ) = __________ J Similarly, we can calculate the heat entering through the oxygen as shown in the following steps: J   H ( O 2 ) = ____________ mol O 2  ___________  ( _____ °C − 25°C ) mol ⋅°C  

H ( O 2 ) = ____________ J Note: The heat capacity of oxygen was obtained from the data in Table 1.6-1 of Geankoplis. The standard enthalpy of reaction at the temperature of 25 °C is given in the problem statement to be:

∆H or,25° C = −116.62

kJ ( __________ mol ) mol

∆H or,25° C = __________ kJ Output items The enthalpies of steam and carbon dioxide can be obtained in a similar way to how we calculated the enthalpy of oxygen entering the reactor. For carbon dioxide and steam, the specific heat values were obtained through linear interpolation of the data from Table 1.6-1 of Geankoplis. J   H ( CO 2 ) = __________ mol CO 2  ____________  ( ______ °C − 25°C ) mol ⋅°C  

H ( CO 2 ) = 8861.5 J J   H ( H 2 O ) = __________ mol H 2 O  __________  ( _______ °C − 25°C ) mol ⋅°C  

H ( H 2 O ) = ___________ J Substituting the calculated enthalpies and the heat of reaction at 25 °C into the general energy balance equation, we get:

∑ H + ( −∆H R


o r,25° C

)+ q = ∑H

P

34


Student View


_________ J + __________ J + ( _____________ J ) + q = 8861.5 J + ________ J The diagram of the combustion of biomass process is showing that the amount of heat released by the reaction is equal to the enthalpy of reaction at 600 °C. Thus, solving for q yields:

 1 kJ  q = ∆H or,600 °C = 8861.5 J + _________ J − _________ J − _________ J − ( __________ J )     1000 J 

∆H or,600 °C = ___________ kJ To obtain the molar heat of combustion at 600 °C, we need to divide this value by the number of moles of biomass. Since we selected a basis of 1 mole of biomass reacting, the molar enthalpy will be given by: ô ∆H = r,600 ° C

∆H or,600 ° C n biomass

=

___________ kJ ___________ mol

ô ∆H = ___________ r,600 ° C


kJ mol

35


Student View


Chapter 2 Principles of Momentum Transfer and Overall Balances In fuel cells, the fuel is usually in gas or liquid phase. Thus, the student must be familiar with the principles of fluid mechanics or momentum transfer, which will be covered in the following problem modules. In later sections of this chapter, situations combining momentum transfer and heat transfer will be illustrated. 2.2-3 Conversion of Pressure to Head of a Fluid 2.3-1 Diffusivity of Hydrogen inside a Fuel Cell 2.5-1 Reynolds Number of Hydrogen flowing into a Fuel Cell 2.6-1 Flow of Hydrogen into Fuel Cells 2.6-3 Velocity of Hydrogen in Bipolar Plate Channel 2.7-1 Energy Balance on Ethanol boiler 2.10-1 Methanol Flow in Fuel Cell 2.10-2 Use of Friction Factor in Laminar Flow 2.10-3 Use of Friction Factor in Turbulent Flow 2.10-4 Trial-and-Error Solution to Calculate Pipe Diameter 2.10-5 Flow of Gas in Line and Pressure Drop 2.10-8 Entry Length for a Fluid in a Rectangular Channel 2.11-1 Compressible Flow of a Gas in a Pipe Line 2.11-2 Maximum Flow for Compressible Flow of a Gas


1


Student View

Principles of Momentum Transfer and Overall Balances

Example 2.2-3: Conversion of Pressure to Head of a Fluid Hydrogen is stored in a compressed gas tank with a volume of 50 L at a pressure of 140 atm and a temperature of 25 °C. What is the pressure in the tank in mm Hg and inches of water? Strategy To determine the pressure of hydrogen as head of a fluid we need to use the equation for calculating pressure of a fluid as a function of height. Solution The equation for pressure as a function of height is shown below. P = ρgh where: ρ = density of fluid g = acceleration due to gravitational force h = head or height of fluid We can solve this equation for the head h to yield: h=

P ρg

To determine the pressure in mm of Hg we need to use the density of mercury, which can be obtained from Table 2-31 Perry’s Chemical Engineers’ Handbook, 8th Edition to be: ρHg = _____________

kg m3

Since the units in the expression for the head of a fluid need to match, the value of g must be in  N  m  2  and the pressure must be converted to Pa  2  . After entering the values into the equation for m  s  h we get:

kg ⋅ m  N    __________ m 2   _________ s 2  140 atm    1 atm 1N        h= kg  m   _____________ 3  9.81 2  m  s  


2


Student View


h = ___________ m Hg

This value can be converted to mm of Hg by multiplying by the conversion factor from m to mm:

 _______________  h = ___________ m Hg    _______________  h = ___________ mm Hg

To calculate the head in inches of water, we need to follow a similar procedure. The difference is that we need to use the density of water instead of the density of mercury, and use a conversion factor to convert meters to inches. Thus, kg ⋅ m  N    _________ m 2   _________ s 2  140 atm    1 atm 1N        h= kg  m   _________ 3  9.81 2  m  s   h = 1446 m H 2O

The density of water was obtained from Table A.2-3 of Geankoplis at a temperature of 25 °C. Converting this value to inches we get:

  1 in H 2 O h = 1446 m H 2 O    _____________ m H 2 O 

h = ___________ in H 2 O


3


Student View


Example 2.3-1: Diffusivity of Hydrogen inside a Fuel Cell Hydrogen in a bipolar plate is diffusing in the anode side of a fuel cell. The current density of a stack mA of 440 fuel cells is 600 2 . cm The hydrogen is entering the fuel cell at a pressure of 2 atm and a temperature of 25 °C. Determine mm 2 the diffusivity of hydrogen through the gas-diffusion layer with a thickness of 100 µm in if the s fuel cell performance is limited by mass transfer. The amount of hydrogen reacted as a function of current is described by the following equation: n H

2

,reacted

=

IN 2F

where: I = current in amperes (A) N = number of cells in the fuel cell stack F = Faraday’s constant = 96485

C mol ⋅ e −

Strategy The diffusivity of hydrogen can be determined using Fick’s Law of diffusion and the definition of the consumption rate of hydrogen in terms of the current.

Solution Fick’s first law relates the diffusive flux to the difference in concentration of a substance. For this problem, Fick’s first law is given by: n H

2

,reacted

= −DH A 2

dC H

2

dx

where: D H = diffusivity of hydrogen through the gas − diffusion layer 2

A = cross − sectional area of the gas − diffusion layer dC H

2

= change in the concentration of hydrogen along the thickness of the gas − diffusion layer dx Daniel López Gaxiola 4 Student View Jason M. Keith



The left hand side of Fick’s first law can be re-written in terms of the current. Hence,

dCH _________ 2 = −DH A 2 _________ dx The problem statement is giving the value of the current density defined as the current of the fuel cell divided by the area. Thus, this equation can be rewritten in terms of the current density I as follows:

dCH _________ 2 = −DH 2 dx _________ Assuming the diffusivity remains constant along the gas-diffusion layer, this equation can be separated and integrated as follows: CH @ x =100µm _________ x=_________ µm 2 = − dx D dC H H ∫C 2 2 _________ ∫ x=0 µm H2 @ x =0

Since all the hydrogen entering the fuel cell is reacting when mass transfer is limiting the fuel cell performance, the concentration of hydrogen in the anode will be given by: CH

2

@ x =100 µm

=0

The concentration of a substance is defined as the number of moles of substance in a volume of solution. Thus, CH = 2

nH

2

V

The concentration of hydrogen entering the channels in the bipolar plate can be obtained using ideal gas law as shown in the following steps: PV = nRT

Solving for the concentration

CH = 2

nH

2

V

=

nH

2

V

from the ideal gas law, we have:

____ _______


5


Student View


Substituting the corresponding quantities in the right side of this equation yields:

CH

2

 101325 Pa  2 atm   1 atm   = Pa ⋅ m3   __________ ( 298.15K )  mol ⋅ K  

CH = __________ 2

mol m3

Substituting the concentration values into the integral equation we get: 0 _____ 1×10−4 m = − dx D dC H H ∫ _________ mol 2 2 _____ ∫ 0 m m3

We can integrate this equation and substitute the values for the current density, number of fuel cells and Faraday’s constant to give: mA    ______ 2  ( ______ cells ) mol cm   (1× 10−4 ) m − 0 m  = − D H  ___________ 3  2 C  m    2  ___________ −  mol ⋅ e  

Finally we can solve for the diffusivity D H to obtain: 2

DH

2

2 mA    1 A   _____ C/s   ________ cm  ______ ______ cells ( )       cm 2  1A 1 m2   1000 mA      1× 10−4 m  = )  ( C  mol   2  __________  _________ 3  mol ⋅ e−  m  

D H = 1.67 × 10 −5 2

m2 s

This diffusivity value can be converted to

mm 2 by multiplying it by a conversion factor from m2 to s

mm2. D H = 1.67 × 10−5 2

m 2  ____________    s  ____________ 

mm 2 D H = _______ 2 s Daniel López Gaxiola Jason M. Keith

6


Student View


Example 2.5-1: Reynolds Number of Hydrogen flowing into a Fuel Cell A compressed gas tank contains hydrogen at room temperature and a pressure of 140 atm. The valve kg on this tank is opened and the hydrogen enters a fuel cell at a rate of 0.75 through a steel pipe hr with an inner diameter of 5.46 mm. Determine the type of pipe connecting the fuel cell to the hydrogen tank and determine if the flow of hydrogen is laminar or turbulent.

Strategy To determine the type of pipe used to connect the hydrogen tank to the fuel cell we need to find the pipe corresponding to the inner diameter given in the problem statement. The flow regime may be determined depending on the value of the Reynolds number.

Solution Appendix A.5 of Geankoplis is showing the properties of different types of standard steel pipes. For 1 the inner diameter of 5.46 mm, the nominal size of the pipe is in. with a Schedule Number of 80. 8 For flow inside a pipe, the Reynolds number is given by: Re =

Dυρ µ

where: D = inner diameter of the pipe in meters (m)

υ = velocity of the fluid inside the pipe in

ρ = density of fluid in

m s

kg m3

µ = viscosity of fluid in

kg m⋅s

The velocity of hydrogen circulating in the pipes can be obtained by dividing the volumetric flow rate of hydrogen by the inner cross-sectional area of the pipe. However, since we are given the mass flow rate, we will have to calculate the volumetric flow rate using the ideal gas law.

= mRT PV M yields: Solving for the volumetric flow rate V Daniel López Gaxiola Jason M. Keith

7


Student View


= mRT V MP Substituting the corresponding quantities into this equation we get: kg   Pa ⋅ m 3   0.75 ________    ( 298 K ) ⋅ hr mol K     = V 101325 Pa   kg  1 kmol  (140 atm )    ______   kmol  1000 moles   1 atm  

Note that conversion factors for the pressure and number of moles were used in order to get the correct units for the volumetric flow rate. 3

= __________ m V hr

To calculate the velocity of hydrogen in the pipe, we also need to determine the cross-sectional area of the pipe. This value can be obtained directly from Table A.5 of Geankoplis. A = _____________ m 2

Now we can obtain the velocity of hydrogen as shown in the following equations. Note that we are multiplying the velocity equation by the conversion factor from hours to seconds, hence to obtain the m velocity in . s

 m3  1 hr ________   hr  ________ s  V υ= = A ____________ m 2

υ = 0.772

m s

The density of hydrogen at the pressure of 140 atm and the temperature of 25 °C (298 K) can be calculated using the ideal gas equation of state: = mRT PV M

Since the density r is equal to dividing the mass flow rate by the volumetric flow rate, we can solve for the density as shown in the following step. ρ=

_______ m = _______ V


8


Student View


We can enter the numeric values into the right side of this equation to get:

( _______ ρ=

 __________   2 kg   1 kmol  atm )      __________   1 kmol   1000 moles  = __________ kg Pa ⋅ m3  m3  8.314 298 K ( )  mol ⋅ K  

The only parameter left to be determined before calculating Reynolds number is the viscosity, which can be obtained from Figure A.3-2 for gases at a pressure of 1 atm. However, for hydrogen at the pressure and temperature conditions in this problem, the viscosity does not depend on pressure. Thus, it is valid to use Figure A.3-2. To obtain the viscosity of hydrogen we need to locate the coordinates for hydrogen in Table A.3-8 and draw a line that passes through these coordinates and the temperature value of 25 °C. Hence, the viscosity will be estimated to be: µ ≈ ____________

kg m ⋅s

Now we can enter the quantities we found for velocity, density and viscosity into the equation for the Reynolds number to yield:

( 5.46 ×10 Re =

−3

m  kg   m )  0.772  _________ 3  s  m   kg ____________ m⋅s

Re = __________

Conclusion:


9


Student View


Example 2.6-1: Flow of Hydrogen into Fuel Cells Hydrogen is exiting the fuel tank in a vehicle through a stainless steel pipe (schedule number 80) L with a nominal diameter of ¼”. The hydrogen is leaving at a flow rate of 446.4 at room hr temperature and at a pressure of 2 atm. This flow rate is equally distributed between the 325 fuel cells required to power this vehicle. The hydrogen is being distributed to the fuel cells by a pipe of 1/16” inner diameter. The hydrogen flow rate entering the channels of the bipolar plate of a fuel cell is equally distributed between the 3 channels of the bipolar plate. (Note: The channels in the bipolar plate will be assumed to be semicircular with an inner diameter of 1/16”)

a) Determine the mass flow rate of hydrogen entering each cell and each one of the channels in the bipolar plate.

Hydrogen Tank

…

n=1

n = 325

Strategy To determine the mass flow rate from the volumetric flow rate we need to use ideal gas law.

Solution We can solve the ideal gas law for the mass flow rate as shown in the following steps: overall = m

PV M overall

RT

Substituting the corresponding quantities into the right hand side of this equation yields:


10


Student View


( 2 atm )  446.4

overall m

L  g  2  hr   mol   = L ⋅ atm    ____________  ( 298.15 K ) mol ⋅ K  

overall = ________ m

g hr

To determine the flow rate of hydrogen entering each fuel cell, we divide the mass flow rate of hydrogen leaving the tank by the total number of fuel cells in the stack. Thus,

fuel cell = m

m n fuel cells

fuel cell = 0.225 m

g hr = ______ cells ________

g hr

This value can by divided by 3 (number of channels on each bipolar plate) to determine the flow of hydrogen to each channel:

channel = m

fuel cell m n channels

g hr = 3 channels

0.225

channel = __________ m

g hr

b) What is the average velocity of the hydrogen leaving the tank? Strategy The velocity of hydrogen can be determined by applying the definition of velocity in terms of the flow rate.

Solution The velocity of a fluid can be determined with the following equation: υ=

V A

First we need to determine the volumetric flow rate of hydrogen to each fuel cell and to each channel in the bipolar plates. This can be done in a similar way to part a) of this problem.


11


Student View


V = fuel cell

L hr = 325 cells 446.4

V n fuel cells

L = ____________ V fuel cell hr

V = channel

V fuel cell n channels

L hr = 3 channels _______

L = _________ V channel hr

Now we need to calculate the cross-sectional areas of different pipes through which hydrogen is circulating. First, the area of the ¼” pipe connected to the fuel tank can be found in Table A.5-1 of Geankoplis to be: A1/4" pipe = ______________ m 2

The cross-sectional area of the pipe entering the fuel cells can be calculated using the equation for the area of a circle, where the diameter will be 1/16”. Hence,

A1/16" pipe

1  π  in  2  πD  16  = =   4  4 

2

A1/16" pipe = _____________ in 2

Converting this value to m2, we have: A1/16" pipe

 ____________ m 2  = ____________ in   1 in 2   2

A1/16" pipe = ____________ m 2

Since the channels on the bipolar plate are assumed to be semicircular, the cross-sectional area of the channel can be calculated as follows:

A channel

 πD 2   4   = 2


12


Student View


Substituting the corresponding quantities into the right hand side of this equation yields: 2

A channel

2  1   _______________ m  π  in    1 in 2  16    = 8

A channel = _______________ m 2 Now we can obtain the velocities in the three different sections, as shown in the following equations:

υH

2

υH

υH

υH

υH

in 1/4" pipe

in fuel cell

2

2

υH

in 1/4" pipe

2

2

L   1 m3   _________    446.4   hr   1000 L   _________   = _______________ m 2

in fuel cell

in channel

2

in channel

= 2.68

m s

L   1 m3   1 hr    ___________     hr   1000 L   3600 s   = ______________ m 2

= ____________

m s

L   1 m 3   1 hr   _________     hr   1000 L   3600 s   = _______________ m 2

= _______

m s

c) Calculate the mass flux of hydrogen circulating through each channel in

kg . m 2 ⋅ hr

Strategy The flux of hydrogen refers to the amount of hydrogen flowing through an area during a period of time.


13


Student View


Solution The flux of hydrogen can be obtained by dividing the mass flow rate of hydrogen entering each channel, by the cross sectional area of the channel. Thus, G=

channel m A channel

Entering numeric values into this equation we get: g  _________  hr  _________  ______________ m 2

_________ G=

G = ___________


kg m 2 ⋅ hr

14


Student View


Example 2.6-3: Velocity of Hydrogen in Bipolar Plate Channel The following figure is showing the flow of hydrogen along the horizontal channel of length L in a bipolar plate.

x =−B x = B y=B y = −B

L

y z x

Hydrogen channels

H2

Reactant air channels The velocity of hydrogen along a square channel in a bipolar plate is given by the following equation:   x  2    y 2  υz = υmax 1 −    1 −      B     B  

Determine the average velocity of the hydrogen flowing through the channel.

Strategy The average velocity can be calculated from the expression of the velocity as a function of the position in the x and y dimensions.

Solution The average velocity can be calculated using Equation 2.6-17 of Geankoplis, shown below: υav =

1 υz dA A ∫A ∫

In Cartesian coordinates dA may be written as dxdy. The cross-sectional area of the channel is obtained by multiplying the dimensions of the channel. Hence, υav =

1

( 2B )

  x 2    y  2  ∫ −B ∫ −B υmax 1 −  B   1 −  B   dxdy    B

2


B

15


Student View


Integrating this equation in the x-direction, we get: υ υav = max2 4B

  ______   ∫ −B  x −  ______  B

B

  y 2  1 −    dy   B   −B 

Evaluating the integrated expression for x yields: υ υav = max2 4B

2   ______    _____      y   ∫ −B B −  ______  −  − ____ −  _____    1 −  B   dy    B

We can reduce this equation to get: υav =

υmax 4B2

 _____    y  ∫ −B _____ −  _____  1 −  B  

υav =

υmax 4B2

2  _____ − _____    y   − 1      dy ∫ −B  3    B  

υav =

2 B  υmax y  − 1     dy _______ ∫ −B   B  

B

2

  dy 

B

Now we can proceed to integrate and evaluate the equation for the average velocity with respect to y as shown in the following steps: B

 ______   υmax  υav =  ____ −   ______   ______   − B

υav =

 _____     ______    υmax     ____ −    −  ____ −    _____    _____     ______   

Reducing this equation we determine the expression for the average velocity of hydrogen in the channels:

υav =

 2B3   υmax  _____ −    _____   ______  

υav =

υmax  _____  ____ υmax  = _____  3  ____

υav =

4υmax 9


16


Student View


Example 2.7-1: Energy Balance on Ethanol Boiler A liquid solution of ethanol and water is entering a boiler before undergoing a reforming reaction for producing hydrogen for proton-exchange membrane fuel cells. A diagram of the process is shown below. Liquid Ethanol/Water Solution m υl = 0.085 s = −13216 kJ H l kg

What is the power required to produce 1323

Ethanol/Water gas mixture m υg = 13.71 s = −10878 kJ H g kg kg of vapor? hr

Strategy The overall energy balance for a steady-state flow system can be used to determine the amount of heat required.

Solution The energy balance equation is given by Equation 2.7-10 of Geankoplis and is shown below: H 2 − H1 +

1 υ2 2 − υ12 ) + g ( z 2 − z1 ) = Q − Ws ( 2α

where: H2 = Enthalpy of the substance exiting the system H1 = Enthalpy of the substance entering the system

α = Kinetic-energy velocity correction factor υ2 = Velocity of the substance exiting the system υ1 = Velocity of the substance entering the system

g = Acceleration due to gravity z 2 − z1 = Difference in height between the outlet and inlet points in the system

Q = Heat added (+) or removed (–) to the system Ws = External work applied by (+) or to (–) the system Daniel López Gaxiola Jason M. Keith

17


Student View


For this problem we will assume turbulent flow, so the value of the kinetic-energy velocity correction factor α is close to 1. The external work term Ws will be neglected in the balance equations as there are no mechanical parts performing work to the system. Since no information is given about the height of the inlet and outlet points in the system, we will assume the values are the same. Thus, the difference z 2 − z1 is equal to zero. After applying these assumptions, the energy balance equation will be given by:

=H −H + 1 ( ______ − ______ ) Q 2 1 2 Where the tilde represents the amount per unit mass. The values given for the mass enthalpies of the ethanol/water mixtures and the velocities of the liquid and gas can be now substituted into this equation to obtain the amount of heat required. Thus,

=H −H + 1 ( ______ − ______ ) Q 2 1 2 2 2 kJ  kJ  1  m  m  Q = ____________ −  −13216  + 13.71  −  ________   kg  kg  2  s   s  

= 2432 kJ Q kg To determine the power required, we need to multiply the heat per kilogram of vapor by the flow rate:

= 2432 kJ  ___________ kg   ______  Q    kg  hr   ______  = ________ kW Q


18


Student View


Example 2.10-1: Methanol Flow in Fuel Cell A solution of 40 wt. % methanol and 60 wt. % water is entering the channels of a bipolar plate in a direct methanol fuel cell. The viscosity and density of this solution are shown below: µ = 1.85 × 10−3 Pa ⋅ s

ρ = 931.5

kg m3

The pressure drop along a rectangular channel is shown by Bahrami et al. [1] to be: ∆P =

µLυ  1 64ε  π  c 2  − 5 tanh     2ε   3 π

where: ∆P = Pressure drop, Pa µ = Viscosity of the fluid, Pa ⋅ s

L = Length of the channel, m

υ = Velocity of the fluid in the channel,

m s

The parameters c and ε are related to the dimensions of the channel as shown in the following figure: Gas-Diffusion Layer

ε= 2c = 1 mm

L = 30 mm

c b

2b = 1 mm

What is the pressure drop in atm and mm Hg along a single channel if the methanol flowing in the fuel cell has a Reynolds number of 300?

Strategy The pressure drop in the channel can be calculated using the equation for the pressure drop given in the problem statement. 1. Bahrami, M., Yovanovich, M. M., Culham, J. R., “Pressure Drop of Fully-Developed, Laminar Flow in Microchannels of Arbitrary Cross Section, Journal of Fluids Engineering, 128, 1036-1044 (2006)


19


Student View


Solution The equation given in the problem statement for the pressure drop depends on the velocity of the fluid in the channel. Since we are not given this value directly, we will have to calculate it from the definition of the Reynolds number. For an open channel, the Reynolds number is defined as follows: Re =

D H υρ µ

where DH is the hydraulic diameter, given by: DH =

4A P

In this equation, A is the cross-sectional area of the channel, and P is the perimeter of the channel in contact with the fluid. Since the fluid in the channel is in contact with the gas diffusion layer, we need to consider all the dimensions of the channel when calculating the wetted perimeter. Hence, the hydraulic diameter can be calculated using the dimensions of the channel as shown in the following step:

DH =

4 ( 2b ⋅ 2c ) ____________

Substituting the values of b and c into this equation yields: DH =

4 (1 mm )

2

 1m    _____________  1000 mm 

D H = 1× 10−3 m

Now we can enter this value for the hydraulic diameter into the equation for Reynolds number and solve for the velocity to get: kg   300 1.85 × 10−3  Re µ m ⋅s   υ= = DH ρ (1×10−3 m )  931.5 mkg3   

υ = __________

m s

The only remaining values that need to be calculated before being able to determine the pressure drop are the parameters b, c and ε : Daniel López Gaxiola Jason M. Keith

20


Student View

Supplemental Material for Transport Process and Separation Process Principles 2b = 2c = ____________

b=c=

ε=

__________  1 m    = ______________ m 2  1000 mm 

c ________________ m = b 5 × 10−4 m

ε = ____ Finally, the pressure drop along the length of the channel of 30 mm can be calculated as follows:

m   −3 kg  1.85 ×10  ( 0.03 m )  ___________  m ⋅s  s   1 atm   ∆P =     π    101325 Pa  64 (1) 2 1 ( ______________ )  − 5 tanh   π  2 (1)    3

∆P = ____________ atm Converting this value to mm Hg, we have:  _____________  ∆P = ___________ atm    _____________ 

∆P = __________ mm Hg


21


Student View


Example 2.10-2: Use of Friction Factor in Laminar Flow An aqueous solution of 40 mol % methanol is flowing along the channels in a bipolar plate of a m direct-methanol fuel cell at a velocity of 0.601 and a temperature of 27 °C. s Bahrami et al.[1] are showing the following equation to determine the friction factor in a rectangular channel: f ⋅ Re A =

12  192ε  π  1 − π5 tanh  2ε   (1 + ε ) ε   

where: Re

A

=

υρ A µ

The parameter ε is a function of the dimensions of the channel, as shown in the following figure. Gas-Diffusion Layer

ε= 2c = 1 mm

c b

L = 30 mm

2b = 1 mm

Use the definition of friction factor to calculate the pressure drop in atm along a single channel.

Strategy The pressure drop in the channel can be calculated using the equation for the friction factor given in Geankoplis.

Solution Equation 2.10-4 of Geankoplis gives the definition of Fanning friction factor:  A∆P    Aw   f= ρυ2 2 1. Bahrami, M., Yovanovich, M. M., Culham, J. R., “Pressure Drop of Fully-Developed, Laminar Flow in Microchannels of Arbitrary Cross Section, Journal of Fluids Engineering, 128, 1036-1044 (2006)


22


Student View


where: A = Cross – sectional area of the channel. Aw = Surface area of the channel being wetted by the fluid. ∆P = Pressure – drop along the channel of length L

ρ = Density of the fluid circulating through the channel υ = Velocity of the fluid in the channel We can start by solving for the pressure drop ∆P from the definition of friction factor, given in Equation 2.10-4 of Geankoplis. ∆P =

___________  ___________    ___________  2 

For the rectangular channel in the bipolar plate, the cross sectional area and the wetted surface area are given by:

A = ( 2b )( 2c ) A w =  2 ( ____________ )  L

Note that for the wetted perimeter we are considering the area of the channel in contact with the gas-diffusion layer. Since b = c, we can rewrite the area and the wetted perimeter as follows and substitute the dimensions of the channel in these equations to get:  2 1 m2 = ___________ m 2 A = ___________ = 4 ( __________ mm )  2  ___________ mm    1 m2  A w = _______ = 8 ( ________ mm )( _______ mm )  = ____________ m 2 6 2  1 × 10 mm  

the value of ε and enter the obtained value into the equation for f ⋅ Re A . Thus, ε=

c 0.5 mm = = ______ b _______ mm


23


Student View


f ⋅ Re A =

12  192  π  1 − π5 tanh  2   (1 + ________ ) _____   

f ⋅ Re A = _________

The Fanning friction factor f can be obtained from this equation but first we need to calculate the Reynolds number Re A using the equation given in the problem statement as shown in the following steps:

υρ Re

A

=

( A) µ

The density of methanol at the temperature of 27 °C can be obtained from Table 2-234 of Perry’s Chemical Engineers’ Handbook, 8th Edition to be: mol L

ρCH OH = 24.486 3

Converting this value to

ρCH OH = 24.486 3

kg , we have: m3

mol  _______ g CH3OH   1 kg   _________ L      L  1 mol CH3OH   1000 g   1 m3 

ρCH OH = ____________ 3

kg m3

Since the methanol entering the fuel cell is diluted in water, the density will depend on the concentration of the methanol solution. Thus, the density can be calculated as shown below: ρ = _________ ρCH OH + _________ ρH 3

O

2

The density of water is given in Table A.2-3 of Geankoplis. Since the value of the density at the temperature of 27 °C, we can use the value in the Table at the temperature closest to 27 °C, which is 25 °C: ρH

O

= _________

2


kg m3

24


Student View


Substituting the individual densities of water and methanol into the equation for the density of the solution ρ yields: kg  kg    ρ = 0.4  ____________ 3  + 0.6  ____________ 3  m  m   

ρ = ____________

kg m3

The viscosity of methanol can be determined using Figure A.3-4 of Geankoplis and the coordinates in Table A.3-12. Thus, µ = ____________

kg m ⋅s

Now we can obtain the value of the Reynolds number Re A as follows: υρ Re

Re

A

A

=

( ) A

µ

m  kg    0.601   ____________ 3  s  m  =

(

____________

____________ m 2

)

kg m⋅s

= ____________

From the value calculated for f ⋅ Re A we can solve for the Fanning friction factor to yield: f=

14.13 14.13 = Re A ____________

f = ____________

Finally we can determine the value of the pressure drop along the channel in the direct-methanol fuel cell by entering all the corresponding values into the equation for ∆P kg  m ____________ 3  0.601  −4 2 ____________ (1.2 × 10 m ) m  s  ∆P = ⋅ 2 ____________ m 2

2

  1 atm ∆P = 942.45 Pa    ____________ Pa 

∆P = ______________ atm


25


Student View


Example 2.10-3: Use of Friction Factor in Turbulent Flow kg and a pressure of hr 2061 kPa. The hydrogen is entering the storage tank through a pipe with an inner diameter of 100 J mm and a length of 10 m. If the maximum friction loss permitted along the pipe is 3.60 , what kg material would you propose for the pipe transporting the hydrogen from the pressure swing adsorption unit to the storage tank?

Pure hydrogen is exiting a pressure swing adsorption unit at a rate of 63.12

The gas leaving the pressure swing adsorption unit has the following properties: µ = 9.769 × 10−6 Pa ⋅ s

ρ = 1.425

kg m3

Strategy To determine what type of pipe to use in this process, we need to determine the parameter ε for the flow conditions in this process.

Solution ε may be obtained from Figure 2.10-3 of Geankoplis. To do this, we need to know D the values of the Reynolds number and the Fanning friction factor.

The parameter

The friction factor f can be calculated from the definition of the friction loss, as shown in the following equation: Ff = 4f

∆L υ2 D 2

Solving for the Fanning friction factor f, we get: f=

_________ _________

The value of the velocity of hydrogen inside the pipe is not given. However, its value can be calculated by dividing the volumetric flow rate of hydrogen by the cross-sectional area of the pipe. Hence, υ=

V A


26


Student View


The volumetric flow rate is calculated by dividing the mass flow rate by the density. The calculation of the volumetric flow rate and the cross-sectional area of the pipe are shown in the following steps: 2

πD 2 π ( 0.1 m ) = = ______________ m 2 A= 4 4 kg 63.12 m  1 hr  hr υ= =   kg  ρA  2  3600 s  1.425 _____________ m ( )   m3  

υ = ___________

m s

Substituting this velocity value into the equation for the friction factor we obtained, we have: f = __________

The other parameter required to locate the parameter

ε in Figure 2.10-3 is the Reynolds number, D

which is obtained as follows: Re =

Dυρ µ

Substituting the corresponding values into this equation yields:

( 0.1 m )  ______ Re =

m  kg   1.425 3  s  m   kg 9.769 × 10−6 m⋅s

Re = ____________ ε in the graph for the Reynolds number and the friction factor. D From Figure 2.10-3, this parameter is estimated to be:

Now we can locate the value of

ε ≈ __________ D


27


Student View


Solving for the pipe roughness ε , we get:

ε ≈ _________ D ≈ _________ ( 0.1 m ) ε ≈ _____________ m

Conclusion:


28


Student View


Example 2.10-4: Trial-and-Error Solution to Calculate Pipe Diameter Hydrogen in a fuel cell vehicle is flowing from the tank to the fuel cell stack through a commercial g steel pipe with a length of 3 m. The hydrogen is being consumed at a rate of 1.17 and is entering s the fuel cell stack at a temperature of 25 °C and a pressure of 2.5 atm. Determine the diameter of the pipe connecting the fuel tank to the fuel cell stack if a head of 240 in Hg is available to compensate for the friction loss.

Strategy From the value of Ff given as head of fluid we can determine the required diameter for feeding hydrogen to the fuel cells.

Solution The friction loss can be calculated by multiplying the head of fluid in m of H2O by the acceleration due to gravity. Thus, m   _________ m H 2O   Ff = 240 in Hg  ____________ 2    s  1 in Hg   Ff = 812.69

J kg

To determine the pipe diameter we have to obtain the diameter from the definition of friction force, given by the following equation: Ff = 4f

∆L υ2 D 2

Solving for the diameter D, we have: D = 4f

∆L υ2 Ff 2

As it can be seen in this equation, we need to obtain the Fanning friction factor f and the velocity υ before being able to calculate the diameter. The friction factor can be obtained from Figure 2.10-3 of Geankoplis. To do this, we need to ε calculate Reynolds number first and obtain the relative roughness of commercial steel. Hence, D ε _____________ m = D D


29


Student View


Before obtaining the velocity of the fluid first we need to convert the mass flow rate to volumetric flow rate. To do this, we need to calculate the density of the fluid using the ideal gas equation of state. The velocity can then be determined by dividing the volumetric flow rate of hydrogen by the cross-sectional area of the pipe as shown in the following steps: 2 g   1 kg    PM kg  mol   1000 g  ρ= = = _________ 3 3 L ⋅ atm RT m  1m  _____________ ( 298.15 K )   mol ⋅ K  1000 L 

( 2.5 atm ) 

g  1 kg  1.17  s  1000 g  m3 =m V = = ___________ kg ρ s _________ 3 m A=

πD 2 4

m3   4  ___________ s  m V υ  = =  πD 2 s  A

υ=

___________ D2

The other parameter needed to obtain the friction factor from Figure 2.10-3 is the Reynolds number, calculated as follows: kg   __________ m  D ( m)  ___________ 3   2 Dυρ D s  m   Re = = kg µ _____________ m⋅s

Re =

169.23 D

The viscosity was obtained from Appendix A.3 of Geankoplis. As we can see, the diameter of the pipe D appears in all the expressions required for determining the friction factor. Therefore, we will select a diameter value and compare the result obtained to the calculated friction force. The initial guess for the diameter will be:


30


Student View


Trial 1

D = 0.010 m

Substituting this value in the equations for Reynolds number, velocity and relative roughness, we get: Re =

υ=

169.23 169.23 = = _____________ D 0.010

__________ __________ m = = __________ 2 2 D (0.010) s

ε ____________ m ____________ m = = = __________ D D (0.010 m)

Locating the values of the relative roughness and the Reynolds number in Figure 2.10-3 we find: f = 0.0085 Substituting this result as well as the velocity and dimensions of the pipe into the equation for the friction force we get: ∆L υ2 Ff = 4f D 2

m  ______   _____ m  s  Ff = 4 ( 0.0085 ) 0.010 m 2 Ff = ______________

2

J kg

It can be seen that this value does not match the calculated friction force. For a second trial, we will select a diameter of 0.020 m. Selecting a higher value for the diameter, will result in an decrease in the velocity of the fluid, thus reducing the value of the friction force. Trial 2

D = 0.020 m

Substituting this value in the equations for Reynolds number, velocity and relative roughness, we get: Re =

169.23 169.23 = = _________ D 0.020


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Student View


υ=

__________ __________ m = = __________ 2 2 D (0.020) s

ε __________ m __________ m = = = __________ D D (0.020 m)

f = _____________ m  __________   __________ m  s  Ff = 4 ( __________ ) 0.020 m 2 Ff = _____________

2

J kg

As we can see, the value obtained for a diameter of 0.020 m yields a friction force of J ____________ . The error between this value and the friction force calculated using the head of kg fluid available is given by: J J − __________ kg kg (100%) J __________ kg

__________ Error % =

Error = __________ %

Conclusion:


32


Student View


Example 2.10-5: Flow of Gas in Line and Pressure Drop Hydrogen at room temperature and standard pressure is entering a fuel cell stack through a smooth pipe with an inner diameter of 1 cm and a length of 3 m. Calculate the pressure of hydrogen in the g fuel tank. The hydrogen consumption rate is 2.053 . s

Strategy The initial pressure of the gas can be determined using the equation for the pressure drop of a gas inside a tube.

Solution We will start by assuming turbulent flow of hydrogen in the pipe. The initial pressure of hydrogen can be calculated using Equation 2.10-10 of Geankoplis shown below: ∆L ⋅ G 2 RT P − P = 4f DM 2 1

2 2

where: P1 = Initial pressure of the fluid in the pipe, Pa P2 = Final pressure of the fluid in the pipe, Pa f = Fanning friction factor ∆L = Length of the tube, m G = Mass flux of the fluid in the pipe,

R = Ideal gas constant,

kg m2 ⋅ s

Pa ⋅ m3 mol ⋅ K

T = Temperature of the fluid, K D = Inside diameter of the pipe, m M = Molecular weight of the fluid,

g mol

The flux of hydrogen may be calculated by dividing the mass flow rate of hydrogen by the crosssectional area of the pipe as shown in the following steps. G=

m A


33


Student View

Principles of Momentum Transfer and Overall Balances 2

πD 2 π ( __________ m ) = = _____________ m 2 A= 4 4 Substituting this value and the mass flow rate into the equation for the mass flux of hydrogen we get: g  1 kg  2.053  s  1000 g  G= _____________ m 2 G = __________

kg m2 ⋅ s

The friction factor can be obtained from Figure 2.10-3 of Geankoplis. To do this, first we need to ε calculate Reynolds number and the relative roughness of the pipe . Thus, D Re =

DG µ

Substituting the corresponding values for the parameters on the right side of this equation yields: kg   0.01 m  __________ 2  m ⋅s   = ____________ Re =  −6 kg   8.8 × 10  m⋅s  

The viscosity of hydrogen was obtained from Appendix A.3 of Geankoplis. We can see that this value for the Reynolds number indicates that the flow of hydrogen to the fuel cell is turbulent. Therefore, the equation we selected for calculating the pressure P1 is valid for this problem. Now with the Reynolds number value we can obtain the friction factor for a smooth pipe to be given by: f = __________ Solving Equation 2.10-10 of Geankoplis for the pressure P1 and substituting all the known quantities into this equation, we can obtain the pressure of hydrogen leaving the fuel tank as shown in the following steps. P1 = 4f

∆L ⋅ G 2 RT + P22 DM


34


Student View


2

kg   Pa ⋅ m3   3 m ⋅  _________ 2   _________ 2 ( 298.15 K ) m ⋅s   mol ⋅ K    101325 Pa    P1 = 4 ( _________ ) + 1 atm    g   1 kg   1 atm    0.01 m 2 ( )    mol   1000 g 

 1 atm  P1 = _____________ Pa    101325 Pa 

P1 = __________ atm


35


Student View


Example 2.10-8: Entry Length for a Fluid in a Rectangular Channel Determine if the velocity profile for the gas flowing through the channels of a fuel cell bipolar plate is fully developed for the following cases:

a) Hydrogen in a Proton-Exchange Membrane Fuel Cell at a temperature of 25 °C with laminar g H2 flow. The fuel consumption rate is of 1.642 . s b) Carbon monoxide in a Solid - Oxide Fuel Cell at a temperature of 800 °C with laminar flow. The kg CO fuel cell is converting carbon monoxide into products at a rate of 0.51 . hr The viscosity of carbon monoxide was found in Appendix A.3-2 to be: µ = 3.8 × 10−5

kg m⋅s

c) Same gases from parts a) and b) with turbulent flow. The dimensions of the channels in the bipolar plate are shown in the following figure: Gas-Diffusion Layer

20 cm

3 mm 5 mm Fuel Gas

Strategy The entry length is the distance required for the establishment of fully-developed velocity profile.

Solution a) To determine if the velocity profile is fully established in the channel of a bipolar plate, we need to calculate the entry length Le, defined by the following equation: Le = 0.0575 Re D

For this problem, the diameter D of the channel will be equal to the hydraulic diameter, i.e. the perimeter of the channel 'wetted' by the fluid flowing through it. Thus,


36


Student View


Le = 0.0575 Re DH

(1)

To determine the entry length, we need to calculate the Reynolds number, as shown in the following steps: Re =

DH G µ

The mass flux of hydrogen G is obtained by dividing the mass flow rate of hydrogen by the crosssectional area of the channel. Hence,

g H 2  1 kg  s  1000 g  m kg G= = = __________ 2 2 A m ⋅s   1m ( ____ mm )( ____ mm )  2   __________ mm  1.642

The hydraulic diameter is calculated using the following equation 2 4A 4 ( ____________ m ) = = _____________ m DH = P 2 ( 0.005 m + 0.003 m )

where: A = Cross-sectional area of the channel P = Perimeter of the channel ‘wetted’ by the fluid. Substituting this value and the mass flux in the equation for Reynolds number we get:

( ________________ m )  ___________ Re =



 −6 kg   8.8 × 10  m ⋅s  

kg   m2 ⋅ s 

= ____________

The viscosity of hydrogen was obtained from Appendix A.3 of Geankoplis. Now we can solve equation (1) for the entry length Le and substitute the calculated values to yield:

Le = 0.0575 Re D H = 0.0575 ( ____________ )( _____________ m ) L e = __________ m

Conclusion:


37


Student View


b) The entry length for the carbon monoxide flowing in the solid-oxide fuel cell will be determined in a similar way to part a) of this problem. Le = 0.0575 Re DH

The calculations of the mass flux and Reynolds number of carbon monoxide are shown in the following steps: kg CO  1 hr  __________   m kg hr  3600 s  G= = = ____________ 2 2 A m ⋅s   1m ( 3 mm )( _____ mm )  2   __________ mm  The hydraulic diameter from part a) is given by: D H = _____________ m

Substituting the hydraulic and the mass flux of CO in the equation for Reynolds number yields:

( _____________ m )  __________ Re =



kg   m2 ⋅ s 

kg ________________ m ⋅s Solving for the entry length, we get:

= __________

Le = 0.0575 Re D H = 0.0575 ( __________ )( ____________ m ) L e = 0.201 m

Conclusion:

c) For turbulent flow, the entry length is relatively independent from Reynolds number and estimated to be: L e = 50D H

Substituting the hydraulic diameter of the channel into this equation yields:

Le = 50 ( ______________ m ) L e = __________ m

Conclusion:


38


Student View


Example 2.11-1: Compressible Flow of a Gas in a Pipe Line A gas mixture of 12.5 mol % ethanol and 87.5 % water is leaving a boiler at a temperature of 400 °C and a pressure of 21.604 atm. The ethanol/water mixture enters a reformer in a large-scale ethanol reforming plant at a rate of kg through a commercial steel pipe with a length of 120 m and an inner diameter of 75 cm. 425.96 s Determine the pressure of the ethanol/water vapor mixture entering the reformer if the viscosity of the gas is 1.705 × 10−5 Pa ⋅ s [2].

Strategy The pressure of the gas entering the reformer can be calculated using the equation for the pressure drop for isothermal compressible flow.

Solution Equation 2.11-9 of Geankoplis can be solved for the pressure P2 at the end of the pipe to yield: P2 2 = P12 −

____________ _________  P1  − ln   DM ____  P2 

where: f = Fanning friction factor ∆L = Length of the pipe, m G = Mass flux of gas in the pipe,

R = Gas constant,

kg m2 ⋅ s

Pa ⋅ m3 mol ⋅ K

D = Inner diameter of the pipe M = Molecular weight of the gas flowing through the pipe,

kg mol

P1 = Pressure at the beginning of the pipe segment, Pa P2 = Pressure at the end of the pipe segment, Pa

2. DOE Hydrogen Program: DOE H2A Analysis Production Case Studies, http://www.hydrogen.energy.gov/ h2a_prod_studies.html. Accessed: July 2010.


39


Student View


To use this equation, the friction factor and the mass flux of the ethanol mixture must be calculated. Before determining the friction factor we need to calculate the relative roughness of the pipe and the Reynolds number. Thus, Re =

Dυρ µ

The velocity of the gas can be calculated with the following equation: υ=

V A

However, to determine the volumetric flow rate, we need to divide the mass flow rate by the density of the fluid. The density of this mixture is calculated using ideal gas law equation of state. ρ=

________ ________

Substituting the corresponding quantities into this equation, after calculating the molecular weight of the gas mixture, we get: M = x Ethanol M Ethanol + x H O M H O 2

M = 0.125

2

 mol H 2 O  mol ethanol  g g   ____   ____  + 0.875 mol mol ethanol  mol  mol H 2 O  

M = _________

g mol

g   1 kg   mol   1000 g   ρ= L ⋅ atm   1 m3   0.08206 673.15 K ( )    1000 L  mol ⋅ K    

( 21.604 atm )  21.5

ρ = __________

kg m3

Now the volumetric flow rate can be determined as follows: kg 425.96 m m3 s = = V = _________ ρ _________ kg s m3 Substituting this value into the equation for the velocity of the fluid yields: Daniel López Gaxiola 40 Jason M. Keith


Student View


m3 _________ V m s υ= = = _________ 2 A  π ( _________ m )  s   4  

The value of the Reynolds number required to determine the friction factor can now be calculated as shown below:

( 0.75 m )  __________

m  kg   _________ 3  s  m   kg    ______________  m ⋅s  

Re =

Re = _______________

The relative roughness

ε of a commercial steel pipe is given by: D

ε ___________ m = = 6 × 10−5 D ___________ m

The friction factor can be obtained from Figure 6-9 of Perry’s Chemical Engineers’ Handbook, 8th Edition to be: f = __________ The only value left to be calculated before being able to solve for the pressure of the gas entering the reformer is the mass flux, defined as: G=

m A

Substituting the mass flow rate and the cross-sectional area of the pipe into this equation, gives:

kg kg 425.96 kg s s G= = = ___________ 2 2 2 m ⋅s  π ( _________ m )  _________ m   4   425.96


41


Student View


Finally we can substitute all the corresponding values into the equation for P2 2 to get: 2

kg   Pa ⋅ m3   4 ( _________ )(120 m )  _________ 2   8.314 2 ( 673.15 ) m ⋅s   mol ⋅ K    101325 Pa    2 P2 =  21.604 atm   − g   1 kg   1 atm    ( 0.75 m )  _________  mol   1000 g   2

kg   Pa ⋅ m3   2  _________ 2   8.314 ( 673.15 )  21.604 atm  101325 Pa    m ⋅s   mol ⋅ K    1 atm   − ln  P2 g   1 kg      _________     mol   1000 g    

 ______________ Pa  P2 2 = ______________ Pa 2 − 4.178 × 1011 Pa 2 − ______________ Pa 2 ln   P2  

This equation can be solved using computer software or trial and error to yield:

P2 ≈ ________________ Pa


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Student View


Example 2.11-2: Maximum Flow for Compressible Flow of a Gas Determine the maximum velocity that can be obtained for the ethanol/water mixture from Example 2.11-1 and compare to the actual velocity of the fluid fed to the ethanol reformer.

Strategy The maximum velocity of the fluid is obtained using the definition of the velocity of sound in an isothermal fluid.

Solution The maximum velocity of the ethanol/water mixture can be determined using Equation 2.11-12 of Geankoplis, as shown below:

υmax =

RT M

Substituting the ideal gas constant, as well as the temperature of the gas in the pipe lines and the molecular weight into this equation yields:

υmax

Pa ⋅ m3   _________ ( 673.15 K )  mol ⋅ K   = g   1 kg    _________  mol   1000 g  

υmax = ___________

m s

The velocity of this gas in the process at the entrance to the ethanol reformer is given by Equation 2.11-13 of Geankoplis:

υ2 =

RTG P2 M

Substituting the pressure found in Example 2.11-1 and the mass flux of ethanol vapor, we find that the velocity is given by: Pa ⋅ m3  kg  _________ ( 673.15 K )  ___________ 2    mol ⋅ K  m ⋅s   υ2 =  g   1 kg  ( _______________ Pa )  _________  mol   1000 g  

υ2 = ___________ Daniel López Gaxiola Jason M. Keith

m s

43


Student View


Chapter 3 Principles of Momentum Transfer and Applications Chapter 3 introduces students to the principles for calculation of power and efficiency of equipment such as compressors, pumps and fans. The following problem modules illustrate the application of this type of equipment to processes for producing hydrogen for fuel cells as well as the derivation of equations for different flow conditions from the general transport equations. 3.1-3 Surface Area in Packed Bed of Cylinders 3.1-4 Pressure Drop and Flow of Gases in Packed Bed 3.2-1 Flow Measurement using a Pitot Tube 3.3-1 NPSH Available for Pump 3.3-2 Calculation of Brake Horsepower of a Pump 3.3-3 Brake-kW Power of a Centrifugal Fan 3.3-4 Compression of Methane 3.8-3 Laminar Flow in a Circular Tube 3.11-1 Dimensionless Groups


1


Student View

Principles of Momentum Transfer and Applications

Example 3.1-3: Surface Area in Packed Bed of Cylinders Natural gas has been proposed as a source of hydrogen for fuel cell vehicle applications because of the existing infrastructure. In a process known as steam reforming, natural gas and steam are reacted into mostly carbon monoxide and hydrogen with some carbon dioxide also produced. There is also excess water in the reformate stream. A water gas shift reactor can be used to convert some of the remaining carbon monoxide into hydrogen according to the reaction: CO + H2O

H2 + CO2

The following figure shows an axisymmetric view of an annular water gas shift reactor which is 8 cm high. In the outer (annular) region, an iron chromium oxide catalyst is present to carry out the water gas shift reaction. A 20 µm thick palladium membrane separates the reaction (outer) zone from the separation (inner) zone.

Gas Flows In

Separation Zone

Reaction Zones Annular Membrane Reactor : Top View (Left) and Side View (Right)

Cylinders of iron chromium oxide catalyst with diameter and length of 0.1 cm are forming a packed lb bed in the reaction zones with a bulk density of 39.75 3 . Determine the void fraction ε, the ft effective diameter of the particles Dp and the hydraulic radius rH for the flow through the packed bed lb if the density of the catalyst is 76 3 . ft Strategy The equations defining the parameters ε, Dp and rH can be used for solving this problem.


2


Student View


Solution First we will calculate the void fraction ε of the packed bed, defined by the following equation: ε=

volume of voids in bed total volume of bed

For simplicity, we will select a basis of 1 ft3 of packed bed. Thus, we can calculate the mass of the bed to be:

lb   m bed = 1 ft 3  _________ 3  = ___________ lb ft  

This mass of packed bed can be used to calculate the volume of the solid cylinders of catalyst, as shown in the following calculation:

Vcatalyst =

_________ lb = _____________ ft 3 lb 76 3 ft

Substituting this volume and the basis of 1 ft3 of packed bed into the equation for ε yields: ε=

1 ft 3 − ___________ ft 3 ______ ft 3

ε = _________ For non-spherical particles, the effective diameter is given by the following equation:

Dp =

6 aV

where a V is the specific area, given by the ratio of the surface area of the catalyst particle to the volume of the particle. Thus, a V is given by:

 π _______  + π ___________ 2 Sp _____   aV = = πD 2 Vp ______ 4 In this equation, the first term in the numerator represents the area of the ends of the cylinder, while the second term is accounting for the area of the walls of the cylinder. We can substitute the values of D and L into this equation to determine the specific area a V as follows:


3


Student View


 π __________  2 + π ( ________ ) 2  Sp ______ 1  _________ cm   aV = =  = 60 2   Vp cm  1 ft  π ( 0.1 cm ) ( __________ ) 4

a V = 1828.8 ft −1 Now we can calculate the effective diameter to be: Dp =

6 1828.8 ft −1

D p = ____________ ft

The hydraulic radius of an object is defined as: rH =

ε a

where a is the ratio of the wetted surface of the particles to the volume of the packed bed. The following equation can be used for calculation of a:

a = a V (1 − ε ) The values of the void fraction and the specific area of the catalyst particles can be entered into this equation to give:

a = 1828.8 ft −1 (1 − __________ ) a = ___________ ft −1

Therefore, the hydraulic radius can now be calculated to yield: rH =

__________ __________ ft −1

rH = 4.98 ×10 −4 ft


4


Student View


Example 3.1-4: Pressure Drop and Flow of Gases in Packed Bed A water-gas shift reactor in a distributed-scale hydrogen plant is producing hydrogen at a rate of lb 65.2 . The reactor consists of a tubular packed bed of 5.25 cm diameter with 31.4 kg of a catalyst h lb with a density of 76 3 . ft The void fraction of the bed is 0.57 and the spherical catalyst pellets have a diameter of 0.1 cm. Determine the pressure drop of the reacting synthesis gas in the packed bed. The synthesis gas is entering the reactor at a pressure of 2066 kPa and has the following properties:

µ = 0.048

lb ft ⋅ h

ρ0 = 0.356

lb ft 3

The differential form of the pressure drop in a packed bed reactor is given by the Ergun equation:

 dP G  1 − φ   150(1 − φ)µ =− + 1.75G      dz ρg c D p  φ   Dp  3

The solution to this differential equation is given by: P = (1 − αW)1/2 Po where: α=

2βo A c (1 − φ)ρc Po

βo =

G(1 − φ) ρo g c D p φ3

150(1 − φ)µ  + 1.75G   Dp  

The first term in the brackets in the equation for βo is dominant for laminar flow and the second term is dominant for turbulent flow. In these equations, the following notation and units are used:  lb  P = Pressure  2   ft 

φ = Void fraction (dimensionless) Daniel López Gaxiola Jason M. Keith

5


Student View


gc = Gravitational constant = 4.17 × 108

lb ⋅ ft h 2 ⋅ lb f

Dp = Diameter of particle in the bed (ft)  lb  µ = Viscosity of gas    ft ⋅ h   lb  ρ0 = Gas density  3   ft   lb  ρc = Density of the catalyst  3   ft 

G=

m  lb  = Mass flux of synthesis gas  2  A  ft ⋅ h 

lb = Mass flow rate of synthesis gas   m h

 lb  υ = Velocity of the gas in the reactor   h

z = Distance down packed bed (ft) A = Cross – sectional area of the reactor (ft2) W = Mass of catalyst in the reactor (lb)

Strategy The Ergun equation can be used for calculating the pressure at the outlet of the packed bed.

Solution First we determine the value of βo. All of the terms in the problem statement are in the appropriate units except for the particle diameter. We have:

 1 ft  D p = 0.1cm   = ______________ ft  30.48cm  Another value we need to calculate to calculate βo is the mass flux of synthesis gas G, given by: m A Daniel López Gaxiola Jason M. Keith

G=

6


Student View


Where the cross sectional area is obtained as follows: 2

π ( ________ cm )   1 ft 2 = ___________ ft 2 A=  2  4 ____________ cm   Entering this value and the mass flow rate into the equation for the mass flux G we get:

lb lb h G= = ______________ 2 2 ____________ ft ft ⋅ h 65.2

Substituting this and the other values into the equation for βo :

βo =

G(1 − φ) ρo g c D p φ3

150(1 − φ)µ  + 1.75G   Dp  

lb (1 − 0.57) 2 ft ⋅ h β = 3 lb  lb ⋅ ft  0.356 3  _______________ 2  ( ____________ ft )( 0.57 ) ft  h ⋅ lb f  lb    +1.75  _______________ 2   ft ⋅ h    _______________

o

βo = 77.91

 lb   150(1 − 0.57)  0.048 ft ⋅ h      ______________ ft 

lb f ft 3

 lb  Note that this term has units of pressure  2f  per unit length (ft). It is also noted that the second  ft  term in the brackets is dominant, suggesting turbulent flow in this industrial reactor. Now we need to determine the value of αW which is needed in the formula for the pressure drop. We first need to obtain the feed pressure, and catalyst weight in the appropriate units. The feed pressure in

lb f ft 2

is:

lb f    _________ in 2   ___________  lb f Po = 2066 kPa    = _____________ 2 ft  ___________ kPa   ___________   


7


Student View


The catalyst weight W in lbm is:

 1 lb  W = 31.4 kg   = ___________ lb  0.454 kg  Thus,

lb   2  77.91 3f  ( _________ lb ) ft  2βo W  αW = = = ________ lb f  A c (1 − φ)ρc Po  lb   2 _____________ ft (1 − 0.57)  76 3  _______________ 2  ft   ft  

The exit pressure can be determined by entering this value into the equation for the pressure ratio P = (1 − αW)1/2 Po Substituting the values of αW and the pressure at the entrance of the packed bed P0 yields: P lb f    ____________ ft 2   

= (________________)1/2

Solving for the pressure at the outlet P, we get: lb  P = (________________)1/2  _____________ 2f ft 

lb f   = 35373 2 ft 

Hence, the pressure drop is given by: lb lb   ___________   ∆P = P − P0 =  35373 2f − ___________ 2f    ft ft   ___________  

∆P = ___________ psi


8


Student View

P : Po


Example 3.2-1: Flow Measurement using a Pitot Tube A Pitot Tube is used for measuring the flow in a circular pipe. To determine the flow of hydrogen at room temperature to a proton-exchange membrane fuel cell, a Pitot tube with a coefficient of 0.84 is used in a pipe with a diameter of 1”. The static-pressure of hydrogen is measured to be 12.34 mm of Hg above atmospheric pressure. Determine the maximum and average velocities and flow rates of hydrogen in the pipe if the reading on the manometer is 0.022 in of Hg. The following figure shows the diagram of Pitot Tube for this problem.

1

2

H2 to Fuel Cell

∆h = 0.022 in Hg

Strategy To solve this problem we need to obtain the velocity value from Bernoulli Equation applied to a Pitot Tube.

Solution The Bernoulli equation applied to a Pitot Tube is given by: υ1 = C p

2 ( ________________ )

___________

where Cp is the value of the Pitot tube coefficient. The viscosity of hydrogen can be obtained from Appendix A.3 of Geankoplis to be. µ = _________________

kg m ⋅s

The density can be calculated using Ideal Gas equation of state as shown in the following steps: kg   atm )  _________  ________ mol   = = RT  −5 _____ ⋅ _____   8.206 × 10 ______ ⋅ _____  ( 298.15 K )  

( _____

ρH

2 @ P = 1 atm


9


Student View


ρH

2

@ P = 1 atm

= _________

kg m3

First we need to convert the manometric static pressure to absolute pressure. To do this, we will use kg the density of mercury as 13533.61 3 , obtained from Table 2-31 of Perry’s Chemical Engineers’ m th Handbook, 8 Edition. Thus,

(

∆Pstatic = ρHg − ρH

2

@ P = 1 atm

) gh

kg kg   m  ∆Pstatic = 13533.61 3 − _________ 3   ___________ 2  ( ____________ m Hg ) m m  s   ∆Pstatic = 1638 Pa

We can add this value to the atmospheric pressure of 101325 Pa to get the absolute pressure to be ______________ Pa. Since the density of hydrogen was calculated at atmospheric pressure, we need to correct the density value for the actual pressure of hydrogen in the pipe. This can be done by multiplying the density at 1 atm of pressure by the pressure ratio. Therefore,

ρH = ρH 2

2

@ P = 1 atm

 P   Patm

ρH = __________ 2

 kg  __________ Pa   = _________ 3   m  101325 Pa  

kg m3

Now we can calculate the pressure difference using the change in the height of mercury in the manometer as shown below:

(

)

∆P = ρHg − ρH g∆h 2

 0.0254 m Hg  kg kg   m  ∆P = 13533.61 3 − __________ 3   ____________ 2  ( __________ in Hg )   m m  s    1 in Hg  ∆P = __________ Pa

Substituting this pressure drop into the equation for the velocity of hydrogen in the tube yields:

υ1 = 0.84

2 ( __________ Pa ) kg __________ 3 m


10


Student View


υ1 = __________

m s

Since the point 1 in the tube is at the center of the tube, where the velocity reaches its maximum value, the velocity we calculated is the maximum velocity. υmax = __________

m s

The average velocity of a fluid in a pipe can be estimated using Figure 2.10-2 of Geankoplis as a function of Reynolds number. The value of Reynolds number using the maximum velocity for this problem is given by:

Re =

Dυmax ρ H

(1 in )  2

µ

0.0254 m  m  kg   ___________   ___________ 3  s  m   1 in  kg _______________ m⋅s

=

Re = 8507

Locating this value in Figure 2.10-2 we can estimate the ratio of the average velocity to the maximum velocity to be: υav ≈ __________ υmax

Solving for the average velocity and entering the value of the maximum velocity into this equation we get: m  υav ≈ ___________ ( υmax ) ≈ ___________  ___________  s  

υav ≈ 27

m s

To calculate the average and maximum flow rates of hydrogen we need to multiply the corresponding velocity value by the cross-sectional area of the pipe. The area of the pipe is calculated as shown below: 2

πD 2 π (1 in )  _______________ m 2  2 = A=   = _______________ m 2 4 4 1 in   Multiplying the cross-sectional area by both velocity values we get:


11


Student View


= _______________ m 2  27 m  V av   s  

= _______________ m V av s

3

= _______________ m 2  ________ m  V max   s  

= _______________ m V max s


3

12


Student View


Example 3.3-1: NPSH Available for Pump Water at a temperature of 25°C is being pumped to a boiler in a distributed-scale steam-methane reforming plant. The water is entering the boiler through a commercial steel pipe with a diameter of 0.1 m and a length of 25 m. The estimated friction factor for the flow conditions in this process is estimated to be 0.0045. Determine the available net positive suction head (NPSH) of the pump if the velocity of the water in m the pipe is of 2.3 s

Strategy This problem can be solved using the equation for the NPSH as a function of the pressure and thermodynamic properties of the fluid in the pipe.

Solution The following equation shows the relation between the NPSH available and the conditions in the system:

g ( NPSH ) A =

P1 − Pvp

ρ

+ gz1 −

υ2 −∑F 2

where: g = Acceleration due to gravitational force = 9.80665

m s2

P1 = Pressure of the fluid before entering the pump (Pa) Pvp = Saturation pressure of fluid at the process temperature (Pa)

 kg  ρ = Density of fluid  3  m  z1 = Difference in height between the pump and the point at pressure P1 (m)

m υ = Velocity of the fluid   s   J 

∑ F = Friction loss in suction line to pump  kg  



Since there is no difference in the height between the pipe and the pump, the term gz1 in the equation for g(NSPH)A is neglected in this problem. Thus, Daniel López Gaxiola Jason M. Keith

13


Student View


g ( NPSH ) A =

P1 − Pvp

ρ

−

υ2 −∑F 2

The density and saturation pressure of water at 25 °C can be found in Geankoplis in Tables A.2-3 and A.2-9 respectively: ρ = _____________

kg m3

Pvp = __________ Pa

The only remaining unknown value in the equation for NSPHA is the friction loss, which can be calculated using the following equation: F = 4f

∆L υ2 D 2

Substituting the corresponding values into this equation, we get: 2

m  2.3   25 m J s ∑ F = 4 ( ___________ ) 0.1 m  2  = ________ kg

Now we can enter all the known quantities into the equation for NSPH to yield: 2

m   9.80665 2  ( NPSH )A s  

m  2.3   (101325 Pa ) − ( __________ Pa ) −  s  − 11.9 J = kg 2 kg __________ 3 m

Solving for the NPSHA:

( NPSH )A

2   m   101325 Pa − __________ Pa  2.3   ( ) ( ) −  s  − 11.9 J  1 = kg m  2 kg   __________ 3 9.80665 2     m s   

( NPSH )A = _________


m

14


Student View


Example 3.3-2: Calculation of Brake Horsepower of a Pump Determine the brake horsepower of the pump from Example 3.3-1 operating at a flow rate of kg for feeding water to a boiler in a Steam-Methane reforming plant. Assume the 683.1 hr characteristic curves of this pump are described by Figure 3.3-3 of Geankoplis.

Strategy The brake horsepower of the pump described in this example can be calculated using the mass flow rate of fluid and the work performed by the pump.

Solution The brake hp of a pump can be calculated using Equation 3.3-2 of Geankoplis:

brake hp =

− WS m

550η

where:

 ft ⋅ lb f  WS = Work performed by the pump    lb  lb = Mass flow rate of fluid   m s η = Efficiency of the pump

First we need to calculate the work performed by the pump, defined as follows:

WS = − H

g gC

In this equation, the value of the head of fluid H is unknown. However, it can be read from the characteristic curves of the pump, shown in Figure 3.3-3. The volumetric flow rate of water required for using Figure 3.3-3 is calculated by multiplying the mass flow rate by the density of water at the process conditions as shown below: kg  ________  hr  ________  m V= = ρ kg  ________________ m3  997.08 3   m  ft 3  683.1


15


Student View


ft 3  ___________ gal  gal V = ___________   = ___________ 3 min  1 ft min 

With this volumetric flow rate, we can estimate the head of fluid and the efficiency of the pump from Figure 3.3-3 to be: H ≈ ___________ ft η ≈ ___________

We can substitute the head of fluid value into the equation for the work WS to yield:

 ft  _____________ 2 hr WS = − ________ ft   _____________ lb ⋅ ft  hr 2 ⋅ lb f  WS = ___________

     

lb f ⋅ ft

lb

The negative value of the work WS indicates that the fluid is performing work over the pump. Now we can enter the values we calculated into the equation for the brake horsepower of the pump to get: lb ⋅ ft    kg   _________  _________  −  ___________ f  683.1     lb   hr   _________   _________  brake hp =  550 ( _________ )

brake hp = ________ hp


16


Student View


Example 3.3-3: Brake-kW Power of a Centrifugal Fan m3 m (measured at 1 atm and 298.15 K) and a velocity of 67.1 enters a min s proton-exchange membrane fuel cell stack through a centrifugal fan. The air is entering the fan at a pressure of 1.009 bar and a temperature of 100°F. What is the discharge pressure of the air if the fan has a brake power of 2.12 kW and an efficiency of 70%?

Air at a flow rate of 2.90

Strategy This problem can be solved by performing a mechanical energy balance on the system, and using the definition of brake power of a centrifugal fan.

Solution The mechanical-energy balance for this problem ( z1 = z 2 , υin = 0 ,

P − P (υ ) WS = 1 2 − out ρ 2

∑ F = 0 ) is given by:

2

 J  WS = Work performed by the centrifugal fan    kg  P1 = Pressure of air at the suction point (Pa) P2 = Pressure of air at the discharge point (Pa)

 kg  ρ = Average density of the air  3  m 

m υout = Velocity of air at the discharge point   s  Before being able to calculate the discharge pressure from this equation, we need to calculate the velocity of air in the outlet, the average density and the work performed by the fan on the fluid. The density of the air entering the fan is calculated using the ideal gas law and the properties of an ideal gas at standard conditions. In the following calculation, the sub-index 1 represents the conditions of the air at the inlet of the fan.

M  T   P1 ρ1 = air  std   ˆ T V   Pstd 1 std


kg air  ___________ kmol  ___________ K    1.009 bar = 3      311.15 K   ___________ bar   ___________ m  kmol 17


Student View


ρ1 = ___________

kg m3

The density of the air leaving through the fan can be determined using the pressure ratio between the suction and discharge points. After substituting the known pressure and density we get: P  P2   kg  1 bar ρ2 = ρ1  2  = ___________ 3    m  1.009 bar   ___________ Pa   P1 

ρ2 = ( ________________ ) P2 Now we can obtain the average density of the air as a function of the discharge pressure P2:

ρ=

ρ1 + ρ2 2

_________

=

kg + ( ________________ ) P2 m3 2

 kg  ρ  3  = _______ + ( ____________ ) P2 m  Substituting this expression into the mechanical-energy balance equation: WS =

P1 − P2 _______ + ( ____________ ) P2

(υ ) − out

2

2

Now we need to calculate the left hand side of this equation. Equation 3.3-2 of Geankoplis defines the brake power of a centrifugal fan as follows: brake kW =

− WS m 1000η

where: kg = Mass flow rate of fluid   m  s  η = Efficiency of the fan

We can solve for the work WS performed by the fan:  ________  WS = _____________    ________ 


18


Student View


can be obtained from the properties of an ideal gas at standard The value of the mass flow rate m conditions. Thus,

= m

________  Tstd  V   ˆ V  ________ 

The sub-index 0 of the temperature indicates the temperature at which the air flow rate was measured. Substituting the corresponding numeric values into this equation yields: m3  1 min   kg  ________    _________  min  60 s   kmol   273.15 K  m=   m3  __________ K     __________ kmol    kg = _____________ m s

Entering this value and the brake power of the fan into the equation for the work WS:

    1000(0.70) J WS = −2.12 kW  = − _____________  kg  _____________ kg  s   Now, after substituting the calculated values, the energy balance equation is given by:  1× 105 Pa  m  1.009 bar  − P2 67.1    J s   1 bar  − _____________ = − kg _______ + ( ____________ ) P2 2 _______________

2

1.009 × 105 Pa − P2 J = kg _______ + ( ____________ ) P2

This equation can be solved for P2 as shown in the following steps: ____________

J  _______ + ( ____________ ) P2  = 1.009 × 105 Pa − P2 kg

_______________ + ( ____________ ) P2 = 1.009 ×105 Pa − P2 ____________ P2 + P2 = (1.009 ×105 ) + 13386.4 P2 =

____________ 0.8673

 __________  P2 = ________________ Pa   = 1.318 bar  __________  Daniel López Gaxiola Jason M. Keith

19


Student View


Example 3.3-4: Compression of Methane A compressor in a steam-methane reforming process for hydrogen production is compressing natural gas at room temperature from a pressure of 1 atm to 21.8 atm before entering the reforming reactor. What percent of power is saved when operating the compressor at isothermal conditions if compared to adiabatic compression?

Strategy The equation for calculating the brake power of a compressor allows us to compare both adiabatic and isothermal compression processes.

Solution The equation for calculating the power required by the compressor is shown below:

brake kW =

− WS m η

and the The value of the work WS in this equation is in kJ. The values of the mass flow rate m efficiency η are not given in the problem statement. However, for calculating the amount of power and η are not required, as it will be shown in the following steps. We need saved, the values of m to calculate the work performed by the compressor at both adiabatic and isothermal conditions. For adiabatic compression, the work WS is given by:

− WS,adiabatic

γ−1   γ γ RT1  P2   =   − 1  γ − 1 M  P1   

The parameter γ represents the heat capacity ratio. For methane, a value of γ = 1.31 is given in Geankoplis. Hence, the work for adiabatic compression is calculated as follows:

− WS,adiabatic

kJ   _____________  _____________   _____________ kmol ⋅ K ( 298.15 K )   21.8 atm  _____________  = − 1     kg 1 atm      _____________    ______  kmol  

− WS,adiabatic = 702.9

kJ kg


20


Student View


For isothermal operation, the work WS is calculated using the following equation: − WS, isothermal =

P _______________ log 2 M P1

Substituting the pressure ratio into this equation and the temperature of methane before entering the compressor yields: − WS, isothermal = _____________

kJ kg

The percent of power saved can be calculated dividing the difference in power between adiabatic and isothermal compression by the power required for adiabatic compression. Thus, Power saved (%) =

( brake kW )adiabatic − ( brake kW )isothermal ×100 ( brake kW )adiabatic

Substituting the equations for the power required by the compressor and the work values in this equation, we get:

 m −W − − WS,isothermal   η  S,adiabatic

(

Power saved (%) =

)

m − WS,adiabatic η

(

× 100

)

kJ kJ    _____________ kg − _____________ kg   × 100 Power saved (%) =  kJ   _____________   kg   Power saved = _____________


21


Student View


Example 3.8-3: Laminar Flow in a Circular Tube An aqueous solution of 40 % methanol is flowing from the fuel reservoir to a stack of directmethanol fuel cells in a mobile phone through a pipe with an inner diameter of 3 mm and a length of 2 cm. Use the equation derived in Geankoplis for laminar flow in a circular tube to determine the pressure drop along the pipe if the maximum Reynolds number for methanol in the pipe is 1850.

Strategy We can use the Hagen-Poiseuille equation to calculate the pressure drop between the methanol reservoir and the fuel cell stack.

Solution The Hagen-Poiseuille equation is derived in Geankoplis for laminar flow in a circular tube, shown below:

P1 − P2 =

32µυav L D2

The viscosity of the methanol solution at 25 ºC can be obtained from Figure A.3-4 to be: µ = _________________

kg m ⋅s

The only parameter we need to calculate before being able to determine the pressure drop is the average velocity. We can obtain the ratio of the average velocity to the maximum velocity from Figure 2.10-2 of Geankoplis at the Reynolds number of 1850. From this figure, it can be seen that for laminar flow the ratio of the velocities remains constant and equal to __________. Thus, υav = ________ υmax

From the definition of Reynolds number we can solve for the maximum velocity to yield: υmax =

_____________ __________

The density of an aqueous solution of 40 % methanol can be obtained from Table 2-109 of Perry’s Chemical Engineers’ Handbook, 7th Edition to be: ρ@20ºC = 934.5


kg m3

22


Student View


Substituting the corresponding quantities into the definition of Reynolds number, we get:

υmax

kg   1850  _______________  m m⋅s   = = _____________ s ( 3 ×10−3 m )  _____________ mkg3   

With this value, we can determine the average velocity from the data read from Figure 2.10-2: m m  υav = _____________  _____________  = 0.6 s  s 

Finally, after entering this velocity and the rest of the values into the Hagen-Poiseuille equation we can determine the pressure drop as shown in the following steps: kg  m  32  _____________  0.6  ( _____________ m ) m ⋅ s  s  P1 − P2 =  2 ( 3 ×10−3 m )

P1 − P2 = ______ Pa


23


Student View


Example 3.11-1: Dimensionless Groups Using dimensionless numbers determine which of the following forces is the most dominant for the flow conditions in a polymer-electrolyte membrane fuel cell and direct-methanol fuel cell: -Inertia Force -Gravity Force -Pressure Force -Viscous Force

The flow conditions in both fuel cells are given in the following set of data. The direct-methanol fuel cell is using an aqueous solution of 40 % methanol as fuel.

Polymer-Electrolyte Membrane Fuel Cell

Direct Methanol Fuel Cell

Pressure: 2.5 atm

Pressure: 1 atm

Temperature: 80 ºC

Temperature: 25 ºC

Velocity of hydrogen: 15

m s

Velocity of methanol: 0.49

Hydraulic diameter

Density: 931.5

of the channel: 1×10 −3 m

m s

kg m3

Hydraulic diameter of

the

channel:

8.57 × 10 −4 m

Strategy The Froude, Euler and Reynolds numbers establish a relation between the forces given in the problem statement and hence can be used for determining the most dominant force for a specified flow conditions.

Solution The definitions of the dimensionless groups that will be used for solving this problem are given in the following equations.


24


Student View


inertia force υ2 N Fr = = gravity force gL N Eu =

pressure force ______ = inertia force ______

N Re =

inertia force Lυρ = viscous force µ

In these equations the parameter L represents the characteristic length. For this problem, the characteristic length will be given by the hydraulic diameter of the channel DH. Thus, inertia force υ2 N Fr = = gravity force gD H N Eu =

pressure force ______ = inertia force ______

N Re =

inertia force _________ = viscous force _________

For comparison of the forces involved in these dimensionless numbers, we will use the reciprocal of the Euler number. By doing this, we can compare the three numbers using the inertia force as reference. Therefore,

1 inertia force _______ = = N Eu pressure force _____ The viscosity of the methanol solution at room temperature is obtained from Appendix A.3 to be: µ = _______________

kg m ⋅s

For hydrogen at a temperature of 80 ºC, the density and viscosity obtained from Table 2-223 of Perry’s Chemical Engineers’ Handbook, 8th Edition and are shown below: µ = 9.9982 × 10−6 ρ = 0.178

kg m⋅s

kg m3

Now we can proceed to calculate the dimensionless numbers for the flow conditions in both types of fuel cell. For the polymer-electrolyte membrane, we can substitute the given parameters into the equations for Froude, Euler and Reynolds as shown in the following steps: Daniel López Gaxiola Jason M. Keith

25


Student View

Principles of Momentum Transfer and Applications 2

m  _______   2 υ s   = N Fr = m gD H   9.80665 2  ( ___________ m ) s  

N Fr = ___________ kg  m ________ 3  ________  1 m  s  = N Eu  101325 Pa  2.5 atm    1 atm 

2

1 = ________________ N Eu

( ___________ m )  ____ N Re =

m  kg   0.178 3  s  m   kg _______________ m ⋅s

N Re = ____________

Conclusion:

Repeating a similar procedure for the direct-methanol fuel cell, we find the following values for the dimensionless numbers: 2

m  ________   2 υ s   = N Fr = m gD H   ________ 2  ( ______________ m ) s  

N Fr = ________


26


Student View


1 = N Eu

kg  m ________  3  m  s  _____________

2

________

1 = ________________ N Eu

( ______________ m )  ________ N Re =

m  kg   ________ 3  s  m   kg ______________ m ⋅s

N Re = ________

Conclusion:


27


Student View


Chapter 4 Principles of Steady - State Heat Transfer Heat transfer is occurring in many chemical and separation processes as a consequence of a temperature difference. In Chapter 4, the following problem modules explain the heat transfer processes involved in fuel cell vehicles and in the processes for producing fuel for fuel cells. 4.1-1 Heat Loss through a Stainless Steel Bipolar Plate 4.3-1 Cooling of a Fuel Cell 4.3-2 Heat Loss from an Insulated Pipe 4.3-3 Heat Loss by Convection and Conduction and Overall U 4.3-4 Heat Generation in a Solid-Oxide Fuel Cell 4.5-1 Heating of Natural Gas in Steam-Methane Reforming Process 4.5-2 Trial-and-Error Solution for Heating of Steam 4.5-3 Heating of Ethanol in Reforming Process 4.5-4 Heat-Transfer Area and Log Mean Temperature Difference 4.5-5 Laminar Heat Transfer and Trial and Error 4.6-3 Heating of Steam by a Bank of Tubes in High-Temperature Electrolysis 4.7-3 Natural Convection in Bipolar Plate Vertical Channel 4.8-2 Steam Condensation in a Fuel Cell 4.9-1 Temperature Correction Factor for a Heat Exchanger 4.9-2 Effectiveness of Heat Exchanger 4.11-1 Radiation in Cylindrical Solid-Oxide Fuel Cell 4.15-1 Cooling Channels in Fuel Cell Bipolar Plates


1


Student View

Principles of Steady-State Heat Transfer

Example 4.1-1: Heat Loss through a Stainless Steel Bipolar Plate Calculate the heat flux through a stainless steel bipolar plate in a polymer-electrolyte membrane fuel cell with a thickness of 4.5 mm. The fuel cell is operating at a temperature of 80 ºC during the summer season in Houghton, Michigan where the temperature is 70 ºF. Strategy The equation for the heat flux obtained from Fourier’s Law can be used to obtain the solution to this problem. Solution Equation 4.1-10 of Geankoplis is defining the heat transfer per unit area as follows:

q k = (T − T ) A x 2 − x1 1 2 We can substitute the values given in the problem statement into this equation, but first we need to convert the temperature outside the fuel cell to ºC: T ( º C) =

T ( º F ) − 32 1.8

=

70º F − 32 1.8

T = _________º C

Entering the temperatures inside and outside the fuel cell stack into the heat transfer equation, as well as the thickness of the bipolar plate represented by x 2 − x1 , we get:

W ________ q m ⋅ K ( ________ K − 294.25 K ) = A __________ m q W = _____________ 2 A m

The thermal conductivity of steel was obtained from Table 4.1-1 of Geankoplis.


2


Student View


Example 4.3-1: Cooling of a Fuel Cell Air at a temperature of 25 °C is being used for cooling a single cell fuel cell. The convective heat W transfer coefficient of the air is 61.2 2 and is capable of removing heat at a rate of 183.6 W. m ⋅ °C What would be the dimensions of the square surface of the fuel cell if its temperature must not exceed 50°C?

Air @ 25 °C

Air @ 25 °C

Strategy The heat transfer rate by convection can be obtained using Newton's Law of Cooling.

Solution The heat flux q when heat is being transferred by forced convection is defined as follows:

q = hA ( TS − T∞ ) where: q = heat transfer rate,

W m2

h = convective heat transfer coefficient,

W m2 ⋅ K

TS = temperature on the surface of the object, °C T∞ = temperature of the air, °C

A = surface area of the fuel cell, m2 Daniel López Gaxiola Jason M. Keith

3


Student View


To determine the dimensions of the surface of the fuel cell, we can substitute the given temperatures and heat transfer rate and solve for the area A to yield:

A=

____ ________ W = ________________ _________ W 50°C − 25°C ( ) m 2 ⋅ °C

A = 0.12 m 2

Since the heat is being removed from the fuel cell through both the left and right faces of the fuel cell, this value of A must be divided by 2. Thus,

A fuel cell =

0.12 m 2  ___________ cm 2    2 1 m2  

A fuel cell = ________ cm 2

The dimensions of a fuel cell with a square surface could be obtained as follows:

L = A fuel cell = ________ cm 2 L = ________ cm

Therefore, for a heat transfer rate of 183.6 W, air at 25 °C can be used to keep the surface area of a ________ cm x ________ cm fuel cell at a temperature of ________ °C.


4


Student View


Example 4.3-2: Heat Loss in Fuel Reforming Applications A pipe made of 308 stainless steel (schedule number 80) with a nominal diameter of 1.5" is carrying methane at a temperature of 400°C in a steam-methane reforming process for producing hydrogen. The pipe is insulated with a layer of glass-fiber with a thickness of 1". Determine the temperature at the interface between the pipe and the glass fiber and the heat loss through the insulated pipe with a length of 15 m. The surface of the insulating material is at a temperature of 25°C. A schematic of the pipe is shown below:

T3

T1

Natural Gas @ 400°C

T2

Strategy The equation for the heat loss through a pipe can be applied to the different layers in the pipe.

Solution The heat loss through the walls of a cylinder is given by: q=

Tin − Tout R

where: Tin = temperature at the inner wall of the pipe, K Tout = temperature at the outer wall of the pipe, K R = resistance of the pipe to the heat transfer through its walls,


5


K W

Student View


In this problem, we need to apply this equation for both the steel pipe and the insulated pipe. The overall heat loss will be obtained using this equation for the insulated steel pipe. The resistance to heat transfer in cylindrical coordinates is calculated with the following equation:

R=

rout − rin kA lm

in this equation: rout = outer radius of the cylinder, m rin = inner radius of the cylinder, m k = thermal conductivity of the material,

W m⋅K

Alm = log mean area, m2 The log mean area of the pipe is defined as: A lm =

A out − A in  ______  ln    ______ 

where Aout and Ain are the outer and inner surface areas of the cylinder, respectively. Applying the equations for resistance and the log mean areas to the steel and the overall pipe we have:

Steel Pipe

Overall

q 1→ 2 =

T1 − T2 R1→ 2

R 1→2 =

r2 − r1 k steel A lm,1→2

q 1→3 =

T1 − T3 R 1→2 + R 2→3

A lm,1→3 = A lm,1→ 2 =

A 2 − A1 A  ln  2   A1 


6


A 3 − A1 A  ln  3   A1 

Student View


To determine the log mean areas we need to look for the radius of the steel pipe in Appendix A.5 of Geankoplis. For the 1.5" pipe: rin = r1 = __________ m rout = r2 = __________m The radius of the pipe including the insulation is obtained by adding the thickness of 1" to the outer radius of the steel pipe. Hence, r3 = __________m + 0.0254 m = __________m With these diameter values and the length of the pipe, the areas A1, A2 and A3 can be calculated as shown in the following steps. A1 = 2πr1L = 2π(0.01905 m)( __________ m) = 1.795 m2 A2 = 2πr2L = 2π(__________m)( __________ m) = __________m2 A3 = 2πr3L = 2π(__________m)( __________ m) = __________m2 The thermal conductivities for the glass fiber and the steel can be found in Appendix A.3 of Geankoplis and are shown below. The conductivity of the glass fiber was selected at the highest temperature available in Table A.3-15. The thermal conductivity of steel was obtained from Table A.3-16. k steel = ____________________ k glass fiber = __________

W m ⋅ °C

W m ⋅ °C

Substituting the values we obtained into the equations for the individual layers yields:

Steel Pipe q 1→ 2 =

R 1→ 2 =

T1 − T2 __________ °C − T2 = °C R1→2 _____________ W °C ____________ m − ____________ m = 1.17 × 10−4 W W 21.6 ____________ m 2 ) ( m ⋅°C

A lm,1→ 2 =

___________ m 2 − ___________ m 2 = 2.025 m 2 2  2.274 m  ln  2   1.795 m 


7


Student View


Overall q 1→3 =

R 2 →3 =

___________ °C − ___________ °C = 2695.6 W °C °C ___________ + 0.139 W W °C ___________ m − 0.02415 m = 0.139 W W ___________ ( ___________ m2 ) m ⋅°C

A lm,2→3 =

___________ m 2 − ___________ m 2 = ___________ m 2 2  4.670 m  ln  2   2.274 m 

Hence, the heat loss through the insulated pipe is 2695.6 W. Since this answer represents the amount of heat lost per unit time, if we assume that the system is at steady state. The heat loss per unit time will be the same in the individual layers. Thus, we can use the equation for the heat loss through the steel pipe to determine the temperature at the steel-glass fiber interface. q 1→ 2 =

______ °C − T2 = 2695.6 W °C _____________ W

Solving for the temperature T2, we get: °C   T2 = ___________ °C − 2695.6 W  _______________  W  T2 = ___________ °C

As it can be seen, the temperature at the pipe - insulator interface is almost the same as the temperature of the inner wall of the steel pipe. This is because most of the heat is lost through the metal pipe due to the high thermal conductivity of steel in comparison to the thermal conductivity of the insulating material.


8


Student View


Example 4.3-3: Heat Loss by Convection and Conduction in a Steam-Methane Reforming Process Natural gas at 400°C is flowing inside a steel pipe with an inner diameter of 1.5 in and an outer diameter of 1.9 in. The pipe is insulated with a layer of glass-fiber with a thickness of 1 in. The btu convective coefficient outside the insulated pipe is 1.23 2 . The temperature on the external ft ⋅ hr ⋅°F surface of the pipe is 43.4°C. Calculate the convective coefficient of natural gas and the overall heat transfer coefficient U based btu on the inside area Ai, if heat is being lost at a rate of 7115 in a pipe with a length of 49.2 ft. hr

Strategy To determine the heat transfer coefficients, we will use the equation for heat loss for a multilayer cylinder.

Solution The heat loss through a cylinder with different layers is defined by the following equation:

q =

Ti − To Ti − To = ∑ R Ri + RA + RB + Ro

where: Ti = Temperature on the internal surface of the pipe To = Temperature on the external surface of the pipe Ri = Convective resistance inside the pipe RA = Conductive resistance through the steel pipe RB = Conductive resistance through the insulation layer Ro = Convective resistance outside the pipe The resistance to heat transfer due to convection is defined as follows: R conv =

1 hA


9


Student View


where: h = Convective heat transfer coefficient A = Area of heat transfer The resistance of a cylinder to heat conduction is calculated as follows: R cond =

rout − rin kA lm

where: rout = Outer radius of the cylinder rin = Inner radius of the cylinder k = thermal conductivity of the material Alm = log mean area of the cylinder We can enter the definitions of the resistances due to conduction and convection into the equation for the heat loss to yield: q =

Ti − To ro − r1 r1 − ri 1 1 + + + h i A i k steel A A,lm k glass −fiber A B,lm h o A o

In this equation: ri = inner radius of the steel pipe =

1.5 in = ____________ in 2

r1 = outer radius of the steel pipe =

1.9 in = ____________ in 2

ro = outer radius of the insulated pipe = ___________ in With these values we can calculate the log mean areas AA,lm and AB,lm and the inner and outer areas of the insulated pipe. Thus,

 1 ft  2 A i = 2πri L = 2π ( 0.75 in )   ( ___________ ft ) = ___________ ft  12 in   1 ft  2 A1 = 2πr1L = 2π ( ___________ in )   ( ___________ ft ) = 24.48 ft  12 in 


10


Student View


 1 ft  2 A o = 2πro L = 2π ( ___________ in )   ( ___________ ft ) = ___________ ft  12 in  A A,lm =

A B,lm =

A1 − A i 24.48 ft 2 − ___________ ft 2 = = 21.80 ft 2 2 A    24.48 ft ln  ln  1  2   ___________ ft   Ai  A o − A1 ___________ ft 2 − 24.48 ft 2 = = ___________ ft 2  Ao   ___________ ft 2  ln  ln    24.48 ft 2    A1 

The temperature of the inner and outer surfaces of the pipe are given in °C and therefore, they must be converted to °F:

Ti (°F) = ( 400°C × 1.8 ) + 32 = ___________ °F To (°F) = ( 43.4°C × 1.8 ) + 32 = ___________ °F Now we can substitute all the values into the heat loss equation and solve for the convective coefficient hi:

1 hi = Ai

 Ti − To ro − r1 r −r 1  − 1 i − −   k steel A A,lm k glass −fiber A B,lm h o A o   q

−1

The thermal conductivity values can be obtained from Appendices A.3-15 and A.3-16 of Geankoplis. However, since the values are given in the SI system they must be converted to the English system. Hence,

k steel = 45

   ___________ m  3600 s    J 1 btu 1 °C btu    = 26   m ⋅ s ⋅ °C  ___________ J   1 ft ft ⋅ hr ⋅°F  1 hr   ___________ °F 

k glass − fiber = ______

   _________ m  3600 s   1 °C  J 1 btu btu    = 0.0317   m ⋅ s ⋅°C  _________ J   1 ft ft ⋅ hr ⋅ °F  1 hr   _____ °F 

The thermal conductivities of steel and glass-fiber were obtained at the highest temperature available in Appendix A.3.


11


Student View


Now we can calculate the convective heat transfer coefficient hi as shown below:    1 ft ( ___________ in − 0.75 in )    1  ___________ in   ___________ °F − 110.1°F − hi = 2 btu btu ___________ ft  7115 26 ___________ ft 2 ) (  hr ft ⋅ hr ⋅°F     1 ft ( ___________ in − 0.95 in )    1  ___________ in  −  − btu btu  0.0317 ___________ ft 2 ) ___________ 2 ___________ ft 2 )  ( ( ft ⋅ hr ⋅ °F ft ⋅ hr ⋅ °F 

1 hr ⋅°F hr ⋅°F hr ⋅°F hr ⋅°F   hi = __________ − 2.95 × 10−5 − __________ − 0.0162 2  ________ ft  btu btu btu btu  h i = __________

−1

btu ft ⋅ hr ⋅°F 2

To calculate the overall heat transfer coefficient U we need to use the equation for the heat loss in terms of U. Thus, q = U i A i (Ti − To )

We can solve this equation for the coefficient U and substitute the corresponding values to get:

btu 7115 q hr = Ui = 2 A i (Ti − To ) ________ ft ( ________ °F − 110.1°F )

Ui = ________

btu ft ⋅ hr ⋅°F 2


12


Student View

−1


Example 4.3-4: Heat Generation in a Tubular Solid-Oxide Fuel Cell A tubular solid-oxide fuel cell with an outer diameter of 2.2 cm and a length of 150 cm is operating mA W at a current density of 202.6 2 . Determine the heat generation rate in 3 if the voltage of the cm m fuel cell is 1 V. Assume that the thickness of the electrodes and electrolyte membrane are small compared to the overall diameter of the fuel cell. The following figure shows a tubular solid-oxide fuel cell:

Cathode interconnection

Electrolyte Cathode Fuel flow

Air Flow

Anode

Strategy The heat generation rate of the fuel cell can be obtained from the power of the fuel cell, which depends on the current and the voltage.

Solution The heat generated by the fuel cell in terms of the power is given by the equation shown below:

q =

P πR 2 L

The power of the fuel cell can be obtained by multiplying the current by the voltage of the fuel cell. Hence, P = IV


13


Student View


Substituting this equation into the equation for the heat generation rate yields: q =

IV πR 2 L

The problem statement is not giving the value of the current. However, if we calculate the crosssurface area of the fuel cell we can determine the value of the current in A: i=

I

Asurface

=

I 2πRL

Solving for the current I and substituting the dimensions of the fuel cell into this equation, we get: mA  1 A   I = πDLi = π ( ________ cm )( ________ cm )  202.6 2   cm  1000 mA   I = ________ A

Entering this value into the heat generation equation we have:

q =

________ A (1 V ) 2

  1 m3  ________ cm  π ________ cm ( )   3  2    __________ cm 

q = 3.68 × 105


W m3

14


Student View


Example 4.5-1: Heating of Natural Gas- in Steam-Methane Reforming Process

Natural gas at a temperature of 310°C is flowing inside a steel pipe Schedule 80 with an inner m3 . The natural gas is being heated by the product of the s W reforming process at 850°C. The convective heat transfer coefficient of the reformate is 1025 2 . m ⋅K Calculate the heat transfer rate in W through a pipe with a length of 7 m. The properties of natural gas are given in the following table.

diameter of 1.5 in at a rate of 7.79 × 10−3

ρ Cp k

µb µw

kg m3 J 3087 kg ⋅ K W 0.0803 m⋅K

7.859

kg m⋅s kg 2.909 ×10−5 m ⋅s

1.942 × 10 −5

Strategy The equation for heat transfer through a pipe will be used to determine the heat flux.

Solution When heat is being transferred through a fluid, the heat flux is given by:

q =

( Tr − Tn )

∑R

where: q = Heat transfer rate in W

Tr = Temperature of the heating medium (reformate), °C or K Tn = Temperature of the fluid inside the pipe (natural gas), °C or K

∑ R = Sum of resistances to heat transfer through the pipe, Daniel López Gaxiola Jason M. Keith

15


°C K or W W

Student View


In this problem, the sum of the resistances is given by the sum of two convective resistances (fluid inside and outside the pipe) and the resistance to heat flow through the steel pipe. Thus, 1

∑R = h A i

+ i

ro − ri 1 + k steel A lm h o A o

The parameters we need to calculate before being able to use the equation for heat transfer in terms of the resistances are: Ai, Alm, Ao and hi. The heat transfer areas are calculated using the inner and outer diameters of the pipe from Appendix A.5.

A i = πDi L = π ( 0.0381 m )( 7 m ) = _________ m 2 A o = πDo L = π ( _________ m )( 7 m ) = _________ m 2 A lm =

A o − A i _________ m 2 − _________ m 2 = = 0.945 m 2 2 A   _________ m  ln  ln  o  2   _________ m   Ai 

The following correlation can be used for calculating the heat transfer coefficient for an aspect ratio (length/diameter) of the pipe greater than 60. 1  µ  k h L = 0.027 N Re 0.8 N Pr 3  b  D  µw 

0.14

where: k = Thermal conductivity of the fluid inside the pipe,

W m⋅K

NRe = Reynolds number NPr = Prandtl number µ b = Viscosity of the fluid in the pipe at the bulk temperature,

kg m⋅s

µ w = Viscosity of the fluid in the pipe at the temperature of the inner wall,

In this problem,

kg m⋅s

L 7m = = _________ . Hence, we can use this correlation to calculate hL. D 0.0381 m


16


Student View


The dimensionless quantities NRe and NPr are defined as follows: N Re =

Dυρ µ

N Pr =

Cpµ k

The velocity of the natural gas is obtained by dividing the volumetric flow rate by the cross-sectional area of the pipe. Thus, after substituting the corresponding quantities into the equations for Reynolds and Prandtl numbers, we have:

N Re

 m3  ______________   s  7.859 kg 0.0381 m   V  m3  ______________  D 2 ρ πr   =  i  = = __________________ kg µ _______________ m⋅s

N Pr =

C pµ

k

3087

=

J  kg   ________________  kg ⋅ K  m ⋅s  = _____________ W ______________ m⋅K

Now we can substitute the dimensionless numbers we just calculated and the properties of the fluid into the equation for the heat transfer coefficient to yield:

kg  W  ______________ 0.0803 1  0.8 m⋅s  m ⋅ K ( ________________ ) ( _______ ) 3 h L = 0.027  kg  0.0381 m  ______________  m⋅s   h L = _____________

0.14

W m2 ⋅ K

The convective heat transfer coefficient hL was obtained using the properties of the fluid inside the pipe. Therefore, the heat transfer coefficient hL is equal to the heat transfer coefficient hi. Substituting the corresponding quantities into the equation for the heat transfer rate q . The thermal conductivity was obtained from Appendix A.3.

q =

( 850°C − 310°C ) 1 0.04826 m − 0.0381 m 1 + + W W W _________ 2 ( _________ m2 ) 45 m ⋅ K ( 0.945 m2 ) _________ m2 ⋅ K ( _________ m2 ) m ⋅K

q = ____________ W Daniel López Gaxiola Jason M. Keith

17


Student View


Example 4.5-2: Trial-and-Error Solution for Heating of Steam Steam at a temperature of 150°C is being heated before entering a steam-methane reforming unit to produce hydrogen for fuel cells. The heating medium is the synthesis gas produced by the steammethane reforming unit at 850 °C. The convective heat transfer coefficient of the syngas is m W 950 2 . The steam is flowing in a 1.5-in schedule 40 steel pipe at a velocity of 20.1 . m ⋅K s Determine the overall coefficient Ui for a pipe with a length of 6.2 m.

Strategy We can determine the convective heat transfer coefficient using the properties of steam at the temperature of the inner wall of the pipe. This temperature will be determined by trial and error.

Solution The overall heat transfer coefficient can be determined from the equation for the heat transfer rate through the pipe:

q = U i A i (To − Ti ) =

To − Ti ∑R

The temperature difference To - Ti can be eliminated from this equation to yield: Ui =

1 ___________

where: 1

∑R = h A i

+ i

ro − ri 1 + kA lm h o A o

The convective coefficient hi can be calculated using the following correlation. 1  µ  k h L = 0.027 N Re 0.8 N Pr 3  b  D  µw 

0.14

The dimensionless quantities in this equation will be determined using the properties of steam at the temperature of the inner wall of the pipe. For the first trial, this temperature will be assumed to be about one-quarter the difference between the temperatures of the steam and the air. Thus, Tw,assumed =

850°C − 150°C + 150°C = __________ °C 4


18


Student View


From Table A.2-12 of Geankoplis we can get the properties of steam at a temperature of 148.9 °C, which is relatively close to the bulk temperature of the steam inside the pipe (150°C). ρ = __________

Cp = 1909

kg m3

µ = µ b = 1.488 × 10−5

J kg ⋅ K

k = __________

kg m ⋅s

N Pr = __________

W m⋅K

The other parameter required to use the correlation for calculating hL is the viscosity of steam at Tw. This can be obtained using linear interpolation from the data in Table A.2-12. Hence, µ w − 2.113 × 10 −5

kg m⋅s

325°C − 315.6°C = 371.1°C − 315.6°C ______________ kg − 2.113 × 10−5 kg m ⋅s m⋅s

Solving for the viscosity µw we get:

µw =

325°C − 315.6°C  kg kg  kg − 2.113 ×10 −5  _______________  + _______________ 371.1°C − 315.6°C  m ⋅s m⋅s  m⋅s

µ w = _______________

kg m⋅s

With the properties of steam, we can now determine the Reynolds number as shown in the following steps. The diameter of the pipe was obtained from Table A.5-1 of Geankoplis

N Re

m  kg   ___________ m  _______   0.525 3  Dυρ s  m   = = = _______________ µ −5 kg 1.488 ×10 m ⋅s

Substituting this value into the correlation for hL we get:

kg W   1.488 × 10−5 ____________  1  0.8 m ⋅s m ⋅ K ( _______________ ) ( 0.95 ) 3 h L = 0.027  kg  0.04089 m  _______________  m ⋅s   h L = ____________

0.14

W m2 ⋅ K

Now we can proceed to calculate the sum of the resistances to heat transfer as follows: Daniel López Gaxiola 19 Student View Jason M. Keith



1

∑R = h A i

+ i

ro − ri 1 + kA lm h o A o

The heat transfer areas are calculated using the inner and outer diameters of the pipe from Appendix A.5.

A i = πDi L = π ( 0.04089 m )( ______ m ) = ___________ m 2 A o = πDo L = π ( 0.04826 m )( ______ m ) = ___________ m 2 A lm =

A o − A i _______ m 2 − _______ m 2 = = 0.866 m 2 2 A   _______ m  ln  ln  o  2   _______ m   Ai 

These values can be entered into the equation for the sum of resistances to yield:

1

∑R = _______

W ( _________ m2 ) m2 ⋅ K

+

0.04826 m − 0.04089 m 1 + W W 45 0.866 m 2 ) 950 2 ( ( _______ m2 ) m⋅K m ⋅K

K

∑ R = _________ W The thermal conductivity value of steel was obtained from Table A.3-16. To determine if the value of Tw we selected is correct, we need to solve for the temperature from the equation for the heat resistance due to the steam in the pipe:

Tw ,calculated − Tb =

Ri ( To − Tb ) ∑R

Solving for Tw and substituting the rest of the values into this equation, we have that:

Tw ,calculated = Tb +

1 h i Ai ∑ R

Tw ,calculated = 150°C +

( To − Tb ) 1

W K  _______ 2 _______ m 2 )  _______  ( m ⋅K W 

(850°C − 150°C )

Tw ,calculated = _________ °C


20


Student View


It can be see that the Tw,assumed does not match Tw,calculated. Hence, we have to repeat the procedure for determining the temperature of the inner wall. The Tw value will affect the value of hL since we have to look for a new value of µw in Appendix A.3. Selecting a higher Tw for the second trial will yield a higher µw, resulting in a lower heat transfer coefficient hL and a higher

∑ R . For the second trial, we will select:

Tw,assumed = 800°C Substituting the viscosity µw into the equation for the heat transfer coefficient hL yields: W  −5 kg  1.488 × 10  m⋅s  m ⋅ K ( ____________ )0.8 ( _____ ) 13 h L = 0.027  kg  0.04089 m  3.95 × 10 −5  m⋅s   W h L = ________ 2 m ⋅K

0.14

__________

For this value of hL we will get the sum of the resistances as follows:

1

∑R = 62.3

W _________ m 2 ) ( 2 m ⋅K

+

0.04826 m − 0.04089 m 1 + W W _______ 0.866 m 2 ) 950 2 _________ m 2 ) ( ( m⋅K m ⋅K

K

∑ R = ________ W Solving for Tw and substituting the rest of the values into this equation, we have that: Tw ,calculated = Tb +

1

hi Ai ∑ R

( To − Tb ) = 150°C +

1

W K  62.3 2 _______ m 2 )  _______  ( m ⋅K W 

( 850°C − 150°C )

Tw ,calculated = __________ °C

The only property that will change for the third trial is the viscosity µw. By changing the temperature again, the effect on the convective coefficient hL will be negligible. Hence, we can use the calculated Tw value of ________°C. Now we can substitute the values of

∑ R at Tw = _________°C and the inner area of the pipe to

obtain the overall heat transfer coefficient Ui as shown in the following steps:

Ui =

1

Ai ∑ R


21


Student View


Ui =

1 K ( ________ m2 )  ________ W  

Ui = 59.8

W m2 ⋅ K


22


Student View


Example 4.5-3: Heating of Ethanol in Reforming Process A vapor mixture of ethanol and water is used in a reforming process to produce hydrogen for protonexchange membrane fuel cells. In a distributed-scale process, the ethanol mixture is flowing at a rate kg of 0.367 and is entering a 1" steel pipe Schedule 40 at a temperature of 210.4°C. Determine the s length of the pipe required if the vapor is exiting at 350°C and the inner wall of the pipe is at a constant temperature of 270°C. The properties of the vapor mixture are summarized in the following table. µ ρ Cp k

1.284 × 10 −5 Pa ⋅ s kg 13.2 3 m J 2211 kg ⋅ K

0.03631

W m⋅K

Strategy To determine the heat transfer area we can use the equation for heat flux through a fluid.

Solution The heat flux when heat is being transferred by a fluid is given by: q = h L ( Tw − T ) A

where Tw is the temperature of the inner wall of the pipe, and T is the bulk temperature of the ethanol and water mixture. Since the problem statement is giving the flow rate and properties of the fluid, the heat transfer rate can be calculated with the equation for sensible heat:

p ( Tout − Tin ) q = mC In this equation, Tin and Tout are the temperatures of the ethanol/water mixture at the inlet and outlet points, respectively. Substituting this equation into the equation for heat flux, we get: p ( Tout − Tin ) mC

______

= ______________________

Solving for the heat transfer area we have: Daniel López Gaxiola Jason M. Keith

23


Student View


p ( Tout − Tin ) mC

A=

______________

To determine the heat convective heat transfer coefficient, the following correlation can be used for a pie with constant wall temperature and if NPe>100 and L/D>60: hL =

k 5.0 + 0.025N Pe 0.8 ) ( D

where: NPe = Peclet number We can calculate Peclet number by multiplying Reynolds number by Prandtl number and thus determine if it is valid to use this correlation. NPe = NReNPr N Re =

N Pr =

Dυρ µ

Cp µ

k

2211

=

J ( ______________ Pa ⋅ s ) kg ⋅ K = ______ W 0.03631 m⋅K

The velocity of the fluid is calculated by dividing the volumetric flow rate by the cross-sectional area of the pipe. The diameter of the pipe is obtained from Appendix A.5 of Geankoplis. m υ= = ρA cross

kg m s = ________ 2 s kg  π ( ________ m )  13.2 3   m  4  ________

Entering this velocity value into the definition of Reynolds number yields:

N Re

m  kg   ________ m  ________  13.2 3  Dυρ s  m   = = = _________________ kg µ −5 1.284 × 10 m ⋅s


24


Student View


Now we can determine the Peclet number to be given by:

N Pe = ____________ ( ______ ) N Pe = ______________

Therefore, the correlation we selected is valid for this problem and the convective heat transfer coefficient is found to be:

W m ⋅ K 5.0 + 0.025 ( _______________ )0.8  hL =  __________ m  0.03631

h L = ___________

W m2 ⋅ K

This value can be entered into the equation for the heat transfer area to yield:

A=

p ( Tout − Tin ) mC h L ( Tw − T )

kg  J  2211 ( _______ °C − 210.4°C )  s  kg ⋅ K  W ________ 2 ( _______ °C − 210.4°C ) m ⋅K

0.367

=

A = __________ m 2

The area for heat transfer is given by: A = πDL

This equation can be solved for the length of the pipe L to give: ________ m 2 L= π ( ___________ m ) L = ________ m


25


Student View


Example 4.5-4: Heat-Transfer Area and Log Mean Temperature Difference kJ in a mid-scale ethanolkg ⋅ °C reforming plant is heated from 210.4 °C to 350 °C. This mixture is flowing at a rate of kg kg 5.18 × 105 . The vapor is being heated by air flowing at a rate of 2.752 × 106 , temperature of day day kJ 560.6 °C and a heat capacity of 1.166 . What type of flow for this heat exchanger will you kg ⋅°C select between countercurrent and parallel flow if the overall heat transfer coefficient is W ?. 92.4 2 m ⋅ °C

An ethanol/water vapor mixture with a heat capacity of 2.23

Strategy To determine which type of flow is more efficient for this process we need to determine the heat transfer area for both types of flow.

Solution The amount of heat gained by the ethanol mixture in terms of the overall heat transfer coefficient is given by the equation shown below: q = U i Ai ∆Tlm

where ∆Tlm is the log mean temperature difference defined as: ∆Tlm =

∆T2 − ∆T1  ∆T  ln  2   ∆T1 

The q equation can be solved for the area Ai to yield: Ai =

q _________

Since we know the inlet and outlet temperatures of the ethanol/water mixture we can calculate the amount of heat gained as shown in the following steps: ethanol C p,ethanol ( Tethanol,out − Tethanol,in ) q = m

q = 5.18 × 105

  kg  kJ  1 day  2.23  ( _______ °C − _______ °C )   day  kg ⋅°C   __________ 


26


Student View


q = __________ kW

To determine the log mean temperature difference required to calculate the heat transfer area we need to obtain the temperature of the air leaving the heat exchanger. We know that the amount of heat gained by the ethanol mixture is being lost by the heating air. Thus, the temperature can be determined as follows: air C p,air ( Tair,in − Tair,out ) q = m

Tair,out = _________ − ____________ = _______ °C −

__________ kW  kg  kJ  1 day 2.752 × 106 1.166    day  kg ⋅ °C  _________ 

Tair,out = _________ °C

Now that we know the inlet and outlet temperatures of both the air and the ethanol mixture, we can calculate the log mean temperature difference. Hence, ∆Tlm =

∆T2 − ∆T1  ∆T  ln  2   ∆T1 

where: ∆T1,countercurrent = Tair,in − Tethanol,out = _________ °C − _________ °C ∆T1,countercurrent = _________ °C ∆T2,countercurrent = Tair,out − Tethanol,in = _________ °C − _________ °C ∆T2,countercurrent = _________ °C

Substituting the values of ∆Ti into the definition of the log mean temperature difference we get: ∆Tlm,countercurrent =

_________ °C − _________ °C  _________ °C  ln    _________ °C 

∆Tlm,countercurrent = _________ °C


27


Student View


For countercurrent flow, we can find the heat transfer area to be given by:

A i,countercurrent

 1000 W  _________ kW    1 kW  = W 92.4 2 ( _________ °C ) m ⋅°C

A i,countercurrent = _________ m 2

To determine the heat transfer area for parallel flow, we can use Figure 4.5-3 of Geankoplis to calculate the log mean temperature difference for parallel flow, which will be given by: ∆Tlm =

∆T2 − ∆T1  ∆T  ln  2   ∆T1 

where: ∆T1,parallel = Tair,out − Tethanol,out = ________ °C − ________ °C ∆T1,parallel = _________ °C ∆T2,parallel = Tair,in − Tethanol,in = ________ °C − ________ °C ∆T2,parallel = _________ °C ∆Tlm,parallel =

_________ °C − _________ °C = _________ °C  _________ °C  ln    _________ °C 

Substituting the log mean temperature difference into the equation for Ai,parallel we have:

A i,parallel

 1000 W  _________ kW    1 kW  = W 92.4 2 ( _________ °C ) m ⋅°C

A i,parallel = _________ m 2 Conclusion:


28


Student View


Example 4.6-3: Heating of Steam by a Bank of Tubes in High-Temperature Electrolysis High-temperature electrolysis is a process for producing hydrogen from water for use in fuel cells. Before entering the electrolysis stack, steam at a pressure of 50 bar is being heated from 650°C to 850°C by a bank of 1" (nominal diameter) commercial steel tubes containing 12 rows normal to the flow and 7 staggered rows in the direction of flow. The length f the tubes is 0.4 m. The heating medium circulating in the tubes is helium coming from a nuclear source and the outer surface of the tubes is at a temperature of 1000°C. Determine the heat-transfer rate to the steam if the m velocity of steam is 16.7 . s A diagram of this heating process is shown below: Sp=0.0418 m

Sn=0.0418 m

Steam

Helium from nuclear source flowing inside the tubes

Strategy To solve this problem we need to calculate the amount of heat transferred by convection. The heat transfer area will depend on the number of tubes.

Solution The amount of heat gained by the steam can be calculated with the following equation:

q = hA ( Tw − Tb ) The bulk temperature of steam Tb is obtained taking the average of the inlet and outlet temperatures: Tb =

850°C + 650°C = _________ °C 2

The temperature of the outer surface of the tube is constant and equal to 1000°C.


29


Student View


To calculate the amount of heat transferred, we need to obtain the heat transfer area of the tubes. The area of a single tube is calculated as follows: A tube = πDL

Substituting the diameter from Appendix A.5 of Geankoplis and the length of 0.4 m into this equation, we can determine the area of a single tube to be:

A tube = π ( _________ m )( ______ m ) A tube = ________ m 2

Since there are 7 columns and 12 rows of tubes, this area must be multiplied by the total number of tubes in the bank. Thus, A = n rows n columns A tube = ( 7 )(12 ) ( _________ m 2 ) A = _________ m 2

Now we need to determine the heat transfer coefficient of steam using the correlation for flow past a bank of tubes, shown in Section 4.6 of Geankoplis. h calculated =

k CN Re m N Pr1/3 D

In this equation, the parameters C and m will depend on the ratio of the distance between the tubes and their outer diameter. For this electrolysis process, the ratio is given by:

Sn Sp _________ m = = = 1.251 D D _________ m Using Table 4.6-2 of Geankoplis, we can find the values of C and m for staggered tubes to be: C = _________ m = _________ The dimensionless parameters NRe and NPr are calculated from the properties of steam at the film temperature, obtained as follows: Tf =

Tw + Tb _________ °C + 750°C = 2 2

Tf = _________ °C


30


Student View


The properties of steam were obtained from Table 2-305 of Perry's Chemical Engineers' Handbook, 8th Edition and are shown below: k = 0.1203

W m⋅K

Cp = 2441.3

J kg ⋅ K

µ = 4.34 × 10−5 Pa ⋅ s

ρ = 9.508

kg m3

Substituting these properties into the definition of Prandtl number we get:

N Pr =

__________________ ( ____________ Pa ⋅ s ) _______________

N Pr = _________

The maximum velocity required to calculate Reynolds number is obtained using the outer diameter of the pipes and the distance between the pipes normal to the direction of flow as shown in the following equation: υmax =

υSn Sn − D

We can enter the velocity of steam, the diameter and the distance between the pipes into this equation to yield:

m ( __________ m ) s = __________ m − __________ m 16.7

υmax

υmax = __________

m s

Now the Reynolds number can be determined as shown in the next steps: N Re =

Dυmax ρ µ


31


Student View


N Re

m  kg   _________ m  ______   9.508 3  s  m   = −5 4.34 × 10 Pa ⋅ s

N Re = ____________

Substituting the dimensionless parameters we calculated and the corresponding values into the equation for hcalculated, we have: h calculated =

k CN Re m N Pr1/3 D

W m ⋅ K ( _______ )( _____________ )0.556 ( ________ )1/3 = __________ m 0.1203

h calculated

h calculated = ____________

W m2 ⋅ K

The value of the heat transfer coefficient that has to be entered into the equation for the heat transfer rate has to be multiplied by a factor that depends on the amount of rows in the direction of flow. For 7 rows and staggered tubes we can find this factor in Table 4.6-3 of Geankoplis to be _______. Thus, the actual value of the heat transferred coefficient will be given by: W   h = ______ h calculated = ______  ____________ 2  m ⋅K  

h = ___________

W m2 ⋅ K

Entering this value into the equation for q , we get: q = ___________

W ( ___________ m2 ) (1000°C − 750°C ) m2 ⋅ K

q = ______________ W


32


Student View


Example 4.7-3: Natural Convection in Bipolar Plate Vertical Channel Hydrogen at standard pressure is flowing by natural convection in the bipolar plate channels of a fuel cell. These channels have a length of 24.94 cm and a thickness and height of 1 mm. Determine the heat transfer rate across the channel if the temperature of the walls is constant and equal to 82 °C. The surface of the gas diffusion layer adjacent to the channel is at a temperature of 85.17 °C. A schematic of the bipolar plate channels is shown in the following figure.

249.4 mm

1 mm 1 mm

Strategy The heat transfer rate by convection can be calculated by using correlations that involve dimensionless groups.

Solution When heat is being transfer by convection, the heat transfer rate q is given by:

q = hA ( T1 − T2 ) The heat transfer area can be calculated from the dimensions of the channel as shown below:

A = 2L ( t + H ) = 2 ( 0.2494 m )( 0.001 m + 0.001 m ) A = ___________ m 2

To determine the heat transfer coefficient, we can use the definition of Nusselt number: N Nu =

hH k

where H is the height of the channel. However, since the Nusselt number is not given, we need to use another correlation in terms of dimensionless groups. In section 4.7 of Geankoplis, multiple correlations are shown as function of the Grashof and Prandtl numbers. We need to select the adequate correlation depending on the value yielded by the product of these dimensionless groups defined as follows: N Gr

H 3ρgβ ( T1 − T2 ) = µ2


33


Student View


In this equation: H = height of the bipolar plate channel, m ρ = density of hydrogen,

kg m3

g = acceleration due to gravity = 9.80665

m s2

β = volumetric coefficient of expansion of hydrogen =

µ = viscosity of hydrogen,

1 , K-1 Tf

kg m⋅s

We can find the properties of hydrogen in Appendix A.3 of Geankoplis at the film temperature given by: Tf =

T1 + T2 ________ °C + ________ °C = 2 2

Tf = ________ °C = ___________ K

The properties of hydrogen at this temperature (shown below) can be substituted into the equation for Grashof number to get: ρ = 0.068

kg m3

µ = 9.92 × 10−6

β=

kg m ⋅s

1 = 2.80 × 10−3 K −1 356.74 K

( 0.001 m )

3

N Gr =

kg   m  −1  0.068 3   __________ 2  ( ___________ K ) ( ________ °C − ________ °C ) m s    2  −6 kg  9.92 × 10   m ⋅s  

N Gr = ________


34


Student View


In Appendix A.3 we can also find the value of Prandtl number for hydrogen to be: N Pr = ________

The product of Grashof and Prandtl numbers can now be obtained as follows:

N Pr N Gr = ________ ( ________ ) = ________ Looking at the correlations in Section 4.7 of Geankoplis, we find that the Nusselt number corresponding to this value of N Pr N Gr is 1. Thus, we can solve for the heat transfer coefficient from the definition of Nusselt number to yield: W   ________  ________  N Nu k m⋅K   h= = H 0.001 m

h = ________

W m2 ⋅ K

Now we can enter the corresponding quantities into the equation for the heat transfer rate to obtain: q = ________

W ______________ m 2 ) ( ________ °C − ________ °C ) ( 2 m ⋅K

q = ________ W


35


Student View


Example 4.8-2: Condensation in Bipolar Plate Channels in Fuel Cells The reaction occurring in a proton-exchange membrane fuel cell is producing water at a rate of kg 7.65 × 10 −5 through each channel on the bipolar plate in the cathode side. This amount of water is s produced as steam at a temperature of 77°C. Determine if the water is condensing in a single channel if the partial pressure of water is 37.91 kPa. The dimensions of the channel are shown in the following figure:

L = 249.4 mm

H = 1 mm t = 1 mm

The Nusselt number for a square tube with constant temperature at the boundaries is 2.98. Frano Barbir in Section 6.5.2 of the book PEM Fuel Cells - Theory and Practice published by Prentice Hall estimates the average temperature in the bipolar plate channels to be 64.1°C.

Strategy In this problem, condensation will occur if the amount of heat removed by convection is higher than the latent heat of condensation of steam.

Solution The amount of heat lost by the steam can be calculated from the equation for convective heat transfer as shown below: q conv = hA∆T

where the change in temperature ∆T is given by: ∆T = T − Tw

in this equation: T = Temperature of steam Tw = Temperature of the walls of the square channel


36


Student View


The convective heat transfer coefficient is obtained from the definition of Nusselt number: N Nu =

hL k

Solving for the convective heat transfer coefficient h and substituting the values of the length L and the thermal conductivity of steam we get:

h=

kN Nu = L

W ( 2.98 ) m⋅K ________ m

________

h = ________

W m2 ⋅ K

The thermal conductivity value used in this equation was obtained from Table 2-305 of Perry's Chemical Engineers' Handbook, 8th Edition. Substituting this value and the heat transfer area into the equation for q conv yields:

A = 4 ( ________ m )( ________ m ) = ______________ m 2 q conv = ________

W ______________ m 2 ) ( 77°C − 64.1°C ) ( 2 m ⋅K

q conv = _____________ W

To determine if the steam is condensing in the fuel cell, we need to compare this amount of heat to the latent heat of vaporization, defined as follows: fg q vap = mh

The heat of vaporization hfg can be obtained from Table A.2-9 of Geankoplis. When condensation is occurring, the saturated vapor pressure is equal to the partial pressure. Hence, we will look for the enthalpy of vaporization at a pressure of 37.91 kPa and substitute into the equation for q vap . q vap = 7.65 × 10 −5

kg  J J  − _____________   _____________ s  kg kg 

q vap = _______ W

Conclusion:


37


Student View


Example 4.9-1: Temperature Correction Factor for a Heat Exchanger The synthesis gas produced in a steam-methane reforming process for hydrogen production is being cooled in a heat exchanger before entering the water-gas shift reaction chamber at 846°C to 600°C. The cooling medium is air entering a heat exchanger at 255.3°C and leaving at 381.3°C. The syngas kg J is flowing at a rate of 1322 and has a heat capacity of 2584 . The air flow is being hr kg ⋅ K distributed through 100 commercial steel pipes with a nominal diameter of 1" Schedule number 40 and a length of 1.2 m. Calculate the mean temperature difference in the exchanger and the overall heat transfer coefficent Uo for the 4 heat exchanger configurations shown in Section 4.9B of Geankoplis.

Strategy The heat transfer rate can be obtained using the definition of sensible heat. This can be used to determine the overall heat transfer coefficient Uo. The mean temperature difference will depend on the type heat exchanger selected.

Solution The heat transfer coefficient can be obtained from the equation shown below: q = U o A o ∆Tm

Solving for Uo, we get: U o = ___________

where: Ao = total outer surface area of the pipes distributing the air flow, m2 ∆Tm = mean temperature difference = FT ∆Tlm , K or °C

The mean temperature difference can be obtained by multiplying a factor FT depending on the heat exchanger type by the log mean temperature difference, defined as: ∆Tlm =

( Thi − Tco ) − ( Tho − Tci ) ( T − Tco ) ln hi ( Tho − Tci )

where: Thi = temperature of the synthesis gas entering the heat exchanger, K or °C Daniel López Gaxiola Jason M. Keith

38


Student View


Tho = temperature of the synthesis gas exiting the heat exchanger, K or °C Tci = temperature of the air entering the heat exchanger, K or °C Tco = temperature of the air exiting the heat exchanger, K or °C The amount of heat removed from the synthesis gas can be calculated as follows:

p ( Thi − Tho ) q = mC Substituting the syngas flow rate, specific heat and the inlet and outlet temperatures yields: q = 1322

kg  1 hr   J   ( ______ °C − ______ °C )    _______ hr  3600 s   kg ⋅ K 

q = ____________ W

We can now determine the log mean temperature difference as shown in the following steps: ∆Tlm =

( ______ °C − ______ °C ) − ( ______ °C − ______ °C ) ( ______ °C − ______ °C ) ln ( ______ °C − ______ °C )

∆Tlm = ________ °C

To calculate the heat transfer area of the tubes, we need to look for the outer diameter of 1" commercial steel pipes in Appendix A.5 of Geankoplis. Hence, the area can be determined as shown below:

A o = πDLn = π ( _________ m )( _____ m )( _____ ) = _________ m 2 In this equation, n is the number of tubes. Now we will proceed to calculate the mean temperature difference for the 4 different types of heat exchanger. For all heat exchanger configurations we need to calculate the parameters Y and Z, defined as: Y=

Tco − Tci ______ °C − 255.3°C = Thi − Tci ______ °C − 255.3°C

Z=

Y = ______


Thi − Tho 846°C − ______ °C = Tco − Tci ______ °C − 255.3°C

Z = ______

39


Student View


For these values of Y and Z, we find the factors FT in Figures 4.9-4 and 4.9-5 and use them to determine the mean temperature differences and overall heat transfer coefficients, given by:

1-2 exchanger

2-4 exchanger

FT = ______

FT = ___

∆Tm = FT ∆Tlm = ______ ( _______ K )

∆Tm = FT ∆Tlm = ____ ( _________ K )

∆Tm = 391.68 K

∆Tm = _________ K

Uo =

q _________ W = A o ∆Tm ________ m 2 ( 391.68 K )

U o = ________

Uo =

W m2 ⋅ K

q _________ W = A o ∆Tm 12.59m 2 ( ________ K )

U o = ________

W m2 ⋅ K

Cross-flow exchanger with shell fluid mixed

Cross-flow exchanger with fluids unmixed

FT = ______

FT = _____

∆Tm = FT ∆Tlm = _____ ( _________ K )

∆Tm = FT ∆Tlm = ______ ( _________ K )

∆Tm = _______ K

∆Tm = _________ K

Uo =

q _________ W = A o ∆Tm ______ m 2 ( _________ K )

U o = ________

Uo =

W m2 ⋅ K


q _________ W = A o ∆Tm _________ m 2 ( _________ K )

U o = 47.10

40


W m2 ⋅ K

Student View


Example 4.9-2: Temperature Correction Factor for a Heat Exchanger kg and a hr temperature of 846°C. This gas is to be cooled to a temperature of 473.86°C before entering a watergas shift reactor to produce additional hydrogen for fuel cells. The cooling medium is air entering at kg J 255.3°C, a flow rate of 6307 and a heat capacity of 1058 . The overall heat transfer hr kg ⋅ K

Synthesis gas is being produced in a steam-methane reforming process at a rate of 1322

W for a heat transfer area of 8.92 m2. Determine the type of flow at which the 2 m ⋅K heat exchanger is operating and the heat transfer rate if the effectiveness is 0.45. The composition of the syngas is shown in the following table:

coefficient is 90

CO H2 H2O

wt. % 20.61 7.18 72.21

Strategy The charts showing the effectiveness of heat exchangers operating at countercurrent flow and parallel flow can be used to determine the type of operation.

Solution The heat transfer rate in a heat exchanger can be calculated as a function of the effectiveness ε as shown in the following equation:

q = εCmin ( THi − TCi ) To determine the value of Cmin, we need to calculate the values of CH and CC. These parameters depend on the flow rate and the heat capacities of the fluids in the heat exchanger. Thus, syngas Cp,syngas CH = m air C p,air CC = m

We can see that the heat capacity of syngas is not given in the problem statement. However, we can use Figure A.3-3 of Geankoplis to determine the heat capacity of each individual component. The heat capacity of the syngas can then be determined by multiplying the mass fraction of each component by its corresponding heat capacity at the film temperature given by:


41


Student View


Tf =

THi + THo 846°C + 473.86°C = 2 2

Tf = ________ °C

Thus, at this temperature value: Cp,CO = __________

Cp,H2 = 14853.2

J kg ⋅ K

J kg ⋅ K

Cp,H2O = __________

J kg ⋅ K

With these individual heat capacities we can determine the heat capacity of syngas as follows:

Cp,syngas = x CO Cp,CO + x H2 Cp,H2 + x H2O Cp,H2O

   J  J  J  Cp,syngas = 0.2061 _________  + _________ 14853.2  + 0.7221 _________  kg ⋅ K  kg ⋅ K  kg ⋅ K    

Cp,syngas = __________

J kg ⋅ K

Now we can substitute the heat capacities and flow rates of air and syngas to obtain CH and CC. Thus, CH = 1322

kg  1 hr   J      __________ hr  3600 s   kg ⋅ K 

CH = __________

CC = 6307

W K

kg  1 hr   J      __________ hr  3600 s   kg ⋅ K 

CC = __________

W K


42


Student View


We can see from the values of CC and CH that the smaller of these values is CH, thus Cmin = CH. Substituting this value into the equation for the heat transfer rate, we get: W  q = __________  __________  ( 846°C − ________ °C ) K  q = __________ kW

Figure 4.9-7 of Geankoplis is showing the effectiveness of a heat exchanger operating at both C countercurrent flow and parallel flow as function of the number of transfer units and the ratio min . C max If we calculate these two values, we can look in this Figure which type of heat exchanger will yield a value of ε = ________ . Thus, W __________ C min K = 0.59 = W C max __________ K

W 90 2 (8.92m2 ) UA m ⋅ K NTU = = W Cmin __________ K NTU = ________

Conclusion:


43


Student View


Example 4.11-1: Radiation in Cylindrical Solid - Oxide Fuel Cell The following figure is a schematic of a cylindrical solid-oxide fuel cell. Jiang et al. [1] developed a thermoelectrical model to estimate the temperature at different parts of this type of fuel cell. The temperatures are estimated to be 1125 K for the air tube and 1200 K for the solid part (membrane electrode assembly). Determine the heat flux due to radiation. Xue et al. [2] estimated the average emissivity of the membrane electrode assembly to be 0.33. The air is being fed to the system through a commercial steel pipe.

Interconnect

Air Feed

Fuel

Fuel electrode Electrolyte membrane Air electrode

Strategy The heat transferred due to radiation can be estimated using the radiation equation for gray bodies given in Section 4.11 of Geankoplis.

Solution The following equation is the definition of heat flux due to radiation:

  1   = σ ( T − T2 ) q"  1 + 1 − 1  ε2   ε1 4 1

4

where:

= Heat flux due to radiation, q"

W m2

1. Jiang, W., Fang, R., Dougal, R. A., Khan, J. A., Journal of Energy Resources Technology, 130, 2008. 2. Xue, X., Tang, J., Sammes, N., Du, Y., Journal of Power Sources, 142, 211 − 222 (2005)


44


Student View


σ = Stefan – Boltzmann constant = 5.676 × 10−8

W m ⋅ K4 2

Ti = Temperature of surface i, K εi = Emissivity of surface i, K We can substitute the given quantities into the equation for heat flux to solve this problem. Hence,

  W  1 4 4   = _______________ 2 4 ( ______ K ) − (1125 K )  q"   1 m ⋅K  + 1 −1  _____   _____ q " = __________

W m2

The emissivity of steel was obtained from Table 5 − 4 of Perry's Chemical Engineers' Handbook, 8th Edition.


45


Student View


4.15-1 Cooling Channels in Fuel Cell Bipolar Plates The following figures show the top and isometric views of a fuel cell bipolar plate with 10 cooling W channels through which air is circulating with a heat transfer coefficient of 400 2 and a m ⋅K temperature of 10°C. The outer walls of the bipolar plates are held at a temperature of 60°C. Determine the steady – state heat loss in one bipolar plate using finite difference numerical methods, with grids 1 mm x 1 mm. The bipolar plates are made of 304 stainless steel. 8 mm

8 mm

6 mm

0.15 m

108 mm

2 mm x 2 mm channels

Strategy To determine the amount of heat removed we need to determine the temperatures at the different nodes, using the Equations in Section 4.15B of Geankoplis.

Solution Since the area surrounding the channel is symmetrical, we can calculate it for one channel and multiply it by the number of times this area is repeated in the whole bipolar plate. If we zoom into the first channel from the left edge of the bipolar plate:

h = 400

W m2 ⋅ K

T∞ = 10°C

The shaded areas in this figure indicate the sets of nodes that will be repeated along the bipolar plate.

Set of Nodes at the Edges of the Bipolar Plate We will start by using the finite difference method on the edges of the bipolar plate. For a grid of 1 mm by 1 mm, the following nodes will be used: T1,1 T1,2 T1,3 T1,4 T1,5 T1,6 T1,7 T1,8 T1,9 T1,10 T2,1 T2,2 T2,3 T2,4 T2,5 T2,6 T2,7 T2,8 T2,9 T2,10 T3,1 T3,2 T3,3 T3,4 T3,5 T3,6 T3,7 T3,8 T3,9 T3,10 T4,1 T4,2 T4,3 T4,4 T4,5 T4,6 T4,7 T4,8 T4,9


46


Student View


The outer nodes are kept at a temperature of 60°C. Thus: T1,1 = T1,2 = T1,3 = T1,4 = T1,5 = T1,6 = T1,7 = T1,8 = T1,9 = T1,10 = T2,1 = T3,1 = T4,1 = _____°C For the first calculation, we will assume the following temperature values (in °C) for the rest of the nodes: T1,1 T1,2 T1,3 T1,4 T1,5 T1,6 T1,7 T1,8 T1,9 T1,10 T2,1 T2,2 T2,3 T2,4 T2,5 T2,6 T2,7 T2,8 T2,9 T2,10 T3,1 T3,2 T3,3 T3,4 T3,5 T3,6 T3,7 T3,8 T3,9 T3,10 T4,1 T4,2 T4,3 T4,4 T4,5 T4,6 T4,7 T4,8 T4,9 60

60

60

60

60

60

60

60

60 60

60

55

55

55

55

55

55

55

55 35

60

50

50

50

50

50

50

50

50 10

60

45

40

35

30

25

20

15

10

We can obtain the first temperature estimation for the interior nodes using Equation 4.15-11 from Geankoplis: Tn – 1,m

q n,m = Tn −1,m + Tn +1,m + Tn,m −1 + Tn,m +1 − 4Tn,m

Tn,m – 1

This equation is applicable to the nodes highlighted below.

Tn,m Tn,m + 1 Tn + 1,m

T1,1 T1,2 T1,3 T1,4 T1,5 T1,6 T1,7 T1,8 T1,9 T1,10 T2,1 T2,2 T2,3 T2,4 T2,5 T2,6 T2,7 T2,8 T2,9 T2,10 T3,1 T3,2 T3,3 T3,4 T3,5 T3,6 T3,7 T3,8 T3,9 T3,10 T4,1 T4,2 T4,3 T4,4 T4,5 T4,6 T4,7 T4,8 T4,9

To start the calculations, we will select node T2,2. We can apply this equation to get: q 2,2 = T1,2 + ______ + T2,1 + ______ − 4T2,2

q 2,2 = ____ + 50 + ____ + 55 − 4 ( ____ ) = ____


47


Student View


Since the heat q2,2 is not equal to zero, the value of T2,2 we assumed is not the temperature at steady state. Setting the value of q2,2 to zero we calculate a new value of T2,2 as shown below: T2,2 = T2,2 =

T1,2 + ____ + T2,1 + ____

4 ____ °C + 50°C + ____ °C + ____ °C = _________ °C 4

This new value of T2,2 will be used to calculate the temperatures at other nodes. Thus, for q2,3: q 2,3 = ____ + T3,3 + ____ + T2,4 − _______ = 60 + 50 + ________ + 55 − _______ = _______ Setting q3,2 = 0 and solving for T3,2 yields: T2,3 =

60 + 50 + _______ + 55 = _______ °C 4

We can repeat the same procedure for all the interior nodes. The first iteration will yield the following temperature values: 60

60

60

60

60

60

60

60

60

60

60

______ ______ ______ ______ ______ ______ ______ ______ 55

60

______ ______ ______ ______ ______ ______ ______ 50

60

45

40

35

30

25

20

15

10

10

Note that we have not done any calculations for the edge nodes. This is because we need additional equations for these nodes, described in the following sections. Section 4.15B-3 of Geankoplis and Section 4.5-3 of Incropera and DeWitt [3] gives the following equations for different boundary conditions. For nodes T2,10, T4,2, T4,3, T4,4, T4,5, T4,6, T4,7, and T4,8, an equation with an adiabatic boundary is needed [3].

3. Incropera, F. P., DeWitt, D. P., Fundamentals of Heat and Mass Transfer, Fourth Edition, John Wiley & Sons, New York (1996).


48


Student View


This equation is applied for nodes with heat conduction from three adjacent nodes with an adjoining adiabat, as shown in the following figures: Tn – 1,m Tn,m – 1

q n,m = 2Tn,m −1 + Tn −1,m + Tn +1,m − 4Tn,m

Tn,m Tn + 1,m

Adiabat

Tn – 1,m

q n,m = 2 _______ + Tn,m −1 + Tn,m +1 − 4 ______ Tn,m – 1 Tn,m Tn,m + 1

Adiabat

Setting qn,m = 0 in these equations, the temperatures are given by: Tn – 1,m

Tn ,m =

Tn,m – 1

2Tn,m −1 + ______ + Tn +1,m

Tn,m Tn + 1,m

___ −

Adiabat Tn – 1,m

Tn ,m =

2 ______ + Tn,m −1 + Tn,m +1 ____

Tn,m – 1 Tn,m Tn,m + 1

Adiabat

Applying this equation to node T4,2 (with q4,2 = 0), we have:

T4,2 =

2 _____ + T4,1 + _____ ___

=

2 ( ______ °C ) + 60°C + 40°C = _____ °C ___

In a similar way, we can find the first estimate of the temperatures at rest of the nodes that follow this equation, highlighted in the grid below: T1,1 T1,2 T1,3 T1,4 T1,5 T1,6 T1,7 T1,8 T1,9 T1,10 T2,1 T2,2 T2,3 T2,4 T2,5 T2,6 T2,7 T2,8 T2,9 T2,10 T3,1 T3,2 T3,3 T3,4 T3,5 T3,6 T3,7 T3,8 T3,9 T3,10 T4,1 T4,2 T4,3 T4,4 T4,5 T4,6 T4,7 T4,8 T4,9


49


Student View


60

60

60

60

60

60

60

60

60

60

60

______ ______ ______ ______ ______ ______ ______ ______ ______

60

______ ______ ______ ______ ______ ______ ______

50

60

______ ______ ______ ______ ______ ______ ______

10

10

To finish the first temperature estimation, we need to determine 3 more nodes: T3,9, T3,10 and T4,9. In nodes T4,9 and T3,10, heat transfer is occurring at an insulated boundary with convection from internal flow and conduction from the two adjacent nodes. The following equation (at steady state) can be used for this case, also given in page 191 of Incropera [3]. The thermal conductivity of stainless steel 304 was obtained from Appendix A.3 of Geankoplis.

Tn ,m

 h ∆x  Tn,m −1 + Tn −1,m +   T∞ k   = h ∆x   2+  k  

Tn – 1,m Tn,m – 1

Adiabat

Tn,m

T∞, h

Substituting the corresponding temperatures into this equation yields:

T3,10

W    400 m 2 ⋅ K ( ______ m )  ____ °C + 45°C +  10°C )  (  h∆x  W T3,9 + _____ +  T∞    ______  k =   = ______ °C m⋅K = h ∆x  W   400 2 ( ______ m )  2+   k  m ⋅K  2 +  W   ______  m⋅K 

Now for T4,9 we have:

T4,9

W    400 m 2 ⋅ K ( ______ m )  31.3°C + ____ °C +  10°C )  (  h ∆x  W T4,8 + T3,9 +  T∞    ______ k    ⋅  m K = = = ______ °C h ∆x  W    400 2 ( ______ m )  2+   k  m ⋅K  2 +   W   ______  m⋅K 


50


Student View


For node T3,9, we can use Equation 4.15-19 of Geankoplis, which corresponds to an interior corner with convection at the boundary: Tn – 1,m 1  h ∆x  ( Tn +1,m + Tn,m+1 ) + Tn,m−1 + Tn −1,m +  k T∞  Tn,m Tn,m+1 2   Tn,m – 1 Tn ,m = h ∆x   T∞, h 3+  k   T n + 1,m

1  h ∆x  T4,9 + ______ ) + T3,8 + ______ +  T∞  ( 2 k   T3,9 = h ∆x   3+  k   W   400 2 ( ______ m )   1 ⋅ m K 10°C )  ( ______ °C + ______ °C ) + 40.56°C + 55°C +  ( W 2   ______ m K  ⋅  T3,9 = W    400 m 2 ⋅ K ( ______ m )  3 +  W   ______  m⋅K 

T3,9 = ______ °C

The following grid shows the temperatures obtained after the first iteration. The highlighted temperatures are the values we just calculated. 60

60

60

60

60

60

60

60

60

60

60

______ ______ ______ ______ ______ ______ ______ ______ ______

60

______ ______ ______ ______ ______ ______ ______ ______ ______

60

______ ______ ______ ______ ______ ______ ______ ______

After completing the first calculation across the grid, we can start a new approximation using the new temperature values. Hence, starting with q2,2 and T2,2, we have: q 2,2 = T1,2 + T3,2 + T2,1 + T2,3 − 4T2,2

q 2,2 = 60 + 52.81 + 60 + ______ − 4 ( ______ ) = ______


51


Student View


Setting the value of q2,2 to zero we calculate a new value of T2,2 as follows: T2,2 =

60 + _______ + 60 + _______ = _______ °C 4

The same procedure used for the first approximation is repeated until the assumed and new temperatures are similar. In this case we will select a tolerance of 0.01. This numerical problem can also be solved using computer software such as Excel or Matlab. The final temperature values in the set of nodes for the edges of the bipolar plate are shown below:

60

60

60

60

60

60

60

60

60

60

______ 59.98 ______ ______ 59.88 ______ ______ 59.55 ______

60

______ ______ 59.93 ______ ______ 59.65 ______ ______ 58.88

60

59.98 ______ ______ 59.86 ______ ______ 59.30 ______

To calculate the total heat lost by the bipolar plate we use Fourier’s Law of Heat Conduction for the interior and exterior nodes. q = kA

∆T ∆T = k∆xL = kL ( ∆T ) ∆x ∆x

This amount must be multiplied by 4, since this set of nodes is repeated 4 times in the fuel cell bipolar plate (4 external corners), as shown in the shaded areas below:


52


Student View


The total heat conducted is the sum of the heat equations applied for all the interior temperature differences. The following figure illustrates the nodes used for the heat loss calculation. The nodes adjacent to the shaded squares were used for determining qinterior, with a direction for heat flow indicated by the arrows. The heat flux for nodes with an adjoining adiabat has to be multiplied by 1 because of symmetry. 2 Symmetry adiabat T2,8 T2,9

T2,10

T3,8 T3,9 T3,10

T4,8

T4,9

Symmetry adiabat

Hence, for the interior nodes we have: q interior,corner = 4kL 0.5 ( T3,10 − T2,10 ) + 0.5 ( T4,9 − T4,8 ) + ( _____ − _____ ) + ( _____ − _____ ) 

W   q interior,corner = 4 16.3  ( _____ m ) 0.5 ( _____ − 59.50 ) + 0.5 ( 58.78 − ______ ) m⋅K   + ( ______ − 59.55 ) + ( 59.02 − ______ )  °C q interior,corner = _______ W

To determine qexterior, we substituted the temperature differences in the nodes adjacent to the shaded squares in the following figure. The direction of heat flow is indicated by the arrows , with a The 1 heat flux for nodes with an adjoining adiabat has to be multiplied by because of symmetry. 2

T2,1 T3,1 T4,1

T1,1 T1,2 T1,3 T1,4 T1,5 T1,6

T1,7 T1,8

T1,9 T1,10

T2,2 T2,3 T2,4 T2,5 T2,6

T2,7 T2,8

T2,9 T2,10

Symmetry adiabat

T3,2 T4,2

Symmetry adiabat


53


Student View


Applying Fourier’s law to the exterior nodes, the heat transfer rate will be given by:

q exterior,corner = 4kL 0.5 ( _____ − T1,10 ) + 0.5 ( T4,2 − _____ ) + ( _____ − T1,9 ) + ( T2,8 − _____ )

+ ( _____ − T1,7 ) + ( T2,6 − _____ ) + ( _____ − T1,5 ) + ( T2,4 − T1,4 ) + ( T2,3 − T1,3 ) + ( T2,2 − T1,2 ) + ( T2,2 − T2,1 ) + ( T3,2 − T3,1 ) 

W   q exterior,corner = 4  16.3  ( ______ m ) 0.5 ( ______ − 60 ) + 0.5 ( ______ − 60 ) + ( ______ − 60 ) m⋅K   + ( ______ − 60 ) + ( ______ − 60 ) + ( ______ − 60 ) + ( ______ − 60 ) + ( ______ − 60 )

+ ( ______ − 60 ) + ( ______ − 60 ) + ( ______ − 60 ) + ( ______ − 60 )  °C

q exterior,corner = ___________ W

Finding the same value as qinterior, corner proves that this system is at steady – state. The heat transfer rate for this set of nodes will be obtained from the average between the heat transfer for the interior and exterior nodes. q corners = __________ W

Set of nodes between cooling channels We need to establish a different nodal network for the spaces between cooling channels. The set of nodes for a 1 mm x 1 mm grid is shown below: T1,11 T1,12 T1,13 T1,14 T1,15 T1,16 T2,11 T2,12 T2,13 T2,14 T2,15 T2,16 T3,11 T3,12 T3,13 T3,14 T3,15 T3,16 T4,12 T4,13 T4,14 T4,15 T4,16

The same equations used for calculating the temperatures at the exterior corners of the bipolar plates will be used again for this set of nodes. The only exception is for node T4,16, which represents the case for heat conduction with two adjoining adiabats. Thus, for this node (setting q4,16 = 0) we have:

Tn ,m =

Tn,m −1 + Tn −1,m

Tn – 1,m

2 Tn,m – 1

T4,16 =

______ + T4,15

Tn,m

Adiabats

2


54


Student View


Following the same procedure described for the first set of nodes in this problem (for exterior corners), or using computer software, we can find the temperatures for the interior nodes to be: T1,11 T1,12 T1,13 T1,14 T1,15 T1,16

60

60

60

60

60

60

T2,11 T2,12 T2,13 T2,14 T2,15 T2,16

59.49 ______ ______ 59.78 ______ ______

T3,11 T3,12 T3,13 T3,14 T3,15 T3,16

______ 58.99 ______ ______ 59.71 ______

T4,12 T4,13 T4,14 T4,15 T4,16

______ 59.26 ______ ______ 59.71

Now we can proceed to calculate the heat transfer rate using Fourier’s law for the temperature difference at the interior and exterior nodes. If we look at the top view of the bipolar plates, we can see that this set of nodes is repeated 4 times in the space between two channels as illustrated below.

Since there are 10 channels, there will be 9 spaces between the cooling channels (see figure below). Therefore the heat transfer rate in a single set of nodes must be multiplied by 36 to obtain the heat transfer in the whole bipolar plate.

In a similar way as we did for the set of nodes for the corners of the bipolar plate, we can determine the heat flow for the set of nodes including the spaces between cooling channels. Thus, q interior,middle = 36kL 0.5 ( T3,11 − T2,11 ) + 0.5 ( _____ − _____ ) + ( _____ − _____ ) + ( T3,12 − T3,13 ) 

W   q interior,middle = 36 16.3  ( _______ m ) 0.5 ( _______ − 59.49 ) + 0.5 ( _______ − 59.26 ) m⋅K   + ( 58.99 − _______ ) + ( 58.99 − _______ )  °C q interior,middle = __________ W

Similarly for exterior nodes, we have:


55


Student View


W   q exterior,middle = 36  16.3  ( _______ m ) 0.5 ( _____ − _____ ) + 0.5 ( _____ − _____ ) + ( T2,12 − T1,12 ) m⋅K   + ( _____ − _____ ) + ( T2,14 − T1,14 ) + ( _____ − _____ ) + ( _____ − _____ ) + ( _____ − _____ ) 

W   q exterior,middle = 36  16.3  ( 0.15 m ) 0.5 ( 59.49 − 60 ) + 0.5 ( _______ − 59.71) + ( _______ − 60 ) m⋅K   + ( _______ − 60 ) + ( 59.78 − 60 ) + ( _______ − 60 ) + ( _______ − _______ )

+ ( 59.71 − _______ )  °C q exterior,middle = ________ W

Taking the average of q interior,middle and q exterior,middle we get: q middle =

________ W − ________ W 2

q middle = ________ W

To obtain the overall heat transfer in the bipolar plate we need to add q corners and q middle : q = ________ W − ________ W q = _________ W


56


Student View


Chapter 5 Principles of Unsteady - State Heat Transfer In this chapter, we will study chemical processes where heat transfer is taking place due to a temperature difference within the system which is changing with time. The following problem modules illustrate different examples where unsteady – state heat transfer processes are occurring in fuel cell vehicles and in the processes for producing fuel for fuel cells. 5.2-1 Cooling of a Cylindrical Solid – Oxide Fuel Cell 5.2-2 Total Amount of Heat in Cooling of a Solid – Oxide Fuel Cell 5.3-2 Heat Conduction in a Fuel Cell Stack 5.3-3 Transient Heat Conduction in a Cylindrical Solid – Oxide Fuel Cell 5.3-4 Two – Dimensional Conduction in a Cylindrical Solid – Oxide Fuel Cell 5.4-1 Unsteady – State Conduction and the Schmidt Numerical Method 5.4-3 Unsteady – State Conduction with Convective Boundary Condition


1


Student View

Principles of Unsteady-State Heat Transfer

Example 5.2-1: Cooling of a Cylindrical Solid – Oxide Fuel Cell A cylindrical solid – oxide fuel cell has an inner radius of 3.1 mm, outer radius of 3.9 mm and a length of 0.2 m. The fuel cell initially at a uniform temperature of 873.15 K enters a medium whose temperature is 60°C. Determine how much time in minutes is required for the fuel cell to be cooled W down to a temperature of 340.23 K if the convective coefficient of the medium is 12 2 . m ⋅K The properties of the membrane electrode assembly for this type of fuel cell are estimated by Xue et al. [1] to be: ρ = 6337.3

kg m3

Cp = 594.3

J kg ⋅ K

k = 2.53

W m⋅K

Strategy This problem can be solved using the simplified equations for systems with negligible internal resistance. Solution The equation for the dimensionless temperature as a function of time is given by:  

h

 

t −  C ρx  T − T∞ = e  p 1 T0 − T∞

In this equation: T∞ = Temperature of the medium T0 = Initial temperature of the fuel cell

x1 = Characteristic dimension of the body However, this equation is only applicable when the Biot number NBi is less than 0.1. Hence, we need to obtain this dimensionless number first, defined as: N Bi = _________

For a cylindrical object, the characteristic length is calculated as follows: x1 =

r 2

1. Xue, X., Tang, J., Sammes, N., Du, Y., Journal of Power Sources, 142, 211 − 222 (2005)


2


Student View


In this case, since the fuel cell is a hollow cylinder, the characteristic dimension will be obtained as shown below:

x1 =

ro − ri = 2

( 3.9 mm − 3.1 mm ) 

1m    1000 mm 

2

= _____________ m

Substituting this parameter into the definition of Biot number, we get: W   12 2  ( _____________ m ) m ⋅K   = _____________ N Bi = W    2.53  m⋅K  

Since this value is less than 0.1, we can proceed to enter the corresponding values into the equation for the dimensionless temperature, to yield:

340.23 K − _________ K =e _________ K − _________ K

  W   12 2   ⋅ m K − t   594.3 J   6337.3 kg ( ______________ m )      kg⋅K   m3   

Solving for the time, we have: t=−

 340.23 K − _________ K  1 ln  −1 ( _______________ s )  _________ K − _________ K 

 _________  t = _________ s    _________  t = _________ min


3


Student View


Example 5.2-2: Total Amount of Heat in Cooling of a Solid – Oxide Fuel Cell Determine the total amount of heat removed from the fuel cell in Problem 5.2-1 after 5 minutes. The characteristic length of the fuel cell is 4 × 10 −4 m .

Strategy The equation for the total amount of heat in J can be used to solve this problem.

Solution The equation for the total amount of heat is given by:  h   t  −  Cp ρx1      Q = C pρV(T0 − T∞ ) 1 − e    

The volume of the cylindrical fuel cell is calculated as follows: V = π ( ______ − ______ ) L V = π [ _______________ − _______________ ] 0.2 m

V = 3.52 × 10 −6 m 3

Substituting the calculated volume and the rest of the values into the equation for the total amount of heat transferred, yields:   W     12 2   ⋅ m K   − ____ s  J  kg   594.3 6337.3 3 ( 4×10−4 m )       J  kg  kg ⋅K  m  3   Q =  594.3   6337.3 3  ( __________ m ) (873.15 K − 333.15 K) 1 − e kg ⋅ K   m         Q = ____________ J


4


Student View


Example 5.3-2: Heat Conduction in a Fuel Cell Stack A stack of 220 fuel cells is initially at a temperature of 353.15 K. This stack enters a room at a temperature of 266.48 K. Assuming all the sides but the front face of the stack are insulated, determine the temperature of the 60th fuel cell after 90 minutes. The thickness of the bipolar plates in this fuel cell stack is 2 mm and it is much thicker than the membrane electrode assembly. The heat W transfer coefficient of the air in this room is 91 2 m ⋅K The properties of the bipolar plates in this fuel cell stack are given below [2]: ρ = 1632

kg m3

Cp = 1414

J kg ⋅ K

k = 24.42

W m⋅K

A schematic of this fuel cell stack is shown in the following figure: ncell = 60

ncell = 220

ncell = 1

T = 266.48 K The sides and back face of the fuel cell stack are insulated

x x1

Strategy The temperature at a point inside the fuel cell stack can be obtained from the charts for unsteady – state heat conduction of a slab.

Solution To use Figure 5.3-5 of Geankoplis to determine the unsteady – state heat conduction in a flat plate, we need to find the parameters X, m and n, defined as: X = ________

m=

k hx1

n = __________

2. King, J.A., Lopez Gaxiola, D., Johnson, B.A., Keith, J.M., Journal of Composite Materials, 44 (7), 839 – 855 (2010).


5


Student View


The characteristic length x1 in these equations corresponds to the distance from the front face to the center of the fuel cell stack, obtained by multiplying the number of fuel cells by the thickness of a bipolar plate, and dividing it by 2. Thus, x1 =

_______ m ( 220 ) 2

x1 = __________ m

The parameter α is the diffusivity of the fuel cell stack defined as: α=

k ______

Substituting the properties of the bipolar plates into this equation, we get:

24.42 α= _________

W m⋅K

kg  J  _________ 3  m  kg ⋅ K 

= _______________

m2 s

To determine the value of n, we need to know the distance at which the 60th fuel cell is located from the center of the stack. This distance can be calculated by multiplying the thickness of a single bipolar plate by (110-60) which is the number of fuel cells from the center.

x = (110 − 60 ) (_________ m) = _________ m Now we can calculate the three values required to use Figure 5.3-5, as shown in the following steps:   60 s  m2  _______________   ( 90 min )   s   1 min  = _________ X= 2 ( _________ m ) 24.42 m=

n=

W m⋅K

W 91 2 ( _________ m ) m ⋅K

= 1.22

_________ m = 0.45 _________ m

From Figure 5.3-5 we can read a value for a dimensionless temperature Y = _______. This value can be used to solve for the temperature in the 60th fuel cell as follows: Daniel López Gaxiola Jason M. Keith

6


Student View


Y=

T1 − T = _________ T1 − T0

where: T1 = Temperature of the cooling medium T0 = Initial temperature of the fuel cell stack. We can enter the corresponding temperatures and solve for the temperature T to get:

T = T1 − _________ ( _______ − _______ ) T = 266.48 K − _________ ( 266.48 K − _________ K ) T = _________ K


7


Student View


Example 5.3-3: Transient Heat Conduction in a Cylindrical Solid – Oxide Fuel Cell A cylindrical solid – oxide fuel cell with a diameter of 3.9 mm and a length of 0.2 m is initially at a temperature of 1150 K. The fuel cell is shut down in a room where the air is at a temperature of 303 K. Calculate the temperature at the center of the fuel cell after 5 minutes, assuming it is insulated on W the flat ends. The heat transfer coefficient of the air is 10 2 . The thermal conductivity and m ⋅K diffusivity of the fuel cell are 2.53

m2 W and 6.72 × 10−7 , respectively. m⋅K s

Strategy The solution to this problem can be found by using the charts for unsteady – state heat conduction in a cylinder.

Solution Since heat is only being transferred through the walls, the fuel cell can be considered as a long cylinder. The parameters n, m and X required to determine the dimensionless temperature Y from Figure 5.3-8 are calculated as shown in the following steps: At the center of the cylinder, x = 0. Thus, the value of n will also be equal to zero: n=

x 0m = x1 ____________ m

n=0 To calculate the value of X, we need to substitute the diffusivity, radius of the fuel cell and the time elapsed, as shown below: X=

αt x12

  60 s  m2  _____________   ( ___ min )   s  1 min    = _________ X= 2 ( _____________ m ) Finally, we can determine the parameter m: m=

k hx1


8


Student View


_______ m=

W m⋅K

W _______ 2 3.9 × 10−3 m ) ( m ⋅K

= ________

For these values, the corresponding Y from Figure 5.3-8 will be equal to ______. Now we can solve for the temperature at the center of the fuel cell from the definition of the dimensionless temperature Y: Y=

T1 − T = ______ _________

T = T1 − __________________

where: T1 = Temperature of cooling medium T0 = Initial temperature of the fuel cell Substituting the temperature values in this equation, we get:

T = 303 K − ______ ( 303 K − ________ K ) T = _________ K


9


Student View


Example 5.3-4: Two – Dimensional Conduction in a Cylindrical Solid – Oxide Fuel Cell Determine the temperature at the center of the solid – oxide fuel cell from Example 5.3-3 now considering heat conduction also occurring through the ends of the cylinder. What does this result indicate?

Strategy The procedure to solve this problem consists of using the charts for unsteady – state heat transfer in a cylinder for both radial and axial directions.

Solution We need to calculate the required dimensionless quantities X, m and n for both radial and axial directions. For the radial direction, these will be given by: x ______ m = x1 _____________ m

n radial =

n radial = ____ m radial =

k hx1

________ m radial =

X radial =

X radial

W m⋅K

W _____ 2 ( 3.9 ×10−3 m ) m ⋅K

= _______

αt x12

  60 s  m2  ______________   ( 5 min )   s   1 min  = ________ = 2 ( ______________ m )

These values will yield a dimensionless temperature Yradial = ______ For heat conduction in the axial direction, we need to calculate the parameters n, m and X and locate them in Figure 5.3-6, applicable for two parallel planes. Thus, n axial =

y 0m = y1 0.1 m

n axial = 0


10


Student View


m axial =

k hy1

______ m axial =

X axial =

X axial

W m⋅K

W ______ 2 ( ______ m ) m ⋅K

= ______

αt x12

 m2  _____________   ( ______ min )( ______ ) s   = = ______ 2 ( ______ m )

Locating these three parameters in Figure 5.3-6 gives a value of Yaxial = ___. Now that we have both Y values for both directions, we can obtain a Y for the overall heat transfer process as follows: Y = Yaxial Yradial = ________

Now we can solve for the temperature at the center of the cylinder to get: Y=

T1 − T = __________ T1 − T0

T = T1 − _________________

T = ______ K − ______ ( ____________________ ) T = _________ K

Conclusion:


11


Student View


Example 5.4-1: Unsteady – State Conduction and the Schmidt Numerical Method A proton – exchange membrane fuel cell stack has a thickness of 0.3 m and is initially at a uniform temperature of 60°C. The front face of the fuel cell stack suddenly exposed to an environmental m2 . s Assuming the convective resistance is negligible and that the back face of the stack is insulated, determine the temperature profile after 24 minutes using the Schmidt numerical method with M = 2 and dividing the fuel cell stack into slices with a thickness of 0.05 m. Follow the special procedure for the first time increment.

temperature of -6.67°C. The bulk thermal diffusivity of the fuel cell stack is 8.69 × 10−6

The next figure illustrates the conditions in this cooling process:

Insulated Face Ta = -6.67 °C

T0 = 60 °C

Strategy The equations we need to use to determine the temperature profile will depend on the boundary conditions.

Solution The number of time steps to use in this problem will be determined from the definition of the parameter M: M=

( ∆x) 2 α∆t

Solving for the time increment ∆t and substituting the corresponding values into this equation, yields: ∆t =

( ∆x) 2 (______ m) 2 = Mα  m2  2  ____________  s  


12


Student View


∆t = _________ s The number of time steps needed for this ∆t is given by:

n time steps =

t (______ min) ( ______ ) = ≈ 10 ∆t ________ s

The front surface of the fuel cell stack corresponds to n =1. In this point, the temperature 1 Ta used for the first time increment is given by: Ta + 0 T1 2

1

Ta =

0

T1 = Initial temperature at point 1.

where:

Ta = Temperature of the environment = -6.67°C

Since there is no convective heat resistance at the interface, for the remaining time increments: T1 = Ta

The general equation for determining the temperature for the slabs n = 2 to 6 is given below: t +∆t

Tn =

t

Tn −1 + t Tn +1 2

We need an additional equation for the insulated face. This is the point where n = 7 and its corresponding equation is given by:

t +∆t

T7 =

( M − 2 )t T7 + 2t T6 M

Now we can proceed to calculate the temperatures for the first time increment. Thus, for n = 1, t+∆t

T1 =

Ta + 0 T1 −6.67°C + _____°C = = ______ °C 2 2

Since the problem indicates we should use the special procedure for the first time increment, this temperature value we just obtained is equal to the ambient temperature at the first time increment: t+∆t

T1 =1 Ta

For n = 2: Daniel López Gaxiola Jason M. Keith

13


Student View


t+∆t

T2 =

T1 + t T3 2

t

The special procedure used at n = 1 for the first time increment also affects this equation: instead of using the temperature of -6.67°C we will use 1 Ta . Hence,

t+∆t

T2 =

1

Ta + t T3 _________ °C + ______ °C = = ________ °C 2 2

The following four slabs corresponding to n = 3 to 6 can be calculated with the general equation for n = 2 to 6, given in previous steps. Substituting the corresponding temperatures into this equation yields: T3 =

t

t +∆t

T4 =

t

t +∆t

T5 =

t

t +∆t

T6 =

t

t +∆t

T2 + t T4 60°C + 60°C = = 60°C 2 2 T3 + t T5 60°C + 60°C = = 60°C 2 2 T4 + t T6 60°C + 60°C = = 60°C 2 2 T5 + t T7 60°C + 60°C = = 60°C 2 2

For n = 7, we use the equation for the insulated face:

t +∆t

T7 =

( 2 − 2 )t T7 + 2t T6 2

= t T6 = 60°C

Now we can proceed to calculate the temperatures from the second to the tenth time increments, as shown in the following steps: For 2∆t: t+2∆t T1

= Ta = ______ °C

t+2∆t T2

=

t + 2 ∆t T3

=

t + 2 ∆t T4

=

t+∆t T1

+ t+∆t T3 2

t+∆t T2

+ t+∆t T4 2

t+∆t T3


+ t+∆t T5 2

=

=

=

______ + 60°C

2

= ______ °C

______ °C + 60°C

2 ______ °C + 60°C

2

= ______ °C

= ______ °C

14


Student View

Supplemental Material for Transport Process and Separation Process Principles t+∆t T4

+ t+∆t T6

______ °C + 60°C

t + 2 ∆t T5

=

t + 2 ∆t T6

=

t + 2 ∆t T7

= t +∆t T6 = ______ °C

2 t+∆t T5

+ t+∆t T7 2

=

2 ______ °C + 60°C

=

2

= ______ °C

= ______ °C

For 3∆t: t+3∆t T1

= Ta = −6.67°C

t+3∆t T2

=

t + 3 ∆t T3

=

t + 3 ∆t T4

=

t + 3 ∆t T5

=

t + 3 ∆t T6

=

t + 3 ∆t T7

= t + 2 ∆t T6 = ____ °C

t+2∆t T1

+ t+2∆t T3 2

t+2∆t T2

+ t+2∆t T4 2

t+2∆t T3

+ t+2∆t T5 2

t+2∆t T4

+ t+2∆t T6 2

t+2∆t T5


+ t+2∆t T7 2

=

=

=

=

=

−6.67°C + ______°C 2

= ______ °C

_______ °C + _______ °C

2 _______ °C + _______ °C

2 _______ °C + _______ °C

2 _______ °C + _______ °C

2

= _______ °C

= _______ °C

= _______ °C

= _______ °C

15


Student View


We can continue to repeat this procedure up to 10∆T. The solutions for the all the temperatures (in °C) are given in the following table: n 1

2

3

4

5

t

60

60

60

______ 60

t+∆t

______ ______ 60

t+2∆t

-6.67

t+3∆t

______ 22.50

t+4∆t

-6.67

______ 39.17

t+5∆t

-6.67

______ ______ 48.54

t+6∆t

-6.67

14.16

t+7∆t

-6.67

______ 29.79

t+8∆t

-6.67

______ ______ 40.73

t+9∆t

-6.67

10.65

t+10∆t

-6.67

______ 24.74

60

6

7

______ 60

______ 60

______ ______ 60

______

______ ______ ______

______ ______ 60

______ 60

______ ______ 60

______

______ ______ ______

______ ______ 53.75

______ 58.96

______ ______ 56.35

______

______ ______ ______

______ ______ 47.76

______ 54.79

______ ______ 51.28

______

We can plot these temperature results to observe the change in temperature within the fuel cell stack: 60 t t+Δt

50

t+2Δt

40 t+3Δt

30 T (°C)

t+4Δt t+5Δt

20

t+6Δt

10 t+7Δt

0

t+8Δt t+9Δt

-10 1

2

3

4

5

6

7

t+10Δt

n


16


Student View


Example 5.4-3: Unsteady – State Conduction with Convective Boundary Condition Determine the temperature profile for the same fuel cell stack as in Example 5.4-1, now with a W W convective coefficient of 13 2 . The bulk thermal conductivity of the fuel cell stack is 20 . m ⋅K m⋅K Use a value of M = 4.

Insulated Face

Ta = -6.67 °C W h = 13 2 m ⋅K

T0 = 60 °C

Strategy In this problem we need to use Schmidt numerical method depending on the boundary conditions.

Solution Before we start applying Schmidt method, we need to determine the number of time increments we need to use. From the definition of the parameter M, we can solve for the time increment ∆t , as shown in the following steps: ( ∆x) 2 M= α∆t

∆t =

( ∆x) 2 (0.05 m) 2 = = ________ s 2 αM  −6 m   8.69 ×10  ( 4) s  

The number of time steps needed for this ∆t is given by:

n time steps

 60 s  (______ min)   t  1 min  ≈ ______ = = ∆t 71.92s

As stated in Section 5.4B of Geankoplis, when the value of M is greater than 3, the value of the environmental temperature Ta will be the same for all time increments. Therefore, Daniel López Gaxiola Jason M. Keith

17


Student View


Ta = −6.67°C

For the node n = 1, corresponding to the front face of the fuel cell stack, the temperature can be calculated using Equation 5.4-7 of Geankoplis: t +∆t

T1 =

1 2N t Ta + [ M − (2N + 2) ]t T1 + 2 t T2 M

{

}

The value of N is a function of the convective heat transfer coefficient and the thermal conductivity of the fuel cell, described by the following equation: N=

h ∆x k

Entering the corresponding values into this equation, yields: ______ N=

W ( ______ m ) m2 ⋅ K = _________ W 20 m⋅K

In order to use Equation 5.4-7, the value of M must satisfy the following constraint:

M ≥ 2N + 2 Substituting numeric quantities into this constraint, we get: 4 ≥ 2(0.0325) + 2

4 ≥ ______ Hence, we can use Equation 5.4-7. From this equation, we can find the temperature

t +∆t

T1 to be:

1 {2 ( ________ )( −6.67°C ) + [ 4 − (________)] ( 60°C ) + 2(60°C)} 4

t +∆t

T1 =

t +∆t

T1 = ________ °C

For the points at n = 2, 3, 4, 5, 6, we use Equation 5.4-2: t +∆t

Tn =

1  t Tn +1 + ( M − 2 ) t Tn + t Tn −1  M

By entering the value of M into this equation, we can get a general equation for t +∆t

t +∆t

Tn :

Tn = ____________________________________


18


Student View


Now we need an equation for the insulated boundary at n = 7, which is Equation 5.4-10 of Geankoplis:

t +∆t

T7 =

t +∆t

T7 =

( M − 2 )t T7 + 2t T6 M t

T7 + t T6 _____

We can now proceed to determine the temperatures for n = 2 to 7 for the first time increment. Thus, t +∆t

T2 = 0.25 ( t T3 + t T1 ) + 0.5t T2

t +∆t

T2 = 0.25 ( ______ °C + 60°C ) + 0.5 ( ______ °C ) = ______ °C

t +∆t

T3 = 0.25 ( t T4 + t T2 ) + 0.5t T3

t +∆t

T3 = 0.25 ( ______ °C + 60°C ) + 0.5 ( ______ °C ) = ______ °C

t +∆t

T4 = 0.25 ( t T5 + t T3 ) + 0.5t T4

t +∆t

T4 = 0.25 ( ______ °C + 60°C ) + 0.5 ( ______ °C ) = ______ °C

t +∆t

T5 = 0.25 ( t T6 + t T4 ) + 0.5t T5

t +∆t

T5 = 0.25 ( ______ °C + 60°C ) + 0.5 ( ______ °C ) = ______ °C

t +∆t

T6 = 0.25 ( t T7 + t T5 ) + 0.5t T6

t +∆t

T6 = 0.25 ( ______ °C + 60°C ) + 0.5 ( ______ °C ) = ______ °C

t +∆t

T7 =

t +∆t

T7 =

t

T7 + t T6 2

( ____ °C ) + ( ____ °C ) = ____ °C 2

For 2∆t: t + 2 ∆t

T1 =

1 [ 0.065t +∆t Ta + _______ t +∆t T1 + 2t +∆t T2 ] 4


19


Student View


1  0.065t +∆t (_______ °C) + _______ t +∆t ( _______ °C ) + 2 t +∆t ( 60°C )  = _______ °C 4

t + 2 ∆t

T1 =

t + 2 ∆t

T2 = 0.25 ( t +∆t T3 + t +∆t T1 ) + 0.5t +∆t T2

t + 2 ∆t

T2 = 0.25 ( _____ °C + 58.92°C ) + 0.5 ( 60°C ) = 59.73°C

t + 2 ∆t

T3 = 0.25 ( t +∆t T4 + t +∆t T2 ) + 0.5t +∆t T3

t + 2 ∆t

T3 = 0.25 ( ____ °C + 60°C ) + 0.5 ( ____ °C ) = ____ °C

t + 2 ∆t

T4 = 0.25 ( t +∆t T5 + t +∆t T3 ) + 0.5t +∆t T4

t + 2 ∆t

T4 = 0.25 ( ____ °C + ____ °C ) + 0.5 ( ____ °C ) = ____ °C

t + 2 ∆t

T5 = 0.25 ( t +∆t T6 + t +∆t T4 ) + 0.5t +∆t T5

t + 2 ∆t

T5 = 0.25 ( ____ °C + ____ °C ) + 0.5 ( ____ °C ) = ____ °C

t + 2 ∆t

T6 = 0.25 ( t +∆t T7 + t +∆t T5 ) + 0.5t +∆t T6

t + 2 ∆t

T6 = 0.25 ( ____ °C + ____ °C ) + 0.5 ( ____ °C ) = ____ °C

t + 2 ∆t

T7 =

t + 2 ∆t

T7 =

t +∆t

T7 + t +∆t T6 2

( ____ °C ) + ( ____ °C ) = ____ °C 2

For 3∆t: t + 3 ∆t

T1 =

1 [0.065t + 2 ∆t Ta + 1.935t + 2 ∆t T1 + 2t + 2 ∆t T2 ] 4

t + 3 ∆t

T1 =

1  0.065t + 2 ∆t (______ °C) + 1.935t + 2 ∆t ( _______ °C ) + 2 t + 2 ∆t ( 59.73°C )  = ______ °C 4

t + 3 ∆t

T2 = 0.25 ( t + 2 ∆t T3 + t + 2 ∆t T1 ) + 0.5t + 2 ∆t T2

t + 3 ∆t

T2 = 0.25 ( 60°C + ______ °C ) + 0.5 ( 59.73°C ) = ______ °C

t + 3 ∆t

T3 = 0.25 ( t +2 ∆t T4 + t + 2 ∆t T2 ) + 0.5t + 2 ∆t T3


20


Student View

Supplemental Material for Transport Process and Separation Process Principles t + 3 ∆t

T3 = 0.25 ( _______ °C + _______ °C ) + 0.5 ( _______ °C ) = _______ °C

t + 3 ∆t

T4 = 0.25 ( t + 2 ∆t T5 + t + 2 ∆t T3 ) + 0.5t + 2 ∆t T4

t + 3 ∆t

T4 = 0.25 ( _______ °C + _______ °C ) + 0.5 ( _______ °C ) = _______ °C

t + 3 ∆t

T5 = 0.25 ( t + 2 ∆t T6 + t + 2 ∆t T4 ) + 0.5t + 2 ∆t T5

t + 3 ∆t

T5 = 0.25 ( _______ °C + _______ °C ) + 0.5 ( _______ °C ) = _______ °C

t + 3 ∆t

T6 = 0.25 ( t + 2 ∆t T7 + t + 2 ∆t T5 ) + 0.5t + 2 ∆t T6

t + 3 ∆t

T6 = 0.25 ( _______ °C + _______ °C ) + 0.5 ( _______ °C ) = _______ °C

t + 3 ∆t

T7 =

t + 3 ∆t

T7 =

t + 2 ∆t

T7 + t + 2 ∆t T6 2

( ____ °C ) + ( ____ °C ) = ____ °C


2

21


Student View


In a similar way, we can continue to perform this calculation for all the 20 time increments. The results for all temperatures (in °C) are shown in the following table: n 1

2

3

4

5

6

7

t

60

60

60

60

60

60

60

t+∆t

_______ _______ _______ 60

t+2∆t

_______ 59.73

t+3∆t

58.00

t+4∆t

_______ _______ _______ 59.98

t+5∆t

_______ 58.99

t+6∆t

57.15

t+7∆t

_______ _______ _______ 59.85

t+8∆t

_______ 58.38

t+9∆t

56.52

t+10∆t

_______ _______ _______ 59.64

t+11∆t

_______ 57.87

t+12∆t

56.00

t+13∆t

_______ _______ _______ 59.40

t+14∆t

_______ 57.42

t+15∆t

55.54

t+16∆t

_______ _______ _______ 59.15

t+17∆t

_______ 57.02

t+18∆t

55.14

t+19∆t

_______ _______ _______ 58.90

t+20∆t

_______ 56.65


_______ _______ 60

_______ _______ 60

_______ 59.93

_______ _______

_______ _______ 60

_______ _______ 60

_______ _______ 60

_______ 59.59

_______ _______

_______ _______ 60

_______ _______

_______ _______ 59.98

_______ _______

_______ _______ 59.93

_______ _______

_______ _______ 59.85

_______

_______ _______ 59.87

_______ _______ 59.54

_______ 58.15

_______

_______ _______ 59.95

_______ _______ 59.70

_______ 58.48

_______

_______ _______ 59.99

_______ _______ 59.84

_______ 58.84

_______

_______ _______ 60

_______ _______ 59.94

_______ 59.21

_______

_______ _______

_______ _______ 59.73

_______

_______ _______ 59.77

_______ _______ 59.36

22


_______ _______

Student View


The graphical representation of these temperatures is shown in the following figure: 60

t

t+Δt

t+2Δt

t+3Δt

t+4Δt

t+5Δt

t+6Δt

t+7Δt

t+8Δt

t+9Δt

t+10Δt

t+11Δt

t+12Δt

t+13Δt

t+14Δt

t+15Δt

t+16Δt

t+17Δt

t+18Δt

t+19Δt

59

T (°C)

58

57

56

55

54 1

2

3

4

5

6

t+20Δt

n


7

23


Student View


Chapter 6 Principles of Mass Transfer Chapter 6 illustrates different situations where the third fundamental transfer process, mass transfer, is occurring in fuel cells or in the chemical processes to produce fuel for fuel cells. The following examples explain in detail how the flux of different species is affecting the performance of fuel cells. 6.1-1 Molecular Diffusion of Water in Air at the Cathode of a Fuel Cell 6.2-1 Equimolar Counterdiffusion at the Cathode Chamber of a Proton – Exchange Membrane Fuel Cell 6.2-5 Diffusivity in Steam-Methane Reforming Process 6.3-1 Diffusion of Methanol in Water in Direct Methanol Fuel Cells 6.3-2 Prediction of Diffusivity of Methanol in Water 6.5-1 Diffusion of Hydrogen through Nafion 6.5-3 Diffusion of Oxygen through Gas Diffusion Layer


1


Student View

Principles of Mass Transfer

Example 6.1-1: Molecular Diffusion of Water in Air at the Cathode of a Fuel Cell Proton – Exchange Membrane Fuel Cells consume hydrogen and oxygen from air to generate electricity, with heat and steam as byproducts. The steam being produced in the cathode side has a partial pressure of 0.22 bar. Some of the steam stays inside the fuel cell to keep the polymer – electrolyte membrane at an adequate humidity. The steam in the bipolar plate channels has a partial pressure of 0.024 bar. The distance between the gas diffusion layer (GDL) and the bipolar plate is 0.729 mm. Determine the diffusive flux from the gas diffusion layer to the bipolar plates of water in the air. The fuel cell is operating at a pressure and temperature of 2 bar and 40°C, respectively. The following schematic illustrates the diffusion process taking place in the fuel cell: GDL

H2 O diffusing in air

Anode

Cathode Electrolyte Membrane

Strategy

We can use Fick’s law of diffusion to determine the steam flux through the air. Solution The equation that describes the flux of a chemical species due to diffusion is Fick’s Law, given by: J*Az = −D AB

dCA dz

This equation can be separated and integrated to yield: z2

J*Az ∫ dz = _________________________ z1

J*Az = ___________________


2


Student View


The point where the steam is being produced will be considered z1 = 0, whereas z2 = 0.729 mm, since the outlet point is 0.729 mm away. The concentrations of the steam CA2 and CA1 will be determined from the ideal gas law, because the fuel cell is operating at relatively low pressure. Thus,

CA1 =

PA1

CA2 =

RT

PA2 RT

Substituting the partial pressures, temperature and ideal gas constant into these equations, we have:

CA1 =

CA2 =

0.22 bar mol = ______ m3  bar ⋅ m3   _______________  ( 313.15 K ) mol ⋅ K   0.024 bar mol = ________ m3  bar ⋅ m 3   ______________  ( 313.15 K ) mol ⋅ K  

Now we can enter the concentrations we calculated into Fick’s law to obtain the flux of steam, as shown below:  mol mol  ________ 3 − ________ 3   m m m   = ________________ 1 m s    0.729 mm − 0 mm )    (  1000 mm   2

J*Az

J*Az = ________

mol m2 ⋅ s

The diffusivity value was obtained from Table 6.2-1 at a temperature of 42°C, which is relatively close to the operating temperature of 40°C.


3


Student View


Example 6.2-1: Equimolar Counterdiffusion at the Cathode Chamber of a Proton – Exchange Membrane Fuel Cell In the diffusion process illustrated in Example 6.1-1, we calculated the flux of water in a fuel cell. For the same fuel cell, determine the flux of air diffusing in the steam.

Strategy In a similar way to example 6.1-1 this problem can be solved using Fick’s Law of Diffusion.

Solution Since we need to determine the flux of air in the steam, we can start from Fick’s Law. Thus,

J*Bz = _______________

We can separate and integrate this equation to get:

J*Bz ( _____________ ) = − D BA ( ______________ )

J*Bz = ____________________

Since the pressure of the system is low, we can use Ideal Gas Law equation of state to calculate the pressures of air, as shown in the following equations: CB1 =

PB1 RT

C B2 =

PB2 RT

We can substitute these equations into Fick’s Law equation to yield:

 __________________  J*Bz = D BA   RT ( z 2 − z1 )  

Since there is only air and steam in the cathode chamber of the fuel cell, we can calculate the partial pressures of air from Dalton’s Law, which states:

P = ∑ Pi Daniel López Gaxiola Jason M. Keith

4


Student View


where: P = Absolute pressure of the system Pi = Partial pressure of component i in the gas mixture Applying this equation for the number of gases in this problem and solving for the partial pressure of air, we have: P = PA + PB PB = P − PA

Substituting this equation into Fick’s law, gives:  __________ − __________   ________________  J*Bz = D BA   = D BA   RT ( z 2 − z1 ) RT ( z 2 − z1 )    

Now we can enter the corresponding quantities into this equation to obtain the solution to this problem:

 m2  ________ bar − ________ bar J*Bz =  2.88 × 10 −5  s  bar ⋅ m3   _______________   ( 313.15 K )( _____________ m ) mol ⋅ K   J*Bz = __________

mol m2 ⋅ s

The negative in this result indicates that the flux is going from point 2 (outlet stream) to point 1 (steam generation).


5


Student View


Example 6.2-5: Diffusion in Steam – Methane Reforming Process Hydrogen for Proton – Exchange Membrane Fuel Cells is being produced from natural gas in a Steam – Methane Reforming Process. The natural gas reacts with steam to produce a mixture of Hydrogen and Carbon Monoxide, known as synthesis gas at a pressure of 20.45 atm and a temperature of 850°C. Determine the diffusivity of hydrogen in CO at these conditions using the Fuller method.

Strategy We can obtain the diffusivity using Fuller method given in Section 6.2E of Geankoplis.

Solution The diffusivity of a gas mixture can be calculated using the following equation: 1/ 2

−7

1.75

1× 10 T D AB =

 1 1  +    MA MB 

1/3 1/3 P ( ∑ υA ) + ( ∑ υB )   

2

For this problem we will select A = hydrogen, and B = carbon monoxide The atomic diffusion volumes of hydrogen and carbon monoxide can be obtained from Table 6.2-2 of Geankoplis. Hence,

∑υ

A

= 7.07

∑υ

B

= ________

We can substitute the process conditions, as well as the molecular weights and volumes of the gas species into the equation for DAB to get: 1/ 2

1× 10 D AB =

−7

1.75

(1123.15 K )

 1 1   ________ + 28   

1/3 1/3 __________ atm ( 7.07 ) + ( ________ )   

D AB = ________________


2

m2 s

6


Student View


Example 6.3-1: Diffusion of Methanol in Water in Direct Methanol Fuel Cells Direct – methanol fuel cells are a type of fuel cells used for portable devices such as laptops and cell phones. The concentration of the methanol solution at the bottom of the reservoir is 32 wt. %. At the top of the tank, located 2 inches away, the concentration is 40 wt. %. Determine the diffusive flux of methanol in water at steady state at a temperature of 15°C. The densities of these two solutions at this kg kg temperature are 950.6 3 for 32 wt. % CH3OH at the bottom of the tank and 937.2 3 for 40 wt. % m m CH3OH at the top.

Strategy The flux of methanol required in this problem can be calculated using the equation for flux of liquids at steady state, given in Section 6.3B of Geankoplis.

Solution The equation for NA for liquid mixtures is given by Equation 6.3-3 of Geankoplis. NA =

D ABCav x A1 − x A2 ________________

(

)

The diffusivity of methanol in water can be obtained from Table 6.3-1 of Geankoplis to be: D AB = ________________

m2 s

The average concentration of methanol can be obtained from the density and molecular weight of the methanol solution at both concentrations, as shown below: ρ1 ρ2 + M1 M 2 Cav = 2

To calculate the molecular weights at points 1 and 2, we use the following equation: M1 =

M2 =

100kg 100kg kg = = _________ wt.% A @ point 1 wt.% B @ point 1  32 68  kgmol + +  kgmol  MA MB  32 18 

100kg 100kg kg = = _________ wt.% A @ point 2 wt.% B @ point 2  ______ ______  kgmol + + kgmol   MA MB ______   32


7


Student View


Now we can substitute the molecular weights to obtain the average concentration:

kg kg 937.2 3 3 m m + kg kg ________ ________ kgmol kgmol kgmol Cav = = ________ 2 m3 950.6

The remaining unknowns are the molar fractions of methanol at points 1 and 2, and the average molar fraction of water in the tank. This can be done as follows:

x A1

wt.% A @ point 1 MA = wt.% A @ point 1 wt.% B @ point 1 + MA MB

x A2

_____________________ ________ = wt.% A @ point 2 wt.% B @ point 2 + ________ MB

x BM =

x B1 + x B2

2

=

(1 − x A1 ) + (__________)

2

Notice that the average molar fraction of B was written in terms of x A1 and x A2 , since only methanol and water are present in the reservoir. Substituting the weight fractions and molecular weights into these equations yields:

x A1

________ 32 = = ________ ________ 68 + 32 ________

x A2 =

x BM =

________ ________

40 60 + ________ ________

= ________

(__________) + (__________) = ________ 2


8


Student View


We can now enter the values we calculated into the equation for the flux of A to get: m2  kgmol   ________  s  m3  NA = ( ________ − ________ )  0.0254 m  ( 0 − 2 in )   ( ________ )  1 in  1.26 × 10 −9

As we can see in this equation, point 1 is at z = 2 in and point 2 is at z = 0, thus: m2  kgmol  1.26 ×10  ________  s  m3  NA = ( ________ − ________ )  0.0254 m  ( −2 in )   ( ________ )  1 in  −9

N A = __________________


kgmol m2 ⋅ s

9


Student View


Example 6.3-2: Prediction of Diffusivity of Methanol in Water Determine the diffusion coefficient of methanol in water using the Wilke – Chang equation at 15 °C. How does this value compare to the experimental diffusion coefficient in Table 6.3-1? Also calculate the diffusion coefficient of methanol in water at the operating temperature of Direct – Methanol Fuel Cells of 50°C.

Strategy Wilke – Chang equation predicts the diffusivity of liquid mixtures, therefore, it can be used to calculate the diffusion coefficient of methanol in water at a given temperature.

Solution Wilke – Chang equation is given in Section 6.3D of Geankoplis: 1/ 2

D AB = 1.173 × 10−16 ( φM B )

T µ B VA 0.6

where: φ = Association parameter of the solvent = 2.6 (This value is given for water in Section 6.3D)

MB = Molecular weight of the solvent

µB = Viscosity of the solvent in Pa ⋅ s VA = Molar volume of the solute at the boiling point (Calculated using values from Table 6.32) First we can calculate the molar volume of methanol. Since the molecular formula is CH3OH, the molar volume is calculated as follows:

 m3   m3   m3  VA = 1 _________  + 4  _________  + 1 0.0074  kgmol   kgmol   kgmol  

VA = _________

m3 kgmol

Substituting the rest of the parameters into Wilke – Change equation, we have: 1/2

D AB = 1.173 × 10−16  2.6 ( _____ ) 


  288.15  0.6   ( ___________ )( _________ ) 

10


Student View


D AB = _____________

m2 s

The viscosity for water was obtained from Appendix A.2 of Geankoplis at 15°C. We can observe this diffusivity value is relatively close to the experimental value in Table 6.3-1 of 1.26 × 10 −9

m2 . s

We can repeat this procedure to determine the diffusivity of methanol in water at the operating temperature of direct – methanol fuel cells. Again, the viscosity of water was obtained from Appendix A.2 at the temperature of 50°C. 1/2

D AB = 1.173 × 10−16  2.6 (18 ) 

D AB = _______________


  323.15  0.6   ( _____________ )( _________ ) 

m2 s

11


Student View


Example 6.5-1: Diffusion of Hydrogen through Nafion Nafion is a polymer manufactured by DuPont that is widely used as a selective membrane in polymer – electrolyte membrane fuel cells. The protons produced from the electrochemical reaction in the anode side of the fuel cell travel through the membrane to the cathode. The Nafion membrane must be humidified to facilitate the flow of protons through it. At a water uptake of 22 (molecules H2O/molecules of sulfonic groups in the membrane), the concentration of protons in the membrane mol was estimated by Spry and Fayer [1] to be 0.54 . Determine the flux of protons through a L membrane with a thickness of 0.175 mm. The diffusivity at a temperature of 30.5°C and a water m2 uptake of 22 was estimated from data obtained by Ochi et al. to be 1.9 × 10 −9 [2]. Assume that the s reaction in the cathode chamber to form water is occurring instantaneously.

Strategy We can use the diffusive flux equation to calculate the flow of protons through the membrane in the fuel cell.

Solution The equation for diffusive flux is given by:

N A = _____________________

Since the problem is stating that the reaction on the cathode chamber is occurring instantaneously, we can assume the concentration CA2 = _________ . We can now calculate the flux by substituting the concentrations, thickness of the membrane and proton diffusivity into the equation shown above, as shown in the following steps: _______________ NA =

m2  mol  − _________    _________ s  L  1 m  ( _________ mm )   1000 mm  

N A = _______________

  

mol m2 ⋅ s

1. Spry, D.B., Fayer, M.D., Journal of Physical Chemistry B, 113, 10210 – 10221 (2009) 2. Ochi, S., Kamishima, O., Mizusaki, J., Kawamura, J., Solid State Ionics, 180, 580 – 584 (2009)


12


Student View


Example 6.5-3: Diffusion of Oxygen through Gas – Diffusion Layer The oxygen in the cathode side of a proton – exchange membrane fuel cell diffuses to the catalyst layer through a gas – diffusion layer (GDL). The GDL has a thickness of 100 µm, and is typically made of a porous material which may be fabricated with either carbon cloth or carbon paper. The GDL has a porosity of 0.5 and a tortuosity of 1.63 [3]. Determine the flux of oxygen through the GDL if its partial pressure is 98.96 kPa and the operating temperature of the fuel cell is 50 °C. Assume there is steam in the pores of the GDL and that the reaction in the catalyst layer in occurring instantaneously.

Strategy The equation defining diffusive flux in porous solids will be used to calculate the flux of oxygen through the GDL.

Solution Section 6.5C of Geankoplis gives the equation for steady state flux in a porous media:

N A = __________________

The diffusivity DAB can be obtained using the Fuller method from Section 6.2E of Geankoplis to be: D AB = __________________

m2 s

The concentrations of oxygen can be obtained from the ideal gas law, as shown below: CA 1 =

_________ kPa  kPa ⋅ m3  _________   ( _________ K ) kgmol ⋅ K  

CA 1 = ___________

kgmol m3

Since we are assuming the reaction in the cathode is occurring instantaneously, the concentration of oxygen at the catalyst layer will be CA 2 = 0

3. Hao, L., Cheng, P., Journal of Power Sources, 186, 104–114 (2009)


13


Student View


Now we can substitute the corresponding values into the equation for diffusive flux to get:

 m2   kgmol  _________  3.04 × 10 −5   _________  s  m3   NA =   _________ ( _________ µm )     N A = _____________


kgmol m2 ⋅ s

14


Student View


Chapter 7 Principles of Unsteady - State and Convective Mass Transfer This chapter covers different situations where mass transfer is taking place, when the conditions are changing with time and where there is a fluid stream that contributes to these mass transfer processes. The following examples illustrate convective and transient mass transfer in fuel cell systems. 7.5-2 Diffusion and Chemical Reaction in the Anode Chamber of a Direct Methanol Fuel Cell 7.5-4 Diffusion of CO2 and O2 through stagnant Nitrogen in a Solid – Oxide Fuel Cell


1


Student View

Principles of Unsteady – State and Convective Mass Transfer

Example 7.5-2 Diffusion and Chemical Reaction in the Anode Chamber of a Direct Methanol Fuel Cell An aqueous 40 mole % methanol solution is entering the anode of a direct – methanol fuel cell. The fuel is diffusing through the gas diffusion layer (GDL) with a thickness of 0.018 cm [1]. The m2 diffusion coefficient of the fuel in the GDL is estimated by García et al. [1] to be 1× 10−9 . The s governing equation for methanol in the GDL is given by: d 2 CA dz 2

=0

with the following boundary conditions: At z = 0 : CA = C b

At z = δ : N A = k1C A The rate constant for the chemical reaction occurring at the catalyst layer located at z = δ is m k1 = 2.8 × 10−6 . Determine the molar fraction of methanol at z = δ and steady state, if the initial s mol concentration Cb of methanol is 500 3 . m

Strategy The molar fraction at the catalyst layer can be obtained by solving the governing differential equation.

Solution We can start by solving the given differential equation given in the problem statement, as shown in the following steps:

∫

d 2CA dz 2

= ∫0

_______ = c1

∫ dC

A

(1)

= c1 ∫ dz

CA = _________________


2


Student View

Supplemental Material for Transport Process and Separation Process Principles 1. García, B.L., Sethuraman, V.A., Weidner, J.W., White, R.E., Dougal, R., Journal of Fuel Cell Science and Technology, 1, 43 – 48 (2004).

Applying the first boundary condition at z = 0, we have: Cb = ________________ c2 = ______

Substituting c2 into Equation 2 yields: CA = _______________

(2)

At z = δ: CA = _______________

(3)

The equation for methanol flux through the GDL is described by Fick’s Law, given by: N A = −D

dC A dz

(4)

At z = δ, the molar flux of methanol is equal to the reaction rate. Therefore: NA

z=δ

= __________

(5)

Since this process is at steady state, we can equal Equations 4 and 5 to get: −D

dC A = _________ dz

(6)

From Equation 1, we have that c1 = _________ . Substituting this into Equation 6 gives: − D ____ = _________

(7)

We can substitute Equation 3 into Equation 7 to solve for c1, as shown in the following steps:

− Dc1 = k1 ( _______________ ) − Dc1 = k1c1δ + k1C b

−c1 ( ______________ ) = ___________

c1 = ________________


(8) 3


Student View


Now we can substitute this equation for c1 into Equation 2 to yield:

CA = ______________ + C b

(9)

Equation 9 can be evaluated at the boundary condition for z = δ to obtain the molar fraction at this point. Hence,

CA = ______________________________

Reducing this equation and writing CA in terms of the molar fraction of methanol, we have:

C ____ =

______ ( D + ______ )

(10)

where C is the overall concentration of the fuel entering the fuel cell. The overall concentration can be obtained by dividing the feed concentration of methanol Cb by the feed molar fraction of 0.4. Substituting numeric values into Equation 10 gives:

mol CH 3OH  mol CH 3OH   m2  ______ ______ _______________    m3 m3 s    _____ =  mol CH 3OH   m2   m 0.4 +  _______________  (1.8 × 10 −4 m )  _______________    mol s   s    ______ xA =

mol  m2  _______________ cm3  s 

  m2   m +  _______________  (1.8 × 10−4 m )  _______________   s   s   

x A = ______


4


Student View


Example 7.5-4: Diffusion of CO2 and O2 through stagnant N2 in a Solid – Oxide Fuel Cell A solid - oxide fuel cell operating at a temperature of 923.15 K and a pressure of 1.9 atm, is producing CO2 from an electrochemical reaction of CO with oxygen from air. The partial pressures of each gas at the gas diffusion layer the bipolar plate channels, located 0.729 mm away, are given in the following table. Label

Gas

Partial Pressure (atm) at Gas – Diffusion Layer

Partial Pressure (atm) at Bipolar Plate

A

CO2

0.47

0.01

B

O2

0.31

0.47

C

N2

1.12

1.42

Determine the molar flux of CO2 and O2 in non – diffusing N2. The diffusion coefficients are given below:

D AB

m2 = 5.84 × 10 s −5

D AC = 5.93 × 10 −5

m2 s

D BC = 7.49 × 10−5

m2 s

Where: A = CO2, B = O2 and C = N2. These diffusion coefficients were estimated using Fuller et al. method described in Section 6.2E of Geankoplis. The following equations describe multicomponent diffusion for two components diffusing in stagnant C [2]:

 PC NA N P + B = ln  2 D AC D BC RT ( z 2 − z1 )  PC1

   

2. Geankoplis, C.J., Mass Transport Phenomena, Holt, Rinehart and Winston Inc., New York, 1972.


5


Student View


1 1 1  1  − D −D N +N N + NB D D BC  AC A B  AB PB2 − A PA2 + AC P 1 1 NB NB  1 − 1  −  D AB DBC DAB D BC  D AB P NA + NB = ln   1 1 RT ( z 2 − z1 )  1 − 1  − NA + NB D AC D BC   DAB D AC N A + N B PB1 − PA1 + P  1 1 1 1 NB NB − −   D AB D BC   D AB DBC

In this problem, the flux will be assumed positive from the bipolar plate to the gas – diffusion layer.

Strategy We can determine the fluxes NA and NB by simultaneously solving the equations given in the problem statement.

Solution First, we can substitute the pressures given and the diffusivity coefficients, as well as the operating conditions of the fuel cell. Thus, NA

5.93 × 10

−5

m

2

NB

+

7.49 × 10

s

−5

m s

2

=

1.9 atm

 _______________ L ⋅ atm  ( 923.15K )( _______________ m )   mol ⋅ K  

 ______ atm    ______ atm 

ln 

We can solve this equation for NB by following the next steps:   1.9 atm  ______ atm  NA =  ln   3  m ⋅ atm   1.12 atm  ______________ 923.15K ______________ m )( )  ( mol ⋅ K     NB m2 − _________________ 2  s −5 m  7.49 × 10  s  N B = 0.612

mol − ____________ m2 ⋅ s

(1)

So far we have obtained one of the simultaneous equations for the diffusion process occurring in this fuel cell. However, in order to solve this problem, we need a second equation, obtained from the equation for N A + N B , as shown below: Daniel López Gaxiola Jason M. Keith

6


Student View

Supplemental Material for Transport Process and Separation Process Principles _____________ NA + NB =

m2 (1.9atm ) s

 m3 ⋅ atm  −4  _____________  ( 923.15 K ) ( 7.29 × 10 m ) mol ⋅ K   1 1  −  m2 m2  _____________ 5.93 × 10−5 s s N A + N B ( ____ atm ) − N A + N B ( 0.01 atm ) +  1 1  NB NB − 2  m2 −5 m  5.84 × 10 _____________ s s × ln   1 1 −  m2 m2  5.84 ×10−5 _____________ s s N A + N B ( 0.31 atm ) − N A + N B ( ____ atm ) +   1 1 NB NB −  2 2 m −5 m  _____________ 7.49 × 10 s s 

In this equation, it can be seen that some terms have the factor

1  − 2 2  m m  5.93 × 10−5 _____________ s (1.9 atm )  s 1 1  − 2  m2 −5 m  5.84 ×10 _____________ s s   1 1 − 2  m2 m  _____________ 7.49 × 10−5 s s (1.9 atm )   1 1 −  2 2 m m  _____________ 7.49 × 10−5 s s  1

NA + NB in common. Thus, this NB

expression can be simplified as follows:

N + NB   ( _________ )( 0.47 atm ) − ( _____ atm )  A + ( ______ )(1.9 atm )   mol N   B N A + N B = ______ 2 × ln   + N N m ⋅s B  ( _________ )( ______ atm ) − ( 0.47 atm )  A + ( ______ )(1.9 atm )  NB  

We can further simplify this equation by removing the molar flow of oxygen in the denominator as shown in the following step: N A + N B = _______

 _______ ( N A + N B ) + __________ N B  mol × ln   2 m ⋅s  −0.449 ( N A + N B ) + __________ N B 

(2)

Now we can substitute Equation 1 into Equation 2 to get:

N A + ( 0.612 − 1.263N A ) = ________

 ________ ( N A + 0.612 − 1.263N A ) + ________ ( 0.612 − 1.263N A )  mol × ln   2 m ⋅s  −0.449 ( N A + 0.612 − 1.263N A ) + ________ ( 0.612 − 1.263N A ) 

Simplifying similar terms and moving all terms to the left side, we have:  ________ N A + 1.096  ________ N A + ln   − ________ = 0  ________ N A + ________ 


7


Student View


This equation can be solved by trial and error or using computer software to obtain the molar flow rate of carbon dioxide to be:

mol m2 ⋅ s

N A = ________________

We can enter this value into Equation 1 to determine the flux of oxygen as shown below: N B = 0.612

mol mol   − 1.263  ________________ 2  2 m ⋅s m ⋅s  

N B = ________

mol m2 ⋅ s

The following figure illustrates the diffusion process occurring in the fuel cell cathode.

NA NB

NC = 0

Gas – Diffusion Layer


Bipolar Plate

8


Student View


Chapter 9 Drying of Process Materials The drying processes discussed in this chapter relate to the removal of water or other organic liquids from solids or fluid substances. The following examples explain different situations where drying and humidity concepts affect the performance of a fuel cell. 9.3-1 Humidity from Vapor – Pressure Data 9.3-2 Use of Humidity Chart 9.3-3 Adiabatic Saturation of Feed Air for Proton – Exchange Membrane Fuel Cells. 9.3-4 Wet Bulb Temperature and Humidity


1


Student View

Drying of Process Materials

Example 9.3-1: Humidity from Vapor – Pressure Data The air in the cathode chamber of a proton – exchange membrane fuel cell is at a temperature of 80°C and a pressure of 1 atm. Determine the partial pressure of water, humidity, saturation humidity, percentage humidity if the air has a relative humidity of 8.55 %. Strategy The definitions of the required parameters can be used to solve this problem. Solution The following definitions can be used to obtain the solution to this problem: H R = 100

HS =

PH2O ______

M H 2O

PHs 2O

M Air P − PHs 2O

H P = 100

H=

H HS

M H 2O

PH 2O

M Air _________

First we can solve for the partial pressure of water from the equation of relative humidity, as this value is given in the problem statement. Thus, PH2O =

______ ( 8.55 )( _______ atm ) = 100 100

PH2O = 0.04 atm

The saturation pressure of water at the temperature of 80°C was obtained from Appendix A.2 of Geankoplis. The partial pressure of water can now be substituted into the Equation for the absolute humidity H to yield:

______ kg H 2 O ( 0.04 atm ) 1 kgmol H= ______ kg Air 1 atm − 0.04 atm 1 kgmol

H = _______


kg H 2 O kg Air

2


Student View


In a similar way, we can use the saturation pressure of 0.468 atm obtained from Appendix A.2 to determine the saturation humidity of the air in the fuel cell, as shown in the following steps: _______ kg H 2 O ( 0.468 atm ) 1 kgmol HS = _______ kg Air 1 atm − _______ atm 1 kgmol

HS = _______

kg H 2 O kg Air

The only remaining value to be calculated is the percentage humidity HP from the ratio of the absolute humidity to the saturation humidity. Therefore, kg H 2 O    0.026 kg Air   H P = 100   _______ kg H 2 O   kg Air   H P = _______ %


3


Student View


Example 9.3-2: Use of Humidity Chart kg H 2 O . Use the kg Air humidity chart to determine the percentage humidity HP, humid volume υH , humid heat CS, and dew point of this air/steam mixture.

The air in a fuel cell has a dry – bulb temperature of 80°C and a humidity of 0.035

Strategy We can locate the given information in the humidity chart to determine the values required to solve this problem.

Solution kg H 2 O at a temperature of kg Air 80°C. From this point we can move horizontally to the left until reaching the saturation line (HP = 100%). The temperature in this point corresponds to the dew point, found to be:

First we locate the point that corresponds to a humidity of H = 0.035

Tsat = _______ °C

At the point we located initially for the dry bulb temperature and absolute humidity, we can read directly the percentage humidity. For the air in the fuel cell, we find that: H P = _______ %

In section 9.3B of Geankoplis, we are given the following equations for the humid heat and volume as a function of the absolute humidity: υH = ( 2.83 × 10−3 ) + ( 4.56 × 10 −3 H )  T CS = ____________________________

m3 kJ kg H 2 O In these equations, υH is in , H is in , T is in K, and CS is in . kg dry air kg Air kg dry air ⋅ K Entering the humidity and the temperature into these equations, we get:    kg H 2 O    υH = ( 2.83 × 10 −3 ) +  4.56 × 10−3  0.035    ( __________ K ) kg Air       υH = __________ Daniel López Gaxiola Jason M. Keith

m3 kg dry air 4


Student View


 kg H 2 O  CS = 1.005 + 1.88  0.035  kg Air   CS = _______

kJ kg dry air ⋅ K

The following figure illustrates how to determine the saturation temperature and percentage humidity from the chart: Percentage Humidity Lines

Adiabatic Saturation Curve 100 %

T = ______°C

Initial

point

corresponding to kg H 2 O T = 80°C and H = 0.035 kg Air

This point is approximately at humidity of 6.25%


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also located a percentage

Student View


Example 9.3-3: Adiabatic Saturation of Feed Air for Proton – Exchange Membrane Fuel Cells. kg H 2 O enters an adiabatic kg Air saturator before being fed to the cathode side of the fuel cell. The air enters the saturator at a dry bulb kg H 2 O temperature of 60 °C and must enter the fuel cell with a humidity of 0.063 . Determine the kg Air final temperature and percent humidity of the air.

Air to be used as reactant in a PEMFC with an initial humidity of 0.057

Strategy Both HP and the temperature of the air entering the fuel cell can be obtained using the humidity chart.

Solution First we need to locate the point that corresponds to the temperature of 60°C and humidity of kg H 2 O 0.057 . Once we located this point in the humidity chart, we move parallel to the adiabatic kg Air saturation curves, until reaching a point where the absolute humidity is 0.063

kg H 2 O . kg Air

Now we are at the point where the air has the conditions required for entering the fuel cell and therefore we can read the temperature and percentage humidity to be: T ≈ ______ °C H P ≈ ______ %

In the following page we can see a chart indicating the method to determine these values.


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Student View


The next chart indicates the procedure followed to determine the temperature and percentage humidity: Percentage Humidity Lines

Adiabatic Saturation Curve

80% 70%

H = 0.063

kg H 2 O kg Air

T = ______°C Initial point corresponding to kg H 2 O T = 60°C and H = 0.057 kg Air

Point corresponding to kg H 2 O H = ________ kg Air Also we can read the percent humidity to be approximately ___ %


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Student View


Example 9.3-4: Wet Bulb Temperature and Humidity Estimate the humidity of the reactant air in a proton – exchange membrane fuel cell if it has a dry and wet bulb temperature of 80 °C and 42.5 °C, respectively.

Strategy The humidity of the air can be determined from the humidity chart using the temperature data given in the problem statement.

Solution The wet bulb temperature corresponds to the temperature of the air when it has 100 % humidity. Hence, we need to move vertically from the temperature axis in the chart until reaching the curve corresponding to 100 % humidity. From this point we move downwards to the right, parallel to the adiabatic saturation curves until we reach the vertical line for T = 80°C. Now we can read the humidity of the air in the fuel cell to be: H = ______

kg H 2 O kg Air

The method to determine the previous humidity value was obtained as follows: Percentage Humidity Lines

Adiabatic Saturation Curve 100%

H = ______

Twet = 42.5 °C Initial point


kg H 2 O kg Air

Tdry = 80 °C

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Student View


Chapter 11 Vapor – Liquid Separation Processes Separation processes in Chemical Engineering are used to transform a mixture of substances into two or more different products. In this chapter, the following problem modules illustrate the principles of chemical processes in which the separation involves the vapor and liquid phases. 11.1-1 Use of Raoult’s Law for Methanol – Water Equilibrium Data 11.3-1 Relative Volatility of Methanol – Water Mixture


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Student View

Vapor – Liquid Separation Processes

Example 11.1-1: Use of Raoult’s Law for Methanol – Water Equilibrium Data A methanol – water mixture is assumed to be in equilibrium with the vapor phase in the reservoir for a Direct – Methanol Fuel Cell. Using the following equilibrium data [1,2] and Raoult’s Law determine the composition of the vapor and liquid phases at 72°C and 101.325 kPa. The vapor pressure data for water was obtained from Appendix A.2 of Geankoplis. T (°C)

PHsat2O (kPa)

sat PCH (kPa) 3OH

64.5 65 66 67.6 69.3 71.2 73.1 75.3 78 81.7 84.4 87.7 89.3 91.2 93.5 96.4 100

24.521 25.03 26.57 28.233 30.328 32.964 35.772 39.109 43.866 50.94 56.577 64.477 68.417 73.598 80.227 89.254 101.325

100.74 102.66 106.73 113.68 121.06 129.41 139.55 151.28 167.02 190.44 209.96 234.53 248.34 264.85 285.04 313.83 351.38

Strategy We need to use Raoult’s Law to obtain the composition of both phases at the given temperature and pressure. Solution The following equation describes Raoult’s Law: yA =

_______ P

1. Wankat, P.C., Separation Process Engineering, Second Edition, Prentice Hall, 2007. 2. Methanex Corporation. "Technical Information & Safe Handling Guide for Methanol” September 2006. Accessed: February 2011. http://www.methanex.com/products/documents/TISH_english.pdf


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Student View


In this problem, A = Methanol and B = Water. As we can see, we cannot yet determine the composition of the vapor phase because we need to calculate the composition of the liquid phase first. This can be done using Equation 11.1-3 of Geankoplis, given by:

PA x A + PB (1 − x A ) = ____ We can use the pressures PA and PB, using linear interpolation of the given data, at the temperature of 72°C. Thus,

PA @ T = 72° C − PA@ T =71.2°C PA@ T =73.1°C − PA @ T = 71.2°C

=

_____ °C − _____ °C 73.1°C − _____ °C

Solving for the pressure PA@ T =72°C we determine the pressure to be: PA@ T =72°C = _______ kPa

Using the same procedure we can determine the pressure of water at 72°C to be: PB@ T = 72°C = ________ kPa

Now we can substitute the pressure values into the equation for the total pressure, and solve for xA to yield:

_______ x A + _______ (1 − x A ) = 101.325 99.53x A = _______

x A = _______

Hence the molar fraction of water is given by: x B = 1 − x A = 1 − _______ x B = _______

Now we can enter the value of xA into Raoult’s Law to determine yA and yB. Thus, yA =

_________ kPa ( _______ ) 101.325 kPa

y A = _______

y B = 1 − y A = 1 − _______ y B = 0.109


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Student View

Vapor – Liquid Separation Processes

Example 11.3-1: Relative Volatility of Water – Methanol Mixture Determine the relative volatility of the methanol – water mixture described in problem 11.1-1 at the temperature of 72°C.

Strategy Section 11.3B of Geankoplis gives an Equation for determining the relative volatility

Solution The relative volatility of a mixture is defined by equation 11.3-3 of Geankoplis, shown below:

α AB =

PA PB

We can substitute the saturation pressures at the temperature of 72°C, which are given in Example 11.1-1. Thus, the relative volatility is found to be: α AB =

_________ kPa _______ kPa

α AB = _______


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Student View


Chapter 12 Liquid – Liquid and Fluid – Solid Separation Processes This chapter includes examples of adsorption processes where one or more components of a gas or liquid stream are adsorbed on the surface of an adsorbent material. The following problem module illustrates pressure swing adsorption process for purifying hydrogen for proton – exchange membrane fuel cells. 12.3-1 Hydrogen Purification in Pressure Swing Adsorption Process


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Student View

Liquid – Liquid and Fluid – Solid Separation Processes

Example 12.3-1 Hydrogen Purification in Ethanol – Reforming Process The dry reformate stream coming from an ethanol reforming process contains 70 mole % H2 and 30 mole % CO2. Lopes et al. [1] study the adsorption of CO2 on a packed bed of activated carbon with a length of 0.267 m. The bed contains 245.6 kg of adsorbent material consisting of activated carbon, 3 −5 m with a particle diameter of 2.9 mm. The average flow rate of the dry reformate is 5 ×10 and has s kg a density of 0.587 3 . m Use the data in the following table to determine the time required to reach the break – point  C  = 0.01 , the time equivalent to the total capacity of the bed, and the time concentration   C0  equivalent to the usable capacity of the bed up to the break – point time. Also calculate the length and capacity of unused bed after the break-point time. What is the saturation capacity of the bed? Breakthrough Concentration of CO2 in the packed bed t(s) 0 27 187 480 600 613 640 667 747 853 973 1067 1200 1360

C/C0 0 0.0063 0.0125 0.0358 0.1391 0.3487 0.4249 0.4884 0.5600 0.6661 0.7168 0.7611 0.8055 0.8561

t(s) 1493 1640 1827 1933 2107 2240 2373 2560 2707 2853 3120 3280 3440 3600

C/C0 0.8814 0.9066 0.9318 0.9444 0.9569 0.9695 0.9757 0.9819 0.9817 0.9879 0.9940 0.9938 1.0000 1.0000

Strategy The break-through and capacity of the packed bed can be determined using the design equations given in Section 12.3D of Geankoplis. Solution First, the break – point time can be obtained from the tabulated data, at the point where

C = 0.01 to C0

be equal to tb = ______ s 1. Lopes, F.V.S., Grande, C.A., Rodrigues, A.E., Chemical Engineering Science, 66, 303 – 317 (2011)


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Student View


The time tt for the total capacity of the bed can be calculated using the integral of the

C curve as a C0

function of time: ∞ C t t = ∫  1 − dt = A1 + A 2 0  C0 

where A1 and A2 are the shaded areas shown in the following figure: 1.0

0.9 0.8

A2

0.7 C/C0

0.6 0.5 0.4 0.3 0.2 0.1 0.0 0

500

1000

1500

2000

2500

3000

3500

4000

t(s)

A1

Cb

As it can be seen in this figure, we can determine the break – through time by calculating the areas A1 and A2. Thus,

A1 = ( _____ s − 0s )(1 − ___ ) A1 = _____ s


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Student View


C/C0

To determine the area A2, we can use the trapezoidal method to calculate the area under the curve and subtract it from the total area A of the rectangle, as shown below: 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0

A2

A3

0

500

1000

1500

2000 t(s)

2500

3000

3500

4000

A1

The area of the rectangle to the right of the break point can be calculated as follows:

A = ( ______ s − _____ s )(1 − ___ )

A = 3413 s The results for the numerical integration using trapezoidal method will yield:

A3 = _________ s Now we can obtain the area A2 to be:

A2 = A − A3 = 3413 s − _________ s A2 = _________ s We can substitute the areas A1 and A2 into the equation for the break-through time tt, to obtain: t t = ______ s + _________ s

t t = _________ s


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Student View


The time equivalent to the available capacity of the bed before the break – point time can be obtained as shown in the following steps: tb  C t u = ∫ 1 − dt = ____ 0  C0 

t u = _____ s The length of the unused bed can be calculated using Equation 12.3-4 of Geankoplis:

 t H UNB = 1 − u  tt

  HT 

where: HT = Total length of the packed bed Substituting the times we calculated and the length of the packed bed into this equation, we get:  _____ s  H UNB = 1 −  ( ______ m )  _________ s 

H UNB = 0.212 m Finally, to determine the saturation capacity of the activated carbon in the bed, we need to obtain the moles of carbon dioxide adsorbed on the bed. The carbon dioxide can be obtained by multiplying the initial concentration of CO2 in the gas by the mass of gas in the time tt of _______ s. Thus,

0.30 Total CO 2 adsorbed =

3 kmol CO 2  kg gas   −5 m  5 × 10   ( ________ s )  0.587  kmol gas  s  m3   kg CO 2   ____   kg gas kmol CO 2   _____ kmol gas

Total CO 2 adsorbed = ________ kg CO 2

The molecular weight of the gas mixture was obtained by multiplying the molar fraction of each component by its corresponding molecular weight and adding the results, as shown below:


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Student View


  kg H 2  kg CO 2  M gas = y H2 M H2 + yCO2 M CO2 = ______  ____  + 0.3  _____  kmol H 2  kmol CO 2    M gas = ________

kg kmol

Now we can divide the total CO2 adsorbed by the mass of activated carbon on the bed. Hence, Saturation Capacity =

________ kg CO 2 245.6 kg adsorbent

Saturation Capacity = _____________


kg CO 2 kg adsorbent

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Student View

Solucionario Procesos de Transporte y Principios de Procesos de Separación 4ta Edicion Christie Geankopolis

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