AIRCONIDITONING INDUSTRIAL PLANT ENGINEERING
“Satisfaction lies in the effort, not the attainment. Full effort is full victory.” - Maha Mahatma tma Gandh Gandhi i
Properties of Moist Air ATMOSPHERIC AIR
Total Atmospheric Presure
= +
Total Enthalpy
H = H + Hv H = m h + mv hv
H = m h + mv hv mv h = h + hv m =
h = h + ωhv hv = h @ t ,v
= +
Properties of Moist Air ATMOSPHERIC AIR Total Enthalpy
h = h + ωh
= +
kJ kg
or
Btu lbm
h ≅ 25 2500.9 + 1.82 t ,kJ/kg h ≅ 10 1060.9 0.9 + 0.4 0.435 t ,Btu/lbm kJ c = 1.005 kg−K Humidity Ratio (Absolute or Specific Humidity),
=
= 0.622
Pv P
= .
kgv
−
kg
or
lbmv lbm
Properties of Moist Air ATMOSPHERIC AIR Relative Humidity, RH
Pv Vv mv Pv RvT RH = = = P V m P RT P = P @
=
@
Properties of Moist Air PROPERTIES IF NOT GIVEN: Enthalpy for saturated water in Btu/lbm (provided tsat is in oF):
= − = . + . Saturation Temperature ( in °C) and Saturation Pressure relation ( in kPa):
= =
. . . . − .
Vapor Pressure (in psi or in Hg):
= −
−
PSYCHROMETRIC CHART
PSYCHROMETRIC CHART
Properties of Moist Air
Problem 1
Calcula ulate the specifi ific volume lume of an air-vapor mixture in cubic meters per kilogram of dry air when the following conditions prevail : t = 30C , w = 0.015 0.015 kg/kg, kg/kg, and Pt = 90 kPa.
Ans:
.
Properties of Moist Air
Problem 2
A sample of air has dry-bulb temperature of 30C and a wet-bulb temperature of 25C . The barometric pres pressu sure re is 101. 101.32 325 5 kPa. kPa. Calc Calcul ula ate the enthalpy of the air if it is adiabatically saturated. saturated. At 25 deg C Pv=3.17 kPa, hg=2547.2 kJ/kg.
Properties of Moist Air
Problem 2
h = h
2
1
25℃
30℃
Ans:
.
Properties of Moist Air
Problem 3
A 4 m x 4 m x 4 m room has a relative humidity ratio of 80%. The The pressure in the room is 120 kPa and tempe empera ratu ture re is 35C (Psat = 5.628 kPa). What is the mass of vapor in the room. Use Rvapor = 0.4615 kNm/kg-K. Ans:
.
Properties of Moist Air
Problem 4
A certain sample of air has a temp temper erat ature ure of 70F (partial (partial pressure pressure of 0.36 psia) and a dew point temperature of 50F . The partial pressure of the water is vapor corresponding to a 50F dew point temperature is 0.178 psia. Determine the relative humidity RH. Ans:
. %
Properties of Moist Air
Problem 5
A coil has an inlet temperature of 60F and outlet of 90F. If the mean temperature of the coil is 110F, find the bypass factor of the coil.
Properties of Moist Air
Problem 5
Bypa By pass ss Fa Fact ctor or (BF) (BF) The inefficiency of the heating coil or cooling coil in not being to heat or cool the incoming air to the temperature of cooling coil. =
− −
Coil Co il Ef Effi ficie cienc ncy y The efficiency of the heating coil or cooling coil to heat or cool the incoming air to the temperature of cooling coil. . = −
Properties of Moist Air
Problem 5
A coil has an inlet temperature of 60F and outlet of 90F. If the mean temperature of the coil is 110F, find the bypass factor of the coil.
Ans:
.
Air Mixing t 1 m1
t 2 m2
Mass balance at mixing point:
= + Energy balance at mixing point:
= + Humidity balance at mixing point:
= + Dry-bulb Dry-bulb temp balance at mixing point: m3 t 3
= + Dew-point temp balance at mixing point:
= +
Air Mixing
Problem 6
The mass of an outside air at 50C in an air conditioning unit is 60 kg. Find the temperature after mixing if the outside air mixed with 40 kg with recirc recircula ulated ted air at 35C.
Air Mixing
Problem 6
t 1=50°C m1=60 kg
t 2=35°C m2=40 kg
m3=100 kg t 3 = ?
Ans:
℃
Air Mixing
Problem 7
In an air con condit ditioning ing sy sys stem, em, if the re-circulated air is three times the outside air and the mass of supply air is 20 kg/s, what is the mass of the outside air?
Air Mixing
Problem 7
Recir Rec ircula culate ted d air
=
Supp Su pply ly ai airr
=
Outs Ou tsid ide e ai airr
=?
Ans:
AC Load Calculations = + =
Total Recirculation Recirculati on System
AC Room 2 1
Q QL = − =
UNIT
= −
AC Load Calculations = −
Introduction of Ventilation Air
= −
AC Room
2 1
Q QL = − =
4
UNIT
3
= +
= −
AC Load Calculations
Problem 8
A room being air conditioned is being held at 25C dry bulb and 50% relative relative humidity humidity.. A flow rate of 5 m3/s of supply air at 15C dry bulb and 80% RH is being delivered to the room to maintain that steady condition at 100 kPa. What is the sensible heat absorbed from the room air in kW?
AC Load Calculations
Problem 8
t , = 25℃ RH = 50% A/C ROOM
t , = 15℃ suppl supply y air air m m Q = 5 s RH = 80%
Ans:
.
AC Load Calculations
Problem 9
If the latent and sensible heat loads are 20 kW and 80 kW respectively, what is the sensible heat ratio?
Ans:
.
AC Load Calculations
Problem 10
In an auditorium maintained at a temperature not to exceed 24C and rela relati tive ve humi humidit dity y not not to exc excee eed d 60%, 60%, a sensible heat load of 132 kW and 78 kg of moisture per hour to be removed. Air is supplied to the auditorium at 18C. How many kilograms of air must be supplied per hour? Ans:
,
AC Load Calculations
Problem 10
t , = 24℃ RH = 60% A/C ROOM
t , = 18℃ suppl supply y air air m =?
=
Ans:
,
AC Load Calculations
Problem 11
An assembly hall was to have an air conditioning unit installed which would be maintained at 26C dry bulb and at 50% RH. The unit delivers air at 15C dry bulb temperature and the calculated sensible heat load is 150 kW and latent heat is 51.3 kW. kW. Twen Twenty ty percent percent by weight weight of extracted extracted air is made up of outside air at 34C dry bulb and 60% RH. While 80% is extracted by the air at 34C dr dry bu bulb an and 60 60% RH RH, wh while 80% is extracted by the air conditioner from the asse assemb mbly ly hall hall.. Deter etermi mine ne the the air air cond condit itio ione ners rs refrigeration capacity in tons of refrigeration and its ventilation load in kW. From psychrometric chart: h3 = 86.5 86.5 kJ/kg, kJ/kg, h2 = 53 kJ/kg Ans:
. ; .
AC Load Calculations
Problem 11
= %
ASSEMBL ASSEMB LY HALL H ALL 26°C db, 50% RH
2 1
℃ Q = 150 kW
QL = 51.3 51.3 kW
4
UNIT
3
= % ℃ %
AC Load Calculations
Problem 11
REFRIGERATION LOAD
2 1
℃
4
UNIT
3
= %
℃ %
Ans:
.
AC Load Calculations
Problem 11
VENTILATION LOAD
2 1
℃
4
UNIT
3
= %
℃ %
Ans:
.
COOLING TOWERS Main Features of Cooling Towers
(Pacific Northwest National Library, 2001)
Air drawn across COOLING falling water Fill located TYPES OF COOLING outside tower TOWERS 1. Natural Draft Cooling Tower •
T O W E R S Air drawn up •
•
Cross flow
•
Counter flow
through falling water Fill located inside tower
COOLING TOWERS TYPES OF COOLING TOWERS 2. Mechanical Draft Cooling Tower (Forced Draft)
COOLING TOWERS TYPES OF COOLING TOWERS 2. Mechanical Draft Cooling Tower Tower (Induced Draft-counter Draft-co unter flow)
COOLING TOWERS TYPES OF COOLING TOWERS 2. Mechanical Draft Cooling Cooli ng Tower Tower (Induced Draft-cross flow) flow )
COOLING TOWERS COOLING TOWER ANALYSIS Hot Water Temperature (In)
1. Range Difference between cooling water inlet and outlet outlet tempera temperatur ture: e: Range = CW inlet temp CW outlet temp
–
e g n a R
= , − ,
(In) to the Tower (Out) from the Tower
Cold Water Temperature (Out) h c a o r p p A
Wet Bulb Temperature Temperature (Ambient)
COOLING TOWERS COOLING TOWER ANALYSIS Hot Water Temperature (In)
2. Approach Difference between cooling tower outlet cold cold wate waterr temp temper era atur ture and ambient wet bulb temperature:
e g n a R
(In) to the Tower (Out) from the Tower
Approach = CW outlet temp Wet bulb temp –
Cold Water Temperature (Out)
= , − ,
h c a o r p p A
Wet Bulb Temperature Temperature (Ambient)
COOLING TOWERS COOLING TOWER ANALYSIS Hot Water Temperature (In)
3. Effectiveness/Efficiency . =
. =
+
, − ,
e g n a R
(In) to the Tower (Out) from the Tower
, − , Cold Water Temperature (Out) h c a o r p p A
Wet Bulb Temperature Temperature (Ambient)
COOLING TOWERS COOLING TOWER ANALYSIS
4. mass balance on CT m = m + = + − = − − = − Make-up water requirement
= − = −
COOLING TOWERS COOLING TOWER ANALYSIS
5. Energy balance on CT (SSSF)
Q w = Q − = −
Cooling Towers
Problem 12
Determine the approximate load on a cooling tower if the entering and leavin aving g temper peratu atures are 96F and 88F, respectively and the flow rate of the water over the tower is 30 gpm.
Cooling Towers
Problem 12
= ℉ 1
=
= ℉ 2
Ans:
.
Cooling Towers
Problem 13
Determine the quantity of water lost by bleed off if the water flow rate over the tower is 30 gpm and the range is 10F. Percent bleeded-off off requi equirred is 0.33%.
Cooling Towers
% =
Problem 13
=
Ans:
.
Ducts
Problem 14
A rectangular duct has a dimensions of 0.25 m by 1 m. Determine the equivalent diameter of the duct.
Ducts
Problem 14
HYDRAULIC DIAMETER or EQUIVALENT DIAMETER
=
=
HYDRAULIC RADIUS or EQUIVALENT RADIUS
=
=
=
Ducts
Problem 14
A rectangular duct has a dimensions of 0.25 m by 1 m. Determine the equivalent diameter of the duct.
Ans:
.
Ducts
Sample Problem
A pipe with a radius of 1.2 m flows partially full as shown. own. What is the appro proxima imate hydraulic radius?
Answer:
.
Ducts
Problem 15
A duct 0.40 m high and 0.80 m wide suspended from the ceiling in a corridor, makes a right angle turn in the horizontal plane. The inner radius is 0.2 m and the outer radius is 1.0 m measured from the same center. The velocity of air in the duct is 10 m/s. Compute the pressure drop in this elbow elbow.. Assu Assuming ming ; f = 0.3, 0.3, = 1.204 kg/m3 and L = 10 m. Ans:
DRYERS Humi Hu mid d Ai Airr
3
= 1
Supply Ai Air
2
HEATER
DRYING CHAMBER
Heated Air
5
Dried Dri ed Pro Product duct
4
Wett Fe We Feed ed
DRYERS Dryer Efficiency
ŋ =
, ,
Supply Ai Air
=
= 1
=
Humi Hu mid d Ai Airr
2
HEATER
= − , =
3
DRYING CHAMBER
Heated Air
= −
5
Dried Dri ed Pro Product duct
4
Wett Fe We Feed ed
DRYERS Bone-dry weight or dry weight Is the final solid weight reached by a hygroscopic hygroscopic substance.
Regain Is the hygroscopic moisture content of a substance expressed as a percentage of the BDW of the material.
=
−
=
Moisture Content = = +
=
DRYERS FEED
PRODUCT
water, | water, |
DRYING
bone-dry, |
bone-dry, |
= +
= +
Note:
Wet basis: Dry basis:
per unit mass of the feed or product. per unit mass of the bone dry material.
Dryers
Problem 16
Copra enters a dryer containing 60% water and 40% of solids and leaves with 5% water and 95% solids. Find the weight of water removed based on each pound of original product.
Dryers
Problem 16 Humi Hu mid d Ai Airr
3
2
DRYING CHAMBER
Heat He ated ed Ai Airr
= . .
% 4
% %
Wet Fe Feed ed
%
= . .
= . .
5
% % = . .
Dried Dri ed Pro Product duct
Ans:
.
Dryers
Problem 17
Wet material containing 215% moisture (dry basis) is to be dried at the rate of 1.5 kg/s in a continuous dryer to give a produ oduct containing 5 % moisture (wet basis). The drying medium consist of air heat eated to 373 K and and containin ning water vapor equivalent to a partial pressure of 1.40 kPa. The air leaves the dryer at 310K and 70% saturated. Calculate how much air will be required to remove the moisture. Ans:
.
Dryers
Problem 17 Humi Hu mid d Ai Airr
= % =
3
= . . = 2
DRYING CHAMBER
Heated Air
5
4
Wett Fe We Feed ed % % ( ( ) ) = . % % = . .
%
= . . % %
Dried Dri ed Pro Product duct
Ans:
.
-Gary Player
