1. INTRODUCTION
1.1
A comp composi osite te wall wall consis consistt of of alte alterna rnativ tivee layer layerss of fir ( 5 cm thick thick ) , alumi aluminum num ( 1 cm thick ), lead ( 1 cm thick ), and corkboard ( 6 cm thick ). The temperature is 60 C of the outside of the fir and 10 C on the outside of the corkboard. Plot the temperatur temperaturee gradient gradient through the wall. Does the temperature temperature profile profile suggest suggest any simplifying assumptions that might be made in subsequent analysis of the wall?
Solution: Thermal Conductivities: k fir = 0.12 W/m.K (Table A.2, Appendix A) k alu alu = 237 W/m.K (Table A.1, Appendix A) k ld ld = 35 W/m.K (Table A.1, Appendix A) k cb W/m.K (Table A.2, Appendix A). cb = 0.04 W/m.K
Ques Questi tion on No. No. 1:
Plot Plot the the temp temper erat atur uree gra gradi dien entt thro throug ugh h the the wall wall.. Answer:
Ques Questi tion on No. No. 2: 2: Does Does the the temp temper erat atur uree profi profile le sugg sugges estt any simp simpli lify fyin ing g assum assumpt ptio ions ns that might be made in subsequent analysis of the wall? Answer: Yes, since the thermal conductivity of aluminum and lead are very high than fir and corkboard, they are considered isothermal. Therefore consider only fir and corkboard. ∆Τ fir + ∆T cb cb = 60 C – 10 C = 50 K q
∆T ∆T = k = k L fir L cb
L fir = 5 cm = 0.05 m Lcb = 6 cm = 0.06 m Then, 1
1. INTRODUCTION
q=
( 0.12 W / m ⋅ K ) ∆T fir ( 0.04 W / m ⋅ K ) ( ∆T cb ) = ( 0.05 m) ( 0.06 m)
3.6∆T fir ∆T cb cb = 3.6∆ Then, 3.6∆Τ fir = 50 K ∆Τ fir + 3.6∆Τ ∆Τ fir = 10.87 K q
= k
10.87 K ∆T 2 = ( 0.12 W / m. K ) = 26.09 W/m L fir 0 . 05 m
Considering all walls: ∆T fir + ∆T alu alu + ∆T ld ld + ∆T cb cb = 60 C – 10 C = 50 K q
= k
∆T ∆T ∆T ∆T = k = k = k L fir L alu L ld L cb
L fir = 5 cm = 0.05 m Lcb = 6 cm = 0.06 m Lalu = 1 cm = 0.01 m Lld = 1 cm = 0.01 m
k L fir ∆T alu = ∆T fir k L alu k L ∆T ld = ∆T fir fir k L ld k L ∆T cb = ∆T fir fir k L cb Then
2
1. INTRODUCTION
q=
( 0.12 W / m ⋅ K ) ∆T fir ( 0.04 W / m ⋅ K ) ( ∆T cb ) = ( 0.05 m) ( 0.06 m)
3.6∆T fir ∆T cb cb = 3.6∆ Then, 3.6∆Τ fir = 50 K ∆Τ fir + 3.6∆Τ ∆Τ fir = 10.87 K q
= k
10.87 K ∆T 2 = ( 0.12 W / m. K ) = 26.09 W/m L fir 0 . 05 m
Considering all walls: ∆T fir + ∆T alu alu + ∆T ld ld + ∆T cb cb = 60 C – 10 C = 50 K q
= k
∆T ∆T ∆T ∆T = k = k = k L fir L alu L ld L cb
L fir = 5 cm = 0.05 m Lcb = 6 cm = 0.06 m Lalu = 1 cm = 0.01 m Lld = 1 cm = 0.01 m
k L fir ∆T alu = ∆T fir k L alu k L ∆T ld = ∆T fir fir k L ld k L ∆T cb = ∆T fir fir k L cb Then
2
1. INTRODUCTION
k 1 1 1 ∆T fir 1 + + + k k k = 50 K L fir L alu L ld L cb 0.12 1 1 1 ∆T fir + + 1 + = 50 237 35 0.04 0.05 0.01 0.01 0.06 ∆T fir = 10.87 K q
= k
10.87 K ∆T 2 = ( 0.12 W / m. K ) = 26.09 W/m L fir 0 . 05 m
There it is equal to simplified solution.
1 .2
Verify Eq Equation (1 (1.15).
Solution: Equation (1.15) dT body
∝T body −T ∞
dt
For verification only Equation (1.3) dU
Q=
dt
= mc
dT dt
Equation (1.16) Q
∝T body −T ∞
Then dT
mc
dt
dT dt
∝T body −T ∞
∝T body −T ∞
Then dT body dt
∝T body −T ∞
where mc is constant.
q = 5000 W/m2 in a 1 cm slab and T = 140 C on the cold side. Tabulate the temperature drop through the slab if it is made of
1.3
• •
Silver Aluminum 3
1. INTRODUCTION
• • • • •
Mild steel (0.5 % carbon) Ice Spruce Insulation (85 % magnesia) Silica aerogel
Indicate which situations would be unreasonable and why. Solution: L = 1 cm = 0.01 m (a) (a) Silv Silver er Sla Slab b
∆T = 5000 W/m2 L Si
q = k
Thermal conductivity of silver at 140 C, 99.99+ % Pure, Table A.1, Appendix A k si = 420 W/m.K
∆T Si = 5000 W/m2 0.01 m
q = ( 420 420 W / m ⋅ K )
∆T SiSi = 0.12 K (b) Alumiu Alumium m Slab Slab q
= k
∆T = 5000 W/m2 L alu
Thermal conductivity of aluminum at 140 C, 99.99+ % Pure, Table A.1, App. A K alu alu = 237.6 W/m.K q
∆T alu 2 = ( 237 237.6 W / m ⋅ K ) 0.01 m = 5000 W/m
∆T alu alu = 0.21 K (c) Mild Mild Steel Steel Slab Slab
∆T = 5000 W/m2 L ms
q = k
Thermal conductivity of mild steel at 140 C, Table A.1, Appendix A K ms ms = 50.4 W/m.K q
∆T ms = ( 50.4 W / m ⋅ K ) = 5000 W/m2 0.01 m
∆T ms ms = 0.992 K (d) (d) Ice Ice Sla Slab b 4
1. INTRODUCTION
∆T = 5000 W/m2 L ice
q = k
Thermal conductivity of ice at 140 C, Table A.1, Appendix A • ice at 0 C, k ice ice = 2.215 W/m.K • Note: there is no ice at 140 C, but continue calculation at 0 C. q
∆T ice 2 = ( 2.215 215 W / m ⋅ K ) 0.01 m = 5000 W/m
∆T ice ice = 22.57 K (e) (e) Spru Spruce ce Sla Slab b
∆T = 5000 W/m2 L Si
q = k
Thermal conductivity of spruce at 140 C, Table A.1, Appendix A K sp = 0.11 W/m.K @ 20 C (available) q
∆T Sp 2 = ( 0.11W / m ⋅ K ) = 5000 W/m 0 . 01 m
∆T Sp Sp = 454.55 K (f) Insula Insulati tion on (85 % Magnes Magnesia) ia)
∆T = 5000 W/m2 L Si
q = k
Thermal conductivity of insulation at 140 C, Table A.1, Appendix A K inin = 0.074 W/m.K @ 150 C (available) q
∆T in 2 = ( 0.074 074 W / m ⋅ K ) 0.01 m = 5000 W/m
∆T SiSi = 675.8 K (g) Silica Silica Aerogel Aerogel Slab
∆T = 5000 W/m2 L Si
q = k
Thermal conductivity of silica aerogel at 140 C, Table A.1, Appendix A k sa = 0.022 W/m.K @ 120 C q
∆T sa 2 = ( 0.022 022 W / m ⋅ K ) 0.01 m = 5000 W/m
∆T sa = 2,273 K
Tabulation: 5
1. INTRODUCTION Slab Silver Aluminum Mild Steel (0.5 % Carbon) Ice Spruce Insulation (85 % Magnesia) Silica Aerogel
Temperature Drop, K 0.12 0.21 0.992 22.57 454.55 675.8 2273
The situation which is unreasonable here is the use of ice as slab at 140 C, since ice will melt at temperature of 0 C and above. That’s it. 1.4
Explain in words why the heat diffusion equation, eq. no. (1.13), shows that in transient conduction the temperature depends on the thermal diffusitivity, α , but we can solve steady conduction problems using just k (as in Example 1.1).
Solution: Equation (1.13) d T − T ref dU dT δ x = ρ cA δ x − Qnet = = ρ cA dt dt dt Answer:
The application of heat diffusion equation eq. no. (1.13) depends on the ∂T thermal diffusivity α as the value of is not equal to zero as it I s under unsteady ∂t state conduction. While in steady conduction depends only on k because the value of dT ∂T ∂2T = 0 for steady state conduction giving = 0 , so q = −k . 2 dx ∂t ∂ x
1.5
A 1-m rod of pure copper 1 cm2 in cross section connects a 200 C thermal reservoir with a 0 C thermal reservoir. The system has already reached steady state. What are the rates of change of entropy of (a) the first reservoir, (b) the second reservoir, (c) the rod, and (d) the whole universe, as a result of the process? Explain whether or not your answer satisfies the Second Law of Thermodynamics.
Solution: Equation (1.9) q = k
∆T L
Thermal conductivity of copper at 100 C, Table A.1, Appendix A 6
1. INTRODUCTION
k = 391 W/m.K L = 1 m ∆T = 200 C – 0 C = 200 K q
200 K 2 = ( 391W / m ⋅ K ) = 78,200 W/m .K 1 m
Q = qA A = 1 cm2 = 1 x 10-4 m2 Q = (78,200 W/m2.K)(1 x 10-4 m2) = 7.82 W
− Qrev
− 7.82 W = - 0.01654 W/K ( 200 + 273 K ) + 7.82 W = = + 0.02864 W/K ( 0 + 273 K )
(a)
S 1
=
(b)
S 2
=
(c)
r S
== 0.0 W/K (see Eq. 1.5, steady state)
(d)
Un S
= S 1 + S 2 = = - 0.01654 W/K + 0.02864 W/K = + 0.0121 W/K
Since
S Un 〉 0 , therefore it satisfied Second Law of Thermodynamics.
1.6
T 1 Qrev T 2
=
Two thermal energy reservoirs at temperatures of 27 C and – 43 C, respectively, are separated by a slab of material 10 cm thick and 930 cm2 in cross-sectional area. The slab has a thermal conductivity of 0.14 W/m.K. The system is operating at steady-state conditions. What are the rates of change of entropy of (a) the higher temperature reservoir, (b) the lower temperature reservoir, (c) the slab, and (d) the whole universe as a result of this process? (e) Does your answer satisfy the Second Law of Thermodynamics?
Solution: Equation (1.9) q = k
∆T L
Thermal conductivity , k = 0.14 W/m.K A = 930 cm2 = 0.093 m2 L = 10 cm = 0.10 m ∆T = 27 C – (- 43 C) = 70 K T 1 = 27 + 273 = 300 K T 2 = -43 + 273 = 230 K q
70 K 2 = ( 0.14 W / m. K ) 0.10 m = 98 W/m 7
1. INTRODUCTION
Q = qA = (98 W/m2)(0.093 m2) = 9.114 W
− Qrev
− 9.114 W ( 300 K ) = - 0.03038 W/K
(a)
1 S
=
(b)
2 S
= Qrev =
(c)
r S
== 0.0 W/K (see Eq. 1.5, steady state)
(d)
Un S
= S 1 + S 2 = = - 0.03038 W/K + 0.03963 W/K = + 0.00925 W/K
Since
S Un 〉 0 , therefore it satisfied Second Law of Thermodynamics.
1.7
T 1 T 2
=
+ 9.114 W ( 230 K )
= + 0.03963 W/K
(a) If the thermal energy reservoirs in Problem 1.6 are suddenly replaced with adiabatic walls, determine the final equilibrium temperature of the slab. (b) What is the entropy change for the slab for this process? (c) Does your answer satisfy the Second Law of Thermodynamics in this instance? Explain. The density of the slab is 26 lb/ft3 and the specific heat 0.65 Btu/lb-F.
Solution:
ρ
c
= ( 26 lb / ft
3
16.018 kg / m3 ) = 416.468 kg/m3 3 1 lb / ft
4186.8 J / kg . K = ( 0.65 Btu / lb. F ) 1 Btu / lb. F = 2721.42 J/kg.K
k = 0.14 W/m.K ∆T = 27 C – (-43 C) = 70 C T 1 = 27 C + 273 = 300 K T 2 = - 43 C + 273 = 230 K A = 0.093 m2 L = 0.10 m (a)
Q
Q T
= ∫
T 2 T 1
= ∫
T 2 T 1
T
dQ T
ρ cVdT
T
T = ρ cV ln T T T (T − T ) = ln T T
ρ cV ( T 2
− T ) 1
2
1
2
1
2
1
8
1. INTRODUCTION
T =
( T − T ) 2
ln
1
T T
( 230 − 300)
=
ln
2
1
∆S =
(b)
Q T
=
230 300
ρ cV ( T 2
= 263.45 K
− T ) 1
T
=
ρ cAL( T 2
− T ) 1
T
( 416.468)( 2721.42 )( 0.093 )( 0.10 )( 230 − 300 )
∆S =
263.45
= - 2801 J/K
(c) This will not satisfy the Second Law of Thermodynamic since this is not a rate of entropy of production of the universe.
1.8
A copper sphere 2.5 cm in diameter has a uniform temperature of 40 C. The sphere is suspended in a slow-moving air stream at 0 C. The air stream produces a convection heat transfer coefficient of 15 W/m 2.K. Radiation can be neglected. Since copper is highly conductive, temperature gradients in the sphere will smooth out rapidly, and its temperature can be taken as uniform throughout the cooling process (i.e., Bi << 1). Write the instantaneous energy balance between the sphere and the surrounding air. Solve this equation and plot the resulting temperatures as a function of time between 40 C and 0 C.
Solution: Energy Balance: Q= Q
dU dt
= −h A(T −T ∞)
dU dt
=
d dt
[ ρ cV (T −T )] = d [ ρ cV (T −T ∞ )] ref
dt
Then: d
[ ρ cV (T −T ∞ ) ] dt d (T −T ∞ ) h A (T −T ∞ ) =− dt ρ cV
− h A( T −T ∞ ) =
ln ( T − T ∞ )
=−
t + C ρ cV
h A
at T (t = 0)
≡ T i,
C = ln (T −T ∞ )
9
1. INTRODUCTION
ln ( T − T ∞ )
=−
t + ln ( T i − T ∞ ) ρ cV h A
T −T ∞ t t = − = − ρ cV T x T i −T ∞ h A ρ cV T x = h A ln
T −T ∞
=e
T i −T ∞
−
t
T x
= 0 C + 273 = 273 K T i = 40 C + 273 = 313 K
T
T x
ρ cV = h A
V =
4 3
3
π r
r = (1/2)(2.5 cm) = 1.25 cm = 0.0125 m 2 A = 4π r 4 3 ρ c π r ρcV ρ cr = 3 2 = T x = h A h ( 4π r ) 3h 2 h = 15 W/m .K
Properties of copper, Table A.1, App. A ρ = 8954 kg/m3 c p = 384 J/kg.K α = 11.57 x 10-5 m2/s2 T x
=
( 8954 kg / m3)( 384 J / kg . K )( 0.0125 m) = 955 sec 3(15 W / m. K )
Then:
T − T ∞
= ( T − T ∞ ) e
t T x
i
T = ( T i − T ∞ ) e
−
t T x
T = ( 313 − 273) e T = 40e
−
−
T = 40e
−
+ T ∞ t 955
+ 273 K
t 955
−
+ 273 K
t 955
C 10
1. INTRODUCTION
where t in seconds Tabulation: Time, t, seconds 0 10 20 40 60 80 100 200 300 400 600 800 1000 5000 10000 100000 1000000
Temperature, T, C 40 39.6 39.2 38.4 37.6 36.9 36.2 32.7 29.6 26.8 22 18 14.7 0.3 0.0 0.0 0.0 0.0
∞
Plot:
11
1. INTRODUCTION
1.9
Determine the total heat transfer in Problem 1.8 as the sphere cools from 40 C to 0 C. Plot the net entropy increase resulting from the cooling process above, ∆S vs T(K).
Solution: T
= 0 C + 273 = 273 K
2 A = 4π r , V =
4 3
3
π r
r = 0.0125 m ρ = 8954 kg/m3 c p = 384 J/kg.K α = 11.57 x 10-5 m2/s2 ∆T = 40 C – 0 C = 40 K Total Heat Transfer: Q = ρ cV ∆T = (8954 kg/m3)(384 J/kg.K)(4/3)(π)(0.0125 m)3(40 K) Q = 1125 J - - - - Answer. Plotting the net-entropy increase:
Equation (1.24)
12
1. INTRODUCTION
∫ T b
∆S =− ρ cV
1 1 T −T ∞
T b 0
b
dT
b
4 ∆S =−(8954)(384 ) π ( 0.0125 ) 3
∫ T b
3
T b 0
1 1 T −T ∞
T T ∆S = −28.13 − ln T − T − ln T T ∞ ∞ b0
b
b0
b
T − T ∆S = −28.13 T ∞ b
b0
T − ln T
b
b0
b
dT
b
T b0 = 40 C = 313 K
∆S = −28.13 T − 313 − ln T 273 313 b
b
T b, C
T b, K
40 35 30 25 20 15 10 5 0
313 308 303 298 293 288 283 278 273
Plot:
13
S 0 0.0622 0.117 0.1642 0.2034 0.2344 0.2569 0.2707 0.2754
1. INTRODUCTION
1.10
A truncated cone 30 cm high is constructed of Portland cement. The diameter at the top is 15 cm and at the bottom is 7.5 cm. The lower surface is maintained at 6 C and the top at 40 C. The outer surface is insulated. Assume one dimensional heat transfer and calculate the rate of heat transfer in watts from top to bottom. To do this, note that the heat transfer, Q, must be the same at every cross section. Write Fourier’s law locally, and integrate between this unknown Q and the known end temperatures.
Solution: T 1 = 40 C
14
1. INTRODUCTION
T 2 = 6 C Q = −kA
D1 − D2
dT dx
=
D1 − D
L x D1 = 15 cm = 0.15 m D2 = 7.5 cm = 0.075 m L = 30 cm = 0.30 m 0.15 m −0.075 m 0.30 m
=
0.15 m − D
x
D = 0.15 m – 0.25 x A =
π
4
D
2
π dT = −k D 4 dx dT π Q = −k ( 0.15 m − 0.25 x ) dx 4 π − Q( 0.15 − 0.25 x ) dx = −k dT 4 2
Q
2
2
Q
m
∫ 0. 3
(0.15 −0.25 x )
−2
dx
0
π =−k dT 4
Thermal Conductivity of Portland Cement, Table A.2, Appendix A. k = 0.70 W/m.K
1 [( 0.15 − 0.25 x) − ] = −( 0.70) π ( 40 − 6) − 0.25 4 π − − Q( 4)[( 0.15 − 0.25( 0.3) ) − ( 0.15) ] = −( 0.70) ( 34) 4 1 − 1 = −( 0.70) π ( 34) Q( 4) 0.075 0.15 4 1 0.3
Q( −1)
0
1
1
Q = -0.70 W – Ans.
1.11.
A hot water heater contains 100 kg of water at 75 C in a 20 C room. Its surface area is 1.3 m2. Select an insulating material, and specify its thickness, to keep the water from cooling more than 3 C / h . (Notice that this problem will be greatly simplified if the temperature drop in the steel casing and the temperature drop in the convective boundary layers are negligible. Can you make such assumptions? Explain.)
15
1. INTRODUCTION
Solution: Specific heat of water at 75 C, Table A.1 , c p = 4194 J/kg.K Q = (100 kg)(4194 J/kg.K)(3 K/hr)(1 hr / 3600 s) Q = 349.5 W A = 1.3 m2 Then: Q = −kA
∆T L
k (1.3)( 75 − 20) L
Q = −349.5 = −
k = 4.89 W/m2.K L
Select Magnesia, 85 % (insulation), Table A.2 k = 0.071 W/m.K L = (0.071 W/m.K) / (4.89 W/m2.K) = 0.01452 m = 1.5 cm Yes, we can make an assumption of neglecting temperature drops as above as the thermal conductivity of steel is much higher than insulation, also negligible temperature drops for thin film boundary. 1.12.
What is the temperature at the left-hand wall shown in Fig. 1.17. Both walls are thin, very large in extent, highly conducting, and thermally black.
Fig. 1.17
Solution: Left: q = h L (T ∞ L Right:
−T ) = 50 (100 – T L) q = h ( T −T ∞ ) = 20 (T r – 20) r
r
L
r
16
1. INTRODUCTION
Equating: q = 50 (100 – T L) = 20 (T r – 20) 5 (100 – T L) = 2 (T r – 20) 100 – T L = 0.4T r – 8 T L = 108 - 0.4T r o C Then; by radiation.
(
q = σ T L σ =
4
− T
r
4
)
5.67040 x 10-8 W/m2.K 4
q = ( 5.67040 × 10−
) (108 − 0.4T + 273) − ( T + 273) = 20( T − 20) q = ( 5.67040 × 10− ) ( 381 − 0.4T ) − ( T + 273) = 20( T − 20) 4
8
4
r
r
4
8
r
r
4
r
r
By trial and error: T r = 42 C (right hand wall) Then T L = 108 – 0.4(42) = 91.2 C (left hand wall)
1.13. Develop S.I. to English conversion factors for: • The thermal diffusivity, α • The heat flux, q • The density, ρ • The Stefan-Boltzmann constant, σ • The view factor, F 1-2 • The molar entropy • The specific heat per unit mass, c In each case, begin with basic dimension J , m, kg , s, C , and check your answer against Appendix B if possible. Solution: (1.) The thermal diffusivity, α Unit of α is m2/s. The conversion factor for English units is: 1 ft 2 3600 s 1= ⋅ 2 h ( 0.3048 m ) 2
1 = 38,750
ft / hr
, checked with Table B.2, o.k. m 2 / s (2.) The heat flux, q
Unit of q is @/m2 or J/s.m2 17
1. INTRODUCTION
The conversion factor for English units is: 2 0.0009478 Btu 3600 s ( 0.3048 m ) ⋅ ⋅ 1= J h ft 2 Btu / h ⋅ ft
2
1 = 0.317
J / s ⋅ m
Btu / h ⋅ ft
2
= 0.317
2
W / m
2
, checked with Table B.2, o.k.
(3.) The density Unit of density ρ is kg/m3 The conversion factor for English units is: 1 lb
1=
0.45359 kg
1 = 0.06243
⋅
( 0.3048 m)
3
3
ft
lb / ft 3
, checked with Table B.2, o.k.
kg / m 3
(4.) The Stefan-Boltzmann constant, σ
σ = 5.6704 x 10-8 W/m2.K 4 = 5.6704 x 10-8 J/m2.s.K 4 The conversion factor for English units is: 0.0009478 Btu
1=
J
1 = 0.0302
⋅
( 0.3048 m ) 2 3600 s ft 2
⋅
h
⋅
K 4
(1.8 F ) 4
Btu / hr . ft 2 . K 4 W / m 2 . K 4
(5.) The view factor F 1-2 The view factor is dimensionless, so there is no nee d for conversion factor. (6.) The molar entropy Unit of molar entropy, S = J/K The conversion factor for English units is. 1=
0.0009478 Btu
J
1 = 0.0005266
⋅
K 1.8 F
Btu / F J / K
(7.) The specific heat per unit mass, c Unit of c is J/kg.K The conversion factor for English units is: 1=
0.0009478 Btu 0.45359 kg
J
⋅
lb
⋅
K 1.8 F
18
1. INTRODUCTION
1 = 0.00023884
1.14.
Btu / lb ⋅ F J / kg ⋅ K
Three infinite, parallel, black, opaque plates, transfer heat by radiation,as shown in Fig. 1.18. Find T 2.
Fig. 1.18
Solution:
( − T ) = σ (T − T )
q = σ T 1
4
4
4
2
2
4
3
T 1 = 100 C + 273 = 373 K T 3 = 0 C + 273 = 273 K T 2
4
1
= [( 373) + ( 273) ] 4
4
2
T 2 = 334.1 K = 61.1 C 1.15.
Four infinite, parallel black, opaque plates transfer heat by radiation, as shown in Fig. 1.19. Find T 2 and T 3.
Fig. 1.19
19
1. INTRODUCTION
Solution:
( − T ) = σ (T − T ) = σ (T − T )
q = σ T 1
4
4
4
2
2
4
3
4
3
4
4
T 1 = 100 C + 273 = 373 K T 4 = 0 C + 273 = 273 K Then: 2T 2
2T 3 T 2
4
4
= T +T 4
1
4
3
= T + T = 2T −T
4
4
4
2
4
4
3
and
(
2 2T 3 4
4T 3
4
4
4
− T ) = T + T 4
4
4
1
− 2T = T + T 4
4
4
1
4
3
4
3
= T + 2T 3T = (373)4 + 2 (273)4 T =317.45 K = 44.45 C 3T 3
4
4
4
1
4
4
3
3
T 2
4
T 2
= 2T −T 4
3
4
4
= 2 (317.45)4 – (273)4
= 348.53 K = 75.53 C
1.16.
Two large, black, horizontal plates are spaced a distance L from one another. The top is warm at a controllable temperature, T h, and the bottom one is cool at a 20
1. INTRODUCTION
specified temperature, T c. A gas separates them. The gas is stationary because it is warm on top and cold on the bottom. Write the equation qrad /qcond = fn ( N , Θ
≡ T T h
), where N is dimensionless group containing σ , k , L, and T c. Plot Ν as a
c
function of Θ for qrad /qcond = 1, 0.8, and 1.2 (and for other values if you wish). Now suppose that you have a system in which L = 10 cm, T c = 100 K, and the gas is hydrogen with an average k of 0.1 W/m.K. Further suppose that you wish to operate in such a way that the conduction and radiation heat fluxes are identical. Identify the operating point on your curve and report the value of T h that you must maintain. Solution:
= σ (T − T 4
qrad
h
4
c
k ( T h
)
− T )
qcond
=
qrad
(T T ) σ L = ⋅ − = ⋅ ( T + T ) (T + T ) k ( T − T ) k
c
L
4
σ L
4
h
2
c
h
qcond
h
qrad
=
qcond
qrad
=
qcond
L σ k σ L
k
c
2
h
c
c
T T ⋅ +1 T +1 T 2
⋅T
3
c
h
h
c
c
⋅ T ⋅ ( Θ + 1) [Θ + 1] = N ⋅ ( Θ + 1) [Θ + 1] 3
2
2
c
where
N =
Θ=
σ LT c
3
k T h T c
N as a function of Θ ;
N =
(1)
qrad qcond
( Θ + 1) ( Θ + 1) 2
qrad qcond
N =
=1 1
( Θ + 1) ( Θ + 1) 2
21
1. INTRODUCTION
(2)
qrad
N =
(3)
= 0.8
qcond
0.8
( Θ + 1) ( Θ + 1) 2
qrad qcond
N =
=1.2 1.2
( Θ + 1) ( Θ + 1) 2
plot of N as a function of Θ:
For the system: L = 10 cm = 0.10 m T c = 100 K k = 0.1 W/m.K For qrad / qcond = 1.0 Then 1 = N ( Θ +1) ( Θ2
+1)
Solving for N :
N =
σ LT c
3
k 22
1. INTRODUCTION
σ = 5.67040 x 10-8 W/m2.K 4 − ( 5.6704 ×10 ) ( 0.10)(100) N = 8
3
= 0.056704
0.10
Then 1 = ( 0.056704)( Θ +1) ( Θ2
+1)
( Θ + 1) ( Θ + 1) = 17.64 2
By trial and error: Θ = 2.145 Then: Th = ΘTc = (2.145)(100 K) = 214.5 K 1.17.
A blackened copper sphere 2 cm in diameter and uniformly at 200 C is introduced into an evacuated black chamber that us maintained at 20 C. Write a differential equation that expresses T(t) for the sphere, assuming lumped thermal capacity. Identify a dimensionless group, analogous to the Biot number, that can be used to tell whether or not the lumped-capacity solution is valid. Show that the lumped-capacity solution is valid. Integrate your differential equation and plot the temperature response for the sphere.
• • • •
Solution: (1) Assuming lumped thermal capacity Q=
dU dt
−σ A(T −T ∞ ) = 4
4
d dr
[ ρ cV (T −T ) ] ref
T = T ( t )
d (T −T ∞ ) dt
=
A −σ (
ρ cV
T
4
−T ∞
4
)
A = 4π r 2 4 3 V = π r 3
d ( T − T ∞ ) dt
− σ ( 4π r ) − 3σ = ( T − T ∞ ) = (T − T ∞ 4 ρ cr ρ c π r 3 2
4
4
4
4
)
3
Differential Equation, T = T ( t ) d ( T −T ∞ ) dt
=
− 3σ ( ρ cr
T 4
−T ∞
4
)
(2) Dimensionless group analogous to the Biot number 23
1. INTRODUCTION
=
Bi
h L k b
Equivalent h , 4 σ (T 4 − T ∞ h = T − T ∞ Biot number equivalent =
h V k b A
=
h r 3k b
(
− T ∞ = 3k ( T − T ∞ ) 4
4
r T
σ
b
(3) Showing that lumped-capacity solution is valid.
r (T
− T ∞ 3k ( T − T ∞ ) 4
σ
Dimensionless group must be << 1 =
4
b
T i = 200 C + 273 = 473 K
= 20 C + 273 = 293 K σ = 5.6704 x 10-8 W/m2.K 4 r = (1/2)(2 cm) = 1 cm = 0.01 m T
For copper: ρ = 8,954 kg/m3 c p= 384 J/kg.K k b = 389 W/m/K @ 200 C, Table A-1, App. A.
r (T
− T ∞ 3k ( T − T ∞ ) 4
4
σ
b
( 5.6704 ×10 − )( 0.01) ( 473 − 293 ) = = 0.00012 << 1 , therefore valid. 3( 389 )( 473 − 293) 8
4
4
(4) Integrating and plotting differential equation. d ( T −T ∞ ) dt dT dt
=
4
ρ cr
− T ∞
∫
− 3σ ρ cr
−3σ T
dT T
=
4
=
(
4
dT 4 T −T ∞ T
4
4
−T ∞ ) 4
− 3σ dt ρ cr
T
4
(T −T ∞ )
=
−3σ t cr ρ
i
Note:
∫
dx x −a 4 4
: 24
1. INTRODUCTION
( x + a ) − ( x − a ) = = ( x − a )( x + a ) 2a ( x − a )( x + a )
1
−a
4
x
4
2
1 x
4
−a
4
=
4
=
4
=
1 x
4
x
4
−a 1
−a 1
=
x 4 − a 4 1
−a
4
x
2
1
=
4
2
2
2
1 2a ( x 1 2
2a 2
2
2
2
2
1
2
2
2
2
2
2
2
2
2
2
1 1 − 1 − x − a x + a 2a ( x + a )
1 3
2
1 − 1 − x − a x + a
1 4a
2
2
1
4a
2
2
− a ) 2a ( x + a ) ( x + a ) − ( x − a) 1 − 2a( x − a ) ( x + a) 2a ( x + a ) 1 1 1 2a( x − a ) − 2a( x + a) − 2a ( x + a )
2
2a
−
2
3
2
2
1
x a + 1 2
2a
4
1 1 x
−a
4
=
4
1 − 1 − x − a x + a
1 4a
3
a
x a + 1 2
2a
3
Then,
∫ x
dx
−a
4
=
4
x − a − 1 x + a 2a
1 4a
ln
3
3
x + C a
Arc tan
Applying: T
∫ ∫ ∫ T i
dT 4
T
−T ∞
4
=
1
T −T ∞ T −T ∞ 1 T T − − − ln ln Arc tan Arc tan T +T 2T T T + T ∞ ∞ ∞ T ∞ ∞ i
3
4T ∞
i
3
i
Substitute values: T
T i
dT 4
T
−T ∞
4
=
T − 293 473 − 293 2 T − Arc tan 473 Arc tan ln − ln − 4( 293) T + 293 473 + 293 4( 293) 293 293
=
T − 293 T + 3.48062 = − 3σ t ln − 2 Arc tan ρ cr 4( 293) T + 293 293
T
T i
T
dT 4
−T ∞
4
1
3
3
1
3
T − 293 − 3( 5.6704 ×10− )t T − 2 Arc tan + 3.48062 = ln 4( 293) T + 293 293 ( 8954)( 384)( 0.01) 1 T − 293 T + 3.48062 = −0.0004978 t ln 2 Arc tan − 4( 293) T + 293 293 8
1
3
3
Tabulation:
25
1. INTRODUCTION T , C 200 182 164 146 128 110 92 74 56 38
T , K 473 455 437 419 401 383 365 347 329 311
Plot :
26
T , s 0 93.5 206.8 346.1 520.9 745.8 1046 1468.5 2119.8 3340.6
1. INTRODUCTION
1.18.
As part of space experiment, a small instrumentation packaged is released from a space vehicle. It can be approximated as a solid aluminum sphere, 4 cm in diameter. The sphere is initially at 30 C and it contains a pressurized hydrogen component that will condense and malfunction at 30 K. If we take the surrounding space to be at 0 K, how long may we expect the implementation package to function properly? Is it legitimate to use the lumped-capacity method in solving the problem? (Hint: See the directions for Problem 1.17).
Solution: Properties of aluminum, Table A.1 ρ = 2707 kg/m3 c p = 905 J/kg.K k b = 237.2 W/m.K @ 30 C From Prob. 1.17, using T i = 30 C + 273 = 303 K T = 0 K T = 30 K σ = 5.6704 x 10-8 W/m2.K r = (1/2)(4 cm) = 2 cm = 0.02 m Check for the legitimacy of lumped-capacity method.
r (T
− T ∞ 3k ( T − T ∞ ) 4
4
σ
b
Then,
∫ ∫
T
dT 4 T −T ∞ 4
T
dT T
T i
4
8
4
= −3σ t ρ cr
−3(5.6704 ×10 −
8
=
4
= 0 K
T
T i
( 5.6704 ×10 − )( 0.02) ( 303 − 0 ) = = 0.000044 << 1 , therefore valid. 3( 237.2 )( 303 − 0)
)t
(2707 )(905)(0.02)
− 1 = − 3( 5.6704 ×10 − )t 3T ( 2707 )( 905)( 0.02) T
8
3
T i
− 3( 5.6704 ×10− )t − = 3T ( 2707)( 905)( 0.02) 3T 1 1 − 3( 5.6704 ×10− )t − = 1
8
1
3
3
i
8
3( 303)
3
3( 30)
3
( 2707)( 905)( 0.02)
1 min 1 hr 1 day 1 week = 5.88 weeks 60 sec s 60 min s 24 hrs 7 days
t = 3,552,427 seconds
27
1. INTRODUCTION
1.19.
Consider heat calculation through the wall as shown in Fig. 1.20. Calculate q and the temperature of the right-hand side of the wall.
Fig. 1.20
Solution: − k ( T 1 − T 2 ) q= L T = 200 C T =0C k = 2 W/m2.K L = 0.5 m h = 3 W/m2.K
= −h( T − T ∞ ) 2
1
q=
− ( 2)( 200 − T ) = −( 3)( T − 0) 2
2
0.50
800 − 4T 2
= 3T
2
T 2 = 114.286 C q = (3)(114.286 – 0) = 343 W/m2. 1.20.
Throughout Chapter 1 we have assumed that the steady temperature distribution in a plane uniform wall is linear. To prove this, simplify the heat diffusion equation to the form appropriate for steady flow. Then integrate it twice and eliminate the two constant using the known outside temperatures T left and T right at x = 0 and x = wall thickness, L.
28
1. INTRODUCTION
Solution: Eq. 1.14, one dimensional heat diffusion equation, ∂2T ρ c∂T 1 ∂T
∂ x
=
2
k ∂t
≡
α
∂t
∂ T = 0 for steady flow. ∂ x 2
Use
2
dT
= C 1 dx T = C 1 x + C 2 at T = T left , x = 0 T left = 0 + C 2 C 2 = T left At T = T right , x = L T right = C 1 L + T left C 1
=
Then,
T =
T right
left
L
T right − T left L
T −T left
=
x
1.21
−T
T right
x + T left
−T
L
left
, therefore linear.
The thermal conductivity in a particular plane wall depends as follows on the wall temperature: k = A + BT , where A and B are constants. The temperatures are T 1 and T 2 on either side of the wall, and its thickness is L. Develop an expression for q.
Solution: dT
q
= −k
q
= −( A + BT )
dx
dT dx
qdx = −( A + BT ) dT
29
1. INTRODUCTION L
q
T 2
∫
( ∫
dx = −
)
A + BT dT
T 1
0
T 2
1 qL = − AT + BT 2 1 qL = − A(T −T ) + B(T −T ) 2 1 − A( T 2 −T 1 ) + B(T 2 2 −T 12 ) 2 q= 2
T 1
2
2
1
2
2
1
L
1.22
Find k for the wall shown in Fig. 1.21. Of what might it be made?
Figure 1.21.
Solution: L = 0.08 m T left − T right = −h(T ∞ − T left ) q = −k L ( 20 − 0) − k = −( 200)(100 − 20) 0.08 k = 64 W/m.K
30
1. INTRODUCTION
From Table A.1, @ 10 C, k = 64 W/m.K This might be Steel, AISI 1010, k = 64.6 W/m.K 1.23
What are T ,i T j, and T r in the wall shown in Fig. 1.22?
Fig. 1.22.
Solution: L1 = 2 cm = 0.02 m k 1 = 2 W/m.C L2 = 6 cm = 0.06 m k 2 = 1 W/m.C L3 = 4 cm = 0.04 m k 3 = 5 W/m.C L4 = 4 cm = 0.04 m k 4 = 4 W/m.C q=
k 1 (100 − T i )
=
L1
k 1 (100 − T i ) L1
( 2)(100 − T ) i
0.02 600 − 6T i
= =
k 2 ( T i − 25) L2
=
k 3 25 − T j L3
=
k 2 ( T i − 25) L2
(1)( T − 25) i
0.06
= T − 25 i
T i =89.29 C
k 1 (100 − T i ) L1
=
k 3 25 − T j
( 2)(100 − 89.29) 0.02 T j = 16.43 C
L3
=
( 5)
25 − T j 0.04
31
k 4 T j
− T
L4
r
1. INTRODUCTION
k 1 (100 − T i ) L1
=
k 4 T j
( 2)(100 − 89.29) 0.02 ( 2)(100 − 89.29) 0.02 T r = 5.72 C
1.24
− T
r
L4
= =
( 4 )(16.43 − T ) r
0.04 ( 4)(16.43 − T r ) 0.04
An aluminum can of beer or soda pop is removed from the refrigerator and set on the table. If h is 13.5 W/m2.K, estimate when the beverage will be at 15 C. Ignore thermal radiation. State all of your other assumptions.
Solution: Properties of aluminum, Table A.1, App. A ρ = 2707 kg/m3 c p = 905 J/kg.K k = 237 W/m.K α = 9.61 x 10-5 m2/s Assume size of can is 50 mm diameter x 100 mm height T i = 0 C, and room at T = 20 C Time constant, π D 2 ρ c L ρ cV 4 = T = h A π D 2 + π DL h 2 T =
ρ cDL
2h ( D + 2 L )
D = 0.05 m L = 0.10 m h = 13.5 W/m2.K T =
( 2707)( 905) ( 0.05)( 0.10) = 648.1 ns 2(13.5) ( 0.5 + 2( 0.10) )
Eq. 1.22. T − T ∞
= e−
t T
T i − T ∞ at T = 15 C
32
1. INTRODUCTION
15 − 20
= e−
t
648.1
0 − 20 t = 898.5 s = 15 minutes
1.25. One large, black wall at 27 C faces another whose surface is 127 C. The gap between the two walls is evacuated. If the second wall is 0.1 m thick and has a thermal conductivity of 17.5 W.m.K, what is its temperature on the back side? (Assume steady state).
Solution:
T3 = temperature on the back side. k ( T 3 − T 2 ) 4 4 q = σ (T 2 − T 1 ) = L L = 0.1 m T 1 = 27 C + 273 = 300 K T 2 = 127 C + 273 = 400 K σ = 5.6704 x 10-8 W/m2.K k = 17.5 W/m.K
q = ( 5.6704 × 10 −8 )( 400 4 − 300 4 ) =
(17.5)( T − 400) 3
0.10
T 3 = 405.67 K = 132.67 C. 1.26. A 1-cm diameter, 1 % carbon steel sphere, initially at 200 C, is cooled by natural convection, with air at 20 C. In this case, h is not independent of temperature. 33
1. INTRODUCTION
Instead, h =3.51(∆t C)1/4 W/m2.K. Plot T sphere as a function of t . Verify the lumpedcapacity assumption. Solution: Properties of 1% carbon steel, Table A.1 ρ = 7801 kg/m3 c p = 473 J/kg.K k = 42 W/m.K α = 1.17 x 10-5 m2/s Verify the lumped-capacity assumption: Bi
≡
h L k
∆t = 200 C – 20 C = 180 C h =3.51(180)1/4 W/m2.K = 12.86 W/m2.K 4 L =
V A
π r 3
3 4π r 2
=
=
r 3
r = (1/2)(1 cm) = 0.005 m L = 0.005 m / 3 = 0.001667 m (12.86 )( 0.001667 ) Bi ≡ = 0.00051 << 1, therefore valid. 42 Use Lumped-Capacity Method T = 20 C T i = 200 C Q=
dU dt
− hA( T − T ∞ ) = or
d ( T −T ∞ ) dt
h
=
dt
=
d ( T − T ∞ )
( T − T ∞ )
5/ 4
(T −T ∞ ) −
5/ 4
dt
− hA( T −T ∞ )
= 3.51( ∆T )
d (T −T ∞ )
d ρ cV T − T ref
ρ cV 1/ 4
W / m 2 . K = 3.51( T −T ∞ )
−3.51 A(T −T ∞ )
1/ 4
5/ 4
ρ cV
= − 3.51 Adt ρ cV
d (T −T ∞ )
= −3.51 Adt ρ cV
34
W / m 2 .K
1. INTRODUCTION
− 3.51 At ( − 4) ( T − T ∞ ) − = ρ cV 1
T
4
T i
1 − At − ( T − T ∞ ) − + = 3.51 ρ cV + 1 5
T
1
4
5
T i
4
1
(T − T ∞ ) A V
=
−
1/ 4
1
( T i − T ∞ )
=
1/ 4
0.8775 At ρ cV
3 r
Then, 1
( T − T ∞ )
−
1/ 4
1
( T − T ∞ )
1/ 4
=
0.8775 3
ρ c
i
( t ) = r
2.6325t
ρ cr
Substitute value, 1
(T − 20 )1/ 4
−
1
( T − 20)
1/ 4
( 200 − 20 ) 1/ 4
=
2.6325t
( 7801)( 473 )( 0.005)
= 0.000143t + 0.273012
1/ 4
(T −20 )1/ 4 =
(T −20)
1
=
1 0.000143t +0.273012 6993 t +1909 4
6993 T = t +1909 + 20 C Tabulation: t,s 0 100 200 300 400 500 600 800 1000 1200 1400 1600 1800 2000
T, C 200 166.8 140.9 120.4 104.7 91 80.4 64.4 53.4 45.6 40 35.8 32.6 30.2
Plot: 35
1. INTRODUCTION
1.27. A 3-cm diameter, black spherical heater is kept at 1100 C. It radiates through an evacuated space to a surrounding spherical shell of Nichrome V. The shell has a 9 cm inside diameter and is 0.3 cm thick. It is black on the inside and is held at 25 C on the outside. Find (a) the temperature of the inner wall of the shell and (b) the heat transfer, Q. (Treat the shell as a plane wall.) Solution: Properties of Nichrome V, Table A.1, Appendix A. ρ = 8,410 kg/m3 c p = 466 J/kg.K k = 10 W/m.K α = 0.26 x 10-5 m2/s
Radiation
Qrad
= σ A (T − T ) 4
1
4
1
2
T 1 = 1100 C = 1373 K T 3 = 25 C + 273 = 298 K σ = 5.6704 x 10-8 W/m2.C Conduction − kA2 ( T 3 Qcond = L
A1
= 4π r
A1
= 4π ( 0.015 )
2
1
− T ) 2
, r 1 = (1/2)(3 cm) = 1.5 cm = 0.015 m 2
m2 36
1. INTRODUCTION
A2
= 4π r
2
2
, r 2 = (1/2)(9 cm) = 4.5 cm = 0.045 m
= 4π ( 0.045 )
2
m2 L = 0.3 cm = 0.003 m A1
Then Qrad
A1 σ
=Q
cond
(T − T ) = kA (T − T ) 4
4
1
2
2
2
3
L
( 5.6704 ×10 )( 4π )( 0.015) [(1373) − T ] = −8
2
4
4
(10)( 4π )( 0.045) ( T − 298) 2
2
2
0.003
(1373) − T = 5.290632 × 10 ( T − 298) 4
4
11
2
2
By trial and error method. (a) Inner Wall Temperature = T 2 = 304.7 K = 31.7 C (b) Heat Transfer, Q
Q = σ A ( T − T ) = ( 5.6704 × 10 − 4
1
1
4
8
2
) ( 4π ) ( 0.015) [ (1373) − ( 304.7 ) ] = 568.4 W 2
4
4
1.28. The sun radiates 650 W/m2 on the surface of a particular lake. At what rate (in mm/hr) would the lake evaporate if all of this energy went to evaporating water? Discuss as many other ways you can think of that this energy can be distributed (h fg for water is 2,257,000 J/kg). Do you suppose much of the 650 W/m2 goes to evaporation? Solution: q = 650 W/m2 = 2,340,000 J/hr.m2 Evaporation rate =
2,340,000 J / hr .m 2,257,000 J / kg
2
= 1.036774 kg/hr.m2
Density of water ρ = 1000 kg/m3 Evaporation rate =
• •
1.036774 kg / hr .m 2 1000 kg / m
3
1000 mm =1.036774 mm/hr 1 m
There are other ways that this energy can be distributed such as cloud barrier, heating up of the lake up to evaporation, haze or atmosphere. Yes, much of the 650 W/m2 goes to evaporation especially on a clear day.
37
1. INTRODUCTION
1.29. It is proposed to make picnic cups, 0.005 m thick, of a new plastic for which k = k o(1 + aT 2), where T is expressed in C, k o = 15 W/m.K, and a = 10-4 C-2. We are concerned with thermal behavior in the extreme case in which T = 100 C in the cup and 0 C outside. Plot T against position in the cup wall and find the heat loss, q. Solution: q
= −k
dT dx
qdx = −k o (1 + aT ) dT 2
T 2
( ∫
q∆ x = k o −
)dT
aT 1+
2
T 1
q∆ x = − k o [ T + aT ] T 3
1
T 2
3
[
1
q∆ x = − k o ( T + aT ) − ( T + aT ) −
q∆ x
(T + 2
1 3
3
= (T + 2
k o
aT 2
3
3
1
2
1 3
3
aT 2
1 3
1
3
1
) −(T + 1
) = (T +
3
1
2
3
aT 1
1 3
1
aT 1
3
]
)
) − q∆ x k o
Solving for q if, T 1 = 100 C T 2 = 0 C ∆ x = 0.005 m q∆ x k o
= (T + 1
q( 0.005)
1 3
3
aT 1
) −(T + 2
1 3
aT 2
3
)
= [(100) + (10 − )(100 ) ] − [( 0) + (10− )( 0 ) ]
0.15 q = 4000 W
3
4
1 3
4
1 3
3
Plotting: Use T 1 = 100 C, a = 10-4 C-2, q = 4000 W, k o = 0.15 W/m.K
(T + 2
1 3
aT 2
∆ x = (
3
) = (T + 1
k o T + aT
3
1
1
3
1
1 3
3
aT 1
) − (T + 2
∆ x =
k o
1 3
aT
2
3
)
q
( 0.15) (100 + (10 ) (100) −4
1
∆ x =
) − q∆ x
3
3
(
− T 2 + 13 aT 2
4000 20
− 0.15 T + (10− )T 1
2
3
4
3
2
4000
38
3
1. INTRODUCTION
Tabulation: T 2, C
x , m 0 0.00136 0.00248 0.00342 0.00424 0.00500
100 80 60 40 20 0
Heat loss , q = 4000 W
1.30. A disc-shaped wafer of diamond 1 lb is the target of a very high intensity laser. The disc is 5 mm in diameter and 1 mm deep. The flat side is pulsed intermittently with 1010 W/m2 of energy for one microsecond. It is then cooled by natural convection from that same side until the next pulse. If h = 10 W/m2.K and T =30 C, plot T disc as a function of time for pulses that are 50 s apart and 100 s apart. (Note that you must determine the temperature the disc reaches before it is pulsed each time.) Solution: Properties of Diamond, Table A.2 ρ = 3250 kg/m3 c p = 510 J/kg.K k b = 1350 W/m.K α = 8.1 x 10-4 m2/s 39
1. INTRODUCTION
L = 1 mm = 0.001 m Bi
=
h L k b
= 10 W/m2.K T =30 C h
Bi
= (10)( 0.001) = 0.0000074 << 1
1350 Therefore lumped capacity solution is valid. t − T −T ∞ T =e T i −T ∞
For 50 s apart, On the first pulse, q = 1010 W/m2
= q A(time) A = π r , r = (1/2)(5 mm) = 2.5 mm = 0.0025 m Q
2
time = 1 µs = 1 x 10-6 s Q = (1010 W/m2)(π)(0.0025)2(1 x 10-6) = 1.9635 W Q = ρ cV ( T i − 30 C ) V = π r 2 L
Q = ( 3250 )( 510 )( π )( 0.0025) ( 0.001)(T i − 30 C ) 2
T i = 90.33 – this is the initial temperature on the first pulse. ∆T = 90.33 C – 30 C = 60.33 C Then: T −T ∞ T i −T ∞
=e
−
t T
Time constant, T = T
ρ cV
h A
=
ρ cL
h
=
( 3250 )( 510 )( 0.001) 10
= 165.75 s
=30 C
For 50 s pulse apart First 50 s, T i = 90.33 C t = 25 s 25 − T − 30 165.75 =e 90.33 − 30 T = 81.88 C 40
1. INTRODUCTION
t = 50 s T − 30
90.33 − 30
=e
50
−
165.75
T = 74.62 C Second 50 s, T i = 60.33 C + 74.62 C = 134.95 C t = 25 s 25 − T − 30 165.75 =e 134.95 − 30 T = 120.26 C t = 50 s T − 30
134.95 − 30
=e
−
50
165.75
T = 107.62 C Third 50 s, T i = 60.33 C + 107.62 C = 167.95 C t = 25 s 25 − T − 30 165.75 =e 167.95 − 30 T = 148.64 C t = 50 s T − 30
167.95 − 30
=e
−
50
165 .75
T = 132.03 C And so on…. Tabulation: st
1 50 s
2nd 50 s
3rd 50 s
t , s 0 25 50 50 75 100 100 125 150
T disc, C 90.33 81.88 74.62 134.95 120.26 107.62 167.95 148.64 132.03
41
1. INTRODUCTION
Plot:
For 100 s pulse apart
First 100 s, T i = 90.33 C t = 50 s 50 − T − 30 165.75 =e 90.33 − 30 T = 74.62 C t = 100 s
42
1. INTRODUCTION
T − 30 90.33 − 30
=e
−
100
165.75
T = 63.00 C Second 100 s, T i = 60.33 C + 63.00 C = 123.33 C t = 50 s − 50 T − 30 = e 165.75 123 .33 − 30 T = 99.03 C t = 100 s T − 30
123.33 − 30
=e
100
−
165.75
T = 81.05 C Third 100 s, T i = 60.33 C + 81.05 C = 141.38 C t = 50 s − 50 T − 30 = e 165.75 141 .38 − 30 T = 148.64 C t = 100 s T − 30
141.38 − 30
=e
−
100
165 .75
T = 112.38 C And so on…. Tabulation: st
1 100 s
2nd 100 s
3rd 100 s
t , s 0 50 100 100 150 200 200 250 300
T disc, C 90.3.3 74.62 63.00 123.33 99.03 81.05 141.38 112.38 90.93
43
1. INTRODUCTION
Plot:
1.31
A 150 W light bulb is roughly a 0.006 m diameter sphere. Its steady surface temperature in room air is 90 C, and h on the outside is 7 W/m2.K. What fraction of the heat transfer from the bulb is by radiation directly from the filament through the glass? (Stae any additional assumptions.)
Solution: Assume black body radiation.
Q = h A( T b Q A
− T ) + σ A(T − T 4
a
b
a
4
)
= h (T −T ) +σ (T −T ) 4
b
a
b
4
a
44
1. INTRODUCTION T b = 90 + 273 = 363 K
= 7 W/m2.K. σ = 5.6704 x 10-8 W/m2.K 4. Q = 150 W but change to 0.150 W as light bulb is very small. It may be a typographical error. h
A = π D
2
= 4π r
2
D = 0.006 m Then: 0.150 π ( 0.006)
1326
= ( 7)( 363 −T ) + (5.6704 ×10− )(363 −T 8
2
4
a
a
4
)
= ( 7) ( 363 − T ) + ( 5.6704 × 10− ) ( 363 − T ) 8
4
4
a
a
By Trial and Error Method: T a = 270.5 K = -2.5 C
Fraction =
Q
σ
=
(T b
4
− T a
4
)
=
h ( T b − T a ) + σ (T b − T a ) ( 5.6704 ×10− )( 363 − 270.5 ) ( 7)( 363 − 270.5) + ( 5.6704 ×10− )( 363 − 270.5 ) A
4
8
4
4
4
8
4
4
Fraction = 0.5126 1.32
How much entropy does the light bulb in Problem 1.31 produce?
Solution:
Un S
1 1 = Q − T T a
1.33
b
1 1 = 0.0001413 W/K = ( 0.15) 270.5 − 363
Air at 20 C flows over one side of a thin metal sheet ( h = 10.6 W/m2.K). Methanol at 87 C flows over the other side ( h = 141 W/m2.K). The metal functions as an electrical resistance heater, releasing 1000 W/m 2. Calculate (a) the heater temperature, (b) the heat transfer from the methanol to the heater, and (c) the heat transfer from the heat of the air.
Solution: (a) q = 1000 W/m2 q = h1 ( T h
− 20 C ) + h ( T −87 C ) (10.6)( T h − 20) + (141)( T h − 87) = 1000 2
h
T h = 88.9 C
45
1. INTRODUCTION
(b) qm
= h (T − 87 ) = (141)( 88.9 − 87 ) 2
h
qm = 267.9 W (c) qa = h1 ( T h − 20) = (10.6)( 88.9 − 20) q a = 730.3 W
1.34
A planar black heater is simultaneously cooled by 20 C air ( h =14.6 W/m2.K) and by radiation to a parallel black wall at 80 C. What is the temperature of the heater if it delivers 9000 W/m2 ?
Solution:
[
q = h ( T − 20) + σ ( T + 273)
4
− ( 80 + 273) ] = 9000 W/m2 4
q = (14.6 )( T − 20) + ( 5.6704 ×10 −8 )[( T + 273)
4
− ( 353) ] = 9000 W/m2 4
By Trial and error method. T = 294.3 C
1.35
An 8-oz. can of beer is taken from a 3 C refrigerator and placed in a 25 C room. The 6.3 cm diameter by 9 cm high can is placed on an insulated surface ( h =7.3 W/m2.K). How long will it take to reach 12 C? Ignore thermal radiation and discuss your other assumption.
Solution: T −T ∞ T i −T ∞
=e
−
t T
Assume aluminum material for the can of beer, Properties of aluminum, Table A.1, Appendix A ρ = 2707 kg/m3 c p = 905 J/kg.K k = 237 W/m.K Then, = 25 C T i = 3 C T = 12 C T
46
1. INTRODUCTION
Time constant: T =
σ cV
h A
π D L 4
V =
2
D = 6.3 cm = 0.063 m
π ( 0.063) ( 0.09) = 2.8055 x 10-4 m3 4 π π A = D ( 2 ) + π DL = ( 0.063) ( 2 ) + π ( 0.063)( 0.09) = 0.02405 m2 4 4 ( 2707 )( 905) ( 2.8055 ×10−4 ) T = = 3915 s ( 7.3)( 0.02405) V =
2
2
2
12 − 25 3 − 25
=e
−
t 3915
t = 2314 sec = 38.6 min 1.36
A resistance heater in the form a thin sheet runs parallel with 3 cm slabs of cast iron on either side of an evacuated cavity. The heater, which releases 8000 W/m2, and the cast iron are very nearly black. The outside surfaces of the cast iron slabs are kept at 10 C. Determine the heater temperature and the inside slab temperatures.
Solution:
q = 8000 W/m2
47
1. INTRODUCTION
Properties of cast iron, Table A.1, Appendix A ρ = 7272 kg/m3 c p = 420 J/kg.K k b = 52 W/m.K L = 3 cm = 0.03 m Inside slab temperature, (10C − T ) q = −k = 8000 W/m2 L (10− T ) q = −( 52 ) 0.03 T = 14.62 C Heater temperature,
( − T ) = 8000 W/m
q = σ T h
4
4
q = ( 5.6704 ×10
−8
2
)[( T + 273) − (14.62 + 273) 4
h
4
] = 8000 W/m
2
T h = 347.2 C
1.37
A black wall at 1200 C radiated to the left side of a parallel slab of type 316 stainless steel, 5 mm thick. The right side of the slab is to be cooled convectively and is not to exceed 0 C. Suggest a convective proceed that will achieve this.
Solution:
1.38
A cooler keeps one side of a 2 cm layer of ice at –10 C. The other side is exposed to air at 15 C. What is h just on the edge of melting? Must h be raised or lowered if melting is to progress? 48
1. INTRODUCTION
Solution: Melting point of ice = 0 C Thermal Conductivity of ice at 0 C = 2.215 W/m.K k ( T 2 − T 1 ) = h ( T 2 − T 3 ) q=− L k ( T 2 − T 1 ) = h ( T 3 − T 2 ) L k ( T 2 − T 1 ) = h ( T 3 − T 2 ) L T 1 = -10 C T 2 = 0 C T 3 = 15 C L = 2 cm = 0.02 m Then, ( 2.215) ( 0 − ( −10) ) = h (15 − 0) 0.02 2 h = 73.83 W/m .K If the melting is to progress the thickness will reduce and h must be raised.
1.39
At what minimum temperature does a black heater deliver its maximum monochromatic emissive power in the visible range? Compare your result with Fig. 10.2.
Solution: Figure 1.15 or Wien’s Law, Eq. (1.29) ( λ T ) eλ =max = 2898 µm.K Minimum visible range, λ = 0.4545 µm Then: (0.4545 µm)(T min) = 2898 µm.K T min = 6376 K From Fig. 10.2 , T = 5900 K 1.40
The local heat transfer coefficient during the laminar flow of fluid over a flat plate of length L is equal to F / x1/2, where F is a function of fluid properties and the flow velocity. How does h compares with h ( x = L). (x is the distance from the leading edge of the plate.) 49
1. INTRODUCTION
1.41
An object is initially at a temperature above that of its surroundings. We have seen that many kinds of convection processes will bring the object into equilibrium with its surroundings. Describe the characteristics of a process that will do so with the least net increase of the entropy of the universe.
Solution:
1 1 ∆S = − ρ cV ∫ T −T ∞ T b T bo
b
dT
b
Determine T b for least net increase of the entropy of the universe. 1 1 − =0 T ∞ T b T b
=T ∞
T − T T ∆S = − ρ cV ∞ − ln ∞ T T ∞ bo
bo
T −T ∞ T ∆S = ρ cV − ln T ∞ T ∞ bo
bo
The characteristic of the process is unsteady state conduction having Biot number increasing from less than one to more than one when reaching equilibrium at T b =T ∞ . 1.42 A 250 C cylindrical copper billet, 4 cm in diameter and 8 cm long is cooled in air at 25 C. The heat transfer coefficient is 5 W/m 2.K Can this be treated as lump-capacity cooling? What is the temperature of the billet after 10 minutes? Solution: Check Biot Number Properties of copper, Table A.1, App. A ρ = 8954 kg/m3 c p = 384 J/kg.K k b = 398 W/m.K Time constant: T =
σ cV
h A
π D L 4
V =
2
D = 4 cm = 0.04 m, L = 8 cm = 0.08 m
π ( 0.04) ( 0.08) = 1.0053 x 10-4 m3 4
V =
2
50
