1. INTRODUCTION 1.1 A composite wall consist of alternative layers of fir ( 5 cm thick ) , aluminum ( 1 cm thick ), lea
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1. INTRODUCTION 1.1
A composite wall consist of alternative layers of fir ( 5 cm thick ) , aluminum ( 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 temperature gradient through the wall. Does the temperature profile suggest any simplifying assumptions that might be made in subsequent analysis of the wall?
Solution: Thermal Conductivities: kfir = 0.12 W/m.K (Table A.2, Appendix A) kalu = 237 W/m.K (Table A.1, Appendix A) kld = 35 W/m.K (Table A.1, Appendix A) kcb = 0.04 W/m.K (Table A.2, Appendix A). Question No. 1:
Plot the temperature gradient through the wall. Answer:
Question No. 2: Does the temperature profile suggest any simplifying assumptions 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.
∆Τfir + ∆Tcb = 60 C – 10 C = 50 K ∆T ∆T q = k = k L fir L cb
Lfir = 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 )( ∆Tcb ) = ( 0.05 m ) ( 0.06 m )
∆Tcb = 3.6∆Tfir Then,
∆Τfir + 3.6∆Τfir = 50 K ∆Τfir = 10.87 K 10.87 K ∆T q = k = ( 0.12 W / m.K ) 0.05 m L fir
=
26.09 W/m2
Considering all walls:
∆Tfir + ∆Talu + ∆Tld + ∆Tcb = 60 C – 10 C = 50 K ∆T ∆T ∆T ∆T q = k = k = k = k L fir L alu L ld L cb
Lfir = 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 ∆Talu = ∆Tfir k L alu
k L fir ∆Tld = ∆T fir k L ld k L fir ∆Tcb = ∆Tfir k L cb
Then
2
1. INTRODUCTION 1 1 1 k = ∆T fir 1 + + + k k k L fir L L L cb alu ld
50 K
0.12 1 1 1 = ∆T fir 1 + + + 237 35 0 . 04 0 . 05 0.01 0.01 0.06
50
∆Tfir = 10.87 K
10.87 K ∆T q = k = ( 0.12 W / m.K ) 0.05 m L fir
=
26.09 W/m2
There it is equal to simplified solution. 1.2
Verify Equation (1.15).
Solution: Equation (1.15) dTbody dt
∝Tbody −T∞
For verification only Equation (1.3) Q=
dU dT = mc dt dt
Equation (1.16) Q ∝Tbody −T∞
Then mc
dT ∝ Tbody −T∞ dt
dT ∝Tbody −T∞ dt
Then dTbody dt
∝Tbody −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) Silver Slab ∆T q = k = 5000 W/m2 L Si
Thermal conductivity of silver at 140 C, 99.99+ % Pure, Table A.1, Appendix A ksi = 420 W/m.K ∆TSi = 5000 W/m2 q = ( 420 W / m ⋅ K ) 0 . 01 m
∆TSi = 0.12 K
(b) Alumium Slab ∆T q = k = 5000 W/m2 L alu
Thermal conductivity of aluminum at 140 C, 99.99+ % Pure, Table A.1, App. A Kalu = 237.6 W/m.K ∆Talu 2 q = ( 237.6 W / m ⋅ K ) 0.01 m = 5000 W/m
∆Talu = 0.21 K
(c) Mild Steel Slab ∆T q = k = 5000 W/m2 L ms
Thermal conductivity of mild steel at 140 C, Table A.1, Appendix A Kms = 50.4 W/m.K ∆Tms 2 q = ( 50.4 W / m ⋅ K ) 0.01 m = 5000 W/m
∆Tms = 0.992 K (d) Ice Slab
4
1. INTRODUCTION
∆T q = k = 5000 W/m2 L ice
Thermal conductivity of ice at 140 C, Table A.1, Appendix A • ice at 0 C, kice = 2.215 W/m.K • Note: there is no ice at 140 C, but continue calculation at 0 C. ∆Tice 2 q = ( 2.215 W / m ⋅ K ) 0.01 m = 5000 W/m
∆Tice = 22.57 K
(e) Spruce Slab ∆T q = k = 5000 W/m2 L Si
Thermal conductivity of spruce at 140 C, Table A.1, Appendix A Ksp = 0.11 W/m.K @ 20 C (available) ∆TSp 2 q = ( 0.11 W / m ⋅ K ) 0.01 m = 5000 W/m
∆TSp = 454.55 K
(f) Insulation (85 % Magnesia) ∆T q = k = 5000 W/m2 L Si
Thermal conductivity of insulation at 140 C, Table A.1, Appendix A Kin = 0.074 W/m.K @ 150 C (available) ∆Tin 2 q = ( 0.074 W / m ⋅ K ) 0.01 m = 5000 W/m
∆TSi = 675.8 K
(g) Silica Aerogel Slab ∆T q = k = 5000 W/m2 L Si
Thermal conductivity of silica aerogel at 140 C, Table A.1, Appendix A ksa = 0.022 W/m.K @ 120 C ∆Tsa 2 q = ( 0.022 W / m ⋅ K ) 0.01 m = 5000 W/m
∆Tsa = 2,273 K Tabulation:
5
1. INTRODUCTION Slab Temperature Drop, K Silver 0.12 Aluminum 0.21 Mild Steel (0.5 % Carbon) 0.992 Ice 22.57 Spruce 454.55 Insulation (85 % Magnesia) 675.8 Silica Aerogel 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 − Tref ) dU dT − Qnet = = ρcA δx = ρcA δx dt dt dt Answer:
The application of heat diffusion equation eq. no. (1.13) depends on the
thermal diffusivity α as the value of
∂T 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 ∂T dT ∂2T = 0 for steady state conduction giving = 0 , so q = −k . 2 ∂t dx ∂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=1m ∆T = 200 C – 0 C = 200 K 200 K q = ( 391 W / m ⋅ K ) 1m
= 78,200 W/m2.K
Q = qA A = 1 cm2 = 1 x 10-4 m2 Q = (78,200 W/m2.K)(1 x 10-4 m2) = 7.82 W (a) (b)
− Qrev − 7.82 W S1 = = T1 ( 200 + 273 K ) = - 0.01654 W/K Q + 7.82 W S 2 = rev = T2 ( 0 + 273 K ) = + 0.02864 W/K
(c)
S r == 0.0 W/K (see Eq. 1.5, steady state)
(d)
S Un = S1 + S 2 = = - 0.01654 W/K + 0.02864 W/K = + 0.0121 W/K
Since
SUn 〉 0 , therefore it satisfied Second Law of Thermodynamics.
1.6
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 cm 2 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 T1 = 27 + 273 = 300 K T2 = -43 + 273 = 230 K 70 K q = ( 0.14 W / m.K ) 0.10 m =
98 W/m2 7
1. INTRODUCTION Q = qA = (98 W/m2)(0.093 m2) = 9.114 W − Qrev − 9.114 W S1 = = T1 ( 300 K ) = - 0.03038 W/K Q + 9.114 W S 2 = rev = T2 ( 230 K ) = + 0.03963 W/K
(a) (b) (c)
S r == 0.0 W/K (see Eq. 1.5, steady state)
(d)
S Un = S1 + S 2 = = - 0.03038 W/K + 0.03963 W/K = + 0.00925 W/K
Since
SUn 〉 0 , therefore it satisfied Second Law of Thermodynamics.
1.7
(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: 16.018 kg / m 3 = 416.468 kg/m3 ρ = ( 26 lb / ft 3 ) 3 1 lb / ft 4186.8 J / kg .K c = ( 0.65 Btu / lb.F ) 1 Btu / lb.F
= 2721.42
k = 0.14 W/m.K ∆T = 27 C – (-43 C) = 70 C T1 = 27 C + 273 = 300 K T2 = - 43 C + 273 = 230 K A = 0.093 m2 L = 0.10 m (a)
Q T dQ = ∫T12 T T
Q T ρcVdT = ∫T12 T T
ρcV (T2 − T1 )
T = ρcV ln 2 T T1 (T2 − T1 ) = ln T2 T T 1
8
J/kg.K
1. INTRODUCTION
− T1 ) ( 230 − 300) = T2 230 = 263.45 K ln ln 300 T 1 Q ρcV ( T2 − T1 ) ρcAL( T2 − T1 ) = (b) ∆S = = T T T T=
(T
2
∆S =
( 416.468)( 2721.42)( 0.093)( 0.10)( 230 − 300) 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/m2.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