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12-1. An air-cooled condenser is to reject 70 kw of heat from a condensing refrigerant to air. The condenser has an air-side area of 210 m^2 and a U value based on this area is 0.037 kW/m^2 .K; it is supplied with 6.6 m3/s of air, which has a density of 1.15 kg/m3. If the condensing temperature is to be limited to 55 C, what is the maximum allowable temperature of inlet air?
Solution: Ao = 210 m^2
Uo = 0.037 kW/m^2 .K q = 70 kw ρ = 1.15 kg/m^3 Condensing Temperature = 55 C w = (6.6 m^3 /s) / (1.15 kg/m^3 ) = 5.739 kg/s cp =1.0 kJ/kg.K
c o
c i
c i c o
t t ln t t
t t t t LMTD
q = UoAoLMTD
9.009K
UA
q LMTD o o
But q = wcp(to - ti)
12.197K
wc
t t q p
o − i= = =
c o
c i
c i c o
t t ln t t
t t t t LMTD
o
i 55 t ln^55 t
55 t
55 t o
i (^) = −
55 - ti = 3.8724(55 - 12.197 - ti) ti = 38.6 C - - - Ans.
12-2. An air-cooled condenser has an expected U value of 30 W/m^2 .K based on the air-side area. The condenser is to transfer 60 kW with an airflow rate of 15 kg/s entering at 35 C. If the condenser temperature is to be 48 C, what is the required air-side area?
Solution: q = UoAoLMTD q = wcp(to - ti)
w = 15 kg/s cp = 1.0 kJ/kg.K
p
o i wc t =t + q
t 35 60 o = +
to = 39 C
c o
c i
o i
t t ln t t
t t LMTD
10.878 K
ln^4835
LMTD 39 35 =
q = UoAoLMTD 60 kw = (30 / 1000)(Ao)(10.878) Ao = 184 m^2 - - - Ans.
12-3. A refrigerant 22 condenser has four water passes and a total of 60 copper tubes that are 14 mm ID and have 2 mm wall thickness. The conductivity of copper is 390 W/m.K. The outside of the tubes is finned so that the ratio of outside to inside area is 1.7. The cooling-water flow through the condenser tubes is 3.8 L/s. (a) Calculate the water-side coefficient if the water us at an average temperature of 30 C, at which temperature k = 0.614 W/m.K, ρ = 996 kg/m^3 , and m = 0.000803 Pa.s. (b) Using a mean condensing coefficient of 1420 W/m^2 .K, calculate the overall heat-transfer coefficient base don the condensing area.
Solution:
(a) Water-side coefficient: Eq. 12-19.
p
k
VD c
k
hD
μ
μ
ρ
D = 14 mm = 0.014 m k = 0.614 W/m.K ρ = 996 kg/m^3 μ = 0.000803 Pa.s cp = 4190 J/kg.K
( )^2
3 3
0.014 m 4 4
3.8 10 m/s V
π
×
−
V = 1.6457 m/s
h 0.
h = 7,313 W/m^2 .K - - - Ans.
(b) Overall heat-transfer coefficient. Eq. 12-8.
o o o o m hiA i
kA
x hA
U A
i i
o m
o o o hA
A
kA
xA h
U
2
1
2 (^2 1) w
w p p
∆p 2 = ( 50 )( 2 )^2
∆∆∆∆p 2 = 200 kPa - - - Ans.
12-5. (a) Compute the fin effectiveness of a bar fin made of aluminum that is 0.12 mm thick and 20 mm long when hf = 28 W/m^2 .K, the base temperature is 4 C, and the air temperature is 20 C. (b) If you are permitted to use twice as much metal for the fin as originally specified in part (a) and you can either double the thickness or double the length, which choice would be preferable in order to transfer the highest rate of heat flow. Why?
Solution:
(a) Aluminum fins k = 202 W/m.K 2y = 0.12 mm = 0.00012 m y = 0.00006 m L = 20 mm = 0.020 m
ky
h M = f
M =
M = 48.1 m-
ML
tanhML η =
ML = (48.1 m-1)(0.020 m) = 0.
η =tanh0. ηηηη = 0.7746 - - - - Ans.
(b) If the fin thickness is doubled.
2y = 0.24 m = 0.00024 m y = 0.00012 m
M =
M = 33.99 m-
ML
tanhML η =
ML = (33.99 m-1)(0.020 m) = 0.
η =tanh0. η = 0.87 > 0. If the length L is doubled L = 40 mm = 0.040 m
M =^28
M = 48.1 m-
ML
η =^ tanhML
ML = (48.1 m-1)(0.040 m) = 1.
η =tanh1. η = 0.498 < 0.
Ans. Therefore double the fin thickness to improve rate of heat flow with an efficiency of 87 % compared to 77.46 %.
12-6. Compute the fin effectiveness of an aluminum rectangular plate fin of a finned air-cooling evaporator if the fins are 0.18 mm thick and mounted on a 16-mm-OD tubes. The tube spacing is 40 mm in the direction of air flow and 45 mm vertically. The air-side coefficient is 55 W/m^2 .K.
Solution: hf = 55 W/m^2 .K
Alumimum Fins, k = 202 W/m.K 2y = 0.00018 mm y = 0.00009 mm
ky
h M = f
M =^55
M = 55 m-1.
Equivalent external radius.
2 2 2 e (^2)
r
= − π
π −
re = 23.94 mm = 0.02394 m ri = 8 mm = 0.008 m (re - ri)M = (0.02394 - 0.008)(55) - 0. re/ri = 23.94 mm / 8 mm = 3 From Fig. 12-8/ Fin Effectiveness = 0.68 - - - Ans.
12-7. What is the UA value of a direct-expansion finned coil evaporator having the following areas: refrigerant side, 15 m^2 ; air-side prime, 13.5 m^2 , and air-side extended, 144 m^2? The refrigerant-side heat-transfer coefficient is 1300 W/m^2 .K, and the air-side coefficient is 48 W/m^2 .K. The fin effectiveness is 0.64.
Solution: η = 0. Ai = 15 m^2 hi = 1300 W/m^2 .K
vertical rows of tubes there are respectively, two, three, four, three, and two tubes. The refrigerant is condensing at a temperature of 52 C and the temperature of the tubes is 44 C.
Solution: Condensing Coefficient: Eq. 12-24. 3 1/ (^2) fg cond (^) tND
g h k h 0.725
μ ∆
ρ
Table A-5 at 52 C. hfg = 370.997 - 251.004 kJ/kg = 119.993 kJ/kg hfg = 199,993 J/kg
ρ = 1 / (0.83179 L/kg) = 1202 kg/m^3
Table 15-5, Liquid Refrigerant 12 μ = 0.000179 PA.s k = 0.05932 W/m.K N = (2 + 3 + 4 + 3 +2) / 5 = 2. ∆t = 52 C - 44 C = 8 K g = 9.81 m/s^2 D = 19 mm = 0.019 m
2 3 1/ cond (^) 0.000174 8 2.8 0. h 0.725 9.81^1202 119,9930.
hcond = 1065 W/m^2 .K - - - Ans.
12-10. A condenser manufacturer quarantees the U value under operating conditions to be 990 W/m^2 .K based on the water-side area. In order to allow for fouling of the tubes, what is the U value required when the condenser leaves the factory?
Solution:
ff i
o o2 o1 h A
A
U
U
Uo1 = 900 W/m^2 .K 1/ hff = 0.000176 m^2 .K/W Ao / Ai ~ 1.
U
o
Uo2 = 1,199 W/m2.K - - - Ans.
12-11. In example 12-3 the temperature difference between the refrigerant vapor and tube was originally assumed to be 5 K in order to compute the condensing coefficient. Check the validity of this assumption.
Solution:
41 0.000180 t3.23 0.
h 0.725 9.81^11092 160,900 0.0779^3 cond
cond (^) t 0. h 2285 ∆
Then,
t U
o
t U
o
t 0.
U 2285
o =∆0.25+ LMTD = 12.33 C
But hcond∆t = UoLMTD
t 0.
t^2285 t
t 0.
t 12.
∆ +
∆t +0.843782∆t0.75^ = 12. ∆t = 8.23 K ∆t max= LMTD = 12.33 K ∆t = 8.23 K to 12.33 K... Ans.
12-12. (a) A Wilson plot is to be constructed for a finned air-cooled condenser by varying the rate of airflow. What should the abscissa of the plot be? (b) A Wilson plot is to be constructed for a shell-and-tube water chiller in which refrigerant evaporates in tubes. The rate of water flow is to be varied for the Wilson plot. What should the abscissa of the plot be?
Solution: (a) Eq. 12-20.
o o f^ (^ p e)^ m hiAi
kA
x h A A
U A
+η
Eq. 12-
h (^) f =38V V in m/s. Varying airflow, the Wilson plot is a graph of 1/Uo versus 1/V0.5. Abscissa is 1/V0.5^ where V is the face velocity in meters per second.
(b) Eq. 12-
o o o m hiA i
kA
x h
U A
Liquid in shell, variation of Eq. 12-13,
h o = ( const)V 0.
Varying water flow, the Wilson plot is a graph of 1/Uo versus 1/V0.6. Abscissa is 1/V0.6^ where V is the face velocity in meters per second.
12-13. The following values were measured on an ammonia condenser.
L
h dx h
L
cv=^ ∫^0 cv
L
dx 4 k t
g h x x
k
h
L 0
3 14 fg 2
cv
μ ∆
ρ
=
^ ∫
μ ∆
ρ
= L
0
-^14
(^14) fg 2 cv (^4) k t x dx
g h L
h k
L
0
(^34)
(^14) fg 2 cv 3 x
4 k t
g h L
h k
μ ∆
ρ
(^34)
(^14) fg 2 cv 3 L
4 k t
g h L
k h
μ ∆
ρ
3 14 fg
(^142) cv (^) k tL
g h k 3
h
μ ∆
ρ
Ans. Eq. 12-23. 3 14 fg 2 cv (^) k tL
g h k h 0.
μ ∆
ρ
12-15. From Fig. 12-21, determine C and b in the equation h = C∆Tb^ applicable to values in the middle of the typical range.
Solution: Use Fig. 12-
Tabulation: Heat-transfer Coefficient Heat flux ∆t W/m^2 .K, h W/m^2 K
400 710 1. 600 1550 2. 800 2820 3. 1000 4170 4. 1500 9000 6.
h = C∆tb
By Curve-Fitting: C = 212. b = 1.
h = 212.8∆∆∆∆t1.08^ - - - Ans.
13-16. Section 12-18 makes the statement that on a graph of the performance of a water chilling evaporator with the coordinates of Fig. 12-23, a curve for a given entering water temperature is a straight line if the heat-transfer
coefficients are constant. prove this statement.
Solution: Use Fig. 12-23.
te = evaporating temperature ta = entering-water temperature (constant) U = heat-transfer coefficient (constant)
= (^) p b− a = a b te 2
t t q wc t t UA
wc (^) p tb−wcpta=0.5UAta+0.5UAtb−UAt e ( ) ( wc 0.5UA)
wc 0.5UAt UAt t p
p a e b (^) −
( ) ( ) (^)
= (^) a p
p a e p (^) wc 0.5UA t
wc 0.5UAt UAt q wc
( ) ( )
wcp 0.5UA
wc 0.5UAt UAt wc 0.5UAt q wcp p a e p a
( )
wc 0.5UA
UAt UAt q wc p
a e p
( a e)
p
p (^) t t wc 0.5UA
wcUA q − −
If U is constant. q = (constant)(ta - te) At constant t 0 , this is a straight line. - - - Ans.
