CHAPTER 12 - CONDENSERS AND EVAPORATORS
12-1.
An air-cooled condenser is to reject 70 kw of heat from a condensing refrigerant to air. The condenser has an
2
2
air-side area of 210 m and a U value based on this area is 0.037 kW/m .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
2
Uo = 0.037 kW/m .K
q = 70 kw
3
ρ = 1.15 kg/m
Condensing Temperature = 55 C
3
3
w = (6.6 m /s) / (1.15 kg/m ) = 5.739 kg/s
cp =1.0 kJ/kg.K
(t c − t i ) − (t c − t o )
(t − t )

ln c i

(t c − t o )
LMTD =
q = UoAoLMTD
q
70
=
= 9.009 K
U o A o (0.037)(210)
But q = wcp(to - ti)
LMTD =
q
70
=
= 12.197 K
wc p (5.739)(11)
to −ti =
(t c − t i ) − (t c − t o )
(t − t )

ln c i

(t c − t o )
LMTD =
9.009 =
12.197
(
)

ln 55 − t i

(55 − t o )
55 − t i
= 3.8724
55 − t o
55 - ti = 3.8724(55 - 12.197 - ti)
ti = 38.6 C - - - Ans.
12-2.
2
An air-cooled condenser has an expected U value of 30 W/m .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
to = ti +
t o = 35 +
q
wc p
60
(15)(1)
Page 1 of 11
CHAPTER 12 - CONDENSERS AND EVAPORATORS
to = 39 C
LMTD =
LMTD =
(t o − t i )
t
−
t )
(

ln c i

(t c − t o )
(39 − 35)
= 10.878 K
(
48
−
35)

ln
(48 − 39)

q = UoAoLMTD
60 kw = (30 / 1000)(Ao)(10.878)
2
Ao = 184 m - - - 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
3
temperature k = 0.614 W/m.K, ρ = 996 kg/m , and m = 0.000803 Pa.s.
2
(b)
Using a mean condensing coefficient of 1420 W/m .K, calculate the overall heat-transfer coefficient
base don the condensing area.
Solution:
(a)
Water-side coefficient:
Eq. 12-19.
 VDρ 
hD

= 0.023
k
 µ 
0.8
 c pµ 


 k 


0.4
D = 14 mm = 0.014 m
k = 0.614 W/m.K
3
ρ = 996 kg/m
µ = 0.000803 Pa.s
cp = 4190 J/kg.K
3.8 × 10 −3 m 3 /s
 60  π 
2
  (0.014 m)
 4  4 
V = 1.6457 m/s
V=
h(0.014 )
 (1.6457)(0.014 )(996) 
= 0.023

0.614
0.000803


0.8
 (4190)(0.00803) 


0.614


2
h = 7,313 W/m .K - - - Ans.
(b)
Overall heat-transfer coefficient.
Eq. 12-8.
1
1
x
1
=
+
+
U o A o h o A o kA m h i A i
A
1
1 xA o
=
+
+ o
U o h o kA m h i A i
Page 2 of 11
0.4
CHAPTER 12 - CONDENSERS AND EVAPORATORS
2
ho = 1420 W/m .K
k = 390 W/m.K
Ao / Ai = 1.7
Am =
1
2
(A o + A i )
A 

A m = 21  A o + o 
1.7 

Ao / Am = 1.25926
x = 2 mm = 0.002 m
2
hi = 7,313 W/m .K
(0.002)(1.2596) + 1.7
1
1
=
+
U o 1420
390
7313
2
Uo = 1060 W/m .K - - - Ans.
12-4.
2
A shell-and-tube condenser has a U value of 800 W/m .K based on the water-side are and a water pressure
drop of 50 kPa. Under this operating condition 40 percent of the heat-transfer resistance is on the water side.
If the water-flow rate is doubled, what will the new U value and the new pressure drop be?
Solution:
U1 = 800 W/m2.K
h1 = Water-side coefficient
1
= 2,000
 1 
(0.40) 
 800 
Eq. 12-13, replace 0.6 by 0.8 for condenser.
0.8
Water-side coefficient = (const)(flow rate)
h1 =
For
w2 / w1 = 2
h2  w 2 

=
h1  w 1 
0.8
0.8
2
h2 = (2000)(2) = 3482.2 W/m .K
Remaining resistance = (0.60)( 1 / 800 ) = 0.00075
New U-Value:
1
1
=
+ 0.00075
U 2 3482.2
2
U2 = 964 W/m .K - - - Ans.
New Pressure Drop:
Eq. 12-11.
Page 3 of 11
CHAPTER 12 - CONDENSERS AND EVAPORATORS
w
∆p 2 = ∆p 1  2
 w1



2
∆p 2 = (50)(2 )2
∆p2 = 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
2
hf = 28 W/m .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
M=
M=
hf
ky
28
(202)(0.00006)
M = 48.1 m
-1
tanhML
ML
-1
ML = (48.1 m )(0.020 m) = 0.962
tanh(0.962)
η=
0.962
η = 0.7746 - - - - Ans.
η=
(b)
If the fin thickness is doubled.
2y = 0.24 m = 0.00024 m
y = 0.00012 m
M=
28
(202)(0.00012)
M = 33.99 m
-1
tanhML
ML
-1
ML = (33.99 m )(0.020 m) = 0.6798
tanh(0.6798)
η=
0.6798
η = 0.87 > 0.7746
If the length L is doubled
L = 40 mm = 0.040 m
η=
Page 4 of 11
CHAPTER 12 - CONDENSERS AND EVAPORATORS
28
M=
(202)(0.00006)
M = 48.1 m
-1
tanhML
ML
-1
ML = (48.1 m )(0.040 m) = 1.924
tanh(1.924 )
η=
1.924
η = 0.498 < 0.7746
η=
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
2
flow and 45 mm vertically. The air-side coefficient is 55 W/m .K.
2
Solution: hf = 55 W/m .K
Alumimum Fins, k = 202 W/m.K
2y = 0.00018 mm
y = 0.00009 mm
M=
hf
ky
M=
55
(202)(0.00009)
-1
M = 55 m .
Equivalent external radius.
2
2

 16  
 16 
π (re )2 −    = (40 )(45) − π 
 2  
2

re = 23.94 mm = 0.02394 m
ri = 8 mm = 0.008 m
(re - ri)M = (0.02394 - 0.008)(55) - 0.88
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,
2
2
2
15 m ; air-side prime, 13.5 m , and air-side extended, 144 m ? The refrigerant-side heat-transfer coefficient
2
2
is 1300 W/m .K, and the air-side coefficient is 48 W/m .K. The fin effectiveness is 0.64.
Solution: η = 0.64
2
Ai = 15 m
2
hi = 1300 W/m .K
Page 5 of 11
CHAPTER 12 - CONDENSERS AND EVAPORATORS
2
hf = 48 W/m .K
Ap = 13.5 m
Ae = 144 m
2
2
Eq. 12-20 neglect tube resistance.
1
1
1
=
+
U o A o h f A p + ηA e
hi A i
(
)
1
1
1
=
+
U o A o (48)(13.5 + 0.64(144)) (1300)(15)
UoAo = 4,025 W/K - - - Ans.
12-8.
A refrigerant 22 system having a refrigerating capacity of 55 kW operates with an evaporating temperature of
5 C and rejects heat to a water-cooled condenser. The compressor is hermetically sealed. The condenser
2
2
has a U value of 450 W/m .K and a heat-transfer area of 18 m and receives a flow rate of cooling water of
3.2 kg/s at a temperature of 30 C. What is the condensing temperature?
Solution: Eq. 12-26.
LMTD =
(t c − t i ) − (t c − t o )
(t − t )

ln c i

(t c − t o )
Heat Rejection:
q = UALMTD = wcp(to - ti)
cp = 4190 J/kg.K


(t o − 30)


q = (450)(18)
 = (3.2)(4190)(t o − 30 )

 ln(t c − 30)
(t c − t o ) 
 
(t − 30)

ln c

(t c − t o ) = 0.60412
tc - 30 = 1.82964 (tc - to)
to = 16.397 + 0.45345 tc - - - Eq. No. 1
Figure 12-12.
At Heat-rejection ratio = 1.2
Condensing Temperature = 36 C
At Heat-rejection ratio = 1.3
Condensing Temperature = 49 C
Heat-rejection ratio = 0.92308 + 0.0076923 tc
q = (0.92308 + 0.0076923t c )(55000) = wc p (t o − t i )
(0.92308 + 0.0076923t c )(55000) = (3.2)(4190)(t o − 30)
33.7865 + 0.031554t c = t o = 16.397 + 0.45345t c
tc = 41.22 C - - - Ans.
12-9.
Calculate the mean condensing heat-transfer coefficient when refrigerant 12 condenses on the outside of the
horizontal tubes in a shell-and-tube condenser. The outside diameter of the tubes is 19 mm, and in the
Page 6 of 11
CHAPTER 12 - CONDENSERS AND EVAPORATORS
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.
1/4
 gρ 2 h fg k 3 

h cond = 0.725
 µ∆tND 


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 = 2.8
∆t = 52 C - 44 C = 8 K
2
g = 9.81 m/s
D = 19 mm = 0.019 m
h cond
 (9.81)(1202)2 (119,993)(0.05932)3
= 0.725

(0.000174)(8)(2.8)(0.019)





1/4
2
hcond = 1065 W/m .K - - - Ans.
12-10.
2
A condenser manufacturer quarantees the U value under operating conditions to be 990 W/m .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:
A
1
1
=
− o
U o2 U o1 h ff A i
2
Uo1 = 900 W/m .K
2
1/ hff = 0.000176 m .K/W
Ao / Ai ~ 1.0
1
1
=
− 0.000176(1)
U o2 900
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:
1
h cond
 (9.81)(1109)2 (160,900)(0.0779)3  4
= 0.725

 (0.000180)∆t (3.23 )(0.016) 
Page 7 of 11
CHAPTER 12 - CONDENSERS AND EVAPORATORS
h cond =
2285
∆t 0.25
Then,
∆t 0.25
1
0.016
=
+ 0.000002735 +
(0.000176) + 0.016 ⋅ 1
Uo
2285
0.014
0.014 6910
1
∆t 0.25
=
+ 0.00036927
Uo
2285
Uo =
2285
0.25
∆t
+ 0.843782
LMTD = 12.33 C
But
hcond∆t = UoLMTD
2285
 2285 


 0.25 ∆t =  0.25
(12.33)
+ 0.843782 
 ∆t

 ∆t
12.33
∆t 0.75 = 0.25
∆t
+ 0.843782
∆t + 0.843782∆t 0.75 = 12.33
∆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.
1
1
x
1
=
+
+
U o A o h f A p + ηA e
kA m h i A i
(
)
Eq. 12-21
h f = 38V 0.5
V in m/s.
0.5
Varying airflow, the Wilson plot is a graph of 1/Uo versus 1/V .
0.5
Abscissa is 1/V where V is the face velocity in meters per second.
(b)
Eq. 12-27
1
1
x
1
=
+
+
U o A o h o kA m h i A i
Liquid in shell, variation of Eq. 12-13,
h o = (const )V 0.6
0.6
Varying water flow, the Wilson plot is a graph of 1/Uo versus 1/V .
0.6
Abscissa is 1/V where V is the face velocity in meters per second.
12-13.
The following values were measured on an ammonia condenser.
Page 8 of 11
CHAPTER 12 - CONDENSERS AND EVAPORATORS
Uo, W/sq m.K
2300
2070
1930
1760
1570
1360
1130
865
V, m/s
1.22
0.975
0.853
0.731
0.61
0.488
0.366
0.244
Water flowed inside the tubes, and the tubes were 51 mm OD and 46 mm ID and had a conductivity of 60
W/m.K. Using a Wilson plot, determine the condensing coefficient.
Solution:
Wilson plot
C2
1
= C 1 + 0.8
Uo
V
Tabulation:
0.8
1/Uo
1/V
0.000434783
0.000483092
0.000518135
0.000568182
0.000636943
0.000735294
0.000884956
0.001156069
0.852928
1.020461
1.13564
1.28489
1.485033
1.775269
2.234679
3.090923
By linear regression:
C1 = 0.000153033
C2 = 0.000325563
But:
C1 =
1 xA o
+
h o kA m
Ao
51
=
= 1.05155
A m (51 + 46) 2
x = (1/2)(51 - 46) = 2.5 mm = 0.0025 m
k = 60 W/m.K
1 (0.0025)(1.05155)
0.000153033 =
+
ho
60
2
ho = 9,156 W/m .K - - - Ans.
12-14.
Develop Eq. (12-23) from Eq. (12-22).
Solution: Eq. 12-22.
2
3
h cv x  gρ h fg x 

=
k
 4µk∆t 
1
4
Page 9 of 11
CHAPTER 12 - CONDENSERS AND EVAPORATORS
L
hcv =
∫h
0
cv dx
L
∫
L
0
 gρ 2 h fg x 3 


 4µk∆t 
L
k
x
hcv =
hcv
2
 k   gρ h fg 

=  
 L   4µk∆t 
hcv
2
 k   gρ h fg 

=  
 L   4µk∆t 
hcv
2
 k   gρ h fg 

=  
 L   4µk∆t 
1
4
1
4
dx
∫
L
x
-1
4 dx
0
1
1
4
 4  3 4 
 x 
0
 3 
4
 4  34
 L
3
L
2
3
4   gρ h fg k 


hcv
 
 3   µk∆tL 
Ans. Eq. 12-23.
1
= 
4
hcv
12-15.
1
4
 gρ 2 h fg k 3 

= 0.943
 µk∆tL 
1
1
4
4
b
From Fig. 12-21, determine C and b in the equation h = C∆T applicable to values in the middle of the typical
range.
Solution: Use Fig. 12-21
Tabulation:
Heat-transfer Coefficient
2
W/m .K, h
400
600
800
1000
1500
h = C∆t
Heat flux
2
W/m
∆t
K
710
1550
2820
4170
9000
1.775
2.583
3.525
4.170
6.000
b
By Curve-Fitting:
C = 212.8
b = 1.08
h = 212.8∆t
13-16.
1.08
- - - Ans.
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
Page 10 of 11
CHAPTER 12 - CONDENSERS AND EVAPORATORS
coefficients are constant. prove this statement.
Solution: Use Fig. 12-23.
te = evaporating temperature
ta = entering-water temperature (constant)
U = heat-transfer coefficient (constant)
t +t

q = wc p (t b − t a ) = UA a b − t e 
2


wc p t b − wc p t a = 0.5UAt a + 0.5UAt b − UAt e
(wc p + 0.5UA )t a − UAt e
(wc p −0.5UA )
 (wc p + 0.5UA )t a − UAt e

q = wc p 
−ta 
(wc p − 0.5UA )


 (wc p + 0.5UA )t a − UAt e − (wc p − 0.5UA )t a 
tb =
q = wc p 
(wcp − 0.5UA )

 UAt a − UAt e 
q = wc p 

 wc p − 0.5UA 
wc p UA
(t a − t e )
q=
wc p − 0.5UA
(


)
If U is constant.
q = (constant)(ta - te)
At constant t0, this is a straight line. - - - Ans.
-000-
Page 11 of 11