(1) (Welty, Rorrer, Foster, 6th Edition International Student Version 20.6)
Saturated steam at atmospheric pressure condenses on the outside surface of a 1-m-long
tube with 150 mm diameter. The surface temperature is maintained at 91 . Evaluate the
condensation rate if the pipe is orientated
a. vertically;
b. horizontally.
Take latent heat of vaporization of steam at atmospheric pressure to be 2250 kJ/kg.
(2) WWWR 21.13
Water flowing at a rate of 4000 kg/h through a 16.5-mm-ID tube enters at 20 . The tube
outside diameter is 19 mm. Saturated atmospheric steam condenses on the outside of the
tube. For a horizontal brass tube 2 m long, evaluate
a.
b.
c.
d.
the convective coefficient on the water side;
the convective coefficient on the condensate side;
the exit water temperature;
the condensation rate.
*You may neglect the temperature drop across the tube wall. Suggested initial guess: T w
= 58 , Ti(out) = 36 .
(3) WWWR 21.14
Saturated steam at atmospheric pressure flows at a rate of 0.042 kg/s/m between two
vertical surfaces maintained at 340 K that are separated by a distance of 1 cm. How tall
may this configuration be if the steam velocity is not to exceed 15 m/s?
(4) (Welty, Rorrer, Foster, 6th Edition International Student Version 20.7)
Saturated steam at 365 K condenses on a 2-cm tube whose surface is maintained at 340
K. Determine the rate of condensation and the heat transfer coefficient for the case of a
1.5-m-long tube oriented
a. vertically;
b. horizontally.
(5) WWWR 21.19
If eight tubes of the size designated in Problem (4) are arranged in a vertical bank and the
flow is assumed laminar, determine
a. the average heat-transfer coefficient for the bank;
b. the heat-transfer coefficient for the first, third and eighth tubes.
(6) WWWR 22.3
An oil having specific heat of 1880
enters a singles pass counterflow heat
exchanger at a rate of 2 kg/s and a temperature of 400 K. It is to be cooled to 350 K.
Water is available to cool the oil at a rate of 2 kg/s and a temperature of 280 K.
Determine the surface area required if the overall heat transfer coefficient is 230
.
(7) WWWR 22.15
Determine the required heat-transfer surface area for a heat exchanger constructed from
10-cm OD tubes. A 95% ethanol solution (cp = 3.810 kJ/kg.K, flowing at 6.93 kg/s is
cooled from 340 K to 312 K by 6.30 kg/s of water that is available at 283 K. The overall
heat transfer coefficient based on outside tube area is 568 W/m2.K. Three different
exchanger configuration are of interest:
a. counterflow, single pass;
b. parallel flow, single pass;
c. crossflow with one tube pass.
(1) (Welty, Rorrer, Foster, 6th Edition International Student Version 20.6)
(a) Vertical Tube (Eq. 21.20)
1/ 4

3


3
  L gk (  L  V )  h fg  8 C L (Tsat  Tw )  


h  0.943 

LT




1/ 4

3


3
6
 961.2(9.81)(0.679) (960)  2.25 10  8 (4206)  (9)  


 0.943 

6
1(297 10 )(9)




 6600W / m 2  K
6600(9)( )(0.15)(1)
m
 0.0125kg / s
2.25 106
(b) Horizontal Tube (Eq. 21-25)
1/ 4

3

3 
  L g (  L  PV )k  h fg  8 C L (Tsat  Tw )  


havg  0.725 

 D(Tsat  TW )




2
 7700W / m  K
(2) Problem 21.13 (WWWR)
Neglecting T across tube
k
0.4
hi  0.023Re0.8
 Eq.20  26
D Pr
Di
14

3

3
  L g  k  h fg  8 c pL T  


ho  0.725 

 DT




20  Tout
For hi, Tin  20c TW  ? Tb 
2
100  TW
For h0, properties evaluated at T f 
2
Toverall Ti T0
q


 R Ri R0
Assuming Tout  36C  309K
TW  58C  331K
Giving
Tb,avg  28C  301K

4  4000 
4m
Rei 



 D   0.0165 863 106  3600 
D


 99000
k=0.611
Pr  5.95
0.611
0.023990000.8 5.950.4
hi 
0.0165
 17200W / m 2  K

3
3

971.8  9.81 971.8 0.673  2.25 106   4194  42   


8


h0  0.725 

6
352 10  0.019  42 






W
 8960
m2  K

1
 5.61 10 4
17200 0.01652
1
R0 
 9.35  10 4
8690 0.0192
Ri 
R  R
i
 R0  14.96  10 4
Ttotal  72K
Ti  27 K
T0  45K

1
4
q
27 45

 48000W
Ri R0

q  mC p (Tout  Tin ) 
4000
 4180  (Tout  Tin )
3600
(Tout  Tin )  10.4 K
 Tout  303.4K TbAVG  298K
Tw  328K =373- T0
*Not far away from original guess
(a)
hH 2 0  17200W 2
m K
(b)
hcondensate  8690W 2
m K
(c)
Tout  303.4K

mcondensate 
48000
 0.0213 kg
6
s
2.25 10
(d)
(3) Problem 21.14 (WWWR)
0.042
(per m)
 0.0717m3 / s
0.586
0.0717
Allowable width =
 0.00478m  0.478cm
(15)(1)
0.478  2  1cm    0.261cm
Film model (Eq. 21-18)




4
k


Tx
4  

3
  L g  (h fg  C L T ) 
8


3
4(0.392)(0.195 10 )(59) x

(59.5)(32.2)(59.5)(993.5)(3600)
Flow rate =
 4  4.425 104 x
x  121,500 ft  37000m
(4) (Welty, Rorrer, Foster, 6th Edition International Student Version 20.7)
1
hvertical
  L gk 3   L  V   h fg  0.68c pL Tsat  Tw    4


 0.943 

L



T

T
 sat w 


1
 971.8  9.80665  0.6733   971.8  0.6288  2278  103  0.68(4194)(365  340)   4




6
1.5

352

10

(365

340)


 0.943
 4464W / m 2  K

m
hAT 4464*  *0.02*1.5* 25

 4.628103 kg / s
h fg
2278*1000
horizontal 
hvertical
D
1.3  
L

m
1
4

4464
1.3(0.02 /1.5)
hAT
 0.0105kg / s
h fg
1
4
 10105W / m2 K
(5) Problem 21.19 (WWWR)
1/ 4
havg

3

3 
  L g (  L  V )h  h fg  8 CPL (Tsat  Tw )  


 0.725 

nD (Tsat  Tw )




1/ 4

3

3
 971*9.81*(971  0.586)0.673  2250000  8  25  4.184  


 0.725 

0.02  0.000352  25




1/ 4
1
 10000   
n
For bank
 ( A)
1/ 4
1
havg  10000     5945W / m2 k
8
For n tubes
q  havg ,n  n  Atube  T
 havg ,n 1 (n  1)  Atube  T  hn * Atube * T
i.e.hn nT  hn 1 (n  1) AT  hn AT
 nth tube
hn  nhn  (n  1)hn1
Top tube => h1  10000W / m2 K
3rd tube :
h2  8408.26W / m2 K
h3  7598W / m2 K
h3  3  7598  2  8408.26  5976.6W / m2 K
8th tube:
h8  5946W / m2 K
h7  6147.3W / m2 K
h8  8  5946  7  6147.3  4532.7W / m2 K
1
n
1/ 4
(6) Problem 22.3 (WWWR)
Oil
Tin = 400K

m = 2 kg/s
Cp = 1880 J/kg-K

q = m Cp  T = 2 (1880) (400-350) = 188000 W
Tw 
q


mCp
188000
 22.5K
24187 
Tw,in = 280K
Tw,out = 302.5 K
400K
350K
U=230
302.5K
Tlm 
A
280K
97.5  70
 83K
ln 97.5
70


q
1888000

 9.85 m2
UTlm 230(83)
Tout = 350K
(7) Problem 22.15 (WWWR)
340K
312K
ETH
H20
311.1K
283K

m ETH = 6.93 kg/s

q = m Cp  T/ ETH = 6.93 (3810) (28)

= m Cp  T/w = 6.30 (4182)  Tw
 Tw = 28.1 K
 TLM  29K
(a) counterflow
A
q 6.93(3810)(28)
 44.9 m2
UTLM
568.29
(b) parallel flow
TLM 
A
(57  0.9)
 13.52 K
ln(57 )
0.9
q
 96.3 m2
UTLM
(c) crossflow

Cmixed = m Cpw = 26350
(water)
shell
(ethanol)
tube

Cunmixed = m Cp = 26403
Y 
Tt , out  Tt ,in 312  340

 0.491
Ts ,in  Tt ,in 283  340

Z
(m C p )tube

(m C p ) shell
F  0.85

26403
1
26350
A
q
44.9

 52.8 m2
U ( FTLM ) 0.85