Rensselaer Polytechnic Institute

ENGR 2250

13-38 Saturated steam at 100 °C condenses in a shell-and-tube heat exchanger (one shell pass, two tube passes) with a surface area
of 0.5 m2 and an overall heat transfer coefficient of 2000 W/m2·K. Water enters at 0.5 kg/sec and 15 °C. Determine:
a. the outlet temperature of the water (in °C)
b. the rate of steam condensation (in kg/sec).
Approach:
We seek the water outlet temperature, which is
equivalent to seeking the heat transfer rate. Therefore,
this is a rating problem, and the preferred approach is
the ε-NTU method. The condensation rate can be
determined from an energy balance on the steam once
the heat transfer rate is calculated.
Assumptions:
1. The overall heat transfer coefficient is uniform
over the heat exchanger.
2. The system is steady.
3. No work is done on or by the steam or liquid water.
4. Potential and kinetic energy effects are negligible.
5. Liquid water is an ideal liquid with a constant
specific heat.
Solution:
a) The governing equation for the ε-NTU method is:
Q Q mc T T ɺ ɺ = = - ε ε max p H in C in ( ɺ )min ( , , )
Because the steam is condensing (constant temperature), Cmin is on the cooling water side and
ε = - - 1 exp( ) NTU and Q m c T T ɺmax w p w H in C in = - ɺ , , , ( )
The cooling water outlet temperature can be obtained from conservation of mass and energy applied to the water. Assuming steady,
no work, negligible potential and kinetic energy effects, and an ideal liquid with constant specific heat so that ∆ = ∆ h c T p :
Q m c T T ɺ = - ɺ w p w C out C in , , , ( )
Combining the above equations and solving for the cooling water outlet temperature:
, ,
,
max
C out C in
w p w
Q
T T
m c
ε
= +
ɺ
ɺ
From Appendix A-6 for water assuming Tout = 45 ºC, Tavg ≈ 30ºC , cp,w = 4.176 kJ/kgK.
Q ɺmax = ( ) 0.5kg s 4.176 kJ sK 100-15 K=177.5kJ s=177.5kW ( )( )
To calculate the NTU, we use the given information:
( )( )
( )( )( )
2000 W m K 0.5m 2 2
0.479
min 0.5kg s 4.176kJ kgK 1000J kJ
UA
NTU
C
= = =
ε = - - = 1 exp 0.479 0.381 ( )
( )( )
( )( )
o o
,
0.381 177.5kJ s
15 C+ = 47.4 C
TC out = 0.5kg s 4.176 kJ kgK Answer
b) The steam condensation rate is obtained from an energy balance on the steam. Using the same assumptions as was used for the
water (except for ideal liquid):
Q mh m Q h ɺ ɺ = → = ɺ ɺ fg fg
From Appendix A-10 at 100 ºC, hfg = 2257 kJ/kg
Q Q ɺ ɺ = = ε max ( ) 0.381 177.5kJ s =67.6 kJ s ( )
( )
67.6kJ s kg
=0.030
2.257 kJ kg