TITLE.PM5

dharm

\M-therm\Th15-3.pm5

HEAT TRANSFER 831

or L =

1009 1

0 0384 2 245

.

.

~

π× × ×

− 17.1 m. (Ans.)
Example 15.22. A feed water heater which supplies hot water to a boiler comprises a shell and
tube heat exchanger with one-shell pass and two-tube passes. One hundred thin-walled tubes each of
20 mm diameter and length of 2 m per pass are used. Under normal operating conditions, water
enters the tubes at 10 kg/s and 17°C and is heated by condensing saturated steam at 1 atm. on the
outer surface of the tubes. The convection coefficient of the saturated steam is 10 kW/m^2 °C. Determine
the water exit temperature.
Use the following properties of water :
cp = 4.18 kJ/kg°C ; μ = 0.596 × 10 –3 Ns/m^2 ; k = 0.635 W/m°C and Pr = 3.93. (M.U.)
Solution. Given : p (number of tube passes) = 2, N (total number of tubes) = 200,
d = 20 mm = 0.02 m ; (length per pass) = 2 m, mm&&wc= = 10 kg/s, tc 1 = 17°C.
Water exit temperature, tc 2 :

m&c=π

4

d^2 × V × ρ × Np

whereVNvelocity of water ; number of tubes per pass

N

p p

== ===

L

N

M

O

Q

P

200

2

100

or 10 =

π

4 × 0.02

(^2) × V × 1000 × 100
∴ V =
10 4
0 02^2 1000 100
×
π×××.
= 0.318 m/s
Condensate out
t = 17 Cc 1 º
θ 1
θ 2
t= ?c 2
Cold fluid (water)
Hot fluid (steam)
t = t = th 1 h 2 sat= 100 Cº
Steam in
Shell
Water out
Water in
Fig. 15.44. One-shell pass and two-tube passes condenser.
Using non-dimensional heat transfer equation to water side, we get
Nu =
hd
k
i = 0.023 (Re)0.8 (Pr)0.33