dharm

\M-therm\Th15-3.pm5

HEAT TRANSFER 829

Overall heat transfer coefficient is given by

11 111

Uh

d

g dh h h

o

is g s

=+ =+ as di ≈ do ...(given)

or U =

hh

hh

g s

g s

×

+

= ×

+

250 600

250 600 = 176.5 W/m

(^2) °C
Total heat transfer rate is given by
Q = U A θm ...(i)
where A = N × (π d L) = N × π × 0.03 × 3 = 0.2827 N m^2
Q = 22.5 × (1 × 10^3 ) × (650 – 377) = 6142.5 × 10^3 W
θm =
θθ
θθ
12
12
12 21
12 2 1
− = −− −
ln ( / ) −−
()()
ln [( ) / ( )]
tt tt
tt tt
hc h c
hc h c

()()
ln [( ) / ( )] ln ( / )
650 350 377 180
650 300 377 180
300 197
300 197
−−−
−−

−
= 244.9°C
Substituting the values is eqn. (i), we get
6142.5 × 10^3 = 176.5 × 0.2827 N × 244.9
or N =
6142.5 10
176.5 0.2827 244.9
×^3
××
= 503 tubes. (Ans.)
Example 15.21. A two-pass surface condenser is required to handle the exhaust from a turbine
developing 15 MW with specific steam consumption of 5 kg/kWh. The condenser vacuum is 660 mm
of Hg when the barometer reads 760 mm of Hg. The mean velocity of water is 3 m/s, water inlet
temperature is 24°C. The condensate is saturated water and outlet temperature of cooling water in
4 °C less than the condensate temperature. The quality of exhaust steam is 0.9 dry. The overall heat
transfer coefficient based on outer area of tubes is 4000 W/m^2 °C. The water tubes are 38.4 mm in
outer diameter and 29.6 mm in inner diameter. Calculate the following :
(i)Mass of cooling water circulated in kg/min,
(ii)Condenser surface area,
(iii)Number of tubes required per pass, and
(iv)Tube length. (P.U.)
Solution. Given : di = 29.6 mm = 0.0296 m ; do = 38.4 mm = 0.0384 m ;
U = 4000 W/m^2 °C ; V = 3 m/s ; tc 1 = 24°C ; x (dryness fraction) = 0.9.
The pressure of the steam in the condenser,
ps = 760 660
760
− × 1.0133 = 0.133 bar
The properties of steam at ps = 0.133 bar, from steam table, are :
tsat = 51°C ; hfg = 2592 kJ/kg
∴ tc 1 = 51 – 4 = 47°C
The steam condensed per minute,
mm&sh(==& ) ()15 1000××^5
60
= 1250 kg/min
(i) Mass of cooling water circulated per minute, m(m)&&wc= :
Heat lost by steam = Heat gained by water
m&h × (x. hfg) = m&c × cpc × (–)ttcc
21