TITLE.PM5

dharm

\M-therm\Th15-3.pm5

828 ENGINEERING THERMODYNAMICS

=

()()

ln [( ) / ( )]

()()

ln [( ) / ( )]

tt tt

tt tt

hc hc

hc hc

12 2 1

12 21

230 65 160 25

230 65 160 25

−− −

−−

= −− −

−−

or θm =

165 135

165 135

−

ln [( / )] = 149.5°C

Also, Q = U A θm

or A =

Q

Uθm

= ×

×

91.35 10

420 149.5

3
= 1.45 m^2. (Ans.)
Example 15.20. Steam enters a counter-flow heat exchanger, dry saturated at 10 bar and
leaves at 35°C. The mass flow of steam is 800 kg/min. The gas enters the heat exchanger at 650°C and
mass flow rate is 1350 kg/min. If the tubes are 30 mm diameter and 3 m long, determine the number
of tubes required. Neglect the resistance offered by metallic tubes. Use the following data :
For steam : tsat = 180°C (at 10 bar) ; cps = 2.71 kJ/kg°C ; hs = 600 W/m^2 °C
For gas : cpg = 1 kJ/kg°C ; hg = 250 W/m^2 °C (P.U.)

Solution. Given : m&s = m&c =

800

60

= 13.33 kg/s ; mm&&g==h

1350

60 = 22.5 kg/s ;

th 1 = 650°C ; tc 1 (= tsat) = 180°C ; tc 2 = 350°C ; d = 30 mm = 0.03 m ; L = 3 m.

Number of tubes required, N :

Heat lost by gases = Heat gained by steam

m&h × cph × (–)tthh 12 = m&c× cpc × (–)ttcc 21

22.5 × 1 × (650 – th 2 ) = 13.33 × 2.71 × (350 – 180)

∴ th 2 = 377°C

( ) Flow arrangement.a

t = 650 Ch 1 º

t = 350 Cc 2 º

Gas

Gas

steam

t = 650 Ch 1 º

t (= 377 C)h 2 º

t (= 180 C)c 1 º

t (= 377 C)h 2 º

Area

Hot fluid (gases)

Cold fluid (steam)

t = 650 Ch 1 º

t = 350 Cc 2 º

t (= 377 C)h 2 º

t (= 180 C)c 1 º

( ) Temperature distribution.b

Fig. 15.42. Counter-flow heat exchanger.