TITLE.PM5

dharm
\M-therm\Th15-3.pm5

830 ENGINEERING THERMODYNAMICS

1250 × (0.9 × 2592) = m&c × 4.187 (47 – 24)

∴ m&c (= m&w) = 30280 kg/min

Water

out Water

in

Condensate out

Steam in (x = 0.9)

Two-pass surface

condenser

t = 24 Cc 1 º

t = 47 Cc 2 º

t = t = th 1 h 2 sat= 51 Cº

Fig. 15.43. A two-pass surface condenser.

(ii)Condenser surface area, A :

Q =

mxh&s×(. )fg

60

= U A θm ...(i)

where, θm =

θθ

θθ

12

12

11 22

11 22

−

=

−− −

ln ( / ) −−

()( )

ln [( ) / ( )]

tt tt

tt tt

hc hc

hc hc

=

(()

ln [( / ( )]

(

ln ( /

51 24) 51 47

51 24) 51 47

27 4)

27 4)

−−−

−−

= − = 12.04°C

Substituting the values in eqn. (i), we get

1250

60

× (0.9 × 2592 × 10^3 ) = 4000 × A × 12.04

or A = 1009.1 m^2
(iii)Number of tubes required per pass, Np :

mdV&wi=××F

HG

I

KJ

π ρ

4

(^2) × N
p
30280
60 4
=π × (0.0296)^2 × 3 × 1000 × Np
or Np =
30280 4
60 0 0296^2 3 1000
×
××(. )××π
= 244.46 say 245
(Total number of tubes required, N = 2Np = 2 × 245 = 490)
(iv)Tube length, L :
A = (πdoL) × (2Np)
1009.1 = (π × 0.0384 × L) × (2 × 245)