Page 158 - DJJ20063- Thermodynamics 1

DJJ20063- Thermodynamics 1

As in example 4.3

h1= 2800 kJ/kg and h2 = 1808 kJ/kg
Also, h3 = hf at 0.035 bar = 112 kJ/kg

Using equation 11.18, with v = vf at 0.035 bar

Pump work = -W34 = f (p4 – p3)
2
= 0.001 x ( 42 – 0.035) x 10
= 4.2 kJ/kg

Using equation 4.14
W12 = h1 – h2 = 2800 – 1808 = 992 kJ/kg

Then using equation 4.20

(h − h ) − (h − h )
 = 1 2 4 3
(h − h 3 ) − (h − h 3 )
R
1
4
( 992 ) ) 2 . 4 ( −
=
( 2800 − 112 ) ) 2 . 4 ( −

= 0.368 or 36.8 %

Using equation 4.10

net work
Work ratio =
gross work

992 - 4.2
=
992

= 0.996

Using equation 4.24
3600
s.s.c. =
(h − h 2 ) − (h − h 3 )
4
1
3600
=
( 992 ) ) 2 . 4 ( −

= 3.64 kg/kW h

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