TITLE.PM5

560 ENGINEERING THERMODYNAMICS

dharm

\M-therm\Th15-2.pm5

(iv)Brake thermal efficiency :

Brake thermal efficiency =

W

hhf

brake

1 − 3

=

225

3037 6 428.8. −

= 0.086 or 8.6%. (Ans.)

(v)Relative efficiency :

Relative efficiency on the basis of indicated work

=

W

hh

W

hh

f

f

indicated

1

1

3

3

−

−

=

W

W

indicated = 281 25

495 8

.

. = 0.567 or 56.7%. (Ans.)

Relative efficiency on the basis of brake work

=

W

hh

W

hh

f

f

indicated

()

()

1

1

3

3

−

−

=

W

W

brake =^225

495 8. = 0.4538 or 45.38%. (Ans.)

Example 12.11. Superheated steam at a pressure of 10 bar and 400°C is supplied to a
steam engine. Adiabatic expansion takes place to release point at 0.9 bar and it exhausts into a
condenser at 0.3 bar. Neglecting clearance determine for a steam flow rate of 1.5 kg/s :
(i)Quality of steam at the end of expansion and the end of constant volume operation.
(ii)Power developed.
(iii)Specific steam consumption.
(iv)Modified Rankine cycle efficiency.
Solution. Fig. 12.14 shows the p-V and T-s diagrams for modified Rankine cycle.
From steam tables :

1. At 10 bar, 400°°°°°C : h 1 = 3263.9 kJ/kg, v 1 = 0.307 m^3 /kg, s 1 = 7.465 kJ/kg K
   2. At 0.9 bar : ts 2 = 96.7°C, hg 2 = 2670.9 kJ/kg, sg 2 = 7.3954 kJ/kg K,
   vg 2 = 1.869 m^3 /kg


2. At 0.3 bar : hf 3 = 289.3 kJ/kg, vg 3 = 5.229 m^3 /kg
   (i) Quality of steam at the end of expansion, Tsup2 :
   For isentropic expansion 1-2, we have
   s 1 = s 2

= sg 2 + cp loge

T

T

sup

s

2

2

7.465 = 7.3954 + 2.1 loge

Tsup 2

()96.7 273+

465 3954

1

F −

HG

I

KJ

= loge

Tsup 2

369 7.

or loge

Tsup 2

369 7.

= 0.0033

Tsup 2

369 7.

= 1.0337 or Tsup 2 = 382 K