TITLE.PM5

552 ENGINEERING THERMODYNAMICS

dharm

\M-therm\Th12-1.pm5

T

s

4

3

5

1

p = 80 bar 1

p = 0.1 bar 2 2

Fig. 12.8

[and hf 4 = 2.008 + hfbgp 2 = 2.008 + 173.88 = 175.89 kJ/kg]

Wturbine = h 1 – h 2 = 3159.3 – 2187.68 = 971.62 kJ/kg

∴ Wnet = 971.62 – 2.008 = 969.61 kJ/kg. (Ans.)

Cycle efficiency, ηcycle:

Q 1 = h 1 – hf 4 = 3159.3 – 175.89 = 2983.41 kJ/kg

∴ ηcycle =

W

Q

net

1

=
969.61
2983.41
= 0.325 or 32.5%. (Ans.)
Example 12.4. A Rankine cycle operates between pressures of 80 bar and 0.1 bar. The
maximum cycle temperature is 600°C. If the steam turbine and condensate pump efficiencies are
0.9 and 0.8 respectively, calculate the specific work and thermal efficiency. Relevant steam table
extract is given below.

p(bar) t(oC) Specific volume (m^3 /kg) Specific enthalpy (kJ/kg) Specific entropy (kJ/kg K)

vf vg hf hfg hg sf sfg sg

0.1 45.84 0.0010103 14.68 191.9 2392.3 2584.2 0.6488 7.5006 8.1494

80 295.1 0.001385 0.0235 1317 1440.5 2757.5 3.2073 2.5351 5.7424

80 bar, 600ºC v 0.486 m^3 /kg

Superheat h 3642 kJ/kg

table s 7.0206 kJ/kgK (GATE, 1998)

Solution. Refer Fig. 12.8

At 80 bar, 600ºC :

h 1 = 3642 kJ / kg ;

s 1 = 7.0206 kJ / kg K.

Since s 1 = s 2 ,

∴ 7.0206 = sf 2 + x 2 sfg 2

= 0.6488 + x 2 × 7.5006

or x 2 = 7. 0.
7.

0206 6488

5006

− = 0.85

Now, h 2 = hf 2 + x 2 hfg 2

= 191.9 + 0.85 × 2392.3

= 2225.36 kJ/kg

Actual turbine work

= ηturbine × ()hh 12 −

= 0.9 (3642 – 2225.36)= 1275 kJ/kg

Pump work = vf()p 2 ()pp 12 −

= 0.0010103 (80 – 0.1) ×

10

10

5

3 kN/m^2 = 8.072 kJ/kg