TITLE.PM5

SECOND LAW OF THERMODYNAMICS AND ENTROPY 251

dharm
/M-therm/th5-2.pm5

Solution. Boiler temperature, T 1 = 160 + 273 = 433 K

Condenser temperature, T 2 = 50 + 273 = 323 K

From steam tables :

Enthalpy of water entering boiler, hf 1 = 687 kJ/kg

Enthalpy of steam leaving boiler, h 2 = 2760 kJ/kg

Condenser pressure = 0.124 × 10^5 N/m^2

Boiler pressure = 6.18 × 10^5 N/m^2 ......(corresponding to 160°C)

Enthalpy of vapour leaving the turbine, h 3 = 2160 kJ/kg

(assuming isentropic expansion)

Enthalpy of water leaving the condenser, hf 4 = 209 kJ/kg

Now Qboiler, Q 1 = h 2 – hf 1 = 2760 – 687 = 2073 kJ/kg

and Qcondenser, Q 2 = hf 4 – h 3 = 209 – 2160 = – 1951 kJ/kg

∴

δQ

T

Q

T

Q

cycle∑ T

=+= +F−

HG

I

KJ

1

1

2

2

2073

433

1951
323
= – 1.25 kJ/kg K
< 0. ...... Proved.
+Example 5.20. In a power plant cycle, the temperature range is 164°C to 51°C, the upper
temperature being maintained in the boiler where heat is received and the lower temperature
being maintained in the condenser where heat is rejected. All other processes in the steady flow
cycle are adiabatic. The specific enthalpies at various points are given in Fig. 5.20.
Verify the Clausius Inequality.

Fig. 5.20

Solution. Temperature maintained in boiler, T 1 = 164 + 273 = 437 K

Temperature maintained in condenser, T 2 = 51 + 273 = 324 K