TITLE.PM5

316 ENGINEERING THERMODYNAMICS

DHARM

M-therm\Th6-1.PM5

(ii) Heat transferred during cooling (constant pressure) process

= m. cp (T 1 – T 0 )

= 8 × 1.005 (650 – 300) = 2814 kJ

Change in entropy during cooling

∆s = mcp loge

T

T

1

0

F

HG

I

KJ

= 8 × 1.005 × loge FH^650300 IK = 6.216 kJ/K

Unavailable energy = T 0 ∆S

= 300 × 6.216 = 1864.8 kJ

Available energy = 2814 – 1864.8 = 949.2 kJ. (Ans.)

Effectiveness, ∈ =

Available energy

Change in available energy

= 1319.2949.2 = 0.719. (Ans.)

+Example 6.4. In a power station, the saturated steam is generated at 200ºC by transfer-

ring the heat from hot gases in a steam boiler. Find the increase in total entropy of the combined

system of gas and water and increase in unavailable energy due to irreversible heat transfer. The

gases are cooled from 1000°C to 500°C and all the heat from gases goes to water. Assume water

enters the boiler at saturated condition and leaves as saturated steam.

Take : cpg (for gas) = 1.0 kJ/kg K, hfg(latent heat of steam at 200°C) = 1940.7 kJ/kg.

Atmospheric temperature = 20°C.

Obtain the results on the basis of 1 kg of water.

Solution. Refer Fig. 6.8.

Temperature of saturation steam = 200 + 273 = 473 K

Initial temperature of gases = 1000 + 273 = 1273 K

Final temperature of gases = 500 + 273 = 773 K

For gases : cpg = 1 kJ/kg K

Latent heat of steam of 200°C

saturation temperature, hfg = 1940.7 kJ/kg

Atmospheric temperature = 20 + 273 = 293 K

Heat lost by gases = Heat gained by 1 kg saturated water when it is converted to steam at

200 °C.

∴ mgcpg (1273 – 773) = 1940.7

[where mg = mass of gases, cpg = specific heat of gas at constant pressure]

i.e., mg =
1940.7
1.0 (1273×−773) = 3.88 kg
Change of entropy of mg kg of gas,

(∆S)g = mg cpg loge FH 1273773 IK

= 3.88 × 1.0 × loge FH 1273773 IK = – 1.935 kJ/K