TITLE.PM5

326 ENGINEERING THERMODYNAMICS

DHARM
M-therm\Th6-2.PM5

Let Tc = common temperature when heat flows between ice and water stops.
Heat lost by water = Heat gained by ice
i.e., 12 × 4.18(300 – Tc) = 4.18(Tc – 273) + 333.5
or 15048 – 50.16Tc = 4.18Tc – 1141.14 + 333.5
or 54.34 Tc = 15855.64
∴ Tc = 291.8 K or 18.8°C.
Change of entropy of water = 12 × 4.18 loge 291.8
300

F

HG

I

KJ

= – 1.39 kJ/K

Change of entropy of ice = 1 × 4.18 loge

291.8

273

F

HG

I

KJ +

333.5

273

= 1.499 kJ/K

Net change of entropy, ∆S = – 1.39 + 1.499 = 0.109 kJ/K

Hence, net increase in entropy = 0.109 kJ/K. (Ans.)

Increase in unavailable energy = T 0 ∆S = 288 × 0.109 = 31.39 kJ. (Ans.)

+Example 6.14. A vapour, in a certain process, while condensing at 400°C, transfers heat

to water at 200°C. The resulting steam is used in a power cycle which rejects heat at 30°C.

What is the fraction of the available energy in the heat transferred from the process vapour

at 400°C that is lost due to the irreversible heat transfer at 200°C?

Solution. Refer Fig. 6.13.

∆′s

∆s

W Increase in

unavailable

energy

s

P N

T

M

Q 1

T

T (673 K) L

1

T (473 K) 1 ′

T (303 K) 0

R Q 1

Fig. 6.13

Temperature of vapour, T 1 = 400 + 273 = 673 K

Temperature of water, T 2 = 200 + 273 = 473 K

Temperature at which heat is rejected, T 0 = 30 + 273 = 303 K.

LMNP (Fig. 6.13) would have been the power cycle, if there was no temperature difference

between the vapour condensing and the vapour evaporating, and the area under NP would have

been the unavailable energy. RTWP is the power cycle when the vapour condenses at 400°C and