TITLE.PM5

324 ENGINEERING THERMODYNAMICS

DHARM
M-therm\Th6-2.PM5

or 20(90 – t) = 30 (t – 30)

∴ t =

20 90 30 30

20 30

×+×

+ = 54°C

Total mass after mixing = 20 + 30 = 50 kg

Available energy of 50 kg of water at 54°C

(A.E.)50 kg = 50 × 4.18 ()log327 283^283

327

−− 283

F

H

I

K

L

NM

O
e QP
= 209 (44 – 40.89) = 649.99 kJ
∴ Decrease in available energy due to mixing
= Total energy before mixing – Total energy after mixing
= 883.65 – 649.99 = 233.66 kJ. (Ans.)

• Example 6.12. In an heat exchanger (parallel flow type) waters enter at 50°C and leaves
  at 70°C while oil (specific gravity = 0.82, specific heat = 2.6 kJ/kg K) enters at 240°C and leaves
  at 90°C. If the surrounding temperature is 27°C determine the loss in availability on the basis of
  one kg of oil per second.
  Solution. Refer Fig. 6.12.

Water

Oil

T = 90°C

(363 K)

02

T = 70°C

(343 K)

W2

T = 240°C

(513 K)

01

T = 50°C

(323 K)

W1

s

Fig. 6.12

Inlet temperature of water, Tw 1 = 50°C = 323 K

Outlet temperature of water, Tw 2 = 70°C = 343 K

Inlet temperature of oil, To 1 = 240°C = 513 K

Outlet temperature of oil, To 2 = 90°C = 363 K

Specific gravity of oil = 0.82

Specific heat of oil = 2.6 kJ/kg K

Surrounding temperature, T 0 = 27 + 273 = 300 K.