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Entropy Change in Heat Exchange of Liquids

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A heat exchanger uses 5000 kg/hr of water to cool hydrocarbon oil from 140^{o}C to 65^{o}C. The oil, flowing at the rate of 2500 kg/hr has an average specific heat of 0.6 kcal/kg.^{o}C. The water enters at 20^{o}C. Determine

- The entropy change of the oil
- Entropy change of water
- The total entropy change as a result of this heat exchange process.

Calculations:

By energy balance, m_{oil}C_{Poil}ΔT_{oil} = m_{water}C_{Pwater}ΔT_{water}

ΔT_{water} = 2500 x 0.6 x (140 - 65) / (5000 x 1) = 22.5^{o}C

Therefore, exit temperature of water = 20 + 22.5 = 42.5^{o}C

Entropy change is given by

ΔS = mC_{P} ln (T_{2}/T_{1})

where m is mass flow rate of stream, T_{1} and T_{2} are respectively initial and final temperatures.

(i) Entropy change of oil:

T_{1} = 140 + 273 = 413 K

T_{2} = 65 + 273 = 338 K

Mass flow rate of oil = 2500 kg/hr = 0.6944 kg/sec

ΔS_{oil} = 0.6944 x 0.6 x ln (338/413) =
-0.0835 kcal/sec.^{o}C

(ii) Entropy change of water:

T_{1} = 20 + 273 = 293 K

T_{2} = 42.5 + 273 = 315.5 K

Mass flow rate of water = 5000 kg/hr = 1.389 kg/sec

ΔS_{water} = 1.389 x 1 x ln (315.5/293) =
0.1028 kcal/sec.^{o}C

(iii) Total entropy change:

Total entropy change is the sum of the
entropy changes of two streams, and = -0.0835 + 0.1028 = 0.0193 kcal/sec.^{o}C

[Index]

Last Modified on: 01-May-2024

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