Entropy Change in Heat Exchange of Liquids

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

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

Calculations:

By energy balance, m_oilC_PoilΔT_oil = m_waterC_PwaterΔT_water

ΔT_water = 2500 x 0.6 x (140 - 65) / (5000 x 1) = 22.5^oC

Therefore, exit temperature of water = 20 + 22.5 = 42.5^oC

Entropy change is given by

ΔS = mC_P ln (T₂/T₁)

where m is mass flow rate of stream, T₁ and T₂ are respectively initial and final temperatures.

(i) Entropy change of oil:

T₁ = 140 + 273 = 413 K

T₂ = 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.^oC

(ii) Entropy change of water:

T₁ = 20 + 273 = 293 K

T₂ = 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.^oC

(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.^oC