Entropy Change in Heat Exchange

An ideal gas C_P = 7R/2 is heated in a steady flow heat exchanger from 70^oC to 190^oC, by another stream of the same ideal gas entering at 320^oC. The flow rates of the two streams are the same.

(i) Calculate ΔS of the two gas streams for countercurrent flow

(ii) What is ΔS_total?

(iii) If heating stream enters at 200^oC, what is ΔS_total?

Calculations:

Basis: 1 mole of gas

Entropy change for the constant pressure heating/cooling process:

ΔS = nC_P ln (T₂/T₁)

Where T₁ and T₂ are respectively initial and final temperatures.

(i) Entropy change of the cold gas stream:

T₁ = 273 + 70 = 343 K

T₂ = 273 + 190 = 463 K

n = 1

Therefore, ΔS = (7R/2) x ln (463/343) = 1.05 R

T₁ = 273 + 320 = 593 K

Since the flow rates of the two streams are same,

ΔT_hotstream = ΔT_coldstream = 190 - 70 = 120 K

T₂ = T₁ - 120 = 593 - 120 = 473 K

n = 1

Therefore, ΔS = (7R/2) x ln (473/593) = -0.79 R

(ii) ΔS_total = 1.05 R + (-0.79R) = 0.26 R

(iii) Heating stream inlet temperature = 200^oC

Therefore, outlet temperature of heating stream = 200 - 120 = 80^oC

ΔS for this stream = (7R/2) x ln (353/473) = -1.024 R

ΔS_total = 1.05 R + (-1.024 R) = 0.026 R