### Entropy Change in Heat Exchange

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

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

(ii) What is ΔStotal?

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

Calculations:

Basis: 1 mole of gas

Entropy change for the constant pressure heating/cooling process:

ΔS = nCP 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

Entropy change of the Hot gas stream:

T1 = 273 + 320 = 593 K

Since the flow rates of the two streams are same,

ΔThotstream = ΔTcoldstream = 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) ΔStotal = 1.05 R + (-0.79R) = 0.26 R

(iii) Heating stream inlet temperature = 200oC

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

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

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