Question

Steam at 700 bar and 600?C is withdrawn from a steam line and adiabatically expanded to 10 bar at a rate of 2 kg/min. What is the temperature of the steam that was expanded, and what is the rate of entropy generation in this process? Solve using 1st and 2nd law of thermodynamics.

Answer 1

The proces is adiabatic so dQ = 0

from first law of thermodynamic

dU = dQ -dW , dQ = 0 ( U = internal enrgy, W = work done , Q = heat

dU = -dW, C_V dT = -PdV , PV= RT , V= RT/P and R = C_P-C_V

Substitute all the values we will get

P^(1-v) T^v = constant, where v = C_P/C_V

_{for steam v = 1.33 from litrature , P1 = 700 bar , T1= 600 oC = 600 oC+273.15 = 873.15 K, P2 = 10 bar ,}

P1^(1-v) T1^v = P2^(1-v) T2^v

Substitute all the values

700bar^(1-1.33) * 873.15K^1.33 = 10bar^(1-1.33)*T2^1.33

T2 = 304.26 K = 304.26K -273.15 = 31.13 oC

Second law of thermodynamic dS = dQ/T in this case dQ = 0

so dS = 0 , hence in case of adiabatic process entropy generation is zero