Question

4.17 A rigid insulated tank is divided into two parts, one that contains 1 kg of steam at 10 bar, 200 °C, and one that contains 1 kg of steam at 20 bar, 800 °C. The partition that separates the two compartments is removed and the system is allowed to reach equilibrium. What is the entropy generation?

This is the problem in the book Fundamentals of Chemical Engineering Thermodynamics 1st edition. In the solution, it says first obtain volume and internal energy at 15.85 bar. Can you tell me how did the answer get 15.85 bar?

Answer 1

State 1:

At P = 10 bar, from the steam table,

Tsat = 179.886 C

Given T = 200 C

Therefore, the steam is superheated.

From the superheated steam table at P = 10 bar and T = 200 C

specific enthalpy H1 = 2828.27 kJ/kg

specific volume V1 = 0.206004 m³/kg

Volume of part 1 v1 = 1 kg x 0.206004 m³/kg = 0.206004 m³

Now,

state 2:

From the superheated steam table at P = 20 bar and T = 800 C

H2 = 4151.59 kJ/kg

V2 = 0.246737 m³/kg

Volume of part 2 v2 = 1 kg x 0.246737 m³/kg = 0.246737 m³

Now let the final state be 3.

By mass balance,

m3 = m1 + m2 = 1 + 1 = 2 kg

By enthalpy balance,

m1H1 + m2H2 = m3H3

.: 1*2828.27 + 1*4151.59 = 2*H3

.: H3 = 3489.93 kJ/kg

Volume of the tank v = v1 + v2 = 0.206004 + 0.246737 = 0.452741 m³

specifc volume at final state = v / m3 = 0.452741 / 2 = 0.2263705 m³/kg

Now look for specifc volume = 0.2263705 m³/kg and H = 3489.93 kJ/kg in the steam table, we get

P = 15.85 bar and T = 513 C