4.17 A rigid insulated tank is divided into two parts, one that contains 1 kg of

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?

Explanation / Answer

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

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