5.57 A closed steel bottle contains carbon dioxide at -20 degrees C, x=20% and t

Question

5.57

A closed steel bottle contains carbon dioxide at -20 degrees C, x=20% and the volume is 0.05 m^3. It has a safety valve that opens at a pressure of 6 MPa. By accident, the bottle is heated until the safety valve opens. Find the temperature and heat transfer when the valve first opens.

Explanation / Answer

Greetings... C.V.: NH3 : m2 = m1 = m ; m(u2 - u1) = 1Q2 - 1W2 Process: constant volume process Þ 1W2 = 0 State 1: v1 = 0.001504 + 0.2 ´ 0.62184 = 0.1259 => m = V/v1 = 0.05/0.1259 = 0.397 kg u1 = 88.76 + 0.2 ´ 1210.7 = 330.9 kJ/kg State 2: P2 , v2 = v1 => superheated vaporC.V.: NH3 : m2 = m1 = m ; m(u2 - u1) = 1Q2 - 1W2 Process: constant volume process Þ 1W2 = 0 State 1: v1 = 0.001504 + 0.2 ´ 0.62184 = 0.1259 => m = V/v1 = 0.05/0.1259 = 0.397 kg u1 = 88.76 + 0.2 ´ 1210.7 = 330.9 kJ/kg State 2: P2 , v2 = v1 => superheated vapor T @ 110°C, u2 = h2 - P2v2 = 1677.6 - 6000 x 0.1259 = 922.2 1Q2 = m(u2 - u1) = 0.397(922.2 - 330.9) = 234.7461 kJ Wish you success!

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