Question

A rigid, 26-L steam cooker is arranged with a pressure relief valve set to release vapor and maintain the pressure once the pressure inside the cooker reaches 150 kPa. Initially, this cooker is filled with water at 175 kPa with a quality of 10 percent. Heat is now added until the quality inside the cooker is 40 percent. Determine the exergy.

Answer 1

The minimum entropy change of the heat-supplying source is -0.87 kJ/K.

Initial entropy of the system

In this case, given the initial conditions, we first use the 10-% quality to compute the initial entropy.

• at initial pressure of 175 kPa

S₁ = 1.485  +  (0.1)(5.6865) = 2.0537 kJ/kg K

Final entropy

The entropy at the final state given the new 40-% quality:

• pressure inside the cooker = 150 kPa

S₂ = 1.4337  +  (0.4)(5.7894) = 3.7495 kJ/kg K

Mass of the steam at specific volume

m₁ = 0.026/(0.001057   +  0.1 x 1.002643) = 0.257 kg

m₂ = 0.026/(0.001053   +  0.4 x 1.158347)  = 0.056 kg

minimum entropy change of the heat-supplying source

ΔS + S₁  - S₂  + S₂m₂  - S₁m₁  -  sfg(m₂  -  m₁)  > 0

ΔS + 2.0537 -  3.7495 + (3.7495 x 0.056)  -   (2.0537 x 0.257)  - 5.6865( 0.056 - 0.257)  >  0

ΔS > -0.87 kJ/K

Thus, the minimum entropy change of the heat-supplying source is -0.87 kJ/K.

Learn more about entropy here: brainly.com/question/6364271

#SPJ1