Question

A cylinder fitted with a movable piston contains water at 3 MPa with 50% quality, at which point the volume is 20 L. The water now expands to 1.2 MPa as a result of receiving 600 kJ of heat from a large source at 300◦C. It is claimed that the water does 124 kJ of work during this process. Is this possible?

Answer 1

Answer:

Yes it is possible

Explanation:

Attached is the solution

Answer 2

Answer:

The process is possible:

Explanation:

We are going to find out if the entropy generated is greater than 0, if it is greater than 0, then the process is feasible. If it is not, the process is not feasible.

$P_{1} = 3 MPa$

$x_{1} = 50 % = 0.5$

$V_{1} = 20 L = 0.02 m^{3}$

$P_{2} = 1.2 MPa$

$T_{H} = 300^{0} C = 573 K$

Received heat energy, $Q_{12} = 600 kJ$

Work done, $W_{12} = 124 kJ$

At state 1, using the steam table:

$T_{1} = T_{s} = 233.9^{0} C\\v_{f1} = 0.001216 m^{3} /kg\\v_{fg1} = 0.06546m^{3} /kg\\u_{f1} = 1004.76 kJ/kg\\u_{fg1} = 1599.34 kJ/kg\\s_{f1} = 2.6456 kJ/kg-K\\s_{fg1} = 3.5412kJ/kg-K$

$v_{1} = v_{f1} + x_{1} * v_{fg1}$

$v_{1} = 0.001216 + 0.5*(0.06546)\\v_{1} = 0.03395 m^{3} /kg$

$M = \frac{V_{1} }{v_{1} } \\M = 0.02/0.03395\\M = 0.5892 kg$

$u_{1} = u_{f1} + x_{1} * u_{fg1}\\u_{1} = 1004.76 + 0.5*1599.34\\u_{1} = 1804.43 kJ/kg$

$s_{1} = s_{f1} + x_{1} * s_{fg1}\\s_{1} = 2.6456 + 0.5*3.5412\\s_{1} = 4.4162 kJ/kg$

$Q_{12} = m(u_{2} - u_{1} ) + W_{12} \\600 = 0.5892(u_{2} -1804.43) + 124$

Solving for u₂

$u_{2} = 2612.3 kJ/kg$

Since P₂ = 1.2 MPa, u₂ = 2612.2 kJ/kg,

then from steam table, T₂ = 200°C, S₂ = 6.5898 kJ/kg-K

The entropy generated will be:

$\triangle S = m(S_{2} -S_{1} ) - \frac{Q_{12} }{T_{H} }\\ \triangle S= 0.5892(6.5898 - 4.4162) - \frac{600 }{573 }\\ \triangle S =0.233 kJ/K$

Since ΔS > 0, this process is possible