## Question

###### Water vapor initially at 240 °C, 1.0 MPa expands in a piston-cylinder assembly isothermally and without...

Water vapor initially at 240 °C, 1.0 MPa expands in a piston-cylinder assembly isothermally and without internal irreversibilities to a final pressure of 0.1 MPa. Evaluate the (a)work done, in kJ/kg, (b) the change of entropy,, in kJ/kg - K and (c) the change of internal energy, in kJ/kg. Assume the system can be treated as a Van der Waals model.. [Given: the specific volume of the water vapor can be determined from the superheated steam table and they are: Vi = 0.2275 m®/kg and V2 = 2.359 m2/kg. The system can be treated as a single-phase condition.) Integration Equations: dx In x-a x-a and X-a X -

## Answers

Given data: Temperature of water vapor, T = 240°C = 513 K. Initial pressure of water vapor, P, = 1 MPa = 1000 kPa. Final pressure of water vapor, P = 0.1 MPa = 100 kPa. Initial specific volume of water vapor, vų = 0.2275 m?/kg. Final specific volume of water vapor, va = 2.359 m /kg. Water vapor undergoes expansion process isothermally (T=C) inside the piston cylinder assemble. It is given that the system can be treated as single phase condition "Van Der Waals” model. Which means, we will the solve the problem by assuming water vapor to be in superheated state (i.e. steam, because steam is a single phase substance) and hence, can be assumed as ideal gas. (a) Calculate the work done by the water vapor in the expansion process as follows: V, Wiso = Pivx In V = 100 kPax 2.359 m /kg x In 2.359 m?/kg 0.2275 m?/kg W = 551.733 kJ/kg ISO (b) Calculate the change in entropy of the system as follows: P AS12 = Cp In R In [because process is isothermal: T1 =T,] P, (？ = 0 -0.4615 kJ/kg-K x In 100 kPa 1000 kPa AS12 = 1.0626 kJ/kg-K (c) Calculate the change in internal energy of the system as follows: For ideal gas, internal energy is a function of temperature only; u = = f(t). And the process of expansion is isothermal. So, u = f(t)= constant. Therefore, Change in internal energy, Auj2 = 0

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