water at 2.0 MPa, 500 °C is brought to saturated vapor in a piston/cylinder with a reversible adiabatic process. find the final temperature and work done during this process.

The final temperature = 93.45°C

The constant volume work done = -568.76 kJ/kg (work is done by the gas)

The constant pressure work done = -756.186 kJ/kg (work is done by the gas)

Explanation:

The parameters given are;

Pressure = 2.0 MPa = 20 Bars

Temperature = 500 °C

Process = Reversible adiabatic (isentropic) process = Constant entropy, s

Therefore;

s₁ = s₂

From the steam tables for super heated steam, s₁ = 7.4335 kJ/(kg·K)

At T₂, we have saturated vapor, hence;

Where:

s₁ = s₂ = 7.4335 kJ/(kg·K), we have;

The final temperature, T₂, is given as follows;          

T₂ = 93.4854 + (7.4339  -7.4335 )/(7.4790 - 7.4339) * (89.9315-93.4854) = 93.45°C

The work done, W in an isentropic process is given as follows

p·dV work,   = m··(T₂ - T₁)

V·dp work, = m··(T₂ - T₁)

The specific heat at constant pressure, , for steam = 1.86 kJ/(kg·K)

R = 0.461 kJ/(kg·K)

∴ = - R = 1.86 - 0.461 = 1.399 kJ/(kg·K)

Hence;

p·dV work = 1.399 * (93.45 - 500 ) = -568.76 kJ/kg

V·dp work = 1.86 * (93.45 - 500 ) = -756.186 kJ/kg

Also work done, , can be expressed as follows;

Where:

k =  / = 1.86/1.399  = 1.33

∴ = 0.461 * (93.45 - 500 )/(1-1.33) = 567.94 kJ/kg done by the gas.

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answer:

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explanation: