A piston/cylinder setup similar to Problem 3.132 contains 0.1 kg of saturated liquid and vapor water at 100 kPa with quality 25%. The mass of the piston is such that a pressure of 600 kPa will float it. The water is heated to 300°C. Find the final pressure, volume, work 1W2 and heat transfer 1Q2.

a) P₂ = 600 KPa

b) V₂ = 0.043442 m³

c) W₂ = 607.2 KJ

d) Q₂ = 793.39 KJ

Explanation:

Since this is a closed system. Therefore,

m₁ = m₂ = m = 0.1 kg

STATE 1:

Consider the initial state of water. From saturated liquid-vapor mixture table, we have:

Quality = x = 25% = 0.25

Pressure = P₁ = 100 KPa

Specific Volume = v₁ = vf + x(vfg) = 0.001043 m³/kg + (0.25)(1.6930 m³/kg)

v₁ = 0.4243 m³/kg

Specific Internal Energy = u₁ = uf + x(ufg) = 417.4 KJ/kg + (0.25)(2088.2KJ/kg)

u₁ = 939.45 KJ/kg

STATE 1':

Now, the water is heated to state 1' until its pressure becomes 600 KPa to lift the piston. Volume remains constant during this process.

STATE 2:

After reaching a pressure of 600 KPa, the water starts to expand at constant pressure, lifting the piston:

P₂ = 600 KPa

T₂ = 300° C

From property table at 600 KPa, Tsat = 158.83° C

Since, T₂ > Tsat. Therefore, water is in super heated state. From super heated water table:

u₂ = 2801.4 KJ/kg

v₂ = 0.43442 m³/kg

So,

a)

Final Pressure = P₂ = 600 KPa

b)

Final Volume = V₂ = mv₂ = (0.1kg)(0.43442 m³/kg)

V₂ = 0.043442 m³

c)

Work Done = W = P₂ΔV = P₂mΔv = mP₂(v₂ - v₁)

W = (0.1 kg)(600000 Pa)(0.43442 m³/kg - 0.4243 m³/kg)

W = 607.2 KJ

d)

Applying energy eqn. or first law of thermodynamics to the system:

Q₂ = m(u₂ - u₁) + W₂

Q₂ = (0.1 kg)(2801.4 KJ/kg - 939.45 KJ/kg) + 607.2 KJ

Q₂ = 793.39 KJ

answer(20) psi;