Question

A 0.5 kg of saturated water vapor at 300°C is heated in a piston–cylinder device. Now the steam expanded reversibly and isothermally to a final pressure of 600 kPa.

1. Determine the heat transferred during this process.

2. Determine the work done during the process.

3. Plot the PV diagram showing all the states and numbers on it.

Answer 1

Using Table A-4 (Saturated water - Temperature table);

• at T₁ = 300^oC and saturated vapor state;

P1 = Psat 8587.9 kPa

V1 = V. 0.021659 m/kg

U1 = ug 2563.6 kJ/kg

Si = Sg = 5.7059 kJ/kg

Using Table A-6 (Superheated water);

• at P₂ = 600 kPa and T₂ = 300^oC;

v2 = 0.43442 m/kg

U2 2801.4 kJ/kg

S2 7.3740 kJ/kg

(a)

Heat transfer during isothermal process;

Q=mT(S2 - 51

Q= (0.5) (300 + 273) (7.3740 – 5.7059)

Q = 477.91 kJ

...(Answer)

(b)

Work done can be given using first law;

Q=W + m(uz - U1

→ 477.91 = W + 0.5(2801.4- 2563.6)

= Ꮃ = 35Ꮽ,01 ᎬᎫ

...(Answer)

(c)

Pressure vs specific volume (P-v) diagram is shown below;

NEV Practice 29 0 8587.9 P(kPa) 600 0.021659 m²/kg 0.43442 m² kg (mes) ده