Question

q1 (25 points). a system consisting of 1.8 kg of water vapour in a piston-cylinder assembly, initially at a volume of is expanded in a constant pressure process to a volume . the system then is compressed isothermally (constant temperature) to a final volume of . during the compression process, the energy transfer by work into the system is 1500 kj. kinetic and potential energy effects are negligible. determine the heat transfer, in kj, for each process

Answer 1

Heat transfer in the constant-pressure expansion: Q1 = 654 Btu & Heat transfer in the isothermal compression: Q2 = -72 Btu

Process 1: Constant-Pressure Expansion

Work Done (W1):

For a constant-pressure process, work is calculated as: W1 = P(V2 - V1)

We need to find the pressure (P).

Use the steam tables to determine the specific volume (v1) at the initial state (350°F, 71.7 ft³).

v1 = 23.9 ft³/lb (from steam tables)

P = mRT/V1

= 3 lb * 0.5956 Btu/lb·°R * (350 + 460)°R / 71.7 ft³

= 45.5 psia

W1 = 45.5 psia * (85.38 ft³ - 71.7 ft³)

= 654 Btu

Heat Transfer (Q1):

Apply the first law of thermodynamics: Q1 - W1 = ΔU1

For an ideal gas, internal energy change (ΔU) depends only on temperature change. Since the process is isothermal, ΔU1 = 0.

Therefore, Q1 = W1 = 654 Btu

Process 2: Isothermal Compression

Heat Transfer (Q2):

For an isothermal process, ΔU2 = 0.

Applying the first law: Q2 - W2 = ΔU2

Q2 = W2 = -72 Btu (negative because work is done on the system)