Thermodynamics Piston-Cylinder Problems

Study Notes

A piston-cylinder device with stops contains steam.
The steam is cooled.
Specific volumes for initial and final states are determined (Table A-6).
Initial state: P₁ = 1 MPa, T₁ = 400°C, v₁ = 0.30661 m³/kg.
Final state: P₂ = 1 MPa, T₂ = 250°C, v₂ = 0.23275 m³/kg.
Pressure is constant during the process.
Boundary work (W) is calculated as: W = mP(v₂ - v₁) = (0.6 kg)(1000 kPa)(0.30661 - 0.23275) m³/kg = 44.3 kJ.

Volume of the cylinder at the final state is 40% of the initial volume.
Final state volume v₂ = 0.12261 m³/kg.
Boundary work (W₁) is calculated as: W₁ = mP(v₁ - 0.4v₁) = (0.6 kg)(1000 kPa)(0.30661 - 0.4 × 0.30661) m³/kg = 110.4 kJ.
Temperature at the final state is determined as T₂ = 151.8°C (Table A-5).
Pressure at the final state is P₂ = 0.5 MPa.

A piston-cylinder device contains air.
Air goes through a three-process cycle.
Properties of air: R = 0.287 kJ/kg∙K, k = 1.4.
Isothermal expansion process: -Initial state: P₁ = 2 MPa, V₁ = 0.01341 m³, T₁ = 350°C -Final state: P₂ = 0.5 MPa, V₂ = 0.05364 m³, T₂ = 350°C -Boundary work (W₁₂): = 37.18 kJ.

Polytropic compression process -Process 2-3: The boundary work W₂₃ -Process3-1: The boundary work W₃₁ = -6.97 kJ
Constant pressure compression process
Net work of the cycle (Wnet) = W₁₂ +W₂₃ + W₃₁ = -4.65 kJ.

Hydrogen gas in a rigid tank is cooled to 300 K.
Final pressure and heat transfer are determined.
Assumptions:
- Hydrogen is an ideal gas (at high temperature and low pressure).
- Stationary tank
- No work interactions.
Properties used:
- R = 4.124 kPa⋅m³/kg⋅K (from Table A-1)
- Cavg = 10.377 kJ/kg⋅K (from Table A-2 at the average temperature).
Final pressure calculated from ideal gas relation: P₂ = 159.1 kPa.
Heat transfer Qout calculated from energy balance: Qout = - m⋅Cavg⋅(T₁-T₂)= -686.2 kJ

Ideal gas in a piston-cylinder device is cooled at constant pressure.
Gas constant and molar mass are determined.
Assumptions:
- No friction between piston and cylinder.
Specific heat ratio is k = 1.667
Gas constant (R) is determined as follows; R= W/(m⋅ΔT) = 2.075 kJ/kg⋅K.
Molar mass (M) is calculated as: M = R/R= 4.007 kg/kmol.
Specific heats (cp and cv) are calculated based on the determined values of R and k.

Helium is compressed by a compressor.
Power input required (W) for mass flow rate of 60 kg/min is to be determined.
Assumptions: -Steady flow -Negligible kinetic/potential energy changes -Ideal gas model with constant specific heats -Properties:cp = 5.1926 kJ/kgK
power input = 872kW

Adiabatic air compressor is powered by a steam turbine.
Net power delivered to the generator is determined.
Assumptions:
steady-flow process.
Kinetic and potential energy changes are negligible.
Adiabatic process (no heat transfer).
Air is an ideal gas with variable specific heats.
Steam and air properties are assumed known and used to compute energy balances for the compressor and turbine.
Net Power = 20,448kW