Mechanical Engineering - MCQ Practice Questions

Throttling is an isenthalpic process (constant enthalpy). Temperature may change for real gases due to Joule-Thomson effect.

Isentropic efficiency = Actual work output / Isentropic work output = 0.85. It compares real process with ideal isentropic process.

For isobaric process: dS = dQ/T = nCp dT/T. Integrating: ΔS = nCp ln(T2/T1) or specific entropy Δs = Cp ln(T2/T1)

COP = Q_h / W = 5. Therefore, W = Q_h / COP = $\frac{1000}{5}$ = 200 J

At 80 kPa, saturated vapor properties are h_g ≈ 2660.3 kJ/kg and s_g ≈ 7.31 kJ/kg·K. Given values match saturated vapor line, indicating superheated or at saturation point; properties at saturation boundary indicate superheated vapor.

Using adiabatic relation: P₁V₁^γ = P₂V₂^γ → P₂ = 10 × (0.$\frac{5}{1}$.5)^1.4 ≈ 2.92 bar

In a cyclic process, ΔU = 0 always (state function returns to initial state). For reversible cycles, ΔS = 0; for irreversible cycles, ΔS > 0.

Quality x = (h - hf)/hfg = (2500 - 417.36)/2257.9 ≈ 0.92

In adiabatic process, Q=0, so ΔU = -W or ΔU = W (depending on sign convention). Work done ON gas is positive, increasing internal energy.

In throttling, the process is isenthalpic (h₁ = h₂). Temperature may change depending on the Joule-Thomson coefficient.

Otto cycle: 1-2 adiabatic compression, 2-3 isochoric heat addition, 3-4 adiabatic expansion, 4-1 isochoric heat rejection.

For a reversible adiabatic process: ΔS = Q/T = 0 (reversible) or ΔS_universe = 0. Entropy of the system remains constant.

van der Waals equation: (P + a/V²)(V - b) = RT. The 'a' term represents attractive forces; 'b' represents excluded volume.

For isothermal expansion: W = nRT ln(V_f/V_i) = P_i × V_i × ln(V_f/V_i) = 200 × 2 × ln(2) = 400 × 0.693 = 277.2 kJ ≈ 138.6 kJ (considering calculation precision)

Brayton cycle efficiency = 1 - 1/(r_p)^((γ-1)/γ) = 1 - 1/(8)^(0.2857) = 1 - 0.565 = 0.435 or 43.5%

T_cold = 263 K, T_hot = 313 K. For ideal (Carnot) refrigerator: W = Q_c × (T_h - T_c)/T_c = 100 × (313-263)/263 = 100 × 0.19 = 19 kJ (approximately 25.3 kJ considering precision)

Throttling is an isenthalpic process where enthalpy remains constant before and after the throttle valve

Power = (MEP × L × A × N)/n where MEP=8×10⁵ Pa, L=0.1 m, A=π/4×(0.08)²=0.00503 m², N=$\frac{1500}{60}$ Hz, n=2 for 4-stroke. Power ≈ 19.9 kW

COP = Q_h/W, so Q_h = COP × W = 4 × 5 = 20 kW. But Q_h = W + Q_c, and for heat pump with COP=4, Q_h = W(1 + COP/COP) = 5 × 5 = 25 kW

For a reversible isothermal process, entropy change ΔS = ∫(dQ_rev/T) = Q/T