(a) A 2 gm quantity of air undergoes the following sequence of quasi-static processes in a piston-cylinder arrangement: (i) An…

(a) Take air as an ideal gas with γ = 1.4 and R = 287 J/kg K. m = 2 g = 0.002 kg, T₁ = 150°C = 423.15 K, P₁ = 5 atm.

State 1: P₁, V₁, T₁. Process 1–2: reversible adiabatic expansion, V₂ = 2V₁. For an adiabatic process, TV^(γ−1) = constant and PV^γ = constant. T₂ = T₁(V₁/V₂)^(γ−1) = T₁/2^(γ−1) P₂ = P₁(V₁/V₂)^γ = P₁/2^γ.

Work done by air in 1–2: W₁₂ = (P₁V₁ − P₂V₂)/(γ−1) = mR(T₁ − T₂)/(γ−1). P₁V₁ = mRT₁ = 0.002 × 287 × 423.15 = 242.888 J.

2^(γ−1) = 2^0.4 = 1.31951, so T₂ = 423.15/1.31951 = 320.69 K. W₁₂ = 0.002 × 287 × (423.15 − 320.69)/(0.4) = 147.02 J.

Process 2–3: constant pressure, volume reduced to initial value V₃ = V₁, so P₃ = P₂. W₂₃ = P₂(V₃ − V₂) = P₂(V₁ − 2V₁) = −P₂V₁ = −P₁V₁/2^γ. 2^1.4 = 2.63902, so W₂₃ = −242.888/2.63902 = −92.04 J.

Process 3–1: constant volume, so W₃₁ = 0.

Net work by air: W_net,by = W₁₂ + W₂₃ + W₃₁ = 147.02 − 92.04 = 54.98 J ≈ 55.0 J. Net work on air = −55.0 J. Thus 55.0 J is delivered by the air to the piston.

(b) Let the flow be steady, one-dimensional, incompressible, adiabatic, with no shaft work. Continuity: A₁V₁ = A₂V₂ = Q. If V₂ > V₁, then A₂ < A₁: the passage is convergent in the forward direction.

Energy equation (Bernoulli): P₁/ρ + V₁²/2 = P₂/ρ + V₂²/2. If V₂ > V₁, then P₂ < P₁: pressure decreases. This is the stated accelerating nozzle.

Can it decelerate? Yes, if the same duct is operated in the reverse direction. Then the fluid enters at the original outlet (area A₂) and leaves at the original inlet (area A₁). The effective inlet area is A₂ and outlet area is A₁, with A₁ > A₂. Continuity gives V_out/V_in = A_in/A_out = A₂/A₁ < 1, so V_out < V_in. Thus the fluid decelerates and the duct acts as a diffuser.

Energy equation then gives P_out − P_in = ρ(V_in² − V_out²)/2 > 0, so pressure rises downstream.

Momentum equation for the control volume: P₁A₁ − P₂A₂ + F_wall = ρQ(V₂ − V₁). When V₂ < V₁, the right side is negative. A divergent wall geometry and the downstream pressure rise provide the wall force required to satisfy momentum.

In differential form, Euler’s equation is ρV dV/dx = −dP/dx. Continuity gives dV/V = −dA/A, hence dV/dx = −(V/A)dA/dx. Therefore, dA/dx > 0 gives dV/dx < 0 and dP/dx > 0, i.e. deceleration in a divergent passage. So the nozzle cannot decelerate the fluid in the stated forward direction, but it can do so when reversed and used as a diffuser with A_out > A_in.

(c) Let x be measured from the insulated surface, so x = 0 is insulated and x = L is maintained at Tw. For one-dimensional steady conduction with uniform heat generation q̇:

d²T/dx² + q̇/k = 0.

Integrate once: dT/dx = −q̇x/k + C₁. At x = 0, the surface is perfectly insulated, so dT/dx = 0. Hence C₁ = 0.

Integrate again: T = −q̇x²/(2k) + C₂. At x = L, T = Tw: Tw = −q̇L²/(2k) + C₂ C₂ = Tw + q̇L²/(2k).

Thus T(x) = Tw + q̇/(2k)(L² − x²).

Maximum temperature occurs at the insulated surface x = 0: T_max = Tw + q̇L²/(2k). Minimum temperature occurs at the cooled surface x = L: T_min = Tw.

Profile:

T(x) ^ | T_max ● at x = 0 | ⋯ | ⋯ | ● Tw at x = L +-------------------------> x 0 L

The curve is parabolic and concave downward.

(d) For an axial flow reaction turbine with constant axial velocity and no losses, let station 1 be rotor inlet and station 2 rotor outlet. Blade speed is u. The inlet absolute velocity V₁ makes angle α with u, so its whirl component is Vw₁ = V₁ cos α. Let Vw₂ be the whirl component at outlet.

Work per unit mass: w = u(Vw₁ − Vw₂).

For constant axial velocity, the degree of reaction is R = 1 − (Vw₁ + Vw₂)/(2u).

Given R = 1/(k+1): 1/(k+1) = 1 − (Vw₁ + Vw₂)/(2u) => Vw₁ + Vw₂ = 2u k/(k+1) => Vw₂ = 2u k/(k+1) − Vw₁.

Then w = u[Vw₁ − (2u k/(k+1) − Vw₁)] = 2u Vw₁ − 2u² k/(k+1).

For maximum utilization, maximise w with respect to u for fixed V₁, α and R: dw/du = 2Vw₁ − 4u k/(k+1) = 0 => u = Vw₁ (k+1)/(2k).

Since Vw₁ = V₁ cos α, u/V₁ = (k+1)/(2k) cos α. At this condition Vw₂ = 0, i.e. the exit absolute velocity is axial.

(e) For laminar flow over a flat plate, the Prandtl number is Pr = ν/α, where ν is momentum diffusivity and α is thermal diffusivity. For laminar forced convection, δ_t/δ_h ≈ Pr^(−1/3), where δ_t is the thermal boundary-layer thickness and δ_h is the hydrodynamic boundary-layer thickness.

• If Pr < 1, then α > ν: heat diffuses faster than momentum, so δ_t > δ_h.
• If Pr = 1, then α = ν, so δ_t = δ_h.
• If Pr > 1, then ν > α: momentum diffuses faster than heat, so δ_t < δ_h.

(e)(i) Pr < 1 u, T∞ → → → → δ_t (thermal) ⋯⋯⋯ δ_h (hydrodynamic) ⋯⋯ plate ________________________________

(e)(ii) Pr = 1 u, T∞ → → → → δ_t = δ_h ⋯⋯⋯ plate ________________________________

(e)(iii) Pr > 1 u, T∞ → → → → δ_h (hydrodynamic) ⋯⋯⋯ δ_t (thermal) ⋯ plate ________________________________

The wall temperature Tw only determines the direction of heat flow; it does not change the relative thicknesses, which are controlled by Pr.