(a) Considering an ideal, isentropic gas flow through a nozzle, show that choking will occur at Mach number (M) = 1. (20…

(a) Choking occurs at M = 1

For isentropic flow of an ideal gas: T₀/T = 1 + [(γ − 1)/2]M², p₀/p = [1 + (γ − 1)M²/2]^(γ/(γ−1)), and V = M√(γRT). The mass flow through area A is ṁ = ρAV = Ap M√(γ/(RT)). Substituting p and T from the stagnation relations, ṁ/A = G = p₀√(γ/(RT₀)) M [1 + (γ − 1)M²/2]^(−(γ+1)/(2(γ−1))).

For fixed p₀ and T₀, G depends only on M. Its maximum is where d(ln G)/dM = 0: 1/M − [(γ + 1)/2] M/[1 + (γ − 1)M²/2] = 0, which gives 1 + (γ − 1)M²/2 = (γ + 1)M²/2, so M² = 1 and M = 1. G rises with M for M < 1 and falls for M > 1, so the maximum is a true one.

Interpretation. The area needed to pass a given ṁ is A = ṁ/G, which is a minimum at M = 1. In a converging nozzle the smallest area is the exit (or throat). Lowering the back pressure raises the throat Mach number until M = 1; after that, lowering it further cannot raise the flow, because pressure disturbances travel at the speed of sound relative to the gas and cannot move upstream against a sonic stream. The throat is therefore choked at M = 1, and the flow is fixed at ṁ_max = 0.0404 p₀A*/√T₀ (for air).

(b) Heated tube: length, mid-tube temperature and pressure drop

D = 0.015 m, u = 1 m/s, T_i = 30°C, T_o = 65°C, T_s = 90°C. Re_D = uD/ν = 0.015/(0.497 × 10⁻⁶) = 3.02 × 10⁴ (turbulent). F = 0.079 Re^(−0.25) = 5.99 × 10⁻³. Nu_D = (F/2)(Re − 1000)Pr/[1 + 12.7(F/2)^½(Pr^(2/3) − 1)] = 152.5. h = Nu k/D = 152.5 × 0.656/0.015 = 6.67 × 10³ W/m²·K.

A_c = πD²/4 = 1.767 × 10⁻⁴ m², ṁ = ρuA_c = 984.4 × 1 × 1.767 × 10⁻⁴ = 0.1740 kg/s. Q = ṁC_p(T_o − T_i) = 0.1740 × 4178 × 35 = 2.54 × 10⁴ W. ΔT_lm = [(90 − 30) − (90 − 65)]/ln(60/25) = 35/0.8755 = 39.98°C. Q = hπDLΔT_lm gives L = 2.54 × 10⁴/(6.67 × 10³ × π × 0.015 × 39.98) = 2.02 m.

Mid-tube temperature. For constant wall temperature, (T_s − T)/(T_s − T_i) = exp(−hπDx/ṁC_p). At x = L the ratio is 25/60, so at x = L/2 it is √(25/60) = 0.6455, and T_mid = 90 − 60 × 0.6455 = 51.3°C.

Pressure drop. F is the Fanning friction factor, so Δp = 4F(L/D)(ρu²/2) = 4 × 5.99 × 10⁻³ × (2.02/0.015) × 984.4 × 1²/2 = 1.59 kPa.

(c) Final pressure after rapid adiabatic filling of a rigid tank

Take the tank as the control volume. For an initially evacuated tank filled adiabatically, with no work, the energy balance is m₂u₂ = m_in h_line. Since m₂ = m_in, u₂ = h_line. For constant specific heats, c_vT₂ = c_pT_line, so T₂ = γT_line = 1.4 × 298.15 = 417.4 K (about 144°C). The pressure at valve closing is 6 MPa.

After storage the tank returns to 298.15 K at constant volume and mass, so p₃ = p₂ × T₃/T₂ = 6 × 298.15/417.4 = 6/1.4 = 4.29 MPa.

The line pressure of 8 MPa and the volume of 150 L do not affect the result.