(a) (i) During a heat transfer process, the entropy change of incompressible substances, such as water, can be determined from ΔS…

(a)(i) Entropy change of a thermal energy reservoir (10 marks)

For an incompressible substance of mass m and average specific heat C, taken from T₁ to T₂ with negligible work, ΔS = m C ln(T₂/T₁), and the heat absorbed is Q = m C (T₂ − T₁).

A thermal energy reservoir (a large lake, the atmosphere) is so large that its mass and heat capacity are very large. It can absorb or give up a finite amount of heat Q with only a tiny change in temperature, so T₂ = T₁ + ΔT with ΔT ≪ T₁. Write x = ΔT/T₁, which is very small. Expanding the logarithm as a series:

ln(T₂/T₁) = ln(1 + x) = x − x²/2 + x³/3 − … ≈ x = (T₂ − T₁)/T₁

Terms of order x² and above are negligible. Therefore

ΔS = m C (T₂ − T₁)/T₁ = Q/T₁

Since T₁ ≈ T₂ = T for the reservoir, this is

ΔS = Q/T

where T is the (constant) reservoir temperature and Q is the heat received by the reservoir (positive) or given up by it (negative). This is the entropy change for a reversible heat transfer at constant temperature, dS = δQ/T.

(a)(ii) Cooling of bread loaves (10 marks)

Data: mass flow of bread = 400 loaves/h × 0.4 kg = 160 kg/h, cooled from 25 °C to −10 °C, with c = 2.90 kJ/kg·°C and latent heat h_fg = 109.15 kJ/kg (freezing). Air enters at −25 °C, with c_p = 1.0 kJ/kg·°C, R = 0.287 kPa·m³/kg·K, and a permitted rise of 5 °C, at 101.3 kPa.

(1) Rate of heat removal from the bread. Each kg of bread gives up sensible heat and latent heat:

• Sensible heat = c ΔT = 2.90 × (25 − (−10)) = 2.90 × 35 = 101.5 kJ/kg.
• Latent heat = 109.15 kJ/kg.
• Total = 210.65 kJ/kg.

Q̇ = 160 × 210.65 = 33 704 kJ/h (about 9.36 kW).

(2) Volume flow rate of air. The heat removed from the bread is gained by the air, so ṁ_air c_p ΔT = Q̇:

ṁ_air = 33 704/(1.0 × 5) = 6740.8 kg/h

The specific volume of air at −25 °C (248.15 K) and 101.3 kPa is v = RT/P = 0.287 × 248.15/101.3 = 0.7031 m³/kg. Hence

V̇ = ṁ_air v = 6740.8 × 0.7031 = 4739 m³/h

(b)(i) Entropy change across a normal shock (10 marks)

For a perfect gas, the Gibbs relation T ds = dh − v dP with dh = c_p dT and v/T = R/P gives

ds = c_p dT/T − R dP/P

The stagnation state of a flow is reached by an isentropic deceleration, so the entropy of the stagnation state equals the entropy of the flow. Integrating between the stagnation states 0x (upstream of the shock) and 0y (downstream), with c_p constant:

s_y − s_x = s₀y − s₀x = c_p ln(T₀y/T₀x) − R ln(P₀y/P₀x)

The flow through the shock is adiabatic with no work, so the stagnation enthalpy is conserved, h₀x = h₀y. Since h₀ = c_p T₀, T₀y = T₀x, and the first term is zero. Therefore

ΔS = s_y − s_x = −R ln(P₀y/P₀x)

Since the shock is irreversible, ΔS > 0, which requires P₀y < P₀x, that is the stagnation pressure falls across the shock. Only this loss of stagnation pressure produces the entropy rise.

(b)(ii) Fanno flow and Rayleigh flow (10 marks)

Both are one-dimensional, steady flows of a perfect gas in a constant-area duct, and in both, the entropy is maximum at M = 1, where the flow chokes. They differ as follows.

• Physical effect: Fanno flow is adiabatic flow with wall friction. Rayleigh flow is frictionless flow with heat addition or removal.
• Conserved quantities: in Fanno flow, mass flux (ρV) and stagnation enthalpy (h₀, so T₀ is constant) are conserved, and momentum is not, because of friction. In Rayleigh flow, mass flux and momentum (the impulse function P + ρV²) are conserved, and stagnation enthalpy is not, since heat is exchanged.
• Curve on the h–s (T–s) plane: the Fanno line is the locus of states of constant G and h₀, and the Rayleigh line the locus of constant G and impulse. Both have a maximum-entropy point at M = 1, with the subsonic branch above it and the supersonic branch below.
• Direction of Mach number: in Fanno flow, friction always drives M towards 1, so subsonic flow accelerates and supersonic flow decelerates, and entropy rises. In Rayleigh flow, heating drives M towards 1 from either side (subsonic flow accelerates, supersonic decelerates), and cooling moves M away from 1.
• Stagnation properties: in Fanno flow T₀ is constant and P₀ falls. In Rayleigh flow T₀ rises with heating and falls with cooling, and P₀ falls with heating.
• Choking limit: Fanno flow reaches M = 1 at the maximum duct length L* (the choked length). Rayleigh flow reaches M = 1 when the heat added reaches its limit (thermal choking).
• Special feature: the Rayleigh line has its maximum temperature on the subsonic side at M = 1/√γ (0.845 for γ = 1.4). The two curves for the same G and h₀ intersect at the two states connected by a normal shock.

(c) Convective heat loss from a wall in a wind (10 marks)

Data: flow length L = 12 m, height 5 m, so the area is A = 60 m². V = 50 km/h = 13.89 m/s, T_∞ = 4 °C, T_s = 10 °C, k = 0.0248 W/m·°C, ν = 1.412 × 10⁻⁵ m²/s and Pr = 0.72.

Reynolds number.

Re_L = V L/ν = 13.89 × 12/(1.412 × 10⁻⁵) = 1.180 × 10⁷

This is above the transition value (5 × 10⁵), so the flow is turbulent over most of the length, and the combined laminar–turbulent correlation is appropriate.

Nusselt number.

Nu = (0.037 Re^0.8 − 871) Pr^0.33 = (0.037 × 4.546 × 10⁵ − 871) × 0.72^0.33

Nu = (16 820 − 871) × 0.8973 = 14 310

Heat transfer coefficient.

h = Nu k/L = 14 310 × 0.0248/12 = 29.6 W/m²·°C

Heat loss.

Q = h A (T_s − T_∞) = 29.57 × 60 × (10 − 4) = 10 650 W ≈ 10.6 kW