(a) (i) What does the NTU of a heat exchanger represent? Is a heat exchanger with a very large NTU necessarily a good one? (10…

(a)(i) Meaning of NTU, and whether a very large NTU is good (10 marks)

What NTU represents. The number of transfer units, NTU = UA/C_min, is the ratio of the heat exchanger's thermal conductance (overall coefficient times area) to the smaller heat capacity rate of the two fluids. It is a dimensionless measure of the heat transfer size of the exchanger: how much heat the exchanger can move per kelvin of temperature difference compared with how much the weaker stream can absorb per kelvin. It also equals the temperature change of the C_min fluid divided by the mean temperature difference between the fluids, so a large NTU means a small mean driving temperature difference is enough to give a large temperature change.

Is a very large NTU good? No, not necessarily.

• Effectiveness rises with NTU but flattens. For a counterflow exchanger with Cᵣ = 1, ε = NTU/(1 + NTU), so ε = 0.5 at NTU = 1, 0.75 at NTU = 3 and only 0.91 at NTU = 10. Beyond about NTU = 3 the extra area buys very little.
• The gain has a ceiling. For a parallel-flow exchanger, ε cannot exceed 1/(1 + Cᵣ), whatever the NTU, and for Cᵣ = 1 ε reaches 1 only as NTU → ∞ even in counterflow.
• A large NTU needs a large area or a long, large exchanger. That raises capital cost, weight, space, fouling area and the tube-side pressure drop, and therefore pumping power.
• So the design is a compromise between effectiveness and cost, and an economically sound design usually has an NTU of about 1–3.

(a)(ii) Effectiveness and flow configuration of the oil-to-oil exchanger (10 marks)

Data: cold oil 18 °C to 50 °C, hot oil 75 °C to 40 °C, equal mass flow rates.

Flow configuration. The cold fluid leaves at 50 °C, which is higher than the hot fluid outlet of 40 °C. In a parallel-flow exchanger both fluids approach the same final temperature from opposite sides, so the cold outlet can never exceed the hot outlet. Therefore the exchanger must be counterflow.

Effectiveness. The temperature changes are ΔTₕ = 75 − 40 = 35 °C and ΔT_c = 50 − 18 = 32 °C. With no heat loss, Cₕ ΔTₕ = C_c ΔT_c, so

Cₕ/C_c = 32/35 = 0.914

The hot stream has the smaller capacity rate, so C_min = Cₕ and Cᵣ = 0.914. The maximum possible heat transfer is C_min (Tₕ,in − T_c,in) = Cₕ (75 − 18) = 57 Cₕ. Then

ε = Q/Q_max = Cₕ (35)/(Cₕ × 57) = 35/57 = 0.614 (about 61%)

Check. The greatest effectiveness a parallel-flow exchanger can give with Cᵣ = 0.914 is 1/(1 + Cᵣ) = 0.52, which is lower than 0.614, and this confirms that the exchanger is counterflow.

(b) Charging of an insulated cylinder with helium (20 marks)

Data: A = 0.5 m², F_ext = 500 kN, V₁ = 0.25 m³, initial state 200 kPa and 100 °C (373.15 K), supply line at 1.2 MPa and 200 °C (473.15 K), T₀ = 300 K, Cₚ = 5.193, Cᵥ = 3.116 and R = 2.0771 kJ/kg·K.

Final pressure. The external force balances a gas pressure of P₂ = F_ext/A = 500/0.5 = 1000 kPa. The initial pressure (200 kPa) is below this, so the piston stays against its stops and the volume remains 0.25 m³ throughout, with no boundary work. The valve is closed when P₂ reaches 1000 kPa.

(i) Final mass. The initial mass is m₁ = P₁V/(RT₁) = 200 × 0.25/(2.0771 × 373.15) = 0.06451 kg.

The system is insulated with no work, so the unsteady energy balance for the filling process is

m₂u₂ − m₁u₁ = (m₂ − m₁) h_in, that is Cᵥ (m₂T₂ − m₁T₁) = (m₂ − m₁) Cₚ T_in

From the gas law at constant volume, m₂T₂ = P₂V/R = 1000 × 0.25/2.0771 = 120.36 kg·K, and m₁T₁ = P₁V/R = 24.07 kg·K. Then

3.116 (120.36 − 24.07) = (m₂ − m₁) × 5.193 × 473.15

300.03 = 2457.0 (m₂ − m₁), so m₂ − m₁ = 0.1221 kg

m₂ = 0.06451 + 0.1221 = 0.1866 kg

(ii) Final temperature.

T₂ = P₂V/(m₂R) = 120.36/0.18662 = 644.9 K

(iii) Entropy generation. The cylinder is insulated, so the entropy generated is the entropy increase of the contents less the entropy carried in by the mass:

S_gen = m₂s₂ − m₁s₁ − (m₂ − m₁) s_in = m₁ (s₂ − s₁) + (m₂ − m₁)(s₂ − s_in)

• s₂ − s₁ = Cₚ ln(T₂/T₁) − R ln(P₂/P₁) = 5.193 ln(644.9/373.15) − 2.0771 ln(1000/200) = 2.8358 − 3.3430 = −0.5014 kJ/kg·K.
• s₂ − s_in = Cₚ ln(T₂/T_in) − R ln(P₂/P_in) = 5.193 ln(644.9/473.15) − 2.0771 ln(1000/1200) = 1.6108 + 0.3787 = 1.9872 kJ/kg·K.

S_gen = 0.06451 × (−0.5014) + 0.1221 × 1.9872 = −0.03235 + 0.24267 = 0.2103 kJ/K

(iv) Irreversibility.

I = T₀ S_gen = 300 × 0.2103 = 63.1 kJ

The entropy generation is positive, as required, and the irreversibility comes from throttling of the helium into the cylinder and mixing of the incoming gas with the gas already inside.

(c) Radiation shape factor and exchange between black surfaces (10 marks)

Definition. The radiation shape factor (view or configuration factor) F₁₂ is the fraction of the radiation leaving surface 1 in all directions that falls directly on surface 2. It depends only on the geometry (sizes, orientations and separation) and not on temperature. For two diffuse surfaces at distance r, with θ₁ and θ₂ the angles between the line joining the elements and the normals to dA₁ and dA₂,

F₁₂ = (1/A₁) ∫_A₁ ∫_A₂ (cos θ₁ cos θ₂/(π r²)) dA₂ dA₁

Net exchange at T₁ and T₂. A black surface emits E_b = σT⁴ and absorbs all incident radiation.

• Energy leaving 1 and reaching 2: Q₁₂ = A₁ F₁₂ σ T₁⁴.
• Energy leaving 2 and reaching 1: Q₂₁ = A₂ F₂₁ σ T₂⁴.

Net exchange from 1 to 2:

Q_net = A₁ F₁₂ σ T₁⁴ − A₂ F₂₁ σ T₂⁴

Reciprocity. If both surfaces are at the same temperature T₁ = T₂ = T, there is no net exchange, Q_net = 0, so

A₁ F₁₂ σ T⁴ = A₂ F₂₁ σ T⁴, that is A₁ F₁₂ = A₂ F₂₁

The shape factors are purely geometric, so this reciprocity relation holds for any temperatures, and the same result follows from the double integral above, which is symmetric in the two surfaces. Using it, the general exchange becomes

Q_net = A₁ F₁₂ σ (T₁⁴ − T₂⁴) = A₂ F₂₁ σ (T₁⁴ − T₂⁴)