(a) The power output of a six-cylinder, four-stroke CI engine is absorbed by a hydraulic dynamometer for which the law is P =…

(a) Performance of a six-cylinder four-stroke CI engine

Brake power: P = WN/20000 = 540 × 2500/20000 = 67.5 kW. Swept volume per cylinder = (π/4)(0.100)²(0.110) = 8.639 × 10⁻⁴ m³. Swept volume rate for 6 cylinders, four-stroke: V̇_s = 6 × 8.639 × 10⁻⁴ × 2500/120 = 0.10799 m³/s.

bmep = P/V̇_s = 67.5/0.10799 = 625 kPa = 6.25 bar. Fuel flow: ṁ_f = (100 × 10⁻⁶/18) × 780 = 4.333 × 10⁻³ kg/s = 15.6 kg/h = 0.26 kg/min. bsfc = 15.6/67.5 = 0.231 kg/kWh (231 g/kWh). Heat input = 4.333 × 10⁻³ × 45 000 = 195 kW, so η_bth = 67.5/195 = 34.6%.

Volumetric efficiency: ρ_a = p/RT = 100/(0.287 × 300) = 1.1614 kg/m³. The air mass that would fill the swept volume is 1.1614 × 0.10799 × 60 = 7.526 kg/min, so η_v = 5.301/7.526 = 70.4%.

Excess air: O₂ needed = (8/3)(0.83) + 8(0.17) = 3.573 kg/kg fuel, so the stoichiometric air is 3.573/0.232 = 15.40 kg/kg fuel. Actual A/F = 5.301/0.26 = 20.39. Excess air = (20.39 − 15.40)/15.40 = 32.4%.

(b) Parsons reaction turbine

(i) Velocity triangles of a stage. The fixed blades give the steam an absolute velocity C₁ at angle α to the blade motion. At the moving-blade inlet, U is the blade speed, C₁ the absolute velocity, V_r1 the relative velocity (C₁ minus U), with whirl C_w1 = C₁ cosα and flow C_f1 = C₁ sinα. At the outlet, V_r2 is the relative velocity leaving at angle β, and the absolute velocity is C₂ = V_r2 + U, with whirl C_w2 and flow C_f2.

(ii) Optimum speed ratio. A Parsons turbine has 50% reaction with identical fixed and moving blades, so the triangles are symmetrical: V_r2 = C₁ and V_r1 = C₂, and the angles are equal (β = α). Whirl at inlet: C_w1 = C₁ cosα. At outlet the relative velocity C₁ has a whirl component C₁ cosα opposing U, so the absolute whirl (in the direction of U) is C_w2 = U − C₁ cosα. Work per kg: W = U(C_w1 − C_w2) = U(2C₁ cosα − U). Exit velocity: C₂² = (U − C₁ cosα)² + (C₁ sinα)² = C₁² + U² − 2UC₁ cosα. Energy supplied = work + exit kinetic energy loss = W + C₂²/2 = C₁²/2 + UC₁ cosα − U²/2. With ρ = U/C₁, the efficiency is η = 2ρ(2cosα − ρ)/(1 + 2ρ cosα − ρ²). Differentiating, dη/dρ = −4(ρ − cosα)/(ρ² − 2ρ cosα − 1)², which is zero at ρ = cosα. The sign change from positive to negative shows a maximum.

(iii) Substituting ρ = cosα: η_max = 2cosα(2cosα − cosα)/(1 + 2cos²α − cos²α) = 2cos²α/(1 + cos²α).

(iv) Triangles at ρ = cosα. Now U = C₁ cosα, so at inlet the relative whirl C₁ cosα − U is zero and V_r1 is purely axial, equal to C₁ sinα. At outlet the absolute whirl U − C₁ cosα is zero and C₂ is purely axial, also C₁ sinα. Both triangles are right-angled. The inlet triangle has U along the base, C₁ at angle α to it, and V_r1 vertical. The outlet triangle has V_r2 = C₁ at angle α to the negative U direction, and C₂ vertical.

(c) Methods of air-conditioning duct design

Velocity reduction method: velocities are assumed for each section and reduced progressively along the run, then area = Q/V. It is simple, but the branches need dampers to balance. Equal friction method: one friction loss per unit length (about 0.8 to 1 Pa/m) is chosen for the whole system and each duct is sized to it. It is the most popular method for low-velocity supply ducts, though branches still need balancing. Static regain method: each section is sized so that the static pressure regained by the fall in velocity equals the friction loss of the next section. This balances the system, lowers fan power and suits high-velocity ducts, though the calculation is longer. Velocity pressure method: all fittings are treated by their loss coefficients in terms of velocity pressure, and the duct is sized for the required total pressure loss. T-method: duct sizes are optimised to minimise the total of initial and operating cost, including fan energy. It needs iteration and is used for large systems.

In all methods dampers are finally used to trim the airflow in each branch.