(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) Brake power from dynamometer law: P = W N / 20000 = 540 × 2500 / 20000 = 67.5 kW.

Swept volume per cylinder: V_s = (π/4) d² L = (π/4)(0.100 m)²(0.110 m) = 0.000275π m³.

Total swept volume rate for a four-stroke, 6-cylinder engine: V̇_s = i V_s N / (2 × 60) = 6 × 0.000275π × 2500 / 120 = 0.107992 m³/s.

Brake mean effective pressure: bmep = BP / V̇_s = 67.5 kW / 0.107992 m³/s = 625.04 kPa = 6.250 bar.

Fuel mass flow rate: ṁ_f = (100 × 10⁻⁶ m³ / 18 s) × 780 kg/m³ = 0.004333 kg/s = 15.6 kg/h.

Brake specific fuel consumption: bsfc = ṁ_f per hour / BP = 15.6 kg/h / 67.5 kW = 0.2311 kg/kWh = 231.1 g/kWh.

Brake thermal efficiency: Heat input = ṁ_f × CV = 0.004333 kg/s × 45000 kJ/kg = 195 kW. η_bt = BP / Heat input = 67.5 / 195 = 0.34615 = 34.62%.

Air density at ambient conditions: ρ_a = P / (R T) = 100 kPa / (0.287 kJ/kg-K × 300 K) = 1.16144 kg/m³.

Theoretical air mass flow if swept volume were filled at ambient conditions: ṁ_th = ρ_a V̇_s × 60 = 1.16144 × 0.107992 × 60 = 7.5256 kg/min.

Volumetric efficiency: η_v = ṁ_actual / ṁ_th = 5.301126 / 7.5256 = 0.7044 = 70.44%.

Stoichiometric air-fuel ratio for C = 0.83, H = 0.17: O₂ required = (8/3)(0.83) + 8(0.17) = 3.5733 kg O₂/kg fuel. Air required = 3.5733 / 0.232 = 15.402 kg air/kg fuel.

Actual air-fuel ratio: A/F = 5.301126 kg/min / 0.26 kg/min = 20.389 kg air/kg fuel.

Percentage excess air: = (20.389 − 15.402) / 15.402 × 100 = 32.38%.

(b)(i) For a reaction stage, the moving-blade inlet and outlet velocity triangles are drawn with U = blade speed, C₁ = absolute inlet velocity, Vr₁ = relative inlet velocity, C₂ = absolute outlet velocity, Vr₂ = relative outlet velocity. Whirl and flow components are Cw = C cos α, Cf = C sin α. Inlet triangle: U horizontal; C₁ at angle α to U; Vr₁ = C₁ − U. Outlet triangle: Vr₂ is relative outlet; C₂ = Vr₂ + U; C₂ at angle β to U. Sketch: `` Inlet: Outlet: C1 Vr2 /| /| Cf1/ |α / |β /__|____ U /__|____ U Vr1 C2 ``

(b)(ii) For a Parsons turbine, degree of reaction R = 1/2, so fixed and moving blades are identical and velocity triangles are symmetrical. Hence Vr₂ = C₁ and Vr₁ = C₂.

Let C₁ be the absolute velocity entering the moving blades, α the fixed-blade angle, and U the blade speed. Whirl at inlet: Cw₁ = C₁ cos α.

At outlet, the relative whirl of Vr₂ is C₁ cos α opposite to U, so absolute whirl: Cw₂ = U − C₁ cos α.

Work done per kg: W = U(Cw₁ − Cw₂) = U[C₁ cos α − (U − C₁ cos α)] W = U(2C₁ cos α − U). ...(1)

Outlet absolute velocity: C₂² = (U − C₁ cos α)² + (C₁ sin α)² C₂² = C₁² + U² − 2U C₁ cos α.

Energy supplied per kg: E = W + C₂²/2 = C₁²/2 + U C₁ cos α − U²/2. ...(2)

Efficiency: η = W/E = [U(2C₁ cos α − U)] / [C₁²/2 + U C₁ cos α − U²/2].

Put ρ = U/C₁. Then: η = [2ρ(2 cos α − ρ)] / [1 + 2ρ cos α − ρ²].

Differentiate with respect to ρ: dη/dρ = 0 ⇒ ρ = cos α.

Thus the optimum blade-to-steam speed ratio for a Parsons reaction turbine is ρ = cos α.

(b)(iii) Substitute ρ = cos α in the efficiency expression: η_max = [2 cos α(2 cos α − cos α)] / [1 + 2 cos² α − cos² α] η_max = 2 cos² α / (1 + cos² α).

Hence proved: η_maximum = (2 cos² α)/(1 + cos² α).

(b)(iv) At ρ_optimum = cos α, U = C₁ cos α. Therefore the inlet relative velocity Vr₁ is axial, of magnitude C₁ sin α. The outlet absolute velocity C₂ is also axial, of magnitude C₁ sin α. The outlet relative velocity Vr₂ = C₁ makes an angle α with the negative direction of U. Thus the velocity triangles are right triangles: Inlet: U horizontal; C₁ at angle α; Vr₁ vertical (axial). Outlet: U horizontal; Vr₂ at angle α to the negative U direction; C₂ vertical (axial).

(c) Various methods of air-conditioning duct design are:

• Velocity reduction method: Assumed air velocities are selected for successive duct sections. Duct area is found from A = Q/V. Friction losses are calculated and branches are balanced by dampers. It is simple but may require extra throttling.

• Equal friction method: A constant friction loss per unit length, usually in Pa/m, is chosen for the whole system or main duct. Duct sizes are selected so that each section has the same friction gradient. Total pressure drop is then found along the critical path. It is widely used for simple low-velocity systems.

• Static regain method: Ducts are sized so that the static pressure regained due to velocity reduction in one section balances the friction loss in the following section. It gives better balance and lower fan power in large systems, but calculations are more involved.

• Velocity pressure method: Both friction and dynamic losses are expressed in terms of velocity pressure. The duct is sized to meet a specified total pressure loss. It is useful for systems with many fittings and transitions.

• T-method: An optimization method in which duct sizes are selected to minimize total owning and operating cost, including fan energy. It requires iterative calculation and is used in large HVAC designs.

• Equivalent resistance method: Branch resistances are converted into equivalent resistances and ducts are sized to equalize pressure drops. It is convenient for balanced multi-branch layouts.

All methods are applied for standard air density, and dampers or balancing devices are finally used to trim the actual airflow.