(a) A shell and tube heat exchanger used in a thermal power plant is designed to condense 500 kg/s of saturated steam entering…

(a) Heat rejected by condensing steam: Q = m_s h_fg = 500 kg/s × 2600 kJ/kg = 1.3×10⁶ kJ/s = 1.3×10⁹ W.

(i) For cooling water, Q = m_w Cₚ ΔT_w. ΔT_w = 45°C − 35°C = 10 K. m_w = Q/(Cₚ ΔT_w) = 1.3×10⁹/(4200 × 10) = 30,952.38 kg/s. Volume flow rate = m_w/ρ = 30,952.38/1000 = 30.952 m³/s. Final: m_w ≈ 3.095×10⁴ kg/s (30.952 m³/s).

(ii) Tube cross-sectional area: A₁ = πD²/4 = π(0.05 m)²/4 = π/1600 = 1.9635×10⁻³ m². N = Volume flow rate/(A₁ × velocity) = 30.952/(1.9635×10⁻³ × 1.0) = 15,763.9. Use whole tubes, so Final: N = 15,764 tubes.

(iii) For a condenser, steam temperature is constant at 60°C. ΔT₁ = 60 − 35 = 25 K, ΔT₂ = 60 − 45 = 15 K. LMTD = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = 10/ln(25/15) = 10/ln(5/3) = 19.576 K. Heat-transfer area: A = Q/(U × LMTD) = 1.3×10⁹/(1500 × 19.576) = 44,271.55 m². Area per tube = πD L. L = A/(NπD) = 44,271.55/(15,764 × π × 0.05) = 17.879 m. Final: length of each tube ≈ 17.88 m.

(iv) Total tube length = N L = 15,764 × 17.879 = 281,841 m. Alternatively, total length = A/(πD) = 44,271.55/(π × 0.05) = 281,841 m. Final: total tube length ≈ 2.818×10⁵ m.

(b) A centrifugal pump is a rotodynamic pump. It uses centrifugal force produced by a rotating impeller to lift and move liquid. Its flow is continuous and steady, and it has no suction or delivery valves. It is suitable for large discharges at low to medium heads. Its discharge falls as head increases and can be regulated by throttling the delivery valve. It normally requires priming, runs at high speed, is compact, and has low maintenance. It is not suitable for very viscous liquids.

A reciprocating pump is a positive-displacement pump. It uses a piston or plunger moving to and fro inside a cylinder. Liquid is admitted through a suction valve and forced out through a delivery valve. Its flow is pulsating, so an air vessel is often used to make delivery smoother. It is suitable for low to medium discharges at very high heads. Its discharge is almost independent of head, except for slip. It is self-priming in many designs, runs at low speed, is bulky, has more wearing parts and valves, and needs higher maintenance. It can handle viscous liquids and must be started with the delivery valve open. Flow is regulated by speed or bypass, not by throttling.

Slip in a reciprocating pump: Theoretical discharge for single acting pump is Q_th = A L N/60, where A is piston area, L stroke, N rpm. Actual discharge Q_act is less than Q_th due to leakage past piston and valves, air ingress, valve lag, incomplete filling, and compressibility. Slip = Q_th − Q_act. Percentage slip = [(Q_th − Q_act)/Q_th] × 100. Normally slip is positive because Q_act < Q_th.

Yes, slip can be negative. It occurs when Q_act > Q_th. This can happen when the suction pipe is long and the pump runs at high speed. The inertia of water in the long suction pipe keeps the suction valve open even after the suction stroke ends, so extra water enters during the beginning of the delivery stroke and is discharged. Thus actual discharge becomes greater than theoretical swept-volume discharge, giving negative slip.

(c) For air, take γ = 1.4, Cₚ = 1.005 kJ/kg·K, R = 0.287 kJ/kg·K. T₁ = 25 + 273.15 = 298.15 K, T₀ = 15 + 273.15 = 288.15 K.

Isentropic compression: T₂ₛ = T₁(P₂/P₁)^((γ−1)/γ) = 298.15(10)^(0.4/1.4) = 298.15(10)^(2/7) = 575.64 K.

First-law efficiency: η_I = (T₂ₛ − T₁)/(T₂ − T₁) = 0.90. T₂ = T₁ + (T₂ₛ − T₁)/0.90 = 298.15 + (575.64 − 298.15)/0.90 = 606.47 K.

Actual work per kg: w = Cₚ(T₂ − T₁) = 1.005(606.47 − 298.15) = 309.86 kJ/kg.

For adiabatic steady flow, entropy generation per kg: s_gen = s₂ − s₁ = Cₚ ln(T₂/T₁) − R ln(P₂/P₁) = 1.005 ln(606.47/298.15) − 0.287 ln 10 = 1.005(0.71006) − 0.66084 = 0.05277 kJ/kg·K.

Irreversibility per kg: i = T₀ s_gen = 288.15 × 0.05277 = 15.20 kJ/kg.

Second-law efficiency: η_II = (w − i)/w = (309.86 − 15.20)/309.86 = 0.9509 = 95.1%.

Final: irreversibility = 15.20 kJ/kg; second-law efficiency = 95.1%.