(a) In a cooling tower used in a thermal power plant, 26,000 kg/s of air enters at DBT = 20°C and relative humidity at 20%. It leaves the cooling tower at 35°C DBT and 80% relative humidity.

I. Find the following :
(i) Total heat added to the air
(i

Question

(a) In a cooling tower used in a thermal power plant, 26,000 kg/s of air enters at DBT = 20°C and relative humidity at 20%. It leaves the cooling tower at 35°C DBT and 80% relative humidity.

I. Find the following :
(i) Total heat added to the air
(ii) Evaporation loss of water
(iii) WBT of the air at inlet and exit
(iv) Change in the volume flow rate of the air in the cooling tower

II. Explain the process on the Psychrometric chart.

Use Psychrometric chart attached at the end. (20 marks)

(b)

A steam power plant operates with a boiler output of 20 kg/s steam at 2 MPa and 600°C. The condenser operates at 50°C, dumping energy into a river that has an average temperature of 20°C. There is an open feed heater with extraction from the turbine at 600 kPa, at its exit is saturated liquid. Find the mass flow rate of the extracted flow (liquid). If the river water should not be heated more than 5°C, how much water should be pumped from the river to the heat exchanger (condenser)?

The steam properties at 2 MPa, 600°C are : h = 3690·14 kJ/kg s = 7·7023 kJ/kg K

At 600 kPa and for s = 7·7023 kJ/kg K, take h = 3270·25 kJ/kg.

Use the Steam Tables given at the end to get other properties. 20 marks

The steam properties at 2 MPa, 600°C are :
h = 3690·14 kJ/kg
s = 7·7023 kJ/kg K

At 600 kPa and for s = 7·7023 kJ/kg K, take h = 3270·25 kJ/kg.

Use the Steam Tables given at the end to get other properties. (20 marks)

(c) IC engine cooling is a complex issue. Discuss in brief various factors affecting the heat transfer from IC engines. (10 marks)

UPSC Answer Check · Accepted Answer

(a) I (i) Assume m_a = 26,000 kg/s dry air, P = 101.325 kPa. At inlet t1 = 20°C, φ1 = 20%: p_s1 = 2.339 kPa, so p_v1 = 0.20 × 2.339 = 0.4678 kPa. W1 = 0.622 p_v1/(P − p_v1) = 0.622 × 0.4678/(101.325 − 0.4678) = 0.002885 kg/kg dry air. At exit t2 = 35°C, φ2 = 80%: p_s2 = 5.628 kPa, p_v2 = 0.80 × 5.628 = 4.5024 kPa. W2 = 0.622 × 4.5024/(101.325 − 4.5024) = 0.028924 kg/kg dry air. h = 1.006t + W(2501 + 1.86t). h1 = 1.006×20 + 0.002885(2501 + 1.86×20) = 27.44 kJ/kg dry air. h2 = 1.006×35 + 0.028924(2501 + 1.86×35) = 109.43 kJ/kg dry air. Total heat added = m_a(h2 − h1) = 26,000(109.43 − 27.44) = 2.13×10^6 kW.

(a) I (ii) Evaporation loss = m_a(W2 − W1) = 26,000(0.028924 − 0.002885) = 677 kg/s.

(a) I (iii) Using the adiabatic-saturation equation W = [(2501 − 2.326t_wb)W_s(t_wb) − 1.006(t − t_wb)]/[2501 + 1.86t − 4.186t_wb], or reading the chart: inlet WBT ≈ 9.3°C; exit WBT ≈ 31.9°C.

(a) I (iv) Specific volume v = 0.287T/(P − p_v), T in K. v1 = 0.287×293.15/(101.325 − 0.4678) = 0.834 m^3/kg dry air. v2 = 0.287×308.15/(101.325 − 4.5024) = 0.913 m^3/kg dry air. Change = m_a(v2 − v1) = 26,000(0.913 − 0.834) = 2.06×10^3 m^3/s increase.

(a) II On the psychrometric chart, state 1 is at 20°C DBT and 20% RH; state 2 is at 35°C DBT and 80% RH. The process line slopes upward to the right: DBT, humidity ratio, enthalpy, wet-bulb temperature and specific volume all increase. Sensible heat transfer warms the air, while latent heat transfer from evaporation raises the moisture content. The line is not a constant wet-bulb line; it lies below the saturation curve. Heat added is the enthalpy difference read on the inclined enthalpy scale, and evaporation loss is the vertical W difference.

(b) At 2 MPa, 600°C: h6 = 3690.14 kJ/kg, s6 = 7.7023 kJ/kg K. At 600 kPa extraction: h_ext = 3270.25 kJ/kg. From steam tables: at 50°C, h_f = 209.33 kJ/kg, h_fg = 2382.7 kJ/kg, s_f = 0.7038 kJ/kg K, s_fg = 7.3715 kJ/kg K, v_f = 0.001012 m^3/kg. At 600 kPa, h_f = 670.56 kJ/kg.

Turbine exhaust entropy s = s6. At 50°C, x = (7.7023 − 0.7038)/7.3715 = 0.9494. h_ex = 209.33 + 0.9494×2382.7 = 2471.5 kJ/kg.

Condensate pump to 600 kPa: w_p = v_f(600 − 12.35) = 0.001012×587.65 = 0.595 kJ/kg. h_cond,in = 209.33 + 0.595 = 209.93 kJ/kg.

Open feedwater heater energy balance: m_ext h_ext + (20 − m_ext)h_cond,in = 20h_f600. Thus m_ext = 20(670.56 − 209.93)/(3270.25 − 209.93) = 20×460.63/3060.32 = 3.01 kg/s.

Condenser duty: remaining flow = 20 − 3.01 = 16.99 kg/s. Q_cond = 16.99(2471.5 − 209.33) = 3.84×10^4 kW. River water: Q_cond = m_w c_p ΔT, c_p = 4.18 kJ/kg K, ΔT = 5°C. m_w = 38,400/(4.18×5) = 1.84×10^3 kg/s.

(c) Factors affecting heat transfer from IC engines include:
Gas side: combustion temperature, flame speed, turbulence and swirl, air-fuel ratio, compression ratio, ignition timing, engine speed and load, residual gas fraction.
Geometry and materials: surface-area-to-volume ratio, cylinder bore/stroke, head/piston/liner design, wall thickness, thermal conductivity, fins and coolant passages.
Coolant side: coolant type (water, glycol, air), specific heat, thermal conductivity, viscosity, flow rate, pressure, coolant temperature, boiling, thermostat and radiator performance.
Deposits and fouling: soot, oil films, scale and corrosion add thermal resistance.
Lubrication and auxiliary: oil flow, piston cooling jets, oil cooler.
Ambient and operation: air temperature, humidity, altitude, transient/steady operation, EGR and turbocharging. These factors control convective and radiative heat transfer to walls, conduction through metal, and convection to the coolant.

(a) In a cooling tower used in a thermal power plant, 26,000 kg/s of air enters at DBT = 20°C and relative humidity at 20%. It…

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