The document contains 10 examples solving thermodynamics problems involving concepts like the first law of thermodynamics, ideal gas law, steady state conditions, heat transfer, work done, efficiency and refrigeration cycles. The last example involves a reversible heat engine and refrigerator operating between different temperature reservoirs. It is determined that the heat rejection to the 40°C reservoir is 5539 kJ.

1. Revision of Thermodynamic Concepts Examples ME0223 SEM-IV Applied Thermodynamics & Heat Engines Applied Thermodynamics & Heat Engines S.Y. B. Tech. ME0223 SEM - IV Production Engineering

2. ME0223 SEM-IV Applied Thermodynamics & Heat Engines Example 1 The e.m.f. in a thermocouple with the test junction at t ⁰ C on gas thermometer scale and reference junction at Ice Point is given by; ε = 0.20 t – 5 X 10 -4 t 2 mV The millivoltmeter is calibrated at Ice and Steam Points. What will this thermocouple read when the gas thermometer reads 50 ⁰ C ? At Ice Point, when t = 0 ⁰ C, ε = 0 mV. At Steam point, when t = 100 ⁰ C, ε = 0.20 X 100 – 5 X 10 -4 X (100) 2 = 15 mV Thus, Δ T = 100 ⁰ C -> Δ ε = 15 mV. At t = 50 ⁰ C, ε = 0.20 X 50 – 5 X 10 -4 X (50) 2 = 8.75 mV Hence, when gas thermometer reads 50 ⁰ C (corresponding to 8.75 mV); the thermocouple will read; 100 15 X 8.75 = 58.33 ⁰ C …ANS

3. Example 2 A water in a tank is pressurised by air and a pressure is measured by a multifluid manometer. The tank is located on a mountain at 1400 m altitude, where the atmospheric pressure is 85.6 kPa. Determine the air pressure in the tank if h 1 = 0.1 m, h 2 = 0.2 m and h 3 = 0.35 m. Take the densities of water, oil and Mercury to be 1000 kg/m 3 , 850 kg/m3 and 13600 kg/m3 respectively. P 1 + h 1 ρ water g + h 2 ρ oil g - h 3 ρ mercury g = Patm P 1 = Patm + g( h 3 ρ mercury - h 1 ρ water - h 2 ρ oil ) = 130 kPa …ANS P 1 = Patm - h 1 ρ water g - h 2 ρ oil g + h 3 ρ mercury g = (85.6)+(9.81)[(13600*0.35)-(1000*0.1)-(850*0.2)] kPa Kg/m 3 m/sec 2 m Kg/m 3 m Kg/m 3 m Oil Air Water h 1 h 2 h 3 1 2 Mercury

4. Example 3 A rigid tank contains a hot fluid that is cooled while stirred by a paddle wheel. Initially, the internal energy of the fluid is 800 kJ. During the cooling process, the fluid loses 500 kJ of heat, and the paddle wheel does 100 kJ of work on the fluid. Determine the final internal energy of the fluid. Tank is stationary -> ∆KE = ∆PE = 0 Thus, ∆E = ∆U Applying the First Law of Thermodynamics (i.e. Energy Balance ); E in – E out = ∆E system W sh,in – Q out = ∆U = U 2 – U 1 U 2 = 400 kJ ….ANS 500 kJ U 1 = 800 kJ U 2 = ? W sh,in = 100 kJ 100 – 500 = U 2 - 800 kJ kJ kJ kJ

5. Example 4 A room is initially at the outdoor temperature of 25 ºC. A large fan that consumes 200 W of electricity is turned on. The Heat transfer rate between the room and the outdoor air is given by Q = UA(T i – T o ) where U = 6 W/m 2 .ºC overall Heat transfer coefficient. A= 30 m 2 exposed surface area of the room. T i – T o are the indoor and outdoor temperatures. Determine the indoor air temperature when steady conditions are established. The electricity consumed by a fan is the Energy input to the room, i.e. 200 W. This increases room air temperature. However, at steady state, rate of Heat transfer from the room = Heat input to the room. Applying the First Law of Thermodynamics (i.e. Energy Balance ); W elect,in = Q out = UA(T i – T o ) E in – E out = dE system / dt = 0 Steady State E in – E out 200 (W) = (6 W/m 2 ºC ).(30 m 2 ).(T i – T o ) T i = 26.1 ºC …ANS

6. ME0223 SEM-IV Applied Thermodynamics & Heat Engines Example 5 A frictionless piston-cylinder device contains 5 kg of steam at 400 kPa and 200 ⁰C. Heat is transferred to the steam until the temperature reaches 250 ⁰C. If the piston is not attached to the shaft, and its mass is constant, determine the Work done by the steam during the Process . Use the data given in FIG. for Sp. Volume. Constant Pressure, Quasi-Static Process. v 2 = 0.59520 m 3 /kg W 1-2 = mP(v 1 – v 2 ) …..V = mv W 1-2 = 121.7 kJ …ANS Pressure (P) Volume (V) P=Const v 1 = 0.53434 m 3 /kg W 1-2 = (5)(400)[(0.59520 – 0.53434) kg kPa m 3 /kg

7. ME0223 SEM-IV Applied Thermodynamics & Heat Engines Example 6 A piston-cylinder device initially contains 0.4 m 3 of air at 100 kPa and 80 ⁰C. The air is now compressed to 0.1 m 3 in such a way that the temperature remains constant. Determine the Work done during this Process. Isothermal, Quasi-Static Process. W 1-2 = -55.5 kJ …ANS Negative sign indicates that the Work is done ON the system. Pressure T 0 = 80 ⁰C = Const. Volume State 2 State 1 P 1 P 2 V 2 V 1 W 1-2 = (100)(0.4) ln 0.4 0.1 kPa m 3

8. ME0223 SEM-IV Applied Thermodynamics & Heat Engines Example 7 An insulated rigid tank initially contains 0.7 kg of Helium at 27 ⁰C and 350 kPa. A paddle wheel with a power rating of 0.015 kW is operated within the tank for 30 min. Determine (a) the final temperature, and (b) the final pressure of the Helium gas. Assume C v = 3.1156 kJ/kg. ⁰C Tank is stationary -> ∆KE = ∆PE = 0 Thus, ∆E = ∆U Applying the Energy Balance ; E in – E out = ∆E system W sh = ∆U = U 2 – U 1 = mC v (T 2 - T 1 ) W sh = (0.015 kW) (30 min) = 27 kJ (a) 27 kJ = (0.7 kg)(3.1156 kJ/kg. ⁰C)(T 2 - 27 ⁰C) T 2 = 39.4 ⁰C ….ANS Applying the Ideal Gas Law ; P2 = 364.5 kPa …ANS He gas m = 0.7 kg T 1 = 27 ⁰C P 1 = 350 kPa W sh 350 kPa (27+273) K = P 2 (39.4+273) K (b)

9. ME0223 SEM-IV Applied Thermodynamics & Heat Engines Example 8 Heat is transferred to a Heat Engine from a furnace at the rate of 80 MW. If the rate of Waste Heat Rejection to a nearby river is 50 MW, determine the Net Power and the Thermal Efficiency of the Heat Engine. Q H = 80 MW and Q L = 50 MW. Net power of the Heat Engine is; W net = Q H – Q L = (80 – 50) MW = 30 MW ….ANS Thermal Efficiency of the Heat Engine is; = 0.375 or 37.5 % ....ANS Furnace W net Q H = 80 kW Q L = 50 MW Heat Engine River η th = Wnet,out Q H 30 MW 80 MW =

10. ME0223 SEM-IV Applied Thermodynamics & Heat Engines Example 9 Heat Loss = 80,000 kJ/hr A Heat Pump is used to used to meet the heating requirements of a house and maintained at 20 ⁰C on a day when the outdoor temperature drops to -2 ⁰C, the house is estimated to lose Heat at the rate of 80,000 kJ/hr. If the Heat Pump is having COP of 2.5, determine (a) Power consumed by the Heat Pump (b) Rate at which Heat is absorbed from the cold outdoor air. = 32,000 kJ/hr = 8.9 kW …ANS The house is to be maintained at 20 ⁰C. Hence the Heat Pump has to deliver the SAME Heat to the house, i.e. 80,000 kJ/hr. House, 20 ⁰C W net = ? Q H = ? Q L = ? Heat Pump COP = 2.5 Outdoor Air, -2 ⁰C (a) W net,in = Q H COP HP 80,000 kJ/hr 2.5 = (b) Q L = Q H – W net,in = (80,000 – 32,000) kJ/hr = 48,000 kJ/hr ….ANS

11. ME0223 SEM-IV Applied Thermodynamics & Heat Engines A reversible Heat Engine operates between two reservoirs at temperatures of 600 ⁰C and 40 ⁰C. the Engine drives a reversible Refrigerator which operates between reservoirs of 40 ⁰C and -20 ⁰C. Heat transfer to the Engine is 2000 kJ and the net Work output of the combined Engine is 360 kJ. Evaluate the Heat transfer to the refrigerant and the net Heat Transfer to the reservoir at 40 ⁰C. Example 10 T 1 = 873 K W 1 Q 1 = 2000 kJ Q 2 = ? Heat Engine Refrigerator T 2 = 313 K Q 4 T 3 = 253 K W 2 W = 360 kJ Q 3 = Q 4 + W 2 η max = 1 - T 2 = 1 - T 1 313 873 = 0.642 W 1 Q 1 = 0.642 W 1 = 0.642 X 2000 = 1284 kJ COP max = T 3 T 2 – T 3 253 313 - 253 = = 4.22 COP max = Q 4 W 2 = 4.22

12. ME0223 SEM-IV Applied Thermodynamics & Heat Engines W 1 – W 2 = W = 360 kJ Example 10…contd W 2 = W 1 - W = 1284 - 360 = 924 kJ Q 4 = COP X W 2 = 4.22 X 924 = 3899 kJ Q 3 = Q 4 + W 2 = 3899 + 924 = 4823 kJ Q 2 = Q 1 – W 1 = 2000 - 1284 = 716 kJ T 1 = 873 K W 1 Q 1 = 2000 kJ Q 2 = ? Heat Engine Refrigerator T 2 = 313 K Q 4 T 3 = 253 K W 2 W = 360 kJ Q 3 = Q 4 + W 2 Thus, Heat Rejection to the 40 ⁰C reservoir = Q 2 + Q 3 = 716 + 4823 = 5539 kJ ….ANS