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ENGR 2213 Thermodynamics

ENGR 2213 Thermodynamics. F. C. Lai School of Aerospace and Mechanical Engineering University of Oklahoma. Energy Analysis for a Control Volume. Conservation of Mass. Total Mass Leaving CV. Total Mass Entering CV. Net Change in Mass within CV. -. =. Steady State. Example 1.

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ENGR 2213 Thermodynamics

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  1. ENGR 2213 Thermodynamics F. C. Lai School of Aerospace and Mechanical Engineering University of Oklahoma

  2. Energy Analysis for a Control Volume Conservation of Mass Total Mass Leaving CV Total Mass Entering CV Net Change in Mass within CV - = Steady State

  3. Example 1 Feedwater Heater: Inlet 1 T1 = 200 ºC, p1 = 700 kPa, Inlet 2 T2 = 40 ºC, p2 = 700 kPa, A2 = 25 cm2 Exit sat. liquid, p3 = 700 kPa, Find Inlet 2 Inlet 1 Exit

  4. Example 1 (continued) Steady State Inlet 2: compressed liquid Table A-4, v2 = 0.001008 m3/kg Exit: saturated liquid Table A-5, v3 = 0.001108 m3/kg

  5. Example 1 (continued) = 54.15 – 40 = 14.15 kg/s

  6. Energy Analysis for a Control Volume Flow work Energy that is necessary for maintaining a continuous flow through a control volume. A cross-sectional area p fluid pressure L width of fluid element F = pA = pAL = pV W = FL

  7. Energy Analysis for a Control Volume Energy carried by a fluid element in a closed system Energy carried by a fluid element in a control volume

  8. Energy Analysis for a Control Volume Conservation of Energy Total Energy Carried by Mass Entering CV Total Energy Carried by Mass Leaving CV Net Change in Energy of CV = - Total Energy Crossing Boundary as Heat and Work +

  9. Steady-Flow Process A process during which a fluid flows through a control volume steadily. ● No properties within the control volume change with time. ● No properties change at the boundaries of the control volume with time. ● The heat and work interactions between a steady- flow system and its surroundings do not change with time.

  10. Steady-Flow Process Conservation of mass Conservation of energy

  11. Steady-Flow Process For single-stream steady-flow process Conservation of mass Conservation of energy

  12. Steady-Flow Devices ● Nozzles and Diffusers A nozzle is used to accelerate the velocity of a fluid in the direction of flow while a diffuser is used to decelerate the flow. The cross-sectional area of a nozzle decreases in the direction of flow while it increases for a diffuser. For nozzles and diffusers,

  13. Example 2 Steam enters an insulated nozzle at a flow rate of 2 kg/s with Ti = 400 ºC, pi = 4 MPa, and It exits at pe = 1.5 MPa with a velocity of Find the cross-sectional area at the exit. Inlet Ti = 400 ºC pi = 4 MPa Exit pe = 1.5 MPa

  14. Example 2 (continued) Inlet, superheated vapor Table A-6, hi = 3213.6 kJ/kg = 2992.5 kJ/kg

  15. Example 2 (continued) Table A-6, he = 2992.5 kJ/kg 1.4 MPa 1.5 MPa 1.6 MPa 250 2927.2 2923.2 2919.2 300 3040.4 3037.6 3034.8 T = 280 ºC v = 0.1627 m3/kg = 0.000489 m2

  16. Steady-Flow Devices ● Turbines A turbine is a device from which work is produced as a result of the expansion of a gas or superheated steam through a set of blades attached to a shaft free to rotate. For turbines,

  17. Example 3 Steam enters a turbine at a flow rate of 4600 kg/h. At the inlet, Ti = 400 ºC, pi = 6 MPa, and At the exit, xe = 0.9, pe = 10 kPa and If the turbine produces a power of 1 MW, find the heat loss from the turbine. Inlet Ti = 400 ºC pi = 6 MPa Exit xe = 0.9 pe = 10 kPa

  18. Example 3 (continued) Inlet: superheated vapor at 6 MPa and 400 ºC Table A-6, hi = 3177.2 kJ/kg Exit: x = 0.9, saturated mixture at 10 kPa Table A-5, hf = 191.83 kJ/kg, hfg = 2392.8 kJ/kg he = hf + xehfg = 191.83 + 0.9 (2392.8) = 2345.4 kJ/kg

  19. Example 3 (continued) he - hi = 2345.4 – 3177.2 = - 831.8 kJ/kg = - 63.1 kW

  20. Steady-Flow Devices ● Compressors and Pumps Compressors and pumps are devices to which work is provided to raise the pressure of a fluid. Compressors → gases Pumps → liquids For compressors, For pumps,

  21. Example 4 Air enters a compressor. At the inlet, Ti = 290 K, pi = 100 kPa, and At the exit, Te = 450 K, pe = 700 kPa and If given that Ai = 0.1 m2 and heat loss at a rate of 3 kW, find the work required for the compressor. Inlet Ti = 290 K pi = 100 kPa Exit Te = 450 K pe = 700 kPa

  22. Example 4 (continued) = 0.72 kg/s Table A-17, at 290 K, hi = 290.16 kJ/kg, at 450 K, he = 451.8 kJ/kg.

  23. Example 4 (continued) = - 119.4 kW

  24. Example 5 A pump steadily draws water at a flow rate of 10 kg/s. At the inlet, Ti = 25 ºC, pi = 100 kPa, and At the exit, Te = 25 ºC, pe = 200 kPa and If the exit is located 50 m above the inlet, find the work required for the pump. Exit Te = 25 ºC pe = 200 kPa Inlet Ti = 25 ºC pi = 100 kPa

  25. Example 5 (continued) he – hi ~ [hf + vf (p – psat)]e - [hf + vf (p – psat)]i = vf (pe – pi) Table A-4, at 25 ºC vf = 0.001003 m3/kg = 0.001003 (200 – 100) = 0.1 kJ/kg g(ze – zi) = 9.8(50)/103 = 0.49 kJ/kg

  26. Example 5 (continued) = 20 (0.1 + 0.75 + 0.49) = 13.4 kW

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