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

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

Example 1 Feedwater Heater: Inlet 1 T 1 = 200 ºC, p 1 = 700 kPa, Inlet 2 T 2 = 40 ºC, p 2 = 700 kPa, A 2 = 25 cm 2 Exit sat. liquid, p 3 = 700 kPa, Find Inlet 1 Inlet 2 Exit

Example 1 (continued) Steady State Inlet 2: compressed liquid Table A-4,v 2 = m 3 /kg Exit: saturated liquid Table A-5,v 3 = m 3 /kg

Example 1 (continued) = – 40 = kg/s

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 W = FL = pAL= pV

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

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

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.

Steady-Flow Process Conservation of mass Conservation of energy

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

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,

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

Example 2 (continued) Inlet, superheated vapor Table A-6, h i = kJ/kg = kJ/kg

Example 2 (continued) Table A-6, h e = kJ/kg 1.4 MPa 1.5 MPa 1.6 MPa T = 280 ºC v = m 3 /kg = m 2

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,

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

Example 3 (continued) Inlet: superheated vapor at 6 MPa and 400 ºC Table A-6, h i = kJ/kg Exit: x = 0.9, saturated mixture at 10 kPa Table A-5, h f = kJ/kg, h fg = kJ/kg h e = h f + x e h fg = (2392.8) = kJ/kg

Example 3 (continued) h e - h i = – = kJ/kg = kW

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

Example 4 Air enters a compressor. At the inlet, T i = 290 K, p i = 100 kPa, and If given that A i = 0.1 m 2 and heat loss at a rate of 3 kW, find the work required for the compressor. Inlet T i = 290 K p i = 100 kPa Exit T e = 450 K p e = 700 kPa At the exit, T e = 450 K, p e = 700 kPa and

Example 4 (continued) Table A-17, at 290 K, h i = kJ/kg, at 450 K, h e = kJ/kg. = 0.72 kg/s

Example 4 (continued) = kW

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

Example 5 (continued) Table A-4, at 25 ºC v f = m 3 /kg h e – h i ~ [h f + v f (p – p sat )] e - [h f + v f (p – p sat )] i = v f (p e – p i ) = (200 – 100) = 0.1 kJ/kg g(z e – z i ) = 9.8(50)/10 3 = 0.49 kJ/kg

Example 5 (continued) = 20 ( ) = 13.4 kW