ENGR 2213 Thermodynamics

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ENGR 2213 Thermodynamics

• F. C. Lai
• School of Aerospace and Mechanical
• Engineering
• University of Oklahoma

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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
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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
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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
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Example 1 (continued)
54.15 40 14.15 kg/s
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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
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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
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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

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Total Energy Crossing Boundary as Heat and Work

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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.
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Steady-Flow Process
Conservation of mass
Conservation of energy
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Steady-Flow Process
For single-stream steady-flow process
Conservation of mass
Conservation of energy
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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,
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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
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Example 2 (continued)
Inlet, superheated vapor Table A-6, hi 3213.6
kJ/kg
2992.5 kJ/kg
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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
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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,
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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
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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
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Example 3 (continued)
he - hi 2345.4 3177.2
- 831.8 kJ/kg
- 63.1 kW
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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,
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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
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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.
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Example 4 (continued)
- 119.4 kW
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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
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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
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Example 5 (continued)
20 (0.1 0.75 0.49)
13.4 kW