ENGR 2213 Thermodynamics F. C. Lai School of Aerospace and Mechanical Engineering University of...
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Transcript of ENGR 2213 Thermodynamics F. C. Lai School of Aerospace and Mechanical Engineering University of...
ENGR 2213 ThermodynamicsENGR 2213 Thermodynamics
F. C. Lai
School of Aerospace and Mechanical
Engineering
University of Oklahoma
Energy Analysis for a Control VolumeEnergy Analysis for a Control Volume
CV i em m m CV
i edm
m mdt
AVm AV
v
Conservation of Mass
Net Change in Mass within CV
Total Mass Entering CV
Total Mass Leaving CV
= -
Steady State i em m
Example 1Example 1
1m 40 kg/ s
33(AV) 0.06 m / s
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 2 3 2m ?, m ? and V ?
Inlet 1 Inlet 2
Exit
Example 1 (continued)Example 1 (continued)
i em m 1 2 3m m m
AVm AV
v
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
33
3
(AV) 0.06m 54.15 kg/ s
v 0.001108
Example 1 (continued)Example 1 (continued)
2 3 1m m m
2 22
2
m v (14.15)(0.001008)V 5.7 m/ s
A 0.0025
= 54.15 – 40 = 14.15 kg/s
Energy Analysis for a Control VolumeEnergy Analysis for a Control Volume
Flow workEnergy that is necessary for maintaining a continuousflow through a control volume.
A cross-sectional areap fluid pressureL width of fluid element
F = pA
W = FL = pAL = pV
Energy Analysis for a Control VolumeEnergy Analysis for a Control Volume
2Ve u pv gz
2
2Ve u gz
2
Energy carried by a fluid element in a closed system
Energy carried by a fluid element in a control volume2V
h gz2
Energy Analysis for a Control VolumeEnergy Analysis for a Control Volume
22CV ei
i ei i e e CV CVdE VV
m h gz m h gz Q Wdt 2 2
Conservation of Energy
Net Change in Energy of CV
Total Energy Carried by MassEntering CV
Total Energy Carried by MassLeaving CV
= -
Total Energy Crossing Boundary as Heat and Work
+
Steady-Flow ProcessSteady-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 ProcessSteady-Flow Process
Conservation of mass
i em m
Conservation of energy 2 2e i
e iCV CV e e i iV V
Q W m h gz m h gz2 2
Steady-Flow ProcessSteady-Flow Process
i em m m
2 2e i
CV CV e i e i(V V )
Q W m (h h ) g(z z )2
Conservation of mass
Conservation of energy
For single-stream steady-flow process
2 2CV CV e i
e i e iQ W (V V )
(h h ) g(z z )m m 2
Steady-Flow DevicesSteady-Flow Devices
CVW 0, PE 0. 2 2
CV e ie i
Q (V V )(h h )
m 2
● Nozzles and DiffusersA 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,
i em m m,
Example 2Example 2
iV 10 m/ s.
eV 665 m/ s.
iV 10 m/ s.
Steam enters an insulated nozzle at a flow rate of 2 kg/s with Ti = 400 ºC, pi = 4 MPa, and
Find the cross-sectional area at the exit.
Inlet Ti = 400 ºCpi = 4 MPa
Exit pe = 1.5 MPa
It exits at pe = 1.5 MPa with a velocity of
eV 665 m/ s.
Example 2 (continued)Example 2 (continued)
2 2i e
e i(V V )
h h2
2 2CV e i
e iQ (V V )
(h h )m 2
ee
e
mvA
V
Inlet, superheated vaporTable A-6, hi = 3213.6 kJ/kg
2 2
3
(10) (665) 13213.6
2 10
= 2992.5 kJ/kg
Example 2 (continued)Example 2 (continued)
ee
e
mvA
V
Table A-6, he = 2992.5 kJ/kg
1.4 MPa 1.5 MPa 1.6 MPa250 2927.2 2923.2 2919.2300 3040.4 3037.6 3034.8
T = 280 ºCv = 0.1627 m3/kg
(2)(0.1627)
665 = 0.000489 m2
Steady-Flow DevicesSteady-Flow Devices
PE 0.
2 2CV CV e i
e iQ W (V V )
(h h )m m 2
● TurbinesA turbine is a device from which work is produced asa result of the expansion of a gas or superheatedsteam through a set of blades attached to a shaft freeto rotate.
For turbines,
i em m m,
Example 3Example 3
iV 10 m/ s.
eV 50 m/ s.
iV 10 m/ s.
Steam enters a turbine at a flow rate of 4600 kg/h.At the inlet, Ti = 400 ºC, pi = 6 MPa, and
If the turbine produces a power of 1 MW, find the heatloss from the turbine.
Inlet Ti = 400 ºCpi = 6 MPa Exit
xe = 0.9pe = 10 kPa
At the exit, xe = 0.9, pe = 10 kPa and
eV 50 m/ s.
Example 3 (continued)Example 3 (continued)2 2e i
CV CV e i(V V )
Q W m (h h )2
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
Example 3 (continued)Example 3 (continued)
2 2 2 2e i
3
V V (50) (10) 11.2 kJ/kg
2 2 10
2 2e i
CV CV e i(V V )
Q W m (h h )2
he - hi = 2345.4 – 3177.2 = - 831.8 kJ/kg
CV4600
Q 1000 ( 831.8 1.2)3600
= - 63.1 kW
Steady-Flow DevicesSteady-Flow Devices
PE 0.
2 2CV CV e i
e i e iQ W (V V )
(h h ) g(z z )m m 2
i em m m,
● Compressors and Pumps
Compressors and pumps are devices to which work is provided to raise the pressure of a fluid.
For compressors,
Compressors → gasesPumps → liquids
For pumps, CVQ 0, PE 0.
Example 4Example 4
iV 6 m/ s.
eV 2 m/ s.
iV 6 m/ s.
Air enters a compressor.At the inlet, Ti = 290 K, pi = 100 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 Kpi = 100 kPa Exit
Te = 450 Kpe = 700 kPa
At the exit, Te = 450 K, pe = 700 kPa and
eV 2 m/ s.
CVQ 3kW
Example 4 (continued)Example 4 (continued)
i i
i
i
A V
RTp
2 2e i
CV CV e i(V V )
Q W m (h h )2
i i
i
A Vm
v
Table A-17, at 290 K, hi = 290.16 kJ/kg, at 450 K, he = 451.8 kJ/kg.
(0.1)(6)(100)
(0.287)(290) = 0.72 kg/s
Example 4 (continued)Example 4 (continued)
2 2i e
CV CV i e(V V )
W Q m (h h )2
2 2
CV 3
6 2 1W 3 (0.72) (290.16 451.8)
2 10
= - 119.4 kW
Example 5Example 5
iV 10 m/ s.
eV 40 m/ s.
iV 10 m/ s.
A pump steadily draws water at a flow rate of 10 kg/s.At the inlet, Ti = 25 ºC, pi = 100 kPa, and
If the exit is located 50 m above the inlet, find the workrequired for the pump.
Inlet Ti = 25 ºCpi = 100 kPa
Exit Te = 25 ºCpe = 200 kPa
At the exit, Te = 25 ºC, pe = 200 kPa and
eV 40 m/ s.
Example 5 (continued)Example 5 (continued)
Table A-4, at 25 ºC
vf = 0.001003 m3/kg
2 2CV CV e i
e i e iQ W (V V )
(h h ) g(z z )m m 2
he – hi ~ [hf + vf (p – psat)]e - [hf + vf (p – psat)]i
= vf (pe – pi)
= 0.001003 (200 – 100) = 0.1 kJ/kg2 2 2 2e i
3
V V (40) (10) 10.75 kJ/kg
2 2 10
g(ze – zi) = 9.8(50)/103 = 0.49 kJ/kg
Example 5 (continued)Example 5 (continued)2 2e i
CV e i e i(V V )
W m (h h ) g(z z )2
= 20 (0.1 + 0.75 + 0.49)
= 13.4 kW