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