Transcript Reflection of Buddhism in Contemporary Cinema

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 = Total Mass Entering CV
Mass within CV
Total Mass
Leaving CV
mCV   mi  me
dmCV
  mi   me
dt
Steady State
AV
m  AV 
v
 mi   me
Example 1
Feedwater Heater:
Inlet 1 T1 = 200 ºC, p1 = 700 kPa, m1  40 kg/ s
Inlet 2 T2 = 40 ºC, p2 = 700 kPa, A2 = 25 cm2
Exit
sat. liquid, p3 = 700 kPa, (AV)3  0.06 m3 / s
Find m2  ?, m3  ? and V2  ?
Inlet 2
Inlet 1
Exit
Example 1 (continued)
 mi   me
AV
m  AV 
v
Steady State
m1  m2  m3
Inlet 2: compressed liquid
Table A-4,
v2 = 0.001008 m3/kg
Exit:
saturated liquid
Table A-5,
v3 = 0.001108 m3/kg
(AV)3
0.06
m3 

 54.15 kg / s
v3
0.001108
Example 1 (continued)
m2  m3  m1 = 54.15 – 40 = 14.15 kg/s
m2 v 2 (14.15)(0.001008)
V2 

 5.7 m / s
A2
0.0025
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
2
V
e u
 gz
2
Energy carried by a fluid element in a control volume
2
2
V
V
 gz
e  u  pv 
 gz  h 
2
2
Energy Analysis for a Control Volume
Conservation of Energy
Total Energy
Total Energy
Net Change in
= Carried by Mass - Carried by Mass
Energy of CV
Entering CV
Leaving CV
Total Energy
+ Crossing Boundary
as Heat and Work




dECV
Ve2
Vi2
  mi  hi 
 gzi    me  he 
 gze   QCV  WCV
dt
2
2




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 steadyflow system and its surroundings do not change
with time.
Steady-Flow Process
Conservation of mass
 mi   me
Conservation of energy
QCV  WCV




Ve2
Vi2
  me  he 
 gze    mi  hi 
 gzi 
2
2




Steady-Flow Process
For single-stream steady-flow process
Conservation of mass
mi  me  m
Conservation of energy


(Ve2  Vi2 )
QCV  WCV  m  (he  hi ) 
 g(ze  zi ) 
2


QCV WCV
(Ve2  Vi2 )

 (he  hi ) 
 g(ze  zi )
m
m
2
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, WCV  0, PE  0.
mi  me  m,
QCV
 (he  hi ) 
m
2
(Ve

2
2
Vi )
Example 2
Steam enters an insulated nozzle at a flow rate of
2 kg/s with Ti = 400 ºC, pi = 4 MPa, and Vi  10 m / s.
It exits at pe = 1.5 MPa with a velocity of Ve  665 m/ s.
Find the cross-sectional area at the exit.
Inlet
Ti = 400 ºC
pi = 4 MPa
Vi  10 m / s.
Exit
pe = 1.5 MPa
Ve  665 m / s.
Example 2 (continued)
mv e
Ae 
Ve
2
2
QCV
(Ve  Vi )
 (he  hi ) 
m
2
(Vi2  Ve2 )
he  hi 
2
Inlet, superheated vapor
Table A-6, hi = 3213.6 kJ/kg
(10)  (665)  1 
 3213.6 

3  = 2992.5 kJ/kg
2
 10 
2
2
Example 2 (continued)
Table A-6, he = 2992.5 kJ/kg
250
300
1.4 MPa 1.5 MPa 1.6 MPa
2927.2 2923.2 2919.2
3040.4 3037.6 3034.8
mv e
Ae 
Ve
(2)(0.1627)

= 0.000489 m2
665
T = 280 ºC
v = 0.1627 m3/kg
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, PE  0.
mi  me  m,
QCV WCV
(Ve2  Vi2 )

 (he  hi ) 
m
m
2
Example 3
Steam enters a turbine at a flow rate of 4600 kg/h.
At the inlet, Ti = 400 ºC, pi = 6 MPa, and Vi  10 m / s.
At the exit, xe = 0.9, pe = 10 kPa and Ve  50 m/ s.
If the turbine produces a power of 1 MW, find the heat
loss from the turbine.
Inlet
Ti = 400 ºC
pi = 6 MPa
Vi  10 m / s.
Exit
xe = 0.9
pe = 10 kPa
Ve  50 m/ s.
Example 3 (continued)
QCV  WCV

(Ve2  Vi2 ) 
 m (he  hi ) 

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)
QCV  WCV

(Ve2  Vi2 ) 
 m (he  hi ) 

2


he - hi = 2345.4 – 3177.2 = - 831.8 kJ/kg
Ve2  Vi2 (50)2  (10)2  1 

 1.2 kJ/ kg

3 
2
2
 10 
4600
QCV  1000 
( 831.8  1.2) = - 63.1 kW
3600
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,
PE  0.
QCV  0, PE  0.
QCV WCV
(Ve2  Vi2 )

 (he  hi ) 
 g(ze  zi )
mi  me  m,
m
m
2
Example 4
Air enters a compressor.
At the inlet, Ti = 290 K, pi = 100 kPa, and Vi  6 m / s.
At the exit, Te = 450 K, pe = 700 kPa and Ve  2 m / s.
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
Vi  6 m / s.
QCV  3kW
Exit
Te = 450 K
pe = 700 kPa
Ve  2 m / s.
Example 4 (continued)
QCV  WCV

(Ve2  Vi2 ) 
 m (he  hi ) 

2


(0.1)(6)(100)
A i Vi
A i Vi

m

= 0.72 kg/s
vi
 RTi  (0.287)(290)


 pi 
Table A-17, at 290 K, hi = 290.16 kJ/kg,
at 450 K, he = 451.8 kJ/kg.
Example 4 (continued)
WCV  QCV
WCV

(Vi2  Ve2 ) 
 m (hi  he ) 

2



 62  22   1  
 3  (0.72) (290.16  451.8)  
  3 

 2   10  
= - 119.4 kW
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 Vi  10 m / s.
At the exit, Te = 25 ºC, pe = 200 kPa and Ve  40 m / s.
If the exit is located 50 m above the inlet, find the work
required for the pump.
Inlet
Ti = 25 ºC
pi = 100 kPa
Vi  10 m / s.
Exit
Te = 25 ºC
pe = 200 kPa
Ve  40 m / s.
Example 5 (continued)
QCV WCV

 (he  hi ) 
m
m
2
(Ve

2
2
Vi )
 g(ze  zi )
he – hi ~ [hf + vf (p – psat)]e - [hf + vf (p – psat)]i
Table A-4, at 25 ºC = vf (pe – pi)
vf = 0.001003 m3/kg
= 0.001003 (200 – 100) = 0.1 kJ/kg
Ve2  Vi2 (40)2  (10)2  1 

 0.75 kJ/ kg

3 
2
2
 10 
g(ze – zi) = 9.8(50)/103 = 0.49 kJ/kg
Example 5 (continued)
 WCV
2
2


(Ve  Vi )
 m (he  hi ) 
 g(ze  zi )
2


= 20 (0.1 + 0.75 + 0.49)
= 13.4 kW