Transcript Reflection of Buddhism in Contemporary Cinema

ENGR 2213 Thermodynamics
F. C. Lai
School of Aerospace and Mechanical
Engineering
University of Oklahoma
Steady-Flow Devices
● Heat Exchangers
Heat exchangers are devices that transfer energy
between fluids at different temperatures.
For heat exchangers, QCV  0, WCV  0,
KE  0, PE  0.
 mi   me ,
 mehe   mh
i i
Classification
► Open Feedwater Heaters (Mixing Chambers)
► Closed Feedwater Heaters
Example 1
Feedwater Heater:
Inlet 1 T1 = 200 ºC, p1 = 700 kPa,
Inlet 2 T2 = 40 ºC, p2 = 700 kPa,
Exit
sat. liquid, p3 = 700 kPa.
Find m2 / m3  ?
Inlet 2
Inlet 1
Exit
Example 1 (continued)
m1  m2  m3
 mi   me
 mh
i i   mehe
m1h1  m2h2  m3h3
 (m1  m2 )h3
 m2 
 m2

h1  
 1 h3
 h2  
 m1 
 m1

m2 h1  h3

m1 h3  h2
Example 1 (continued)
Inlet 1: Superheated vapor
Table A-6,
h1 = 2844.8 kJ/kg
Inlet 2: compressed liquid
Table A-4, 40 ºC h2 ~ [hf + vf (p – psat)]2
psat = 7.384 kPa
= 167.57 + 0.001008 (700 – 7.384)
vf = 0.001008 m /kg
= 168.27 kJ/kg
hf = 167.57 kJ/kg
Exit:
saturated liquid
Table A-5,
h3 = 697.22 kJ/kg
Example 1 (continued)
m2 h1  h3

m1 h3  h2
2844.8  697.22

 4.06
697.22  168.27
Steady-Flow Devices
● Throttling Devices
Throttling devices are flow-restricting devices (valves,
porous plugs) used to reduce the pressure of a gas or
a liquid.
Unlike turbines, they produce a pressure drop without
producing any work.
The pressure drop in the fluid is often accompanied
by a change in temperature. The temperature change
depends on the Joule-Thomson coefficient.
Steady-Flow Devices
● Throttling Devices
Joule-Thomson Coefficient
 T 
1 
 v  
      v  T   
cp 
 T p 
 p h
µ < 0, T increases
µ = 0, T no change
µ > 0, T decreases
For an ideal throttling process, QCV  0, WCV  0,
KE  0, PE  0.
hi = he
For ideal gases,
Ti = Te
Example 2
A throttling calorimeter is a device for determining the
quality of a two-phase liquid-vapor mixture.
Inlet 1 sat. mixture, p1 = 2 MPa,
Exit
T2 = 150 ºC, p2 = 100 kPa,
Find the quality of water at the inlet.
Sat. mixture
p1 = 2 MPa
T2 = 150 ºC
p2 = 100 kPa
Example 2 (continued)
Assuming an ideal throttling process,
hi = he
Inlet saturated mixture at 2 MPa
Table A-5 hf = 908.79 kJ/kg
hg = 2799.5 kJ/kg
Exit superheated vapor
Table A-6 he = 2776.4 kJ/kg
hi  hf
hi = hf + xi (hg – hf)
xi 
hg  hf
2776.4  908.79
= 0.988

2799.5  908.79
Uniform-Flow Process
Examples for uniform-flow (unsteady-flow) processes:
► Charge or discharging a vessel,
► Inflate tires or balloons,
► Cooking with a regular or pressure cooker,
● Any instant during the process, the state of the
control volume is uniform.
● The fluid properties may differ from one inlet (or
exit) to another, but the fluid flow at an inlet (exit) is
uniform and steady.
Uniform-Flow Process
Conservation of Mass
mCV   mi  me
Conservation of Energy
ECV




Ve2
Vi2
  mi  hi 
 gzi    me  he 
 gze   QCV  WCV
2
2




QCV  WCV




Ve2
Vi2
  me  he 
 gze    mi  hi 
 gzi 
2
2




+ (m2u2 – m1u1)CV
Example 3
A rigid, insulated tank which is initially evacuated is
connected through a valve to a supply line that carries
steam at 1 MPa and 300 ºC. Now the valve is opened
and steam is allowed to flow slowly into the tank until
the pressure reaches 1 MPa at which point the valve
is closed. Find the temperature of the steam inside
the tank.
Example 3 (continued)
p1 = 1 MPa
T1 = 300 ºC
Example 3 (continued)
Charging process
m1 = 0,
initially evacuated
Uniform-Flow Process
me = 0,
no outlet
m2 = mi
QCV – WCV = Σ mehe - Σ mihi + (m2u2 – m1u1 )CV
insulated
rigid
no outlet
mihi = m2u2
initially evacuated
hi = u2
Example 3 (continued)
Inlet: superheated vapor at 1MPa and 300 ºC
Table A-6, hi = 3051.2 kJ/kg
Final state: p2 = 1MPa and u2 = 3051.2 kJ/kg
Table A-6, T2 = 456.2 ºC
There is an increase in temperature of 156.2 ºC
due to the conversion of flow work to sensible
internal energy.
Example 4
A 0.2-m3 rigid tank equipped with a pressure regulator
contained steam at 2 MPa and 300 ºC. When heated,
the regulator keeps the steam pressure constant by
letting out some steam. But, the temperature inside
rises. Find the amount of heat transfer when the
temperature inside the tank reaches 500 ºC.
Example 4 (continued)
Discharging Process
Q
Uniform-Flow Process
p1 = 2 MPa
T1 = 300 ºC
Example 4 (continued)
No inlet, one outlet
me = (m1 – m2)CV
Initial state: p1 = 2MPa and T1 = 300 ºC
Table A-6, v1 = 0.1255 m3/kg
u1 = 2772.6 kJ/kg
h1 = 3023.5 kJ/kg
Final state: p2 = 2MPa and T2 = 500 ºC
Table A-6, v2 = 0.1757 m3/kg
u2 = 3116.2 kJ/kg
h2 = 3467.6 kJ/kg
Example 4 (continued)
V
0.2
m1 

 1.594 kg
v1 0.1255
V
0.2
m2 

 1.138 kg
v 2 0.1757
me = (m1 – m2)CV = 1.594 – 1.138 = 0.456 kg
QCV – WCV = Σ mehe - Σ mihi + (m2u2 – m1u1 )CV
rigid
no inlet
QCV = mehe + (m2u2 – m1u1 )CV
Example 4 (continued)
h1  h2
he 
2
3023.5  3467.6

 3245.6 kJ / kg
2
QCV = mehe + (m2u2 – m1u1 )CV
= (0.456)(3245.6) + (1.138)(3116.2)
- (1.594)(2772.6)
= 606.7 kJ
Example 5
A 0.3-m3 rigid tank is filled with saturated liquid water
200 ºC. A valve at the bottom of the tank is opened,
and liquid is withdrawn from the tank. Heat is
transferred to the water such that the temperature in
the tank remains constant. Determine the amount of
heat transfer by the time one-half of the water (in mass)
has been withdrawn.
Example 5 (continued)
Discharging Process
Uniform-Flow Process
Sat. liquid
T1 = 200 ºC
Q
Example 5 (continued)
No inlet, one outlet
me = (m1 – m2)CV
Initial state: Sat. liquid at T1 = 200 ºC
Table A-4, v1, u1, h1
Final state: Sat. mixture at T2 = 200 ºC
m2 = ½ m1, v2 = 2v1, find x2
Table A-4, v2, u2, h2
Exit:
Sat. liquid at T2 = 200 ºC
Table A-4, he