Transcript Repaso4

Thermodynamics Review
Craig Bradshaw, Lambert Fellow and PhD
Candidate
Slides provided by Prof. S. F. Son, and M.
Mathison
Adapted from Prof. G. A. Risha and other
sources such as Kaplan AEC Education
Thermodynamics is the transformation
of energy
THE FIRST LAW OF THERMODYNAMICS
• Energy cannot be created or destroyed,
but transformed into different forms
• Applies to systems classified as either
closed or open
Thermodynamic Systems
Energy Balance: Finite Time
E stored  E in  E out  E gen
Energy Balance: Rate
E
 E  E  E
stored
in
out
gen
Energy Transfer by
Work and Heat:
Qout< 0
Win < 0
Qin > 0
Wout > 0
Engines burn fuel to put heat
into the system and produce
work (out of the system)
What is your system?
Closed (fixed mass) or open (fixed volume)
Fixed Mass: Closed System
dEcv  
 Q W
dt
ΔE stored  Q  W
ΔU  ΔKE  ΔPE  Q  W
Note: Closed systems are a subset of open systems
Systems and Sign Convention
Consider a system that contains a lightbulb powered by
electricity. Which of the following is true?
a) Positive work, positive heat transfer
b) Positive work, negative heat transfer
c) Negative work, positive heat transfer
d) Negative work, negative heat transfer
dEcv
 Q  W  0
Q  W
+ dt
Which of the following would be considered a closed system?
a) A pump
b) A tire
c) A turbine
d) A jet nozzle
Closed System: Work
W   Fdx
P
Boundary Work:
Wboundary work   pdV  m  pdv
Note that if
volume is constant,
W=0
Constant Pressure:
W  pV  pV2  V1   mpv  pv 2  v1 
Polytropic:
pv n  constant
p 2 V2  p1V1
V2
W
if n  1, W  p1V1ln
if n  1
1 n
V1
Work: Special Cases (Ideal Gas)
Ideal Gas Law:  pV  mRT
 pv  RT
Constant Temperature:  pv  constant
 v2 
 p1 
W  mRTln    mRTln  
 v1 
 p2 
Isentropic: s2 = s1  pv k  constant
p 2 V2  p1V1 mRT1   p 2 
W

1   
1 k
k  1   p1 

specific heat ratio,
k
cp
cv
k 1
k




Closed System Work
When using the previous equations for boundary work, it must
be assumed that a quasi-equilibrium process exists. If a
quasi-equilibruim process exists, we have assumed
a) The pressure at any instant to be everywhere constant.
b) An isothermal process.
c) The heat transfer to be small.
d) The boundary motion to be infinitesimally small.
e) That no friction exists.
Control Volume: Open System
h in 
Energy entering
the control volume
in the forms of
kinetic energy,
potential energy,
and enthalpy
h in 
Vin2
 gz in
2
Energy exiting the
control volume in the
forms of kinetic energy,
potential energy, and
enthalpy (h = u + pv)
Vin2
 gz in
2
2




dE CV  
Vout
Vin2
 in  h in 
 out  h out 
 Q  W  m
 gz in    m
 gz out 
dt
2
2




dE CV
0
At steady-state,
dt
Nozzles and Diffusers
Assumptions: (1) adiabatic, Q = 0
(2) no volume changes, W = 0
(3) steady-state, d/dt = 0
(4) change in potential energy negligible
dE CV  
QW
dt



Vin2

 in  h in 
m

gz
in  

2



2


Vout

 out h out 
m
 gz out 

2


2
Vout
Vin2
h in 
 h out 
2
2
or
2
Vin2 Vout
h  h out - h in 

2
2
Nozzles and Diffusers
h-s Diagram for Nozzle
Nozzle Efficiency:
Compares the performance
of a real nozzle or diffuser
to the performance of an
ideal, isentropic nozzle or
diffuser operating between
the same pressures
 nozzle 


h actual
V

h ideal
V
2
outlet
2
outlet,s
V
V
2
2
2
inlet
2
inlet
Actual process
Isentropic process
with same initial state
and final pressure
Turbines, Pumps, and Compressors
Assumptions: (1) adiabatic, Q = 0
(2) change in potential energy negligible
(3) steady-state, d/dt = 0
(4) change in kinetic energy negligible
dE CV  
QW
dt



Vin2

 in  h in 
m

gz
in  

2



W
 h in  h out

m

2


Vout

 out h out 
m
 gz out 

2


Turbines, Pumps, and Compressors
Turbine and compressor/pump efficiencies both compare the performance of
an actual device to the performance of an ideal, isentropic device operating
between the same pressures
• Turbine Efficiency: In turbines,
the actual power generation will be
less than the ideal power generation
h actual h inlet  h outlet actual
t 

h ideal
hinlet  houtlet,s s
isentropic
• Pump/Compressor Efficiency:
In pumps/compressors, the actual
power consumption will be greater
than the ideal power consumption
isentropic
h outlet  h inlet,s s
h ideal
c 

h actual h outlet  h inlet actual
Throttling Valves
Assumptions: (1) adiabatic, Q = 0
(2) no volume changes, W = 0
(3) steady-state, d/dt = 0
(4) change in potential energy negligible
(5) change in kinetic energy negligible
dE CV  
QW
dt



Vin2

 in  h in 
m

gz
in  

2



2


Vout

 out  h out 
m

gz
out 

2


h in  h out
Boilers, Condensers, and Evaporators
Assumptions: (1) constant volume, W = 0
(2) change in potential energy negligible
(3) steady-state, d/dt = 0
(4) change in kinetic energy negligible
dE CV  
QW
dt



Vin2


m in  h in 
 gz in  
2



2


Vout


m out  h out 
 gz out 
2



Q
 h out  h in

m
Heat Exchangers
Assumptions: (1) constant volume, W = 0
(2) change in potential energy negligible
(3) steady-state, d/dt = 0
(4) change in kinetic energy negligible
(5) adiabatic, Q = 0
dE CV  
QW
dt



Vin2

 in  h in 
m

gz
in  

2



2


Vout

 out  h out 
m

gz
out 

2


m
in h in

m
out h out
Example Problem 1
Given: Gas initially at 1 MPa and 150 oC receives 7.2 MJ of work while 1.5 kW of
heat are removed from the system. Calculate the internal energy change for
the system over a period of one hour.
Analysis:
negligible
E  U  KE  PE  Q  W
U  Q  W
Work done ON system: W = -7.2 MJ
  -1.5 kW
 OUT of the system: Q
Heat
  -1.5kW(3600s)  -5.4MJ
Q t  Q
U  5.4 MJ  (7.2 MJ)
U  1.8 MJ
Example Problem 2
Given: Calculate the power required to compress 10 kg/s of air from 1 atm and 37 oC
to 2 atm and 707 oC.
For low pressure air:
T = 310 K; h = 290.4 kJ/kg
T = 980 K; h = 1023 kJ/kg
Analysis:
dE CV  
QW
dt



Vin2

 in  h in 
m

gz
in  

2



2


Vout

 out  h out 
m

gz
out 

2


 m
 (hin  h out )
W
  10
W
kg
(290.4
s
  7326
W
kJ
s
 1023)
kJ
kg
 7326 kW
  7326 kW
W
c
Done ON system
Example Problem 3
P
Given: A gas goes through the following thermodynamic
B
cycle.
A to B: constant-temperature compression
B to C: constant-volume cooling
C to A: constant-pressure expansion
The pressure and volume at state C are 1.4 bar and
0.028 m3, respectively. The net work from C to A is
10.5 kJ.
What is the net work from one complete cycle A-B-C?
Analysis:
Wcyc  WAB  WBC  WCA
V  0

B
C


A
B
A

Wcyc  pdV  pdV  pdV  pdV

C
A
C
V
Example Problem 3

B

Wcyc  pdV  pdV  10.5 kJ
P
B
A
Constant Temperature:
 VB 

W  RTln 
 VA 
A
C
Need volume @ A, constant pressure A to C
W  pA VA  VC 
Compression, W<0
WCA
 VA  VC 
pA
2
WCA
10500
J
1
bar
1
Pa
m
1Nm
3
VA  VC 
 0.028 m 




5
pA
1.4 bar 1x10 Pa 1 N
1J

VA  0.103 m3
V
Example Problem 3
P
Constant Temperature:
B
WAB
 VB 

 RTln 
 VA 
C
Ideal gas and constant mass
WAB
A
V


 VB 
1x10 5 Pa 1 N
1J
 0.028 
3
  1.4 bar
 pVA ln 
0.103
m
ln


2
1 bar 1 Pa  m
1 N  m  0.103 
 VA 
WA B  -18.8 kJ
Wcyc  18.8 kJ  10.5 kJ
Wcyc  8.3 kJ
PROPERTIES OF PURE SUBSTANCES
• Use of Steam Tables
• Use of R-134a Tables, NH3
Tables, and P-h Diagrams
Pure Substances
• What is a pure substance?
• Materials with unchanging chemical composition
• Three common phases of pure substances:
• Liquid, vapor, and solid
• Also consider liquid-vapor mixtures, which are a
combination of liquid and vapor where pressure and
temperature are not independent
• State Postulate: Two independent, intensive
properties are required to fix a state
– Intensive: Independent of system size (usually per mass)
– Extensive: Dependent on size of system
T-v Phase Diagram
Diagram courtesy of Jerry M. Seitzman, 2001.
T-v Phase Diagram
@ Constant Pressure
T > Tsat  “superheated vapor”
T = Tsat  “two-phase liquid-vapor”
T < Tsat  “compressed liquid”
P-v Phase Diagram
@ Constant Temperature
P < Psat  “superheated vapor”
P = Psat  “two-phase liquid-vapor”
P > Psat  “compressed liquid”
Common Properties
A set of properties that completely describes a system,
specifies the state of the system.
• Mass, m
– Extensive
– Units [kg]
• Volume, V
– Extensive
– Units [m3]
• Pressure, P
– Intensive
– Absolute units [Pa, psia]
– Gage, Pgage = Pabs - Patm
• Temperature, T
– Absolute units [K or oR]
– T[K] = T[oC] + 273.15
– T[oR] = T[oF] + 459.67
• Specific Volume, v
– Intensive
– Units [m3/kg]
– v = 1/r = V/m
• Internal energy, u
– Intensive
– Units [kJ/kg]
– U = m*u [kJ]
• Enthalpy, h
– Intensive
– Units [kJ/kg]
– h = u + Pv; H = U + P V
– H = m*h
Properties [cont’d]
• Entropy, s
– S, extensive [kJ/K]
– s, intensive [kJ/kg-K]
• Enthalpy
–
–
–
–
H, extensive [kJ] 
H, intensive [kJ/kg]
H = U + PV
h = u + Pv]
• Gibbs’ Function
–
–
–
–
G, extensive [kJ]
g, intensive [kJ/kg]
g = h - Ts = u + Pv - Ts
G = H - TS = U + PV – TS
s 

Q rev
T
Saturated Liquid Vapor Mixture (SLVM)
Quality:
mass of vapor
x
m vap
m liq  m vap

mg
Quality is a function
of the horizontal
distances on P-v
and T-v diagrams
mf  mg
mass of liquid
h mix  h f  xh fg  h f  x h g  h f 
s mix  s f  xs fg  s f  x s g  s f 
u mix  u f  x u g  u f 
v mix  v f  x v g  v f 
x
u mix  u f h mix  h f s mix  s f v mix  v f



ug  uf
hg  hf
sg  sf
vg  vf
Example
• The Purdue Physics department did a
demonstration of collapsing a barrel by
boiling a small amount of water for
several minutes, then sealing the barrel
and letting it cool (see images below).
Room temperature is 25oC. What is the
maximum pressure difference (between
the outside and inside of the barrel)
possible that could be realized if the
barrel were strong enough NOT to
collapse?
•
•
Assume saturated vapor at atmospheric conditions. Table A-4, for
p=101.42 kPa, T=100oC, yields v1 = vg =1.6720 m3/kg
At state 2: v2 = v1 =1.6720 m3/kg
– The MAXIMUM pressure difference possible would be for the
water vapor to be cooled to room temperature. At room
temperature (25oC), vg = 45.3 m3/kg and vf = 0.001003 m3/kg.
– Therefore, the water is a SLVM at 1.6720 m3/kg and 25oC. The
SLVM is at the saturation pressure at 25oC, which is 3.1698 kPa.
– For this process, ΔP = 101.42 – 3.1698 kPa.
•
Thus, the maximum pressure difference is 98.25 kPa.
IDEAL GASES
• Equations of State
• Enthalpy and Internal Energy
Changes
Ideal Gas Equation of State
P  pressure [Pa]
V  volume [m3]
m  mass [kg]
R  gas constant [J/kg-K]
T  temperature [K]
r  density [kg/m3]
v  specific volume [m3/kg]
n  number of moles [mol]
Ru  universal gas constant [J/kmol-K]
MWgas  molecular weight [kg/kmol]
Ru = 8.314 kJ/kmol-K
= 8314 J/kmol-K
= 1.986 cal/mol-K
PV  mRT
Pv  RT
P  ρRT
PV  nR u T
Ru
R
MWgas
Ideal Gas Relationships
Specific Heats:
constant volume specific
heat [kJ/kg-K]
cp – cv = R
gas constant [kJ/kg-K]
constant pressure specific
heat [kJ/kg-K]
Internal Energy:
u = cvT [kJ/kg]
Enthalpy:
(Assuming
constant
specific heat)
h = cpT [kJ/kg]
Entropy:
T2 
p 2 
s  c p   R ln  
T1 
p1 
(Assuming
constant
specific heat)
T2 
v 2 
s  c v   R ln  
T1 
v1 
Ideal Gas Relationships
Constant Temperature:
p1v1  p2 v2
Constant Pressure:
T1 v1

T2 v 2
Constant Volume:
T1 p1

T2 p 2
Isentropic Ideal Gas Relationships
Isentropic with constant specific heats:
p1v1k  p 2 v k2
k = specific heat ratio
k
T1  v 2 
  
T2  v1 
k 1
T1  p 2 
  
T2  p1 
1k
k
cp
cv
Non-ideal Gases
• Gases behave ideally at low pressures and
high temperatures
• What is meant by low pressure and high
temperature?
– Compressibility factor quantifies the deviation of
a pure substance from ideal gas behavior at a
given temperature and pressure
– Z = 1 (ideal gas)
– For a real gas, Z > 1 or Z < 1
– For Z very close to unity, we can assume ideal gas
behavior, most of the time
Compressibility Factor
• Reduced pressure and temperature:
– PR = reduced pressure; TR = reduced temperature
– Pcr = critical pressure; Tcr = critical temperature
• Principle of Corresponding States: The
compressibility factor is approximately the same
at the same PR and TR for all fluids
– Therefore, a generalized compressibility chart can be
used to find the compressibility factor of any fluid
Generalized Compressibility Chart
Sonntag, Borgnakke,
and Van Wylen,
Fundamentals of
Thermodynamics, 5th
Edition, page 763.
Property Evaluation
• STEP 1: ALWAYS start in the saturated or “two-phase liquidvapor” tables
• STEP 2: Identify two, independent, intensive properties
– Pick one property and find its location; compare the second using the
following rules:
– If the properties are pressure and temperature, and
• Tsat < T or Psat > P, then “superheated vapor”
• Tsat > T or Psat < P, then “compressed liquid” or “sub-cooled”
• Tsat = T or Psat = P, then “liquid-vapor” and need another property
– If one of the properties is not pressure or temperature
• Let b be your property
• b > bg, then “superheated vapor”
• b < bf , then “compressed liquid” or “sub-cooled”
• bf< b < bg, then “liquid-vapor”
Property Evaluation
• STEP 3: Evaluate the remaining properties
– If “superheated vapor,” then go to superheated tables
– If “compressed liquid” or “sub-cooled,” then go to compressed liquid
tables
• If data does not exist, assume the following:
v = vf
h = hf
u = uf
s = sf
where the saturated liquid property is evaluated at the given
temperature since pressure does not impact liquids that much
– If “liquid-vapor,” then continue using the “two-phase liquid-vapor” tables
and find quality
Saturated Mixture Table
Superheated Steam
Superheated Steam [cont’d]
Mollier Diagram (P vs h)
Example 1
Given: Steam at 2.0 kPa is saturated at 17.5 oC. In what state will the steam be at
40 oC if the pressure is 2.0 kPa?
Analysis:
@ P = 2.0 kPa = 0.02 bar, and T = 40oC
Psat > P  “superheated vapor”
Example 2
Given: What is the change in internal energy of air (assumed to be an ideal gas)
cooled from 550 oC to 100 oC (assume constant cv= 0.718 kJ/kg-K) ?
Analysis:
“ideal gas”  u = cvT = cv (T2-T1)
given cv= 0.718 kJ/kg-K
u = 0.718 kJ/kg-K (550-100)K
u = 323.1 kJ/kg
Example 3
Given: A boy on the beach holds a spherical balloon filled with air. At 10:00 a.m., the
temperature on the beach is 20 oC and the balloon has a diameter of 30 cm.
Two hours later, the balloon diameter is 30.5 cm. Assuming the air is an ideal
gas and that no air was lost or added, what is the temperature on the beach
at noon?
Analysis:
“ideal gas”  PV  mRT
constant mass, constant pressure, and same gas,
V1 V2
T1V2 T1D32

 T2 
 3
T1 T2
V1
D1
T2


20  273 K(30.5 cm)3

(30 cm)3
 308 K  35 o C
Example 4
Given: Steam initially at 1 MPa and 200 oC expands in a turbine to 40 oC and 83%
quality. What is the change in entropy?
Analysis:
@ P1=1 MPa = 10 bar, and T1=200 oC >> Psat > P  “superheated vapor”
@ T=40 oC and x2 = 0.83  “two-phase liquid-vapor”
Example 4 - Cont’d
Given: Steam initially at 1 MPa and 200 oC expands in a turbine to 40 oC and 83%
quality. What is the change in entropy?
Analysis:
s1 = 6.6940 kJ/kg-K (from superheated table for 1 MPa, 200 oC)
@ 40 oC, s2,f = 0.5725 kJ/kg-K; s2,g = 8.257 kJ/kg-K
s2 = s2,f +x2(s2,g-s2,f)
= 0.5725+0.83*(8.257- 0.5725)
s2 = 6.9506 kJ/kg-K
s2 - s1 = (6.9506 – 6.694) kJ/kg-K
s2 - s1 = 0.2566 kJ/kg-K
Example 5
Given: A 3 kg mixture of water and water vapor at 70 oC is held at constant pressure
while heat is added. The enthalpy of the water increases by 50 kJ/kg. What
is the change in entropy?
Heat flow = enthalpy change since
Analysis:
s 

Q rev
T
Q
50 kJ/kg
 s 

To (70  273)K
s = 0.1458 kJ/kg-K
integrate

Q  H
Example 6
Given: Through an isentropic process, a piston compresses 2 kg of an ideal gas at
150 kPa and 35 oC in a cylinder to a pressure of 300 kPa. The specific heat
of the gas for constant pressure processes is 5 kJ/kg-K; for constant volume
processes, the specific heat is 3 kJ/kg-K. What is the final temperature of the
gas?
Analysis:
“ideal gas and constant entropy” 
cp = 5 kJ/kg-K; cv = 3 kJ/kg-K
T2 
T1  p 2 
  
T2  p1 
k
cp
cv

1k
k
 T2 
5
 1.667
3
35  273 K  406.44 K  133.44 o C
 300 


150


1-1 .6 6 7
1 .6 6 7
T1
 p2 
 
 p1 
1-k
k
Example 7
Given: 3 kg of steam with a quality of 30% has a pressure of 12.056 bar. At that
pressure, the specific volume of a saturated fluid is vf = 1.5289 cm3/g. The
specific volume of the vapor is vg = 14.1889 cm3/g. What is the specific
volume of the steam?
Analysis:
vf = 1.5289 cm3/g
vg = 14.1889 cm3/
vmix = vf +x(vg-vf)
= 1.5289 +0.30*(14.1889 - 1.5289) cm3/g
vmix = 5.3269 cm3/g
Second Law
• THE SECOND LAW OF
THERMODYNAMICS
• ENTROPY
Thermo in a Nutshell
1. There is a game. (0th law, thermal
equilibrium)
2. The best you can do is tie. (1st Law,
energy balance)
3. You can never tie. (2nd Law, entropy
increase principal)
4. You can’t quit. (3rd Law, absolute zero)
2nd Law of Thermodynamics
Generally, there are two ways to state the
second law of thermodynamics:
(1) Kelvin-Planck: It is impossible to build a cyclic engine
that will have a thermal efficiency of 100%
* All work can be converted to heat, but all heat CANNOT
be converted into work
* Maximum possible thermal efficiency for a heat engines
is the Carnot cycle efficiency
(2) Clausius Statement: It is impossible to devise a cycle
such that its only effect is the transfer of heat from a
low-temperature body to a high-temperature body
* An input of work is always required for refrigeration cycles
Entropy Changes
Net entropy must ALWAYS increase:
• The equality applies for reversible processes
T2
dQ
S  
T
T1
• The inequality applies for irreversible processes,
in which entropy production contributes to entropy change
For a constant temperature, reversible process, entropy
change is given by:
Q
S  S2  S1 
To
Isentropic process, entropy change is given by:
S  S2 S1  0
Adiabatic process,
entropy change is given by:

S  S2 S1  0
discharging energy to reservoir
temperature of the reservoir
Entropy Changes
Evaluating the property of entropy:
• For incompressible solids and liquids, constant specific
heats,
 T2 
s 2  s1  cln  
 T1 
• For ideal gases with constant specific heats,
 T2 
v2
s 2  s1  c v ln    Rln
v1
 T1 
 T2 
p2
s 2  s1  c p ln    Rln
p1
 T1 
Problem 4
Given: 1 kg of steam is initially at 400 oC and 800 kPa. The steam expands
adiabatically to 200 oC and 400 kPa in a closed process, performing 450 kJ
of work, given the following information. Which law does this process
violate: (zeroth law, first law, second law, first and second law)?
at 400 oC and 800 kPa
u = 2959.7 kJ/kg
h = 3267.1 kJ/kg
s = 7.5716 kJ/kg-K
at 200 oC and 400 kPa
u = 2646.8 kJ/kg
h = 2860.5 kJ/kg
s = 7.1706 kJ/kg
Analysis:
(1) Zeroth law not applicable, does not deal with thermal equilibrium
(2) Check first law
negligible
E  U  KE  PE  Q  W
Problem 4 Cont’d
at 400 oC and 800 kPa
u = 2959.7 kJ/kg
h = 3267.1 kJ/kg
s = 7.5716 kJ/kg-K
at 200 oC and 400 kPa
u = 2646.8 kJ/kg
h = 2860.5 kJ/kg
s = 7.1706 kJ/kg
U  Q  W
Q W
Q
W
u  
  u 
m m
m
m
W > 0 done
BY system
Q
 2646.8 kJ kg  2959.7 kJ kg  450 kJ kg
m

Q
 137 .1 kJ kg 
m
0
NOT adiabatic Q = 0, as stated
Therefore, violates 1st Law
Problem 4 Cont’d
at 400 oC and 800 kPa
s = 7.5716 kJ/kg-K
at 200 oC and 400 kPa
s = 7.1706 kJ/kg
(3) Check 2nd Law
s  0 for adiabatic process
s  7.1706 kJ kg-K  7.5716 kJ kg-K
s  -0.401 kJ kg-K < 0


violates 2nd Law

Violates both 1st and 2nd Laws
CYCLES
• Rankine Cycle (Steam)
• Vapor Compression Cycle
(Refrigeration)
• Otto Cycle (Gasoline
Engine)
Basic Cycles
Power Generation
(Heat Engine)
Refrigeration, AirConditioning, Heat Pump
Basic Cycles: Definitions
Cycle: Series of processes that eventually brings the
system back to its original state
Power Cycle:
• A cycle that uses heat energy to do work on the surroundings
• Performance is measured by thermal efficiency, which is the
ratio of useful work output to the energy input required to run
the cycle:
work done BY system
Wnet Wout  Win
th 

Q in
Q in
Q net Q in  Q out


Q in
Q in
work done ON system
heat OUT OF system
heat INTO system
Carnot Cycle
Carnot Cycle:
(1) Theoretical implementation
of a cycle without any
irreversibilities
(2) Work output is the maximum
possible for any heat engine
due to reversible processes
A
B
D
C
Processes
A to B: isothermal expansion of saturated liquid to saturated vapor
B to C: isentropic expansion of vapor (Q = 0; s = 0)
C to D: isothermal compression of vapor
D to A: isentropic compression of vapor (Q = 0; s = 0)
Carnot: P-v and T-s Diagrams
A
B
B
A
D
C
D
Carnot Efficiency:
th,Carnot
TH  TL
TL

 1
TH
TH
C
cold reservoir
temperature
hot reservoir
temperature
Problem 1
Given: A Carnot engine receives 100 kJ of heat from a hot reservoir at 370 oC and
rejects 37 kJ of heat. Determine the temperature of the cold reservoir.
Analysis:
 th 
Q net Q H  Q L 100 kJ  37 kJ


Qin
QH
100 kJ
th  0.63 or 63%

Use absolute
temperatures!
TH  TL
T
 1 L
TH
TH
TL  TH  TH th,Carnot

QH
W
th,Carnot 
TL  (370  273)  370  273 0.63

TH
TL = 237.91 K = -35.09 oC
QL
TL
Problem 2
Given: What is the maximum thermal efficiency possible for a power cycle operating
between 600 oC and 110 oC?
Analysis:
Carnot yields maximum possible efficiency
TL
TH
110  273K
1
600  273K
th,Carnot  1
th,Carnot



th,Carnot  0.561 or 56.1%
Problem 3
Given: A heat pump takes heat from groundwater at 7 oC and maintains a room at
21oC. What is the maximum coefficient of performance possible for this heat
pump?
upper limit of heat pump is set
by a Carnot heat pump cycle
Analysis:
COPheat pump
TH
QH
W
QL
TL
QH
QH
TH



W Q H  Q L TH  TL

21  273 K
COPheat pump 
21  273 K  7  273 K
COPheat pump  21
How does a refrigerator COP differ?
Rankine Cycle
Rankine Cycle:
• Vapor-power cycle commonly used
in power plants with water as the
working fluid
• Efficiency is ratio of useful output to
required input:
Wnet h1  h 2   h 4  h 3 
 th 

h1  h 4 
Q in
Processes
1 to 2: Isentropic expansion of the working fluid through the turbine from saturated
vapor at state 1 to the condenser pressure (Q = 0; s = 0)
2 to 3: Heat transfer from the working fluid as it flows at constant pressure through
the condenser with saturated liquid at state 3
3 to 4: Isentropic compression in the pump to state 4 in the compressed liquid
region. (Q = 0; s = 0)
4 to 1: Heat transfer to the working fluid as it flows at constant pressure through the
boiler to complete the cycle
Rankine-Superheat
Superheat: Used to increase
the temperature of the working
fluid entering the low-pressure
turbine to reduce the wear on
turbine blades (point 3)
Combustion Power Cycles
• Differ from vapor cycles since they cannot return
to their initial conditions
• Due to significant complexities with computing
mixtures of fuel and air, combustion power cycles
are often analyzed as air-standard cycles
– Air-Standard Otto Cycle: Hypothetical closed system
using air as the working fluid to simplify the chemistry
due to combustion
Air-Standard Otto Cycle
Air-Standard Otto Cycle: Hypothetical closed system using air as the
working fluid to simplify the chemistry due to combustion
Processes
1 to 2:
2 to 3:
3 to 4:
4 to 1:
Isentropic compression of the working fluid (Q = 0; s = 0)
Constant volume heat addition
Isentropic expansion of the working fluid (Q = 0; s = 0)
Constant volume heat rejection
Air-Standard Otto Cycle: Performance
Compression Ratio:
V1 V4 Vb
rv 


V2 V3 Va
compression from A to B
expansion from C to D

Ideal Thermal
Efficiency:
specific heat ratio
th  1 rv1k
compression ratio

Refrigeration Cycles
• In refrigeration cycles, heat is transferred from a
low-temperature area (i.e. inside the refrigerator) to
a high-temperature area (e.g., in the kitchen)
• Since heat spontaneously flows only from high to low
temperature areas, work is required to force heat transfer
• Opposite of heat engines
• Systems:
• Refrigerator: Heat is removed from air inside
• Air conditioner: Heat is removed from air in an occupied
space
• Heat Pump: Heat is supplied to air in an occupied space
• Chiller: Heat is removed from water
Refrigeration Cycles: Performance
• In refrigeration cycles, the coefficient of
performance (COP) is used in place of thermal
efficiency to measure performance
• The COP is always the ratio of useful energy
transfer to the work input
QL
COPrefrigerat or,AC 
W
COPheat pump
QH

W
heat transfer from lowtemperature reservoir
work input
heat transfer from hightemperature reservoir
MISCELLANEOUS
• Mixture of Gases
• Heat Transfer
– Conduction
– Convection
– Radiation
Modes of Heat Transfer
• All heat transfer driven by
temperature difference
• Heat transfer occurs via 3 modes
– conduction (solid/fluid; molecular
collision)
• depends on temperature gradient
– convection (fluid; motion)
• depends on temperature difference
– radiation (no medium needed –
electromagnetic waves)
• depends on T4
Conductive Heat Transfer
•
Conduction heat transfer
– Transfer of energy by putting a hotter object in contact with a cooler
system)
Convective Heat Transfer
• Convection heat transfer
– transfer of energy between a solid surface and a moving fluid
Newton’s law of cooling,
h = heat transfer coefficient.
Convective Heat Transfer
• The “wind chill”
factor is an example
• The air flow makes a
given temperature
“feel” (higher heat
loss) like a lower
temperature
Radiant Heat Transfer
• Radiation heat transfer
– transfer of energy due to emission and absorption of
electromagnetic radiation
Comfort in a Room
• At the same air temperature in a
room in a house it “feels” cooler in
the winter than the summer
– Why?
Thermal Circuit Model
• A model used often to calculate the heat transfer through a
1-D system is called the thermal circuit model
• In this model, each layer is replaced by an equivalent
resistor called the thermal resistance
– For conduction,
– For convection,
Mass and Mole Fraction
Mass Fraction: ratio of component’s mass to
the total mass of the mixture
mi
wi 
m tot
mass fraction of ith species
m
mass of ith species
m
i
total mass of mixture
w
i
1
Mole Fraction: ratio of component’s moles to
the total moles of the mixture
liquid-phase
gas-phase
moles of ith species
Ni
yi  x i 
N tot
mole fraction of ith species
x
i
1
y
total moles of mixture
i
1
Mass and Mole Fraction [cont’d]
total mass of mixture
Molecular Weight of Mixture:
MWmix
mass fraction of ith species
m
 tot 
N tot
 x MW 
i
total moles of mixture
Mass to Mole Fraction Conversion:
xi 
wi
MWi
wi
MWi

Mole to Mass Fraction Conversion:
wi 
x i MWi 
x i MWi 

i
molecular weight
of ith species
mole fraction of ith species
Non-Reacting Ideal Gases
mass of ith species
Partial Pressure:
mi R i T
pi 
 xip
V
partial pressure
of ith species
total pressure
of mixture
mole fraction of ith species
Partial Volume:
partial volume
of ith species
Vi 
mi R i T
p
Amagat’s Law: total volume is the sum of partial volumes
V
Mole Fraction:
xi 
p i Vi

p V
V
i
mass fraction
u
w u
internal energy
i i
h
enthalpy
w h
i i
assume adiabatic
and reversible
s
entropy
w s
i i
The End
Appendices
A: SI Prefixes
B: Constants
C: Unit Conversions
Appendix A: SI Prefixes
Appendix B: Constants
Appendix B: Constants [cont’d]
Appendix C: Unit Conversions
Appendix C: Unit Conversions [cont’d]
Appendix C: Unit Conversions [cont’d]
Extra Examples
Liquid/Vapor Tables & Charts
Solution
Solution (continued)
Liquid/Vapor Tables & Charts
Solution
Solution (continued)
Enthalpy
Solution
Solution (continued)
Solution (continued)
Closed Systems
Solution
Steady Flow/Steady State
Solution
Reversible Processes
Solution
Solution (continued)
Reversible Processes
Solution
Three-Process Cycles
Solution
Solution (continued)
Otto Cycle
Solution
Solution (continued)
Solution (continued)
Reverse Cycles
Solution
Solution (continued)
Non-Condensable Gas Mixtures
Solution