### Lecture 05: Chap 2 Review

```Chapter 2: Review
Lecture 05:
Chapter 2 Review
Quiz Today?
Main Concepts of Chap 2.
• Work done by compressed gas system:
• Heat Transfer
W 
 pdV
▫ Conduction
▫ Convection
• 1st law of Thermodynamics:  E  Q in  W out
• Power cycle, Refrigeration cycle, Heat Pump cycle
• Thermal Efficiency and Coefficient of Performance.
• Read Chap 3: Sections 1-5
Homework Assignment:
From Chap 2: 46, 59, 82,97
3
Recall Chapter 2 concepts:
1st Law of Thermodynamics:
 E system   P E   K E   U  Q in  W o u t
system
Qin
ΔE
for most Thermodynamics applications:
 U  Q in  W o u t
Wout
ΔKE = ΔPE = 0 then
4
Work done by Gas System
W > 0 : Expansion of Gas
W < 0 : Compression of Gas
W 

V
f
PdV
Vi
= Area under the p-V graph
over process
Run
Animation
5
Heat Transfer
Conduction:
Q   A
where A is area
κ is thermal conductivity
dT
dx
Convection:
Q  h c A  Tb  T f
Q   A T
4
b

where A is area
hc is the convection coefficient
Tb -Tf is the difference between the
body and free steam fluid temp.
where Tb is absolute surface temperature
ε is emissivity of the surface
σ is Stefan-Boltzmann constant
A is surface area
Thermodynamic Cycle:
 E  Q cycle  W cycle = 0
P
S1
Clockwise around the cycle:
Work is done by the system.
S2
Power Cycle: W cycle > 0
Counter clockwise around the cycle:
Work is done on the system.
S4
S3
Refrigeration Cycle: W cycle < 0
v
Sec 2.6: Energy Analysis of Cycles
Cycle Models:
Power cycle:
7
Effectiveness 
R esult
Effort
Refrigeration cycle:
Heat Pump Cycle
Q in  W cycle  Q out
W cycle  Q in  Q out
W cycle  Q in  Q out
W cycle
Q in
Q out

 
W cycle
Q in

Q in  Q out
Q in

Q in
Q out  Q in
 
W cycle

Q out
Q out  Q in
8
Concept Questions:
True or False:
a) In principle, expansion work can be evaluated using ∫pdV for both
actual and quasi-equilibrium expansion processes. False
b) The change in the internal energy of a system between two states is
the change in the total energy of the system between the two states
less the change of the system’s kinetic and gravitation potential
energies between these states.
True
c) The change in gravitational potential energy of a 2 lb mass whose
elevation decreases by 40 ft where g = 32.2 ft/s2 is -2576 ft-lbf . True
d) The rate of heat transfer from a hot baked potato to the ambient
air is greater with forced convection than natural convection. True
9
Example 3 (Problem 2.91):
A heat pump maintains a dwelling at 68oF. When operating steadily, the
power input to the heat pump is 5 hp, and the heat pump receives energy
by heat transfer from 55oF well water at a rate of 500 BTU/min.
a) Determine the COP.
b) Evaluating electricity at \$0.10 per kW-hr, determine the cost of
electricity in a month when the heat pump operates for 300 hr.
10
11
Example (2.70): A gas is contained in a vertical piston-cylinder assembly by a
piston weighing 1000 lbf and having a face area of 12 in2. The atmosphere exerts a
pressure of 14.7 psi on the top of the piston. An electrical resistor transfers energy
to the gas in the amount of 5 BTU as the elevation
Patm=14.7 psi
of the piston increases by 2 ft. The piston and
cylinder are poor thermal conductors and friction
h = 2 ft
can be neglected. Determine the change in internal
Apiston = 12 in2
energy of the gas, in BTU, assuming it is the only
Wpiston = 1000 lbf
significant internal energy change of any component
present.
Welec= - 5 BTU
12
13
Example (2.83): A power cycle has a thermal efficiency of 40% and
generates electricity at a rate of 100 MW. The electricity is valued at \$0.08
per kWh. Based on the cost of fuel, the cost to supply Qin is \$4.50 per GJ.
For 8000 hours of operation annually, determine, in \$
(a) the value of electricity generated per year
(b) the waste heat returned to the environment
(c) the annual fuel cost
(d) Is operation profitable?
Qin
Fuel
Wout =100 MW
Air
Qout
14
15
Example Problem (2.63)
A gas is compressed in a piston cylinder assembly form p1 = 2 bar to p2 = 8 bar,
V2 = 0.02 m3 in a process during which the relation between pressure and
volume is pV1.3 = constant. The mass of the gas is 0.2 kg . If the specific internal
energy of the gas increase by 50 kJ/kg during the process, deter the heat
transfer in kJ. KE and PE changes are negligible.
16
17
End of Lecture 05:
• Solutions to example problems follow
18
Example 3 (Problem 2.91) Page 1 of 2:
A heat pump maintains a dwelling at 68oF. When operating steadily, the
power input to the heat pump is 5 hp, and the heat pump receives energy
by heat transfer from 55oF well water at a rate of 500 BTU/min.
a) Determine the COP.
b) Evaluating electricity at \$0.10 per kW-hr, determine the cost of
electricity in a month when the heat pump operates for 300 hr.
Principle: COP for heat pump (written in terms of power)
Q in  W in  Q out
 
Q out
W in
W in  5 h p
where:
and
Q in  5 0 0 B tu / m in
therefore:
 
Q out
W in

Q in  W in
W in

Q in
W in
1 
500 btu / m in
1hp
60 m in
5 hp
2545 btu / h
1hr
 1  3.36
19
Example 3 (Problem 2.91)…page 2 of 2:
A heat pump maintains a dwelling at 68oF. When operating steadily, the
power input to the heat pump is 5 hp, and the heat pump receives energy
by heat transfer from 55oF well water at a rate of 500 BTU/min.
a) Determine the COP.
b) Evaluating electricity at \$0.10 per kW-hr, determine the cost of
electricity in a month when the heat pump operate for 300 hr.
Principle: Cost = Cost of energy * Power * time
where:
W in  5 h p
\$ / E  \$0.10 / kW  hr
 t  300 hr
therefore:
C\$ 
\$
E n erg y
 W in  t 
\$ 0 .1 0
kW  h r
 5hp  300hr
1 kW
1 .3 4 1h p
 \$ 1 1 1 .8 6
20
Example (2.70): Page 1 of 4
A gas is contained in a vertical piston-cylinder assembly by a piston weighing 1000
lbf and having a face area of 12 in2. The atmosphere exerts a pressure of 14.7 psi on
the top of the piston. An electrical resistor transfers energy to the gas in the
amount of 5 BTU as the elevation
of the piston increases by 2 ft. The piston and
Patm=14.7 psi
cylinder are poor thermal conductors and friction
can be neglected. Determine the change in internal
h = 2 ft
energy of the gas, in BTU, assuming it is the only
Apiston = 12 in2
significant internal energy change of any component
Wpiston = 1000 lbf
present.
Solution: Apply the 1st law of thermodynamics
 PE   K E   U  Q in  W out
 U  Q in  W o u t   P E   K E
Welec= - 5 BTU
21
Example Problem (2.70) …page 2 of 4
where
mg = 1000 lbf
A = 12 in2
Δh = 2 ft
Welec_in = 5 BTU
Because of the statement “poor thermal conductors”, it can be assumed that
this is an adiabatic process (Q = 0) and we will also assume that the process
occurs as a slow quasi-equilibrium process in which case the kinetic energy
terms will also be small (ΔKE = 0). Finally, since the piston floats on the
contained gas, the outside atmospheric pressure maintains a constant
pressure on the cylinder…so this is a constant pressure process (isobaric)
therefore:
KE  0
Q in  0
 P E  m g  h  (1 0 0 0 lb f )( 2 ft )  2 0 0 0 ft  lb f
W elect   5 B T U
(neg. since its put into the system)
V2
W PV 

V1
p d V  p (V 2  V1 )
(for constant pressure)
22
Example Problem (2.70) …page 3 of 4
Ftop=patm A
For equilibrium:
W=1000lbf
F  0
Fb o tto m  Fto p  W  0
p A  p a tm A  W  0
p  p a tm 
W
Fbottom=p A
 1 4 .7 lb f / in 
2
A
1 0 0 0 lb f
1 2 in
2
 9 8 .0 3 p si
and the increase in Volume:
V 2  V1  A  h
V 2  V1  1 2 in ( 2 ft )
2
1 2 in
 2 8 8 in
3
1 ft
therefore the work done by the gas was positive work by the system
V2
W PV 

V1
p d V  p (V 2  V1 )  (9 8 .2 lb f / in )( 2 8 8 in )
2
3
1 ft
1 2 in
 2 3 5 7 lb f  ft
23
Example Problem (2.70) …page 4 of 4
Returning to the 1st law:
 U  Q in  W out   PE   K E
 U  0  (W P V  W elec )   P E  0
 U   ( 2 3 5 7 ft  lb f  5 B T U
7 7 8 ft  lb f
1B T U
)  ( 2 0 0 0 ft  lb f )
 U   4 6 7 ft  lb f
 U   4 6 7 ft  lb f
1B T U
7 7 8 ft  lb f
  0 .6 0 B T U
24
Example (2.70): A gas is contained in a vertical piston-cylinder assembly by a
piston weighing 1000 lbf and having a face area of 12 in2. The atmosphere exerts a
pressure of 14.7 psi on the top of the piston. An electrical resistor transfers energy
to the gas in the amount of 5 BTU as the elevation
Patm=14.7 psi
of the piston increases by 2 ft. The piston and
cylinder are poor thermal conductors and friction
h = 2 ft
can be neglected. Determine the change in internal
Apiston = 12 in2
energy of the gas, in BTU, assuming it is the only
Wpiston = 1000 lbf
significant internal energy change of any component
present.
Welec= - 5 BTU
25
Example (2.83): Page 1 of 3
A power cycle has a thermal efficiency of 40% and generates electricity at
a rate of 100 MW. The electricity is valued at \$0.08 per kWh. Based on
the cost of fuel, the cost to supply Qin is \$4.50 per GJ. For 8000 hours of
operation annually, determine, in \$
(a) the value of electricity generated per year
(b) the waste heat returned to the environment
(c) the annual fuel cost
(d) Is operation profitable?
Qin
Fuel
Wout =100 MW
Air
Qout
26
Example (2.83): Page 2 of 3
A power cycle has a thermal efficiency of 40% and generates electricity at
a rate of 100 MW. The electricity is valued at \$0.08 per kWh. Based on
the cost of fuel, the cost to supply Qin is \$4.50 per GJ. For 8000 hours of
operation annually, determine, in \$
(a) the value of electricity generated per year and
R evenue   100 M W

10 kW  8000 hr   \$0.08  \$64  10



M W  year   kW  hr 
year
3
(b) next find the heat generated
1 0 0 M W
W out
W out
 

Q in 

Q in

0 .4 0

6
 250 M W
and the heat returned to the environment
Q in  W out  Q out

Q out  Q in  W out
 (250  100)M W  150 M W
27
Example (2.83): page 3 of 3
A power cycle has a thermal efficiency of 40% and generates electricity at
a rate of 100 MW. The electricity is valued at \$0.08 per kWh. Based on
the cost of fuel, the cost to supply Qin is \$4.50 per GJ. For 8000 hours of
operation annually, determine, in \$
(c) Cost of fuel?
C ost fuel
\$32.4  10
 \$4.50  G J s  8000 hr  3600 s
  250 M W  




3
year
 G J  10 M W  year  hr
(d) Is operation profitable?
Profit = Revenue – Costs
Profit

\$ 64  10
year
6

\$ 32 . 4  10
year
6

\$ 31 . 6  10
6
year
So, this could be profitable, but the calculation
ignore other costs such as capital and labor.
6
28
Example Problem (2.63) page 1 of 4
A gas is compressed in a piston cylinder assembly form p1 = 2 bar to p2 = 8 bar,
V2 = 0.02 m3 in a process during which the relation between pressure and
volume is pV1.3 = constant. The mass of the gas is 0.2 kg . If the specific internal
energy of the gas increase by 50 kJ/kg during the process, deter the heat
transfer in kJ. KE and PE changes are negligible.
Solution: starting with the 1st Law of Thermodynamics
 P E   K E   U  Q in  W o u t
where: ΔKE=0
ΔPE = 0
p1 = 2 bar
also:
p2 = 8 bar
ΔU/m = 50 kJ/kg
V1 = ?
m = 0.2 kg
V2 = 0.02 m3
pV1.3 = constant
therefore:
1 .3
p1V1
 p 2V 2
1 .3
 p 
V1   2 
 p1 
1 / 1 .3
 8 bar 
V2  

 2 bar 
1 / 1 .3
0 .0 2 m  0 .0 5 8 1 m
3
3
29
Example Problem (2.63)…page 2 of 4
A gas is compressed in a piston cylinder assembly form p1 = 2 bar to p2 = 8 bar,
V2 = 0.02 m3 in a process during which the relation between pressure and
volume is pV1.3 = constant. The mass of the gas is 0.2 kg . If the specific internal
energy of the gas increase by 50 kJ/kg during the process, deter the heat
transfer in kJ. KE and PE changes are negligible.
Solution continued:
also:p V
 p 2V 2
1 .3
1 .3
 (8 b a r )(0 .0 2 m )
3
p  0 .0 4 9 5 / V
1 .3
1 .3
 0 .0 4 9 5 b a r  m
 (0 .0 4 9 5 b a r  m
3 .9
)V
3 .9
 1 .3
therefore:
1 .3
p1V1
 p 2V 2
1 .3
 p 
V1   2 
 p1 
1 / 1 .3
 8 bar 
V2  

 2 bar 
1 / 1 .3
0 .0 2 m  0 .0 5 8 1 m
3
3
30
Example Problem (2.63)…page 3 of 4
so work done is:
V2
W 

V1
V2
PdV 

((0 .0 4 9 5 b a r  m
3 .9
0.0495 bar  m
3.9
)V
 1 .3
)dV 
0 .0 4 9 5 b a r  m
 0 .3
V1


 0 .3


 V1
3.9
 3.224 m  0.9  1.177 m  0.9 


 0.3

V
 (0.02 m 3 )  0.3  (0.581 m 3 )  0.3 


 0.3
0.0495 bar  m
V2
3 .9
0.0495 bar  m
3.9
 3.224 m  0.9  1.177 m  0.9 


 0.3
  0.338 bar  m
3
100 kN / m
1 bar
2
1kJ
1kN  m
  33.8 kJ
31
Example Problem (2.63)…page 4 of 4
Internal Energy is given as
 U  (  U / m )( m )  (5 0 kJ / kg )(0 .2 kg )  1 0 kJ
Finally back at the 1st Law:
 PE   K E   U  Q in  W out
gives
Q in   P E   K E   U  W o u t
 0  0  (1 0 kJ )  (  3 3 .7 8 kJ )
  2 3 .7 8 kJ
```