#### Transcript document

```ME 200 L5: Energy Accounting
Conservation of Energy for a Closed System
First Law of Thermodynamics
Spring 2014 MWF 1030-1120 AM
J. P. Gore, Reilly University Chair Professor
[email protected]
Gatewood Wing 3166, 765 494 0061
Office Hours: MWF 1130-1230
TAs: Robert Kapaku [email protected]
Dong Han [email protected]
Summary of L1 – L4
►Kinetic Energy and Potential Energy are macroscale mechanical energies of a mass.
►Internal Energy is an extensive property of a working
substance and is defined by composition, T, & P.
►Internal Energy + Kinetic Energy + Potential Energy
may transform amongst themselves and also
change by heat transfer and work interactions,
all subject to Laws of Thermodynamics!
►Heat Transfer is a result of temperature difference
►Work Transfer is all modes of energy transfer that
are not a result of temperature difference.
Mechanical Work Example
• An object of mass 80 lb (or
36.29 kg), initially at rest,
experiences a constant
horizontal acceleration of 12
ft/s2 (or 3.6576 m/s2) due to the
action of a resultant force that is
applied for 6.5 s. Determine the
work of the resultant force, in
ft-lbf, in Btu, in J and kJ.
• Given
–
–
–
–
m = 80 lb or 36.29 kg
V1 = 0 ft/s or 0 m/s
a = 12 ft/s2 or 3.6576 m/s2
t = 6.5 s
• Find
– W in ft-lbf and Btu?
– And W in J and kJ
• Sketch
R
m
• Assumptions
z
– The 80 lb (36.29 kg) mass is
the system.
– Motion is horizontal, so the
system experiences no
change in potential energy.
– The horizontal acceleration
is constant.
x
• Basic Equations
W 
a
dV
dt

1
m V22  V12
2

V2  V1  a t 2  t1 


3
Mechanical Work Example
• Solution
R
m
z
x
SI System
V2  0  3.6576m / s 2 6.5s  0
V2  23.77 m s
V2  V1  a t 2  t1 

 0
W

1
m V22  V12
2

British System
V2  0  12 ft s 2 6.5s  0
V2  78 ft s
1
1lbf
2
 7560 ft  lbf
80lb  78 ft s 
2
32.2lb  ft s 2
1Btu
W  7560 ft  lbf
 9.71Btu
778lbf  ft
W
(1) Work does not depend on the units we use to measure it! Different
numerical values are assigned to identical work in different measuring
systems!
(2) See Conversion factors on inside cover of book.
10.255 kJ = 9.71 Btu because 1Btu=1.0551 kJ or 1 kJ = 0.9478 Btu
4
Sign Convention
• Our sign convention for work is easy to
remember:
– Work done by a system is considered to be
useful to mankind, so is defined to be positive.
– Therefore work done by the accelerator on the
particle is positive.
– Of course, if the work for “the system” defined
as “the particle” is to be calculated, then it is
negative but equal in magnitude to the work
done by the accelerator.
5
Piston Cylinder Systems
• Piston cylinder systems are widely used
• There function is to transfer expansion and
compression energy change into linear
motion and eventually rotary motion.
Applications: I. C. engines, hydraulic jacks,
bicycle air-pump, balloon inflator etc.
Stroke=Crank circle diameter. The cylinder
length must clear the end to end motion of
the connecting rod. The clearance volume defines
the compression ratio. The BDC volume defines the
cylinder capacity. The pressure, temperature and volume
within the cylinder are related and determine power output.
Expansion and Compression Work
W  Fdx
F  PA Adx  dV
W  PdV
The above derivation is applicable to
experimental pressure volume traces (see Fig. 2.5 in text)
as well as theoretical approximations to
processes for defining the system behavior.
Work is a path function and can not be evaluated by just knowing the end states 1 and 2.
Also, in writing the above equations, the assumption that the pressure in the cylinder is
uniform through out the volume has been made. This makes the work a quasi-steady
approximation to reality. None the less, this approximation has been found to be very
useful in industry.
Expansion and Compression Work
W  Fgasdx  PgasA p dx
2
1W2   PgasdV
1
If gas volume decreases, work is negative and is done on the gas.
If gas volume increases, work is done by the gas on the piston and
hence on the connecting rod and the crank shaft etc.
Practice these derivations
2
P  Const.1W2  Pgas  dV  Pgas(V2  V1)
1
2 PV
V
PV  Const.1W2   ( 1 1 )dV  P1V1 ln 2
V
V1
1
2 PV n
P V  PV
n
PV  Const.1W2   ( 1 1 )dV  2 2 1 1
n
1 n
1 V
Examples of work functions involving
different processes: Shaft Work
W  ss2 F  d s
1
W  Fds
Fds d
W  d
W  t
W  
τ – torque
9
Examples of work functions involving different
processes: Spring Work
W  ss2 F  d s
1
Wspring  Fdx
F  kx

1
Wspring  k x22  x12
2

k – spring constant
xi – displacement from equilibrium
10
Examples of work functions involving different
processes: Work done by “flowing” electrons
Electric Power
W   i
i –
R ––
electric current (amp)
potential difference (V)
resistance (ohms)
Ohm’s Law
  Ri
11
1. Torsion of a solid bar
See eq. 2.18
2. Stretching of a liquid film
See eq. 2.19
3. Charging of Electrolytic cell, Electric Field, Magnetic Field
Work done by electromotive force
Work done by dielectric in a uniform electric field
Work done by magnetic material in a field
Learning Outcomes for Lecture 5
►Apply closed system energy balances,
observing sign conventions for work and
heat transfer.
►Conduct energy analyses of components
undergoing thermodynamic processes.
Closed System Energy Balance
►Energy is an extensive property that
includes the internal energy, the kinetic energy
and the gravitational potential energy.
►For closed systems, energy is transferred in
and out across the system boundary by two
means only: by work and by heat.
►Energy is conserved. This is the first law of
thermodynamics.
Closed System Transient Energy Balance
dE dKE dPE dU



dt
dt
dt
dt
Time rate of change of energy
contains kinetic, potential and
internal energy
dKE dPE dU  


 Q W
dt
dt
dt
time rate of change of
net rate at which
the energy contained
energy enters via heat
within the system at
transfer at time t
time t
time rate form of
the energy balance
net rate at which
energy is being
transferred by work
at time t
15
Closed System Transient Energy Balance
►The time rate form of the closed system energy
balance is
dE  
 Q W
(Eq. 2.37)
dt
►The rate form expressed in words is
time rate of change
of the energy
contained within
the system at
time t
net rate at which
energy is being
transferred in
by heat transfer
at time t
net rate at which
energy is being
transferred out
by work at
time t
Change in Energy of a System
►The changes in energy of a system from state 1
to state 2 consist of internal, kinetic and potential
energy changes.
E2 – E1 = (U2 – U1) + (KE2 – KE1) + (PE2 – PE1) (Eq. 2.27a)
(Eq. 2.27b)
E = U + KE + PE
►Energy at state 1 or state 2 or any other state is
defined in reference to a standard state.
►Definition of energy at all states must have
identical standard base state.
►Changes in the energy of a system between
states, defined with identical standard state
have significance.
Closed System Energy Balance
►The energy concepts introduced thus far are
summarized in words as follows:
change in the amount
of energy contained
within a system
during some time
interval
net amount of energy
transferred in across
the system boundary by
heat transfer during
the time interval
net amount of energy
transferred out across
the system boundary
by work during the
time interval
►Using previously defined symbols, this can be
expressed as: E2 – E1 = Q – W (Eq. 2.35a)
►Alternatively, KE + PE + U = Q – W (Eq. 2.35b)
In Eqs. 2.35, a minus sign appears before W because
energy transfer by work from the system to the surrounding is
taken as positive.
Example Problem
Imagine a party at a college location as sketched below. Bob goes to the refrigerator
door to get a soda…
Music
speakers
A/C
Vent
Well
insulated
party
room
Electrical
supply
cable
Refrigerator
(fridge) door open
Door
locked
Example 1
An electric generator coupled to a
windmill produces an average
power of 15 kW. The power is used
to charge a storage battery. Heat
transfer from the battery to the
surroundings occurs at a constant
rate of 1.8 kW. For 8 h of
operation, determine the total
amount of energy stored in the
battery, in kJ.
Given
W = -15 kW
Q = -1.8 kW
Δt = 8 h
Assumptions
The battery is a closed system.
The work and heat transfer rates
are constant.
Find: ΔE in kJ?
System
W = ?15 kW
storage
battery
Q = ?1.8 kW
Δt = 8 h
20
Example 1
Basic Equations
E  Q  W
Q  tt 2 Q dt
Q  Q t
W  tt 2 W dt
W  W t
1
1
1kJ s 3600s
W  W t   15kW 8h
1kW
1h
W  4.32105 kJ
E  51,800   4.32  105   3.8  105 kJ
Solution
1kJ s 3600s

Q  Qt   1.8kW 8h
1kW
1h
Q  51,800kJ
21
Example 2
An electric motor draws a current of
10 amp with a voltage of 110 V.
The output shaft develops a
torque of 10.2 N-m and a
rotational speed of 1000 RPM.
determine for the motor, each in
kW.
the electric power required.
the power developed by the output
shaft.
the rate of heat transfer.
• Find
– Welectric in kW?
– Wshaft in kW?
– Q in kW?
Sketch
I = 10
amp
V = 110 V
+
motor
-
τ = 10.2 N-m
ω = 1000 RPM
Given
I = 10 amp
V = 110 V
τ = 10.2 N-m
ω = 1000 RPM
• Assumptions
– The motor is a closed system.
– The system is at steady state.
• Basic Equations
dE  
 Q W
dt
W shaft  
Welectric  I
22
Example 2
• Given
–
–
–
–
• Basic Equations
I = 10 amp
V = 110 V
τ = 10.2 N-m
ω = 1000 RPM
dE  
 Q W
dt
Welectric  VI
W shaft  
• Solution
1Watt am p 1kW
Welectric  110V 10am p
1volt
103W

W
 1.1kW
• Find
electric
– Welectric in kW?
– Wshaft in kW?
– Q in kW?
rev  2 rad 1 min
1kW

W shaft  10.2 N  m1000

min  rev
60s 103 N  m s

W shaft  1.07kW
0
• Sketch
I = 10 amp
V = 110 V
+
motor
-
τ = 10.2 N-m
ω = 1000 RPM
dE  
Q  W
 Q W
dt
Q  Welectric  W shaft
Q  1.1kW  1.07kW
Q  0.03kW
23
Example 3
A gas within a piston-cylinder assembly (undergoes a
thermodynamic cycle consisting of) three processes:
– Process 1-2: Constant volume, V = 0.028 m3, U2 – U1 = 26.4 kJ.
– Process 2-3: Expansion with pV = constant, U3 = U2.
– Process 3-1: Constant pressure, p = 1.4 bar, W31 = -10.5 kJ.
There are no significant changes in kinetic or potential
energy.
1.
2.
3.
4.
Sketch the cycle on a p-V diagram.
Calculate the net work for the cycle, in kJ.
Calculate the heat transfer for process 2-3, in kJ.
Calculate the heat transfer for process 3-1, in kJ.
24
Example 3
• Find
–
–
–
–
• Assumptions
p-V diagram
Wnet = ? in kJ
Q23 = ? in kJ
Q31 = ? in kJ
• System
gas
• Given
– 1-2: V = 0.028 m3, U2 – U1 =
26.4 kJ
– 2-3: pV = constant, U3 = U2
– 3-1: p = 1.4 bar, W31 = -10.5 kJ
– The gas is the closed system.
– For the system, ΔKE = ΔPE
= 0.
– Volume change is the only
work mode.
• Basic Equations
E  Q  W
E  KE  PE  U
W 
2
1
pd 
25
Example 3
•
Solution
Wcycle  W12  W23  W31
0
W12  VV2 pdV
1
W12  0
W23  VV3 pdV
2
p
c
V
c
dV
V
dV  c VV3
 cln V 3
V2
2 V
2 V
V
V
W23  c ln 3  p3V3 ln 3
V2
V2
W23  VV3
W31  VV1 pdV  p V1  V3 
3
V3  V1 
W31
p
26
Example 3
•
Solution
W
V3  V1  31
p
 10.5kJ
1bar
103 N  m
V3  0.028m 
1.4bar 105 N m2
1kJ
3
V3  0.103m3
V
W23  p3V3 ln 3
V2

3

W23  1.4bar 0.103m ln


0.103m3
0.028m3
105 N m2
0.103m3
1kJ
3
W23  1.4bar
0.103m ln
1bar
0.028m3 103 N  m
W23  18.78kJ
27
Example 3
•
Solution
Wcycle  W12  W23  W31
Wcycle  0  18.78   10.5   kJ
Wcycle  8.28kJ
0
0
0
KE  PE  U  Q23  W23
Q23  18.78kJ
Q23  W23
0
Q31  U1  U3   W31
Q31  26.4kJ  10.5kJ 
Q31  36.9kJ
0
KE  PE  U  Q31  W31
Q31  U1  U3   W31
0
U2  U1  U3  U2   U1  U3   0
U1  U3   U2  U1
U1  U3  26.4kJ
28
```