Lecture 02: Work and Energy

Report
Chapter 2
Lecture 02:
Work and Energy
Today’s Objectives:
•
•
•
•
•
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Be able to distinguish between work and energy.
Be able to calculate Kinetic Energy
Be able to calculate Potential Energy
Be able to calculate Work done by an acting force
Be able to calculate Power
Be able to calculate Work done by gases undergoing changes
of state.
• Be able to explain the concept of Internal Enegy
Reading Assignment:
• Read Chap 2. Sections 1-5
Homework Assignment:
From Chap 2: Problems 6, 20, 24, 32
Sec 2.1: Reviewing Mechanical Concepts of Energy
Work, Heat, and Energy
Energy is conserved, but can be converted to different types
Ways to Transfer Energy Into or Out of A System
Work – transfers by applying a force and causing a
displacement of the point of application of the force.
Mechanical Waves – allow a disturbance to propagate
through a medium.
Heat – is driven by a temperature difference between two
regions in space.
Matter Transfer – matter physically crosses the boundary
of the system, carrying energy with it.
Electrical Transmission – transfer is by electric current.
Electromagnetic Radiation – energy is transferred by
electromagnetic waves
W = PE = KE = U
3
Sec 2.2: Broadening Our Understanding of Work
Thermodynamics “Work”
Physics definition of work is W = F s
But, in thermodynamics often we are working with
fluids (non-solids), so we need a broader definition.
“Work is done by a system on
its surroundings if the sole
effect on everything external
to the system could have
been done by raising (or
dropping) a weight.”
4
Sec 2.2: Broadening Our Understanding of Work
Joule’s Experiment (1845-Salford, England)
Joule dropped a known mass and measured the change in
temperature of the water.
The experiment was conducted in the basement of his family’s
brewery, where there was a constant ambient temperature.
The friction of the water molecules rubbing together caused the
temperature to increase.
Joule’s Equipment - Manchester
5
Sec 2.2.1: Sign Convention
6
Work Sign Convention
W > 0 : Work done BY the system
W < 0 : Work done ON the system
Sign is not inherently important, but this is the convention.
Vf
W   PdV
Vi
W < 0 : Work done ON the system
(System is compressed)
W > 0 : Work done BY the system
(System expands)
Sec 2.2 .2: Power
Power = Rate of Energy Transfer
Book’s convention: Dot above symbol represents the rate.
The rate of work can be expressed as
7
Sec 2.2 : Work
8
Work Properties
Work is NOT a property of a system like V, T, or E.
Work occurs when the system undergoes a process.
A differential of a property is exact.
V  
Vf
Vi
dV  V f  Vi
The differential of work depends upon the path.
W 
Vf
V
Vi
pdV
Sec 2.2 : Work
9
But, work depends on the process.
For “Bobby” work
depends on the path,
since friction is a nonconservative force.
Sec 2.2 : Work
So, we need to have a PV relationship for
the process.
The process of changing volume is NOT
necessarily in equilibrium.
- He balloon popping, gas does not
instantly mix with air
- Gas cylinder rupture, pressure inside is
higher then outside for some time, t
For this class, we will used an idealized process, that are
completely reversible. We call this type of process
- quasi-equilbirium
- quasi-static
10
Sec 2.2 : Work
11
V
V
V
Non-quasi-static Process
Consider a box initially divided in half.
- Initially, one is filled with gas, the other a vacuum.
- The divider is then removed.
- The gas takes some time to fill the new volume.
During that time, there are different local values for P
in the volume. There is also likely some heat
generated, as the process is irreversible.
Thermodynamics ≠ Kinetics
Sec 2.2 : Work
12
V
V
V
V
V
V
V
V
Quasi-static
Now we move the wall slowly, such that the gas is able to
adjust instantly. This is a reversible quasi-static process.
13
V
Example (2.34): Air contained within a piston-cylinder assembly
undergoes three processes in series. Evaluate W.
Process 1-2: Compression at constant
pressure from p1=10 psi, V1=4.0 ft3 to state 2
Process 2-3: Constant volume heating to state 3, where p3=50 psi
Process 3-1: Expansion to the initial state, during which the p-V
relationship is pV = constant.
50
P psi
10
 
V ft
3
4
14
Sec 2.3: Broadening Our Understanding of Energy
15
Energy
Physics energy types
Kinetic Energy: Energy of objects in motion
Potential Energy: Energy of objects in a field (g,E,B)
Internal Energy
Spring
Chemical
Pressure
Pressure can be a form of energy if P> Patm
Thus, the general energy
equation is
E  PE  KE  U
P
Patm
16
Example Problem (2.37) A 10 V battery supplies a constant current of
0.5 amp to a resistance for 30 min.
a) Determine the resistance, in ohms.
b) For the battery, determine the amount
of energy transfer to work, in kJ.
Solution:
17
Example Problem (2.31)
Air contained within a piston-cylinder assembly is slowly heated. As
shown in Fig P2.31, during the process the pressure first varies linearly
with volume and then remains constant. Determine the total work in
kJ.
P (kPa)
Solution:
2
150
100
3
1
50
0.030 0.045
0.070
V (m3)
18
End of Lecture 02
• Slides that follow show solutions to Example
problems.
19
Example (2.34): Air contained within a piston-cylinder assembly
undergoes three processes in series. Evaluate W.
Process 1-2: Compression at constant pressure from P1=10 psi, V1=4.0 ft3
to state 2
3
50
Process 1-2: Isobaric Process
W12   PdV  P V2  V1 
V2
P psi
V1
since:
10
1
2
 
V ft
3
4
PV
1 1  PV
3 3
then : V2  V3 
and
V2  V3
P1
V1
P3
 10 psi 
3
3
V2  
4
ft

0.8
ft



50
psi


W12  10 psi0.8  4 ft
3 144  ft 2
in2
BTU
778  ftlb f
 5.92BTU
20
Example (2.34): Air contained within a piston-cylinder assembly
undergoes three processes in series. Evaluate W.
Process 2-3: Constant volume heating to state 3, where P3=50 psi
Process 3-1: Expansion to the initial state, during which the P-V
relationship is PV = constant.
50
3
Process 2-3: Isovolumetric Process
P psi
10
2
1
 
V ft
3
4
V23  0  W23  0
21
Example (2.34): Air contained within a piston-cylinder assembly
undergoes three processes in series. Evaluate W.
Process 2-3: Constant volume heating to state 3, where P3=50 psi
Process 3-1: Expansion to the initial state, during which the P-V
relationship is PV = constant.
50
3
Process 3-1: Isothermic Process
P psi
V1
W31  
V3
10
2
1
 
V ft
3
C
PdV  
dV  C ln
V3 V
V1
W31  P1V1 ln
4
W31  10 psi   4 ft
3
 
V1
V3
 
V1
V3
 ln 
4 ft 3
0.8 ft
3

144  f 2t
in
2
BTU
 778 ft lb f
 11.9 BTU
22
Example Problem (2.37) A 10 V battery supplies a constant current of
0.5 amp to a resistance for 30 min.
a) Determine the resistance, in ohms.
b) For the battery., determine the amount
of energy transfer to work, in kJ.
Solution: Electrical Work Principle
Pelec  VI
therefore: V = 10 V
Welec  Pelec t
I = 0.5 A
Δt=30 min
Pelec  VI  (10V )(0.5 A)  5 W
then
Welec  (5W )(30 min)
60 s 1 J / s 1 kJ
1 min 1 W 1000 J
Welec  0.15 kJ
23
Example Problem (2.31)
Air contained within a piston-cylinder assembly is slowly heated. As shown in
the figure, during the process the pressure first varies linearly with volume
and then remains constant. Determine the total work in kJ.
P (kPa)
2
3
Solution:
150
VB
1
WA B  p dV
100

VA
Conceptually, this represents
the area of the P-V plot
underneath the process lines.
W12 
50
V2
 p dV  A
PV _ trapezoid
V1
0.030 0.045
W23 
0.070
V (m3)
V3
 p dV  A
PV _ rectangle
V2
1
 (100  150)(0.045  0.030) kPa  m3
2
2
1 kJ
3 1kN / m
 1.875 kPa  m
 1.875 kJ
1kPa 1 kN  m
 (150)(0.070  0.045) kPa  m3
1kN / m 2 1 kJ
 3.75 kPa  m
 3.75 kJ
1kPa 1 kN  m
3
Wtotal  W13  W12  W23  1.875  3.75  5.625 kJ

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