1. Redo due Thursday, Dec. 4th Presentations for Exam 3 (or 1 or 2) ???
Announcements
3rd exam
Redo due Thursday, Dec. 4th
Presentations for Exam 3 (or 1 or 2) ???
if ?? when?? how many??
Final exam, Wednesday, Dec. 10th
Extra problem solving sessions??
Physics colloquium Thursday, Dec. 4th at 4 PM
Professor Scott Wollenwebber, WFU School of Medicine -- "Positron Emission Tomography: From Basic Physics to Functional Images"
Today’s lecture – analysis of thermodynamic processes
Efficiency of a thermodynamic process
Carnot, Otto, Diesel processes
Notion of entropy
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PHY Lecture 21

2. Review of thermodynamic ideas
“First law” of thermodynamics: DEint = Q - W
For an ideal gas: PV = nRT
Special cases: Isovolumetric (V=constant) W = 0
Isobaric (P=constant)
Isothermal process (T=constant) DEint = 0
Adiabatic process (Q = 0)
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PHY Lecture 21

3. Extra credit:
Show that the work done by an ideal gas which has an initial pressure Pi and initial volume Vi when it expands adiabatically to a volume Vf is given by:
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4. Examples process by an ideal gas:
P (1.013 x 105) Pa Vi Vf Pi Pf A B C D
A®B
B®C
C®D
D®A Q W
Pf(Vf-Vi)
-Pi(Vf-Vi)
DEint
Efficiency as an engine:
e = Wnet/ Qinput
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5. Otto cycle:
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6. P d c b e
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7. P Otto cycle QAB=0 QCD=0 Example: r=5, g=1.4 e=0.475 11/13/2018
PHY Lecture 21

8. Example: VD/VC=5, VA/VB=15, g=1.4 e=0.558
Diesel cycle
QAB=0
QCD=0 D A V
Example: VD/VC=5, VA/VB=15, g= e=0.558
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PHY Lecture 21

9. Stirling engine A P Th D Example: B Th = 3Tc VB=VC=5VA=5VD g = 1.3 Tc
A®B
B®C
C®D
D®A Q W
DEint P Th D
Example:
Th = 3Tc VB=VC=5VA=5VD g = 1.3 B Tc C
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10. Carnot process
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11. Carnot cycle A®B B®C C®D D®A Q W DEint 11/13/2018W
DEint
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12. Carnot cycle
QBC=0
QDA=0
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13. Examples
Efficiency of a Carnot engine operating between the temperatures of Tc=0oC and Th=100oC:
 For a Carnot engine, it is clear that we cannot achieve e=100%; not possible to completely transform heat into work. It is possible to show that the Carnot engine is the most efficient that one can construct between the two operating temperatures Tc and Th.
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14. Carnot cycle for cooling and heating
“coefficient of performance”
COPheating=|Qh/W|=Th/(Th-Tc)
COPcooling=|Qc/W|=Tc/(Th-Tc)
Example: Suppose that on a cold winter day, a heat pump has a compressor which brings outdoor air at Tc=-3oC into a room at Th=22oC. What is the COP?
COP=295.15/25=11.8
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15. More about Carnot cycle
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16. Carnot cycle shown in a T-S diagram:
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17. Other examples of entropy calculations: Ideal gas:
Isovolumetric process: Isobaric process:
Melting of solid having mass m and latent heat L at melting temperature TM:
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18. Peer instruction question:
Consider the “square cycle” shown below. What can you say about the entropy change in each cycle:
P (1.013 x 105) Pa Vi Vf Pi Pf A B C D
SABCDA= 0
SABCDA> 0
SABCDA< 0
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