1. Ch19 Heat Engines and Refrigerators
講者： 許永昌 老師

2. Contents Turning Heat into Work Energy transfer diagrams
Heat engine: Thermal Efficiency
Refrigerators:
Coefficient of performance.
Practical applications:
Ideal-gas Heat Engines
Ideal-gas Refrigerators
The Limits of Efficiency
The Carnot Cycle.

3. Turning Heat into Work (請預讀P567)
Example:
The gas do positive work on the piston.
W< (work done by the environment on the system)
Ws=-W (work done by the system)
W+Q=DEth.
Q=Ws+DEth.
Based on the 2nd Law
Tflame > Tf > Ti.
System
Emech+Eth Q W

4. Energy Transfer diagram (請預讀P568)
Three diagrams:
Phase diagrams: tell us the phase of a matter
PV diagram for ideal gas:
tell us the change of state of the described system
If n is provided, we can get P, V, T and W.
If CV is provided, we can get DEth and Q.
Energy Transfer diagram for all substance:
Emphasize the ideas about the transfer and transformation of energy for a cyclic process.
Cyclic process: A system will return to its initial state at the end of process. P V

5. Energy Transfer diagram (continue)
Energy reservoir:
An object or a part of the environment so large that its temperature does not change when heat is transferred between the system and the reservoir.
Hot reservoir: TH > Ts.
Cold reservoir: TC < Ts.
Where Ts is the temperature of the system and it is changeable during the process.
QH & QC:
Because we already use the arrow to describe the direction of energy transfers, therefore, QH & QC >0 in this diagram .

6. Energy Transfer Diagram (continue)
Properties:
The process described in this diagram must be a cyclic process for reuse (不然就不是引擎了).
Work can be transformed into heat with 100% efficiency.
There are no perfect engines that turn heat into work with 100% efficiency.
Stop to think: In Ch17, we get Q=Wout of an isothermal process, i.e. the heat is 100% transformed into work. Is the 3rd property wrong? Th Th Tc Tc
No perfect engine

7. Heat engine: Thermal Efficiency (請預讀P570~P572)
It must be a closed-cycle device for reuse.
DEth=0.
Because of the 1st Law of thermodynamics (energy conservation), we get QH=Wout+QC.
Thermal efficiency:
h~0.1~0.5 for real engine.

8. A Heat-Engine Example Description: PV diagram:
Energy Transfer diagram:
12 2 1
23
31
Stop to think:
How to calculate its h?
TH & TC?

9. Refrigerators Similar to the heat engine.
It pumps energy from the cold reservoir to the hot one.
The coefficient of performance (COP):
Stop to think:
Since the 2nd Law says that heat is not spontaneously transferred from a colder to a hotter object, how can we transfer energy from the cold reservoir to the hot reservoir?
Adiabatic
process
Th’ Th Tc
Tc’

10. No perfect Heat Engine Proof: Suppose we had a perfect engine
We could use its output to provide the work input to the refrigerator.
The combined system becomes a perfect refrigerator .
It conflicts with the 2nd Law of thermodynamics.
In fact, we can get the upper limit of the efficiency.

11. Homework
Student Workbook
2, 3, 4, 5

12. Ideal-Gas Heat Engines(請預讀P575~P579)
Strategy:
Model:
Identify each process in the cycle.
Visualize:
Draw the PV diagram of the cycle.
Solve:
Use the ideal-gas law to complete your knowledge of n, P, V, and T at one point in the cycle.
Calculate Q, Ws, and DEth for each process by W+Q=DEth.
Add just the positive values of Q to find QH.
Wout=SiWi.
Calculate the quantities you need to complete the solution.
Assess:
Is (DEth)net=0?
Do all the signs of Ws and Q make sense?
Does h have a reasonable value?

13. Ideal-gas summary Table 19.1 Table 19.2 Process Gas Law Work W Heat Q
這些表上除了PV=nRT外，都是各位可以推得的，請務必會推導。
這些乃是Ch 16~Ch 18的重點。
Table 19.1
Table 19.2
Process
Gas Law
Work W
Heat Q
DEth
Isochoric
DV=0
Q=nCVDT
DEth=Q
Isobaric
DP=0
-PDV
Q=nCpDT
DEth=Q+W
Isothermal
DT=0
nRTln(Vf/Vi)
Q=-W
DEth=0
Adiabatic
nCVDT
Q=0
DEth=W
Any
PV=nRT
Area under curve
Monatomic
Diatomic
Eth
3/2*nRT
5/2*nRT CV
3/2*R
5/2*R CP
7/2*R g
5/3~1.67
7/5=1.4

14. Work done during one full cycle
For a heat engine, the gas must go around the PV trajectory in a clockwise direction for Wout to be positive.
Besides, a refrigerator uses a counterclockwise(ccw) cycle.

15. Stop to Think (請預讀P577) Stop to think 19.3
What is the thermal efficiency of this heat engine?
0.10
0.50
0.25 4
Can’t tell without knowing QC.
P(Pa)
QH=4000J
40,000
20,000
V(m3)
0.1
0.2

16. Some Examples of Heat Engines
Otto Cycle:
Diesel Cycle:
Brayton Cycle:

17. The Brayton Cycle (請預讀P578~P579)
12 & 34: adiabatic
PVg=constant.
23 & 41: isobaric
Thermal Efficiency:

18. The conditions of TH & TC (請預讀P580)
Key concept: The heat is always transferred from a hotter object to a colder object.
Heat Engine:
THMax(T of Q>0 processes).
TCMin(T of Q<0 processes).
Refrigerator:
THMin(T of Q<0 processes).
TCMax(T of Q>0 processes).

19. Homework
Student Workbook
8, 9, 10, 14

20. Ideal-Gas Refrigerators (請預讀P579~P581)
To make a Brayton refrigerator you must both reverse the cycle and change the hot and cold reservoirs.
Example 19.3 is a very nice example.
TH T2=108oC
TCT4=-23oC

21. Homework
Student Workbook 15

22. The Limits of Efficiency (請預讀P582~P588)
很多課本都是由此推導出entropy的概念，可見得有多重要。
What is a perfectly reversible engine?
It has the maximum efficiency.
Q:Compare to what?
A perfectly reversible engine must use only two types of processes.
Q: How about the other processes?
The Carnot Cycle and its efficiency.
Entropy.

23. What is a perfectly reversible engine?
A device that can be operated as either a heat engine or a refrigerator between the same two energy reservoirs and with the same energy transfers, with only their direction changed.
A perfectly reversible engine has maximum efficiency.

24. A perfectly reversible engine must use only two types of processes (請預讀P584)
Adiabatic Process:
Frictionless mechanical interactions with no heat transfer (Q=0).
Isothermal Process:
During this process, the heat should be transferred infinitely slowly.  reversible.
Any engine that uses only these two types of processes is called a Carnot engine.

25. The Carnot Cycle (請預讀P585~P588)
Any Carnot engine operating between TH and TC must have exactly the same efficiency as any other Carnot engine operating between the same two energy reservoirs.
The Carnot Cycle is an ideal-gas cycle.
Find the thermal efficiency Wout/QH:
Known: TH and TC.
Ideal gas: PV=nRT.
Adiabatic Process:
TVg-1=C=constant.
Q=0.
W=DE-Q=nCVDT.
Isothermal Process:
W=-PdV=-nRTD(lnV)
Q=DE-W=-W.
We get 5 10 15 20 30 40 50 60 70 3 QH 4 2 QC 1

26. The maximum efficiency
The thermal efficiency:
The coefficient of performance:
dQrevTdS.
S is entropy and it is a state variable.

27. Homework Student Workbook: Student Textbook:
17, 18.
Student Textbook:
19.15, 19.19, 19.43, 19.61
請自行製作terms and notation 的卡片。