Lecture 15

• heat engines and refrigerators using ideal gas as working substance

• Brayton cycle

Refrigerators• closed cycle uses external work to

remove heat from cold reservoir and exhaust heat to hot reservoir (2nd law does not allow spontaneous): e.g. air-conditioner or kitchen...make air that is cooler than environment even colder

• exhaust more heat than removed from inside (cool room by leaving refrigerator door open?)

• coefficient of performance:

• perfect refrigerator ( ) forbidden by 2nd law (informal statement # 3): real refrigerator uses work ( )

!Eth = 0 (cyclical) : QH = QC + Win

Win = 0; K =!K <!

Example

• 1.0 L of 20 degree Celsius water is placed in a refrigerator. The refrigerator’s motor must supply an extra 8.0 W power to chill the water to 5 degree Celsius in 1.0 hr. What is the refrigerator’s coefficient of performance?

No perfect Heat Engine ( )• connect perfect engine to refrigerator: no net work for 2

combined, but heat transferred from cold to hot (not allowed by 2nd law)

• informal statement # 4: no perfect heat engine, must waste heat...

(! = WoutQH

>! 1 from energy conservation)

! = WoutQH

< 1

! = 1

Example of Proof by Contradiction

• want to prove statement “A” (e.g., there exists a perfect engine) is not true

• assume A is true, find a violation of basic law (e.g. ,2nd law of thermodynamics) assumption is incorrect, A is not true

• Using only energy conservation and heat not transferred from cold to hot, deduce heat engines and refrigerators exist; must use closed-cycle processes; no perfect...

• upper limit on ?!, K

Unanswered questions

Summary

Ideal gas Heat Engines

Ideal gas summary I

• closed cycle trajectory: clockwise for Wout > 0Wout = Wexpand ! |Wcompress| = area inside closed curve

Ideal gas summary II• depends only on T

• identify each process, draw pV diagram

• use ideal gas law to know n, p, V, T at one point

• use ideal gas law and equations for specific processes for p, V, T at beginning/end of each process

• calculate for each process

• by adding `s: confirm by area within curve

• add positive values of Q to find

• check:

Strategy for heat engine problems

(!Eth)net = 0, ! < 1, signs of Ws and Q...

QH

WsWout

Q, Ws and !Eth

Eth

Brayton cycle (heat engine)• adiabatic compression (1 2): raises T;

isobaric expansion (2 3): raises T further, heat by fuel; adiabatic expansion (3 4): spins turbine, T still high; isobaric compression (4 1): heat transferred to cooling fluid

• Thermal efficiency:Process 2! 3 (isobaric):QH = nCP (T3 " T2)Process 4! 1 (isobaric):QC = |Q41| = nCP (T4 " T1)# !B = 1" T4!T1

T3!T2Use pV = nRT and pV ! = constant (adiabatic)to give p(1!!)/!T = constant

# T1 = T2

!p2p1

"(1!!)/!= T2r

(1!!)/!p ,

where rp $ pmax

pminand T4 = T3r

(1!!)/!p

# !B = 1" 1

r(!!1)/!p

(increases with rp)

TH ! T3; TC " T1

! = 1! QC

QH

Brayton cycle (refrigerator)

• heat engine backward, ccw in pV: low-T heat exchanger is “refrigerator”

• sign of W reversed, area inside curve is : used to extract from cold reservoir and exhaust to hot...

• gas T lower than (1 4), higher than (3 2) gas must reach by adiabatic expansion, by adiabatic compression

QH

Win

QC

TC TH

T3(> TH)T1(< TC)

Comparison of Brayton cycle heat engine and refrigerator

• Brayton cycle refrigerator is not simply heat engine run backward, must change hot and cold reservoir: heat transferred into cold reservoir for heat engine , from cold reservoir in refrigerator ; heat transferred from hot reservoir for heat engine , into hot reservoir for refrigerator

• heat engine: heat transfer from hot to cold is spontaneous, extract useful work in this process via system...

• refrigerator: heat transfer from cold to hot not spontaneous, make it happen by doing work via system...

(TC ! T1)(TC ! T4)

(TH ! T3)(TH ! T2)