Lecture 17 heat engines and refrigerators
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1) Heat engines operate in a thermodynamic cycle to absorb heat (Q>0) from a hot reservoir and use it to do mechanical work (W>0), dumping excess heat (QC<0) into a cold reservoir. 2) The efficiency of a heat engine is defined as the work done (W) divided by the total heat absorbed (QH), and is always between 0 and 1. 3) Refrigerators and heat pumps operate using the same thermodynamic cycles as heat engines but with the goal of maximizing heat transfer (QC for refrigerators, QH for heat pumps) for a given expenditure of work (W). Their performance is quantified by coefficients of performance K that can be greater than 1
Lecture 17 heat engines and refrigerators
- 1. Lecture 17 Heat engines and refrigerators.
- 2. Heat engine = device with a working substance (eg. gas) that operates in a thermodynamic cycle. In each cycle, the net result is that the system absorbs heat (Q > 0) and does work (W > 0). Examples: - Car engine: burns fuel, heats air inside piston. Piston expands, does mechanical work to move car - Animal: burns “food” to be able to move
- 3. Hot and cold reservoirs Stages of the cycle –Absorb heat from hot reservoir (QH) –Perform mechanical work (W ) –Dump excess heat into cold reservoir (QC < 0) Reservoir = large body whose temperature does not change when it absorbs or releases heat.
- 4. Energy flow Working substance in engine completes a cycle, so ΔU = 0: (Q H + QC ) −W = 0 W = QH + QC = QH − QC This relation follows naturally from the diagram (QH “splits”). Draw it every time!
- 6. Limitations We are not saying that you can absorb 10 J of heat from a hot source (a burning fuel) and produce 10 J of mechanical work... You can absorb 10 J of heat from a hot source (a burning fuel) and produce 7 J of mechanical work and release 3 J into a cold source (cooling system). … so at the end you absorbed 10 J but used (= converted to work) only 7 J. (We’ll see later that it is impossible to make QH = W, or QC = 0)
- 7. Efficiency Efficiency = what you use what you pay for For a heat engine: e = W QH 0<e <1 Example: A heat engine does 30 J of work and exhausts 70 J by heat transfer. What is the efficiency of the engine? W = 30 J QC = 70 J ⇒ QC = −70 J QH = W − QC = 100 J e= W = 0.3 (or 30%) QH
- 8. ACT: Two engines Two engines 1 and 2 with efficiencies e1 and e2 work in series as shown. Let e be the efficiency of the combination. Which of the following is true? e1 = A. e > e1 + e2 B. e = e1 + e2 e2 = C. e < e1 + e2 W1 TH Q1 Q1 W2 Q2 W1 e1 Q2 e= W1 +W2 Q1 = W1 Q1 = Q2 +W1 > Q2 Q1 + W2 Q1 < W1 Q1 1 1 < Q1 Q2 + W2 Q2 e2 W2 Q3 TC
- 9. DEMO: Stirling engine The Stirling engine 1 2 ΔV Gas warms up 2: isotherm c a 4: isotherm Hot gas hot water 100°C d 3: isochoric 1: isochoric b 3 hot water 100°C 4 ΔV Gas cools down Cold gas Room temperature 20°C Room temperature 20°C
- 10. Internal combustion engines The heat source (fuel combustion) is inside the engine and mixed with the working substance (air) - Otto (4 stroke gasoline) - Diesel Note: No real cold and hot reservoirs.
- 11. Otto cycle Idealization of the four-stroke gasoline engine Start QC
- 12. Intake: • mix of air and fuel enter • at patm • n increase QC Compression: Adiabatic compression: • temperature increase • no heat exchange • work done on the gas (small because of small pressure) QC
- 13. Combustion: Heating at constant volume • No work QC Power stroke: Adiabatic expansion •Temperature decrease • No heat exchange •Work done by the gas (large because of large pressure) QC
- 14. Heat reject: When piston at the bottom, very fast cooling, i.e. at constant volume • Excess heat absorbed by water jacket • Valve opens → Pressure drops to patm QC Exhaust: n decrease QC
- 15. Compression ratio r = Vmax Vmin compression ratio Compression: T2V2γ −1 =TV1 γ −1 1 γ −1 T2 =T1r γ −1 γ −1 T3 =T4r γ −1 =T1 ( rV2 ) Expansion: TV3γ −1 =T4V4γ −1 3 =T4 ( rV3 ) QC
- 16. Efficiency of the Otto cycle QH = nCV (T3 −T2 ) > 0 QC = nCV (T1 −T4 ) <0 W QH + QC e= = QH QH = T2 =T1r γ −1 T3 =T4 r γ −1 = = QC T3 −T2 +T1 −T4 T3 −T2 T4r γ −1 −T1r γ −1 +T1 −T4 T4r γ −1 −T1r γ −1 (T 4 ( −T1 ) r γ −1 − 1 (T 4 −T1 ) r γ −1 ) e =1− 1 r γ −1
- 17. In-class example: Otto’s engine efficiency Two idealized Otto cycles have a compression ratio of 5 and 10, respectively. What is the ratio of their efficiencies? Take the gas mixture to be a diatomic gas. e ( r = 10 ) e ( r = 5) A. 1.27 =? Diatomic gas: B. 1.33 C. 1.50 D. 1.67 E. 2.00 e ( r = 10 ) e ( r = 5) 7 R CP 7 γ = = 2 = = 1.4 CV 5 5 R 2 1− = 1 101.4 −1 1− 1 51.4 −1 = 1.27
- 18. Efficiency of Otto Cycle 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 Monatomic Diatomic Nonlinear' 1 5 9 13 17 Conmpression Ratio (V2/V1) Compression Why not simply use a higher compression ratio? • V2 big huge, heavy engine • V1 small temp. gets too high premature ignition need to use octane in gas to raise combustion temperature
- 19. Real four-stroke engine The Otto cycle is an idealization: • assumes ideal gas • neglects friction, turbulence, loss of heat to walls For r = 8 and γ = 1.4 (air), e = 0.56 Realistic cycle of 4-stroke engine e ~ 0.3
- 20. Diesel engine • Intake and compression happen without fuel. • Fuel is injected after compression, and keeps pressure constant. • Compression rate r is 15-20 → Larger temperatures → Fuel ignites spontaneously • Ideal efficiency e ~ 0.65-0.70
- 21. Refrigerators • Absorb heat from cold reservoir (QC > 0) • Work done on engine (W < 0) • Dump heat into hot reservoir (QH > 0) Energy balance: W = QH + QC W = QH − QC (We want as much QC while paying for the smallest possible W .) Coefficient of performance (refrigerator) K = QC W 0 <K < ∞
- 22. Heat pumps A very efficient way to warm a house: bring heat from the colder outside. Inside of house TH Heat pump This time we are interested in QH : Coefficient of performance (heat pump) K = QH W 1<K < ∞ Outside of house TC Same energy diagram as refrigerator