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PHYSICS 231 Lecture 31: Engines and fridges

PHYSICS 231 Lecture 31: Engines and fridges. Remco Zegers Question hours: Thursday 12:00-13:00 & 17:15-18:15 Helproom. Metabolism. U=Q+W. Work done (negative). Heat transfer: Negative body temperature < room temperature. Change in internal energy: Must be increased: Food!. t. t.

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PHYSICS 231 Lecture 31: Engines and fridges

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  1. PHYSICS 231Lecture 31: Engines and fridges Remco Zegers Question hours: Thursday 12:00-13:00 & 17:15-18:15 Helproom PHY 231

  2. Metabolism U=Q+W Work done (negative) Heat transfer: Negative body temperature < room temperature Change in internal energy: Must be increased: Food! PHY 231

  3. t t t Metabolic rate U = Q + W Metabolic rate: rate in which food and oxygen are transformed into internal energy (to balance losses due to heat loss and work done). |W/t| Body’s efficiency: |U/t| PHY 231

  4. Body’s efficiency U/t~oxygen use rate can be measured W/t can be measured PHY 231

  5. Types of processes A: Isovolumetric V=0 B: Adiabatic Q=0 C: Isothermal T=0 D: Isobaric P=0 PV/T=constant First law of thermo- Dynamics: U=Q+W PHY 231

  6. Isovolumetric processes (line A) • V=0 • W=0 (area under the curve is zero) • U=Q (Use U=W+Q, with W=0) • In case of ideal gas: • U=3/2nRT • if P then T (PV/T=constant) • so U=negative Q=negative • (Heat is extracted from the gas) • if P then T (PV/T=constant) • so U=positive Q=positive • (Heat is added to the gas) p v PHY 231

  7. Adiabatic process (line B) p • Q=0 • No heat is added/extracted from the • system. • U=W (Use U=W+Q, with Q=0) • In case of ideal gas: • U=3/2nRT • if T • U=negative W=negative • (The gas has done work) • if T • U=positive W=positive • (Work is done on the gas) v PHY 231

  8. isothermal processes p • T=0 • The temperature is not changed • Q=-W (Use U=W+Q, with U=0) • if V • W=positive Q=negative • (Work is done on the gas • and energy extracted) • if V • W=negative Q=positive • (Work is done by the gas • and energy added) v PHY 231

  9. isobaric process p • P=0 • Use U=W+Q • In case of ideal gas: • W=-PV & U=3/2nRT • if V then T (PV/T=constant) • W: positive (work done on gas) • U: negative • Q: negative (heat extracted) • if V then T (PV/T=constant) • W: negative (work done by gas) • U: positive • Q: positive (heat added) v PHY 231

  10. Cyclic processes The system returns to its original state. Therefore, the internal energy must be the same after completion of the cycle (U=0) PHY 231

  11. W=-Area under P-V diagram =-[(50-10)*10-3]*[(1.0-0.0)*105] -½[(50-10)*10-3]*[(5.0-1.0)]*105= =4000+8000 W=-12000 J Cyclic Process, step by step 1 Process A-B. Negative work is done on the gas: (the gas is doing positive work). U=3/2nRT=3/2(PBVB-PAVA)= = 1.5*[(1E+5)(50E-03)-(5E+5)(10E-03)]=0 The internal energy has not changed U=Q+W so Q=U-W=12000 J: Heat that was added to the system was used to do the work! PHY 231

  12. W=-Area under P-V diagram =-[(10-50)*10-3*(1.0-0.0)*105]= W=4000 J Work was done on the gas Cyclic process, step by step 2 Process B-C U=3/2nRT=3/2(PcVc-PbVb)= =1.5[(1E+5)(10E-3)-(1E+5)(50E-3)]=-6000 J The internal energy has decreased by 6000 J U=Q+W so Q=U-W=-6000-4000 J=-10000 J 10000 J of energy has been transferred out of the system. PHY 231

  13. Cyclic process, step by step 3 Process C-A W=-Area under P-V diagram W=0 J No work was done on/by the gas. U=3/2nRT=3/2(PcVc-PbVb)= =1.5[(5E+5)(10E-3)-(1E+5)(10E-3)]=+6000 J The internal energy has increased by 6000 J U=Q+W so Q=U-W=6000-0 J=6000 J 6000 J of energy has been transferred into the system. PHY 231

  14. -AREA Summary of the process A-B B-C C-A PHY 231

  15. What if the process was done in the reverse way? Net work was performed on the gas and heat extracted from the gas. We have built a heat pump! (A fridge) What did we do? The gas performed net work (8000 J) while heat was supplied (8000 J): We have built an engine! PHY 231

  16. More general engine turns water to steam heat reservoir Th W=|Qh|-|Qc| efficiency: W/|Qh| e=1-|Qh|/|Qc| Qh work is done the steam moves the piston work engine W Qc The efficiency is determined by how much of the heat you supply to the engine is turned into work instead of being lost as waste. the steam is condensed coldreservoirTc PHY 231

  17. Reverse direction: the fridge heat is expelled to outside heat reservoir Th Qh work is done a piston compresses the coolant work engine W Qc the fridge is cooled coldreservoirTc PHY 231

  18. The 2nd law of thermodynamics 1st law: U=Q+W In a cyclic process (U=0) Q=W: we cannot do more work than the amount of energy (heat) that we put inside 2nd law: It is impossible to construct an engine that, operating in a cycle produces no other effect than the absorption of energy from a reservoir and the performance of an equal amount of work: we cannot get 100% efficiency What is the most efficient engine we can make given a heat and a cold reservoir? PHY 231

  19. AB isothermal expansion BC adiabatic expansion W-, T- W-, T- W+, Q- Tcold DA adiabatic compression CD isothermal compression Carnot engine W-, Q+ Thot PHY 231

  20. Carnot cycle inverse Carnot cycle Work done by engine: Weng Weng=Qhot-Qcold efficiency: 1-Tcold/Thot A heat engine or a fridge! By doing work we can transport heat PHY 231

  21. Next lecture Entropy and examples PHY 231

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