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Lecture 15 Heat Engines Review & Examples

p p b p a. Hot reservoir at T h. b a. Q h. Heat pump. W. c d. Heat leak. Q c. adiabats. Cold reservoir at T c. V. V 1 V 2. Lecture 15 Heat Engines Review & Examples. Traditional thermodynamic entropy: S = k ln W = k s

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Lecture 15 Heat Engines Review & Examples

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  1. p pb pa Hot reservoir at Th b a Qh Heatpump W c d Heatleak Qc adiabats Cold reservoir at Tc V V1 V2 Lecture 15Heat EnginesReview & Examples

  2. Traditional thermodynamic entropy: S = k lnW = ks We want to calculate S from macrostate information (p, V, T, U, N, etc.)Start with the definition of temperature in terms of entropy: The entropy changes when T changes: (We’re keeping V and N fixed.) If CV is constant: ReviewEntropy in Macroscopic Systems

  3. Two blocks, each with heat capacity* C = 1 J/K are initially at different temperatures, T1 = 250 K, T2 = 350 K. They are then placed into contact, and eventually reach a final temperature of 300 K. (Why?) What can you say about the total change in entropy DStot?a) DStot < 0 b) DStot = 0 c) DStot > 0 T1 = 250K T2 = 350K Tf = 300 K T1 T2 Tf Tf Two masses each with heat capacity C = 1J/K* ACT 1: An Irreversible Process *For a solid, C = CV = Cp, to good approximation, since DV  0.

  4. Two blocks, each with heat capacity* C = 1 J/K are initially at different temperatures, T1 = 250 K, T2 = 350 K. They are then placed into contact, and eventually reach a final temperature of 300 K. (Why?) What can you say about the total change in entropy DStot?a) DStot < 0 b) DStot = 0 c) DStot > 0 T1 = 250K T2 = 350K Tf = 300 K T1 T2 Tf Tf Two masses each with heat capacity C = 1J/K* Solution This is an irreversible process, so there must be a net increase in entropy.Let’s calculate DS: The positive term is slightly bigger than the negative term. *For a solid, C = CV = Cp, to good approximation, because DV  0.

  5. Example Process To analyze heat engines, we need to be able to calculate DU, DT, W, Q, etc. for the processes that they use. How much heat is absorbed by 3 moles of helium when it expands from Vi = 10 liters to Vf = 20 liters and the temperature is kept at a constant 350 K? What are the initial and final pressures?

  6. Solution To analyze heat engines, we need to be able to calculate DU, DT, W, Q, etc. for the processes that they use. How much heat is absorbed by 3 moles of helium when it expands from Vi = 10 liters to Vf = 20 liters and the temperature is kept at a constant 350 K? What are the initial and final pressures? Q = Wby The 1st law. For an ideal gas, DT = 0 DU = 0. Positive Q means heat flows into the gas. Wby = nRT ln(Vf/Vi) = 6048 J An expanding gas does work. pi = nRT/Vi = 8.72105 Pa Use the ideal gas law, pV = nRTpf = pi/2 = 4.36105 Pa Where is the heat coming from?In order to keep the gas at a constant temperature, it must be put in contact with a large object (a “heat reservoir”) having that temperature. The reservoir supplies heat to the gas (or absorbs heat, if necessary) in order to keep the gas temperature constant.Very often, we will not show the reservoir in the diagram. However, whenever we talk about a system being kept at a specific temperature, a reservoir is implied.

  7. Example Process (2) Suppose a mole of a diatomic gas, such as O2, is compressed adiabatically so the final volume is half the initial volume. The starting state is Vi = 1 liter, Ti = 300 K. What are the final temperature and pressure?

  8. Solution Suppose a mole of a diatomic gas, such as O2, is compressed adiabatically so the final volume is half the initial volume. The starting state is Vi = 1 liter, Ti = 300 K. What are the final temperature and pressure? Equation relating p and V for adiabatic process in a-ideal gas g is the ratio of Cp/CV for our diatomic gas (=(a+1)/a). Solve for pf. We need to express it in terms of things we know. Use the ideal gas law to calculate the final temperature. Alternative: Use the equation relating T and V for an adiabatic process to get the final temperature. a = 5/2

  9. Helpful Hints in Dealing with Engines and Fridges Sketch the process (see figures below). Define Qh and Qc and Wby (or Won) as positive and show directions of flow. Determine which Q is given. Write the First Law of Thermodynamics (FLT). We considered three configurations of Carnot cycles: Tc Th Tc Th Th Tc Qh Qh Qh Qleak= Qh Won Won Wby Qc Qc Qc Qleak= QC Heat pump:We pay for Won, we want Qh. Qh = Qc + Won = KWon Carnot: K = Th/(Th - Tc) Refrigerator:We pay for Won, we want Qc. Qc = Qh - Won = KWon Carnot: K = Tc/(Th - Tc) Engine:We pay for Qh, we want Wby. Wby = Qh - Qc = eQh Carnot: e = 1 - Tc/Th This has large ewhen Th - Tc is large. These both have large K when Th - Tc is small.

  10. Hot reservoir at Th Qh Engine W Qc Cold reservoir at Tc ACT 2: Entropy Change in Heat Pump Consider a Carnot heat pump. 1) What is the sign of the entropy change of the hot reservoir during one cycle? a) DSh < 0 b) DSh = 0 c) DSh > 0 2) What is the sign of the entropy change of the cold reservoir? a) DSc < 0 b) DSc = 0 c) DSc > 0 3) Compare the magnitudes of the two changes. a) |DSc| < |DSh| b) |DSc| = |DSh| c) |DSc| > |DSh|

  11. Hot reservoir at Th Qh Engine W Qc Cold reservoir at Tc Solution Consider a Carnot heat pump. 1) What is the sign of the entropy change of the hot reservoir during one cycle? a) DSh < 0 b) DSh = 0 c) DSh > 0 2) What is the sign of the entropy change of the cold reservoir? a) DSc < 0 b) DSc = 0 c) DSc > 0 3) Compare the magnitudes of the two changes. a) |DSc| < |DSh| b) |DSc| = |DSh| c) |DSc| > |DSh| Energy (heat) is entering the hot reservoir, so the number of available microstates is increasing.

  12. Hot reservoir at Th Qh Engine W Qc Cold reservoir at Tc Solution Consider a Carnot heat pump. 1) What is the sign of the entropy change of the hot reservoir during one cycle? a) DSh < 0 b) DSh = 0 c) DSh > 0 2) What is the sign of the entropy change of the cold reservoir? a) DSc < 0 b) DSc = 0 c) DSc > 0 3) Compare the magnitudes of the two changes. a) |DSc| < |DSh| b) |DSc| = |DSh| c) |DSc| > |DSh| Energy (heat) is entering the hot reservoir, so the number of available microstates is increasing. Energy (heat) is leaving the cold reservoir.

  13. Hot reservoir at Th Qh Engine W Qc Cold reservoir at Tc Solution Consider a Carnot heat pump. 1) What is the sign of the entropy change of the hot reservoir during one cycle? a) DSh < 0 b) DSh = 0 c) DSh > 0 2) What is the sign of the entropy change of the cold reservoir? a) DSc < 0 b) DSc = 0 c) DSc > 0 3) Compare the magnitudes of the two changes. a) |DSc| < |DSh| b) |DSc| = |DSh| c) |DSc| > |DSh| Energy (heat) is entering the hot reservoir, so the number of available microstates is increasing. Energy (heat) is leaving the cold reservoir. It’s a reversible cycle, so DStot = 0. Therefore, the two entropy changesmust cancel. Remember that the entropy of the “engine” itself does not change.

  14. p pb pa b a Combustion exhaust / intake adiabats c d V V1 V2 Example: Gasoline Engine It’s not really a heat engine because the input energy is via fuel injected directly into the engine, not via heat flow. There is no obvious hot reservoir. However, one can still calculate work and energy input for particular gas types. We can treat the gasoline engine as an Otto cycle: bc and da are nearly adiabatic processes, because the pistons move too quickly for much heat to flow.

  15. p pb pa b a c d adiabats V V1 V2 Solution Calculate the efficiency:

  16. p pb pa b a c d adiabats V V1 V2 Solution Write it in terms of volume instead of temperature. We know the volume of the cylinders. Compression ratio  V2/V1 10g = 1.4 (diatomic gas)  e = 60% (in reality about 30%, due to friction etc.)

  17. Solution Why not simply use a higher compression ratio?  If V2 big, we need a huge, heavy engine (OK for fixed installations).  If V1 small, the temperature gets too high, causing premature ignition. High compression engines need high octane gas, which has a higher combustion temperature.

  18. Free Energy Example Suppose we have a liter of water at T = 100° C. What is its free energy, if the environment is T = 20° C? Verify the result by calculating the amount of work we could obtain.Remember that cH2O = 4186 J/kg.K.

  19. Solution Suppose we have a liter of water at T = 100° C. What is its free energy, if the environment is T = 20° C? Verify the result by calculating the amount of work we could obtain.Remember that cH2O = 4186 J/kg.K. DF = DU - TDS, where T is the temperature of the environment. DU = mcDT = 1 kg * 4186 J/kg.K * 80 K = 3.349105 J. DS = mc ln(TH2O/Tenv) = 1011 J/K DF = 3.87104 J Remember to measure temperature in Kelvin. Otherwise, you’ll get the entropy wrong.

  20. Solution Suppose we have a liter of water at T = 100° C. What is its free energy, if the environment is T = 20° C? Verify the result by calculating the amount of work we could obtain.Remember that cH2O = 4186 J/kg.K. If we run a Carnot engine, the efficiency at a given water temperature is: E(T) = 1 - T/Tenv. So, for each small decrease in water temperature, we get this much work out of the engine: dW = eQ = -emc dT Thus, the total work obtained as T drops from 100° C to 20° C is:

  21. Beware !!!Maximum Stot Does Not Always Mean Minimum Fsys When we introduced the Helmholtz free energy, F, last lecture (see the “hot brick” discussion), we assumed that the volume of the brick was constant. If the volume weren’t constant, then the brick could gain or lose energy (and entropy) by contracting or expanding. That would change the analysis. It is very common that the pressure, not the volume, is constant (e.g., if our system is a gas at constant atmospheric pressure). In this situation, we use a different form of free energy, called “Gibbs free energy”: G = U+pV-TS. The pV term takes into account the work that is done during volume changes. We won’t use Gibbs free energy in this course, but it important to be aware that the calculation of free energy depends on the situation.

  22. Supplement: Gibbs Free Energy Most phase transitions are observed under constant p,T conditions, not constant-V,T. Unless the stuff is in a closed vessel, it’s open to the air (thus, at atmospheric pressure). In this case, the entropy of the environment changes not only when the system energy changes but also when its volume changes. The reservoir is (by assumption) in equilibrium at fixed T and p. So, as heat flows, the change of the reservoir’s entropy is: The change in the total entropy is thus: where: DVR = -DV and DUR = -DUbecause total volume and total energy are constant. Variables U, V, and S are of the system. Fixed p and T are of the reservoir. In these conditions, maximizing Stot means minimizing the system’s G.

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