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Heat Pumps and Refrigerators • Available Work and Free Energy • Work from Hot and Cold Bricks

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Heat Pumps and Refrigerators • Available Work and Free Energy • Work from Hot and Cold Bricks Lecture 14 Heat Pumps, Refrigerators, and Bricks! • Pumping Heat: Heat pumps and Refrigerators • Available Work and Free Energy • Work from Hot and Cold Bricks Reading for this Lecture: Reading for Lecture 16: Elements Ch 10 Elements Ch 11 Lecture 14, p 1 Run the Engine in Reverse The Carnot cycle is reversible (each step is reversible): p Qh Reverse all Hot reservoir at Th the arrows 2 Qh 1 Th Engine W Qc 3 Q 4 Tc c Cold reservoir at T V V V V c b a c d V When the engine runs in reverse: Heat is transferred from cold to hot by action of work on the engine . Note that heat never flows spontaneously from cold to hot; the cold gas is being heated by adiabatic compression (process 3). Qc / Qh = Tc / Th is still true. (Note: Qh, Q c, and W are still positive!) Lecture 14, p 2 Refrigerators and Heat Pumps Refrigerators and heat pumps are heat engines running in reverse. How do we measure their performance? It depends on what you want to accomplish. Hot reservoir at Th Refrigerator: Qh We want to keep the food cold (Q c). We pay for W (the electric motor in the fridge). Engine W So, the coefficient of performance , K is: It’s not called “efficiency”. Q Q Q T c K ≡c = c = c Cold reservoir at Tc W Qh− Q c T h − T c Heat pump: We want to keep the house warm (Qh). We pay for W (the electric motor in the garden). The coefficient of performance , K is: Q Q T K ≡h = h = h W Qh− Q c T h − T c Lecture 14, p 3 Helpful Hints in Dealing with Engines and Fridges Sketch the process (see figures below). Define Qh and Q c and Wby (or W on ) 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: T T T h h h Q = Q Qh Qh Qh leak h Wby Won Won Q Q Q Tc c Tc c Tc c Qleak = Q C Engine: Refrigerator: Heat pump: We pay for Qh, We pay for W on , We pay for W on , we want Wby . we want Q c. we want Qh. Wby = Qh -Qc = εQh Qc = Qh -Won = ΚWon Qh = Q c + W on = ΚWon Carnot: ε = 1 - Tc/T h Carnot: Κ = Tc/(T h -Tc) Carnot: Κ = Th/(T h -Tc) This has large ε These both have large Κ when T -T is small. when T -T is large. h c h c Lecture 14, p 4 Act 1: Refrigerator There is a 70 W heat leak (the insulation is not perfect) from a room at temperature 22 °C into an ideal refrigerator. How much electrical power is needed to keep the refrigerator at -10 ° C? Assume Carnot performance. a) < 70 W b) = 70 W c) > 70 W Lecture 14, p 5 Act 1: Refrigerator There is a 70 W heat leak (the insulation is not perfect) from a room at temperature 22 ° C into an ideal refrigerator. How much electrical power is needed to keep the refrigerator at -10 ° C? Assume Carnot performance. Hint: Q must exactly compensate for the heat leak. c Hot reservoir at Th a) < 70 W Qh b) = 70 W Heat Fridge W c) > 70 W leak Qc Cold reservoir at Tc Lecture 14, p 6 Solution There is a 70 W heat leak (the insulation is not perfect) from a room at temperature 22 ° C into an ideal refrigerator. How much electrical power is needed to keep the refrigerator at -10 ° C? Assume Carnot performance. Hint: Q must exactly compensate for the heat leak. c Hot reservoir at Th Qh The coefficient of performance is: Heat Fridge W Qh T h leak WQQQ=−=hcc( −= 1) Q c ( − 1) Qc Qc T c Cold reservoir at Tc We need Q c = 70 J each second. 295 Therefore we need W≡70( − 1) J / s 263 This result illustrates an unintuitive property of refrigerators and heat The motor power is 8.5 Watts. pumps: When Th -Tc is small, they Watt = Joule/second. pump more heat than the work you pay for. Lecture 14, p 7 Supplement:Supplement: PeltierPeltier coolercooler Driving a current (~amps) through generates a temperature difference. 20-50 ˚C typical Not so common – they’re more costly, take a lot of power, and you still have to get rid of the heat! But....no moving parts to break. How’s it work Electrons pushed from electron-deficit material (p-type) to electron-rich material (n-type); they slow down, cooling the top connector. Similarly, they heat up in going from n to p-type (bottom connector). Despite the radically different construction, this heat pump must obey exactly the same limits on efficiency as the gas-based pumps, because these limits are based on the 1 st and 2 nd laws, not any details. Physics 213: Lecture 10, Pg 8 The Limits of Cooling Q T The maximum efficiency is ε = C ≤ C W TH −TC Refrigerators work less well as Th -Tc becomes large. The colder you try to go, the less efficient the refrigerator gets. The limit as T C goes to zero is zero efficiency ! Since heat leaks will not disappear as the object is cooled, you need to supply more work the colder it gets. The integral of the power required diverges as Tc → 0. Therefore you cannot cool a system to absolute zero. Lecture 14, p 9 Exercise: Heat Pump Suppose that the heat flow out of your 20 ° C home in the winter is 7 kW. If the temperature outside is -15 ° C, how much power would an ideal heat pump require to maintain a constant inside temperature? Hot reservoir at Th Q a) W < 7 kW Heat h b) W = 7 kW leak Heat W c) W > 7 kW pump Qc Cold reservoir at Tc Lecture 14, p 10 Solution Suppose that the heat flow out of your 20 ° C home in the winter is 7 kW. If the temperature outside is -15 ° C, how much power would an ideal heat pump require to maintain a constant inside temperature? a) W < 7 kW This is what you’d Hot reservoir at Th need with an electric Q b) W = 7 kW blanket or furnace. Heat h c) W > 7 kW leak Heat W pump The coefficient of performance is: Qc Q T c c Cold reservoir at Tc WQQQ=−=hch(1 − ) = Q h (1 − ) Qh T h We need Qh = 7000 J each second. 258 Beware: Therefore we need W≡7000(1 − ) = 836 J / s 293 Real heat pumps are not nearly ideal, so the advantage is smaller. The electric company must supply 836 Watts, much less than the 7 kW that a furnace would require! Lecture 14, p 11 ACT 2: Work from a hot brick Brick, 400K We saw that the efficiency of any heat engine is given by ε = 1 - Qc/Qh. Q Heat a brick to 400 K . Connect it to a Carnot Engine. What h Wby is the average efficiency if the cold reservoir is 300 K ? The Q brick has a constant heat capacity of C = 1 J/K . C 300 K a) ε < 25% b) ε = 25% c) ε > 25% Lecture 14, p 12 Solution Brick, 400 K We saw that the efficiency of any heat engine is given by ε = 1 - Qc/Qh. Q Heat a brick to 400 K . Connect it to a Carnot Engine. h Wby What is the average efficiency if the cold reservoir is 300 K ? Q The brick has a constant heat capacity of C = 1 J/K . C 300 K a) ε < 25% b) ε = 25% c) ε > 25% Did you use: ε = 1 - Tc/T h ? If so, you missed that the brick is cooling (it’s not a constant T reservoir). Therefore, the efficiency (which begins at 25%) drops as the brick cools. We must integrate: (dQ h = -CdT) T T c Th T W=− CT()1 −C dTC = 1 − C dT by ∫T ∫ T Th T C Th This is an interesting result. =CTT()hcc −− TCln =−∆+∆ UTS c Let’s discuss it. Tc Lecture 14, p 13 Available Work and Free Energy We just found that the work that the engine can do as the brick cools from its initial temperature to Tc is: Wby=−∆ Ubrick + TS c ∆ brick The form of this result is useful enough that we define a new quantity, the “free energy” of the brick: In the ACT, Tenv was Tc. Fbrick ≡ Ubrick - Tenv Sbrick With this definition, Wby = -∆Fbrick . This is the best we can do. In general , Wby will be smaller: Wby ≤ -∆Fbrick = Fi –Ff Free energy tells how much work can be extracted. It is useful, because it is almost entirely a property of the brick. Only the temperature of the environment is important. Lecture 14, p 14 ACT 3: Work from a cold brick? We obtained work from a hot brick, initially at 400K. If instead the brick were initially at 200K , could we still do work in our 300K environment? a) Yes. Brick, 400 K b) No, you can’t have Th < Tc. Qh c) No, we would have to put work in. Wby QC Hint: Fill in the diagram: 300 K Lecture 14, p 15 Solution We obtained work from a hot brick, initially at 400K.
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