1 / 6

Heat Engines, Refrigerators, and Heat Pumps

high temperature infinite thermal reservoir at T H. q H > 0. w < 0. engine (system). q L < 0. low temperature infinite thermal reservoir at T L. Heat Engines, Refrigerators, and Heat Pumps

wyatt
Download Presentation

Heat Engines, Refrigerators, and Heat Pumps

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. high temperature infinite thermal reservoir at TH qH > 0 w < 0 engine (system) qL < 0 low temperature infinite thermal reservoir at TL Heat Engines, Refrigerators, and Heat Pumps Science owes more to the steam engine than the steam engine owes to science. An infinite thermal reservoiris a heat source or sink so large in extent that the addition or removal of heat from this reservoir does not change its temperature. If an ice cube is tossed into a swimming pool, would the swimming pool act as infinite thermal reservoir? A heat engineextracts heat from a high temperature infinite thermal reservoirs, transforms some of that heat into useful work, and discards the remaining waste heat into low temperature infinite thermal reservoirs. An idealized heat engine operating between a single high temperature reservoir at TH and a low temperature reservoir at TL is diagramed below: All signs are defined relative to the system, the engine. 20.1

  2. What is the algebraic relation that exists between w, qH, and qL for this heat engine? Since the purpose of this engine is to convert the heat extracted to useful work the heat engine’s efficiency is defined as: heat engine efficiency  - w / qH What is the efficiency of a heat engine which extracts +1000 kcal of heat from a high temperature reservoir and dumps -732 kcal of waste heat into a low temperature reservoir? efficiency =- w / qH = - (- qH - qL) / qH = - [- (+1000 kcal) - (-732 kcal)] / (+1000 kcal) = 0.268 20.2

  3. high temperature infinite thermal reservoir at TH qH < 0 w > 0 engine (system) qL > 0 low temperature infinite thermal reservoir at TL Operated in reverse with work supplied to it, the heat engine can act as either a refrigerator or a heat pump. Acting as arefrigeratorthe supplied work is used to extract heat from a low temperature thermal reservoir, the refrigerator. The relevant performance factor is the ratio of the heat removed to the work supplied: performance factor for a refrigerator  qL / w What is the performance factor for a refrigerator which extracts +732 kcal of heat from a low temperature reservoir and dumps -1000 kcal of waste heat into a high temperature reservoir? 20.3

  4. Acting as a heat pump the supplied work is used to pump heat from a low temperature thermal reservoir, e.g., the ground, to a high temperature reservoir, e.g. a building. The relevant performance factor is the ratio of the heat delivered to the high temperature reservoir to the work supplied: performance factor for a heat pump  - qH / w What is the performance factor for a heat pump which uses +268 kcal of work to pump -1000 kcal of heat into a high temperature reservoir? 20.4

  5. high temperature infinite thermal reservoir at TH qH = -1340 kcal qH = +1000 kcal | w | = 268 kcal heat pump heat engine qL = -732 kcal qL = +1072 kcal low temperature infinite thermal reservoir at TL Now suppose we couple a heat engine and a heat pump operating between the same two thermal reservoirs. The heat engine operating with an efficiency of 0.268 extracts +1000 kcal of heat from the high temperature reservoir, converts some of this heat to -268 kcal of useful work that is supplied to the heat pump, and dumps the remaining -732 kcal of waste heat into the low temperature reservoir. Assume that the heat pump operates with a lower efficiency of 0.200. The heat pump will then pump: qH = - w / 0.200 = - (+268 kcal) / 0.200 = - 1340 kcal of heat into the high temperature reservoir, while extracting: qL = - w - qH = - (+268 kcal) - (-1340 kcal) = + 1072 kcal from the low temperature reservoir. 20.5

  6. The net effect of coupling the heat engine and heat pump operating at different efficiencies between the same two thermal reservoirs is to extract: | 1072 kcal | - | - 732 kcal | = + 340 kcal of heat from the low temperature reservoir and pump: | -1340 kcal | - | + 1000 kcal | = + 340 kcal of heat into the high temperature reservoir. If the above analysis were correct, you should be able to couple your stove to your refrigerator and operate them both without ever having to hook up to the power company!!! No device has ever been constructed which is capable of pumping heat from a cold reservoir to a hot reservoir in an isolated cyclic process and this is another statement of the2nd Law of Thermodynamics. What would be the net heat pumped from the cold reservoir to the hot reservoir, if the efficiencies of the heat engine and heat pump were the same? An alternative statement of the 2nd Law of Thermodynamics is the all heat engines operating between the same two thermal reservoirs must have the same efficiencies. 20.6

More Related