Heat Engine A heat engine is a device

Heat Engine  A heat engine is a device

Heat Engine A heat engine is a device that absorbs heat (Q) and uses it to do useful work (W) on the surroundings when operating in a cycle. Sources of heat include the combustion of coal, petroleum or carbohydrates and nuclear reactions. Working substance: the matter inside the heat engine that undergoes addition or rejection of heat and that does work on the surroundings. Examples include air and water vapour (steam). In a cycle, the working substance is in the same thermodynamic state at the end as at the start. HE2 Thermal Physics Heat Engine Hot Body (source of heat) Q1

W E Q2 Cold Body (absorbs heat) HE2 Thermal Physics Example of a Heat Engine Open system HE2 Thermal Physics Internal Combustion Engine d a

HE2 Thermal Physics Comparison of Otto and Diesel Cycles Work per cycle = Area inside combustion Q=0 Q=0 HE2 Thermal Physics Nuclear Power Plant: A Very Large Heat Engine HE2 Thermal Physics Heat Engine Hot Body

(source of heat) Q1 W E Q2 Cold Body (absorbs heat) HE2 Thermal Physics Efficiency of a Heat Engine Efficiency, = Work out/Heat in: W Q1

Apply First Law to the working substance: U = Q1 Q2 W But in a cycle, U = 0 Thus, W = Q1 Q2. Substituting: W Q1 Q2 Q2 1 Q1 Q1 Q1 Lesson: is maximum when Q2 is minimum. HE2 Thermal Physics The Stirling Engine

Closed system See: http://www.animatedengines.com/ltdstirling.shtml Operates between two bodies with (small) different temperatures. Can use stray heat HE2 Thermal Physics The Stirling Cycle TH >TC isothermal Heat in (TH - TC ) is proportional to the amount of work

that is done in a cycle. = air temp =hot water 2 isothermal Heat out HE2 Thermal Physics HE2 Thermal Physics Carnot Cycle Hot Reservoir T1 Q1 W

C Q2 Cold Reservoir T2 HE2 Thermal Physics Carnot Cycle Pressure a nRT1 P= V Q1 b

Q=0 nRT2 P= V T1 Q=0 d Q2 P= const . V

c T2 Volume HE2 Thermal Physics Carnot Cycle Pressure a nRT1 P= V Q1 b W

Q=0 nRT2 P= V d Q2 T1 Q=0 P= const .

V c T2 Volume HE2 Thermal Physics Carnot Cycle From a to b: isothermal, so that U = 0 and Q = - W Thus, Q1 = +nRT1ln(Vb/Va) (+ve quantity) From b to c: adiabatic, Q = 0, so that TV-1 is constant. 1 Thus, T1Vb-1 = T2Vc-1 or T1 Vc T2 Vb

Similarly, from c to d: isothermal, so that U = 0 and Q = - W Thus, Q2 = +nRT2ln(Vd/Vc) = -nRT2ln(Vc/Vd) (-ve) Similarly, d to a: adiabatic, Q = 0, so that TV-1 is constant. Thus, T2Vd-1 = T1Va-1 or 1 T1 Vd T2 Va HE2 Thermal Physics Carnot Cycle We see that: Which means that Now also: But as the volume ratios are equal:

T1 Vc T2 Vb 1 V d Va 1 Vc Vb Vd Va

Q1 nRT1 ln(Vb / Va ) T1 ln(Vb / Va ) Q2 nRT2 ln(Vc / Vd ) T2 ln(Vc / Vd ) Q1 T1 Q2 T2 This is an important result. Temperature can be defined (on the absolute (Kelvin) scale) in terms of the heat flows in a Carnot HE2 Thermal Physics Cycle. Whats Special about a Carnot Cycle? (1) Heat is transferred to/from only two reservoirs at fixed temperatures, T1 and T2 - not at a variety of temperatures. (2) Heat transfer is the most efficient possible because the temperature of the working substance equals the temperature

of the reservoirs. No heat is wasted in flowing from hot to cold. (3) The cycle uses an adiabatic process to raise and lower the temperature of the working substance. No heat is wasted in heating up the working substance. (4) Carnot cycles are reversible. Not all cycles are! HE2 Thermal Physics Whats Special about a Carnot Cycle? (5) The Carnot theorem states that the Carnot cycle (or any reversible cycle) is the most efficient cycle possible. The Carnot cycle defines an upper limit to the efficiency of a cycle. Recall that for any cycle, the efficiency of a heat engine is given as: W Q2 E = =1 Q1

Q1 For an engine using a Carnot cycle, the efficiency is also equal to: T2 C = 1 T1 Where T1 and T2 are the temperatures of the hot and cold reservoirs, respectively, in degrees Kelvin. As T2 > 0, c is always <1. HE2 Thermal Physics Efficiency of a Stirling Engine Question: What is the maximum possible efficiency of a Stirling engine operating between room temperature (25 C) and boiling water (100 C)? Maximum efficiency would be achieved by a Carnot cycle operating between reservoirs at T1 = 373 K and T2 = 298 K.

298 W c = 1 = 0.20 = 373 Q1 Question: What is the maximum possible efficiency of a Stirling engine operating between room temperature (25 C) and ice (0 C)? 273 W c = 1 = 0.08 = 298 Q1 HE2 Thermal Physics

Kelvin-Planck Statement of the Second Law of Thermodynamics It is impossible to construct a device that - operating in a cycle - will produce no other effect than the extraction of heat from a single body and the performance of an equivalent amount of work OrA cyclical engine cannot convert heat from a single body completely into work. Some heat must be rejected at a lower temperature. Thus, efficiency, < 1! HE2 Thermal Physics Heat Engine Hot Body (source of heat) Q1 W= -Q1

E Q2 = 0 Cold Body (absorbs heat) HE2 Thermal Physics Heat Engine Hot Body (source of heat) Q 1= 0 POSSIBLE! E W Examples: friction

creating heat; isothermal compression of ideal gas Q2 = W Cold Body (absorbs heat) HE2 Thermal Physics Refrigerator: A heat engine operating in reverse Hot Body Q1 Refrigerator Efficiency: heat out Q2 R

work in W W E Q2 Cold Body HE2 Thermal Physics Refrigerator Efficiency R heat out Q2 work in W First Law tells us that Q2 + W -Q1 = 0. Thus, W = Q1 Q2

Q2 R Q1 Q2 For a Carnot refrigerator, the efficiency is: 1 Q1 Q2 Q1 T1 T1 T2 1 1 c

Q2 Q2 T2 T2 R R c T 2 T1 T2 Efficiency is usually >1! The smaller the T difference, the more efficient is the refrigerator. HE2 Thermal Physics

Clausius Statement of the Second Law of Thermodynamics (applies to refrigerators) It is impossible to construct a device that - operating in a cycle - will produce no other effect than heat transfer from a colder body to hotter body. OrHeat cannot flow from a cold body to a hotter body by itself. Work has to be done in the process. HE2 Thermal Physics HE2 Thermal Physics Efficiency of a Heat Pump The purpose of a heat pump is to extract heat from a cold body (such as the River Thames) and pump it to a hot body (such as an office building). The efficiency is defined as the amount of heat pumped in to the hot

body per the amount of work done: Q C hp 1 W The First Law tells us that W = Q1-Q2 So, substituting, we find: C hp Q1 T 1 1 Q1 Q2 T1 T2 1 T2 / T1 hp is always > 1! For maximum , T2 should be T1 (just slightly less).

HE2 Thermal Physics

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