Скачать презентацию PHYS 1001 Physics 1 REGULAR Module 2 Thermal Скачать презентацию PHYS 1001 Physics 1 REGULAR Module 2 Thermal

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PHYS 1001 Physics 1 REGULAR Module 2 Thermal Physics 1 HEAT ENGINE ap 06/p PHYS 1001 Physics 1 REGULAR Module 2 Thermal Physics 1 HEAT ENGINE ap 06/p 1/thermal/pt. G_engines. ppt

2 HEAT ENGINES AND REFRIGERATORS Heat engines: Entropy (§ 20. 1 p 673 § 2 HEAT ENGINES AND REFRIGERATORS Heat engines: Entropy (§ 20. 1 p 673 § 20. 2 p 675 § 20. 5 p 682) Heat engines: Carnot cycle (§ 20. 6 p 684 § 20. 7 p 690) Internal Combustion Engines: Otto & Diesel Cycles (§ 20. 3 p 678) Refrigerators (§ 20. 4 p 680) References: University Physics 12 th ed Young & Freedman

3 Heat engine: device that transforms heat partly into work (mechanical energy) by a 3 Heat engine: device that transforms heat partly into work (mechanical energy) by a working substance undergoing a cyclic process.

4 W>0 |QH| = |W|+|QC| petrol engine: fuel + air petrol engine - hot 4 W>0 |QH| = |W|+|QC| petrol engine: fuel + air petrol engine - hot exhaust gases + cooling system

All heat engines absorb heat QH from a source at a relatively high temperature All heat engines absorb heat QH from a source at a relatively high temperature (hot reservoir TH), perform some work W and reject some heat QC at a lower temperature (cold reservoir TC). First Law for a cyclic process: U = 0 = Qnet – W Qnet = |QH| - |QC| = W Thermal efficiency, e represents the fraction of QH that is converted to useful work. 5

Problem (Y & F Example 20. 1) A large truck is travelling at 88 Problem (Y & F Example 20. 1) A large truck is travelling at 88 km. h-1. The engine takes in 10 000 J of heat and delivers 2 000 J of mechanical energy per cycle. The heat is obtained by burning petrol (heat of combustion Lcomb = 5. 0 107 J. kg-1). The engine undergoes 25 cycles per second. Density of petrol = 700 kg. m-3 (a) What is thermal efficiency of the heat engine? (b) How much heat is discarded each cycle? (c) How much petrol is burnt per hour (kg. h-1)? (d) What is the petrol consumption of the truck in L/100 km? 6

7 Solution ISEE v = 88 km. h-1 QH = 1. 0 104 J/cycle 7 Solution ISEE v = 88 km. h-1 QH = 1. 0 104 J/cycle W = 2 103 J/cycle Lcomb = 5. 0 107 J. kg-1 f = 25 cycles. s-1 (a) e=? e = W / QH = (2 103) / (1. 0 104) = 0. 20 = 20% (b) QC = ? J |QH| - |QC| = W sensible |QC| = 8. 0 103 J/cycle

(c) 8 QH = 1 104 J/cycle QH / t = (25)(1 104) J. (c) 8 QH = 1 104 J/cycle QH / t = (25)(1 104) J. s-1 = 2. 5 105 J. s-1 QH/ t = ( m/ t) Lcomb LC = 5. 0 107 J. kg-1 m/ t = ( QH/ t )/ Lcomb = (2. 5 105 ) / (5. 0 107) = 5 10 -3 kg. s-1 m/ t = (5 10 -3)(60) = 18 kg. h-1 (d) fuel consumption = ? L/100 km v = 88 km. h-1 v = d/ t t = ? h d = 100 km t = d / v = 100 / 88 h = 1. 1364 h mass used traveling 100 km m = ( m/ t) t = (18)(1. 1364) = 20. 455 kg =m. V 103 L = 1 m 3 volume used in traveling 100 km V = m / = (20. 455 / 700) m 3 = 0. 0292 m 3 = 29. 2 L petrol consumption = 29. 2 L/100 km

9 Hot Reservoir QH W Engine Surroundings QC= 0 Can a heat engine be 9 Hot Reservoir QH W Engine Surroundings QC= 0 Can a heat engine be 100% efficient in converting heat into mechanical work ? Why does this engine violate the Second Law ?

10 ENTROPY considerations: S(engine) = 0 cyclic process S(surrounding) = 0 no heat transfer 10 ENTROPY considerations: S(engine) = 0 cyclic process S(surrounding) = 0 no heat transfer to surroundings S(hot reservoir) < 0 heat removed S(total) < 0 violates Second Law All heat engines (heat to work): efficiency, e < 1

11 Hot reservoir QH W Engine QC Cold reservoir surroundings 11 Hot reservoir QH W Engine QC Cold reservoir surroundings

12 S(hot reservoir) = - |QH|/ TH S(cold reservoir) = + |QC|/ TC S(engine) 12 S(hot reservoir) = - |QH|/ TH S(cold reservoir) = + |QC|/ TC S(engine) = 0 (cyclic process) S(surroundings) = 0 (no heat transfer to surroundings) S(total) = - |QH|/ TH + |QC|/ TC Useful work can only be done if S(total) 0 | QH| / TH | QC | / TC

13 An ideal engine e. g. Carnot Cycle S = 0 | QH / 13 An ideal engine e. g. Carnot Cycle S = 0 | QH / TH | = | QC / TC | e = W / QH Sadi Carnot (1824) Frenchman W = |QH| - |QC| e = (|QH| - |QC| / QH)= 1 - |QC| / |QH| e = 1 – TC / TH < 1 This is the absolute max efficiency of a heat engine NOTE: All heat (QH & QC) exchanges occurred isothermally in calculating the efficiency e = 1 – TC / TH

14 CARNOT CYCLE Heat engine with the maximum possible efficiency consistent with 2 nd 14 CARNOT CYCLE Heat engine with the maximum possible efficiency consistent with 2 nd law. All thermal processes in the cycle must be reversible - all heat transfer must occur isothermally because conversion of work to heat is irreversible. When the temperature of the working substances changes, there must be zero heat exchange – adiabatic process. Carnot cycle – consists of two reversible isotherms and two reversible adiabats

15 Carnot Cycle P QH adiabatic isothermals 3 Diagram not to scale, adiabats are 15 Carnot Cycle P QH adiabatic isothermals 3 Diagram not to scale, adiabats are much steeper than shown 4 2 W 1 QC released to surroundings NOTE: All heat (QH & QC) exchanges occurred isothermally in calculating the efficiency e = 1 – TC / TH V

QC 16 1 and 2: “cold” 3 and 4: “hot” QH p 3 4 QC 16 1 and 2: “cold” 3 and 4: “hot” QH p 3 4 2 1 V 1 2 Isothermal compression 2 3 Adiabatic compression 3 4 Isothermal expansion 4 1 Adiabatic expansion All energy exchanges are reversible – there are no non-recoverable energy losses

Proof: e = 1 - |QC| / |QH| = 1 - TC / TH Proof: e = 1 - |QC| / |QH| = 1 - TC / TH Isothermal change Q = n R T ln(Vf /Vi) p 3 1 2: Isothermal compression heat QC rejected to sink at constant temperature TC. |QC | = n R TC ln(V 2 / V 1) 4 2 e = 1 - |QC| / |QH| = 1 – TC ln(V 2 / V 1) / TH ln(V 4 / V 3) 1 V 3 4: Isothermal expansion heat QH supplied from source at constant temperature |QH | = n R TH ln(V 4 / V 3) Adiabatic change Q = 0 Ti. Vi -1 = Tf. Vf -1 17 1 and 2: “cold” 3 and 4: “hot” (Eq. 1)

18 2 3: Adiabatic expansion 4 1: Adiabatic expansion In practice, the Carnot cycle 18 2 3: Adiabatic expansion 4 1: Adiabatic expansion In practice, the Carnot cycle cannot be used for a heat engine because the slopes of the adiabatic and isothermal lines are very similar and the net work output (area enclosed by p. V diagram) is too small to overcome friction & other losses in a real engine. e = 1 - TC ln(V 1 / V 2) / TH ln(V 4 / V 3) (Eq. 1) Efficiency of Carnot engine - max possible efficiency for a heat engine operating between TC and TH

19 TC = 25 o. C The strength & hardness of metals decreases rapidly 19 TC = 25 o. C The strength & hardness of metals decreases rapidly above 750 o. C “Re-author” Y&F Example 20. 3

20 Diathermal wall: A highly thermally conducting wall. 20 Diathermal wall: A highly thermally conducting wall.

21 CAR ENGINE The four–stroke OTTO cycle of a conventional petrol engine 21 CAR ENGINE The four–stroke OTTO cycle of a conventional petrol engine

The four–stroke OTTO cycle of a conventional petrol engine intake stroke compression stroke ignition The four–stroke OTTO cycle of a conventional petrol engine intake stroke compression stroke ignition power stroke 22 exhaust stroke intake stroke: isobaric expansion compression stroke: adiabatic compression ignition: isochoric heating of gas power stroke: adiabatic expansion of gas exhaust stroke: isochoric cooling of gas / isobaric compression

23 Intake 1 Power 3 Compression 2 Exhaust 4 23 Intake 1 Power 3 Compression 2 Exhaust 4

Otto Cycle P 24 IGNITION fuel combustion isothermals adiabatic 3 QH 2 power stroke Otto Cycle P 24 IGNITION fuel combustion isothermals adiabatic 3 QH 2 power stroke compression stroke W P o 5 V 2 4 1 intake stroke exhaust stroke QC cooling of exhaust gases released to surroundings V V 1 = r V 2 r = compression ratio

25 Otto Cycle P 3 isothermals adiabatic QH 2 4 Po 1 5 QC 25 Otto Cycle P 3 isothermals adiabatic QH 2 4 Po 1 5 QC released to surroundings V V 2 V 1 5 1: inlet stroke volume increases as piston moves down creating a partial vacuum to aid air/fuel entering cylinder via the open inlet valve.

26 Otto Cycle P 3 isothermals adiabatic QH 2 4 Po 1 5 QC 26 Otto Cycle P 3 isothermals adiabatic QH 2 4 Po 1 5 QC released to surroundings V V 2 V 1 1 2: compression stroke inlet valve closes piston moves up compressing the air/fuel mixture adiabatically.

27 Otto Cycle P 3 isothermals adiabatic QH 2 4 Po 1 5 QC 27 Otto Cycle P 3 isothermals adiabatic QH 2 4 Po 1 5 QC released to surroundings V V 2 V 1 2 3: ignition – spark plug fires igniting mixture - constant volume combustion.

28 Otto Cycle P isothermals adiabatic 3 QH 2 4 Po 1 5 QC 28 Otto Cycle P isothermals adiabatic 3 QH 2 4 Po 1 5 QC released to surroundings V V 2 V 1 3 4: expansion or power stroke – heated gas expands adiabatically as the piston is pushed down doing work (Vmax = r Vmin). Compression ratio, r

29 Otto Cycle P 3 isothermals adiabatic QH 2 4 Po 1 5 QC 29 Otto Cycle P 3 isothermals adiabatic QH 2 4 Po 1 5 QC released to surroundings V V 2 V 1 4 1 start of Exhaust stroke – outlet valve opens and mixture expelled at constant volume then 1 5

30 Otto Cycle P 3 isothermals adiabatic QH 2 4 Po 1 5 QC 30 Otto Cycle P 3 isothermals adiabatic QH 2 4 Po 1 5 QC released to surroundings V V 2 V 1 1 5: Exhaust stroke – piston moves up producing a compression at constant pressure, Po (atmospheric pressure).

31 Otto cycle – standard petrol engine (4 stroke) Idealized model of thermodynamic processes 31 Otto cycle – standard petrol engine (4 stroke) Idealized model of thermodynamic processes in a typical car engine. For compression ratio, r ~ 8 and = 1. 4 (air) TH (peak) ~ 1800 °C TC (base) ~ 50 °C e ~ 56% (ideal engine) e ~ 35% (real engine). Efficiency increases with larger r engine operates at higher temperatures pre–ignition knocking sound and engine can be damaged. Octane rating – measure of anti-knocking – premium petrol r ~ 12. In practice, the same air does not enter the engine again, but since an equivalent amount of air does enter, we may consider the process as cyclic.

p Comparison of theoretical and actual p. V diagrams for the four-stroke Otto Cycle p Comparison of theoretical and actual p. V diagrams for the four-stroke Otto Cycle engine. 32 V

33 Diesel Cycle 5 to 1: intake stroke isobaric expansion P 1 to 2: 33 Diesel Cycle 5 to 1: intake stroke isobaric expansion P 1 to 2: compression stroke abiabatic compression 2 3 2 to 3: ignition isobaric heating P o 3 to 4: power stroke adiabatic expansion 4 to 1: exhaust stroke (start) isochoric cooling 1 to 5: exhaust stroke (finish) isobaric compression adiabatic isothermals QH 4 5 V 2 1 V 1 Rudolf Diesel QC released to surroundings V

34 Diesel Cycle P fuel ignition QH 2 3 compression stroke P o isothermals 34 Diesel Cycle P fuel ignition QH 2 3 compression stroke P o isothermals adiabatic power stroke 4 5 1 QC released to cooling of exhaust gases surroundings V V 2 V 1

35 35

DIESEL CYCLE No fuel in the cylinder at beginning of compression stroke. At the DIESEL CYCLE No fuel in the cylinder at beginning of compression stroke. At the end of the adiabatic compression high temperatures are reached and then fuel is injected fast enough to keep the pressure constant. The injected fuel because of the high temperatures ignites spontaneously without the need for spark plugs. Diesel engines operate at higher temperatures than petrol engines, hence more efficient. For r ~ 18 and = 1. 4 (air) TH (peak) ~ 2000 °C e ~ 68% (ideal engine) Petrol engine TC (base) ~ 50 °C real efficiency ~ 40 % ~ 56% (ideal engine) real efficiency ~ 35 % 36

Diesel engines 37 • Heavier (higher compression ratios), lower power to weight ratio. • Diesel engines 37 • Heavier (higher compression ratios), lower power to weight ratio. • Harder to start. • More efficient than petrol engines (higher compression ratios). • No pre-ignition of fuel since no fuel in cylinder during most of the compression • They need no carburettor or ignition system, but the fuel-injection system requires expensive high-precision machining. • Use cheaper fuels less refined heavy oils – fuel does need to be vaporized in carburettor, fuel less volatile hence safer from fire or explosion. • Diesel cycle – can control amount of injected fuel per cycle – less fuel used at low speeds.

38 http: //people. bath. ac. uk/ccsshb/12 cyl/ These engines were designed primarily for very 38 http: //people. bath. ac. uk/ccsshb/12 cyl/ These engines were designed primarily for very large container ships.

39 Problem For theoretical Otto cycle, calculate: (a) max cycle temperature (b) work per 39 Problem For theoretical Otto cycle, calculate: (a) max cycle temperature (b) work per kilogram of fuel (c) Efficiency (d) Max efficiency Carnot engine working between same temperatures Engine characteristics: compression ratio = 8: 1 cp = 1. 005 k. J. kg-1. K-1 c. V = 0. 718 k. J. kg-1. K-1 Inlet conditions p = 97. 5 k. Pa and T = 50 o. C Heat supplied = 950 k. J. kg-1

Solution Identify / Setup cp = 1. 005 k. J. kg-1. K-1 c. V Solution Identify / Setup cp = 1. 005 k. J. kg-1. K-1 c. V = 0. 718 k. J. kg-1. K-1 V 1/ V 2 = 8 p 1 = 97. 5 103 Pa T 1 = 50 o. C = 323 K QH = 950 k. J. kg-1 TH = T 3 = ? K W = ? k. J. kg-1 e = ? e. Carnot = ? 40 e = W / |QH| e. Carnot = 1 – TC / TH isothermals adiabatic 3 QH 2 4 Po 1 5 QC released to surroundings V V 2 Adiabatic change Q = 0 T V -1 = constant Qp = m cp T QV = m c. V T Otto Cycle P V 1 = cp / c. V = 1. 005 / 0. 718 = 1. 4 W = |QH| - |QC|

Adiabatic compression Execute -1 T 2 / T 1 = (V 1/V 2) T Adiabatic compression Execute -1 T 2 / T 1 = (V 1/V 2) T 2 = T 1(V 1 / V 2) -1 = (323)(8)0. 4 K = 742 K Constant volume heating QH = m c. V (T 3 – T 2) T 3 = (QH/m)/c. V +T 2 = (950/0. 718 + 742) K = 2065 K = 1792 o. C Adiabatic expansion T 4 / T 3 = (V 3 /V 4) -1 T 4 = T 3(1/8)0. 4 = 899 K Constant volume heat rejection QC = m c. V (T 4 – T 3) QC/m = (0. 718)(899 - 323) k. J. kg-1 = 414 k. J. kg-1 W = |QH| - |QC| = (950 – 414) k. J. kg-1 = 536 k. J. kg-1 e = W / |QH| = 536 / 950 = 0. 56 (real engine < 0. 4) e. Carnot = 1 – TC / TH = 1 – 323 / 2067 = 0. 84 41

Calculate the above quantities for the diesel cycle with a compression ratio = 20 Calculate the above quantities for the diesel cycle with a compression ratio = 20 42 Diesel Cycle P adiabatic isothermals QH 2 3 4 P o 5 V 2 1 V 1 QC released to surroundings V

Adiabatic compression V 1 / V 2 = 20 T 2 / T 1 Adiabatic compression V 1 / V 2 = 20 T 2 / T 1 = (V 1/V 2) -1 T 2 = T 1(V 1 / V 2) -1 = (323)(20)0. 4 K = 1071 K 43 Isobaric heating QH = m cp (T 3 -T 2) QH/m = cp (T 3 –T 2) T 3 = T 2 + (1/cp)(QH/m) = 1071 +(1/1. 005)(950) K = 2016 K Isobaric heating / Adiabatic expansion V 2 / T 2 = V 3 / T 3 V 3 / V 2 = T 3 / T 2 V 4 / V 2 = 20 V 2 = V 4 / 20 V 3 / V 4 = (1/20)(T 3/T 2) = (1/20)(2016/1071) = 0. 0941 T 3 V 3 -1 = T 4 V 4 -1 T 4 = T 3 (V 3/V 4) -1 = (2016)(0. 0941)0. 4 = 783 K Isochoric cooling QC = m cv (T 4 – T 1) QC/m = cv (T 4 – T 1) = (0. 718)(783 – 323) K = 330. 3 k. J. kg-1 W = QH – QC = (950 – 330) k. J = 620 k. J. kg-1 e = W / QH = 620 / 950 = 0. 65 e = 1 - QC/QH = 1 – 330/950 = 0. 65 The work output and efficiency are considerably higher than for Otto Cycle

Example Consider two engines, the details of which are given in the following diagrams. Example Consider two engines, the details of which are given in the following diagrams. For both engines, calculate the heat flow to the cold reservoir and the changes in entropy of the hot reservoir, cold reservoir and engine. Which engine violates the Second Law? What is the efficiency of the working engine? 44

Solution First Law: U = Qnet – W Engine: cyclic process U = 0 Solution First Law: U = Qnet – W Engine: cyclic process U = 0 Qnet = W |QH| - |QC| |QC | = |QH| - W Engine 1: |QC| = 1000 - 200 = 800 J Engine 2: |QC| = 1000 - 300 = 700 J 45

 S(total) = - |QH|/ TH + |QC|/ TC Engine 1: S = (- S(total) = - |QH|/ TH + |QC|/ TC Engine 1: S = (- 2. 5 + 2. 7) J. K-1 = + 0. 2 J. K-1 > 0 Second Law validated Engine 2: S = (- 2. 5 + 2. 3) J. K-1 = - 0. 2 J. K-1 < 0 Second Law not validated Engine 1 is the working engine efficiency, e = (work out / energy input) 100 = (200 / 1000)(100) = 20 % 46

Semester 1, 2007 Examination question (5 marks) A hybrid petrol-engine car has a higher Semester 1, 2007 Examination question (5 marks) A hybrid petrol-engine car has a higher efficiency than a petrol-only car because it recovers some of the energy that would normally be lost as heat to the surrounding environment during breaking. (a) If the efficiency of a typical petrol-only car engine is 20%, what efficiency could be achieved if the amount of heat loss during breaking is halved? (b) Is it possible to recover all the energy lost as heat during braking and convert it into mechanical energy? Explain your answer. 47

Solution Identify / Setup 48 efficiency Second Law of Thermodynamics 100% of heat can Solution Identify / Setup 48 efficiency Second Law of Thermodynamics 100% of heat can not be transformed into mechanical energy e<1 Execute (a) Reduce heat loss by half (b) Would require |QC| = 0, this would be a violation of the Second Law

49 What is a heat pump ? Better buy this quick: 500 % efficiency 49 What is a heat pump ? Better buy this quick: 500 % efficiency

50 evaporator gas absorbs heat from interior of frig. |QH| = |Qc| + |W| 50 evaporator gas absorbs heat from interior of frig. |QH| = |Qc| + |W| cold compressor heats gas by compression hot condenser gas to liquid (high pressure) expansion value rapid expansion: liquid to gas, sudden large drop in temp (~35 o. C)

51 The compressor compresses the gas (e. g. ammonia). The compressed gas heats up 51 The compressor compresses the gas (e. g. ammonia). The compressed gas heats up as it is pressurized (orange). The gas represents the working substance eg ammonia and the compressor driven by an electric motor does work W. The condenser coils at the back of the refrigerator let the hot ammonia gas dissipate its heat QH. The ammonia gas condenses into ammonia liquid (dark blue) at high pressure gas (gas liquid). The high-pressure ammonia liquid flows through the expansion valve The liquid ammonia immediately boils and vaporizes (light blue), its temperature dropping to about – 35 °C by the expansion. This makes the inside of the refrigerator cold by absorption of heat QC. The cold ammonia gas enters the compressor and the cycle repeats.

Refrigeration Cycle Heat engine operating in reverse – it takes heat from a cold Refrigeration Cycle Heat engine operating in reverse – it takes heat from a cold place and gives it off at a warmer place, this requires a net input of work. |QH| = |QC| + |W| Best refrigerator – one that removes the greatest amount of heat |QC| from inside the refrigerator for the least expenditure of work |W| coefficient of performance, K (higher K value, better the refrigerator) K = what we want / what we pay for K = extraction of max heat from cold reservoir / least amount of work 52

p 53 QH W QC V p 53 QH W QC V

54 |QH| = |Qc| + |W| 54 |QH| = |Qc| + |W|

55 |QH| = |Qc| + |W| 55 |QH| = |Qc| + |W|

56 |QH| = |Qc| + |W| 56 |QH| = |Qc| + |W|

PHYS 1001 2009 Exam Question 10 In the figure above, cylinder A is separated PHYS 1001 2009 Exam Question 10 In the figure above, cylinder A is separated from cylinder B by an adiabatic piston which is freely movable. In cylinder A there is 0. 010 mole of an ideal monatomic gas with an initial temperature 300 K and a volume of 1. 0 x 10 -4 m 3. In cylinder B, there is 0. 010 mole of the same ideal monatomic gas with the same initial temperature and the same volume as the gas in cylinder A. Suppose that heat is allowed to flow slowly to the gas in cylinder A, and that the gas in cylinder B undergoes thermodynamic process of adiabatic compression. Heat flows into cylinder A until finally the gas in cylinder B is compressed to a volume of 0. 5 x 10 -5 m 3. Assume that CV = 12. 47 J. mol-1. K-1 and the ratio of heat capacities is = 1. 67. (a) Calculate the final temperature and pressure of the ideal monatomic gas in cylinder B after the adiabatic compression. (b) How much work does the gas in cylinder B do during the compression? Explain the meaning of the sign of the work. (c) What is the final temperature of the gas in the cylinder A? (Hint: the pressure exerted by the gas in cylinder A on the piston is equal to that exerted by the gas in cylinder B on the piston. ) (d) How much heat flows to the gas in the cylinder A during the process? 57

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