Presentation on theme: "PHYSICS: FUN EXCITING SIMPLE"— Presentation transcript:
1 PHYSICS: FUN EXCITING SIMPLE Physics 1 REGULARModule 2 Thermal PhysicsHEAT ENGINEPHYSICS: FUN EXCITING SIMPLEap06/p1/thermal/ptG_engines.ppt
2 HEAT ENGINES AND REFRIGERATORS Heat engines: Entropy (§20.1 p §20.2 p675 §20.5 p682)Heat engines: Carnot cycle (§20.6 p §20.7 p690)Internal Combustion Engines: Otto & Diesel Cycles (§20.3 p678)Refrigerators (§20.4 p680)References: University Physics 12th ed Young & Freedman
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 + airpetrol engine - hot exhaust gases + cooling system
5 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| = WThermal efficiency, e represents the fraction of QH that is converted to useful work.
6 Problem (Y & F Example 20.1)A large truck is travelling at 88 km.h-1. The engine takes in J of heat and delivers J of mechanical energy per cycle. The heat is obtained by burning petrol (heat of combustion Lcomb = 5.0107 J.kg-1). The engine undergoes 25 cycles per second.Density of petrol = 700 kg.m-3(a) What is the 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?
7 Solution I S E Ev = 88 km.h QH = 1.0104 J/cycle W = 2103 J/cycleLcomb = 5.0107 J.kg-1 f = 25 cycles.s-1(a)e = ?e = W / QH = (2103) / (1.0104) = 0.20 = 20% sensible(b)QC = ? J |QH| - |QC| = W |QC| = 8.0103 J/cycle
8 (c)QH = 1104 J/cycle QH /t = (25)(1104) J.s-1 = 2.5105 J.s LC = 5.0107 J.kg-1QH/t = (m/t) Lcomb m/t = (QH/t )/ Lcomb = (2.5105 ) / (5.0107) = 510-3 kg.s-1m/t = (510-3)(60)(60) = 18 kg.h-1(d)fuel consumption = ? L/100 kmv = 88 km.h-1 v = d/t t = ? hd = 100 kmt = d / v = 100 / 88 h = hmass used traveling 100 km m = (m/t)t = (18)(1.1364) = kg= m V L = 1 m3volume used in traveling 100 km V = m / = ( / 700) m3 = m3 = 29.2 Lpetrol consumption = L/100 km
9 Hot Reservoir QH W Engine QC= 0 Surroundings 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) = cyclic processS(surrounding) = 0 no heat transfer to surroundingsS(hot reservoir) < 0 heat removedS(total) < violates Second LawAll heat engines (heat to work): efficiency, e < 1
11 Hot reservoirQHWEnginesurroundingsQCCold reservoir
12 S(hot reservoir) = - |QH|/ TH S(cold reservoir) = + |QC|/ TCS(engine) = 0 (cyclic process)S(surroundings) = 0 (no heat transfer to surroundings)S(total) = - |QH|/ TH + |QC|/ TCUseful work can only be done if S(total) 0| QH| / TH | QC | / TC
13 An ideal engine e.g. Carnot Cycle S = 0 | QH / TH | = | QC / TC | e = W / QH W = |QH| - |QC|e = (|QH| - |QC| / QH)= 1 - |QC| / |QH|e = 1 – TC / TH < 1Sadi Carnot (1824) FrenchmanThis is the absolute max efficiency of a heat engineNOTE: All heat (QH & QC) exchanges occurred isothermally in calculating the efficiency e = 1 – TC / TH
14 CARNOT CYCLEHeat engine with the maximum possible efficiency consistent with 2nd 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 Q W Q V adiabatic isothermals 3 Diagram not to 4 scale, adiabatsareWmuch steeper than2shown1QCreleased tosurroundingsVNOTE: All heat (QH & QC) exchanges occurred isothermally in calculating the efficiency e = 1 – TC / TH
16 QC QH p V 1 and 2: “cold” 3 and 4: “hot” 1 2Isothermalcompression2 3Adiabaticcompression3 4Isothermalexpansion4 1AdiabaticexpansionAll energy exchanges are reversible – there are no non-recoverable energy losses
17 Proof: e = 1 - |QC| / |QH| = 1 - TC / TH Isothermal change Q = n R T ln(Vf /Vi)1 2: Isothermal compressionheat QC rejected to sink at constanttemperature TC.|QC | = n R TC ln(V2 / V1)3 4: Isothermal expansionheat QH supplied from source at constant temperature|QH | = n R TH ln(V4 / V3)e = 1 - |QC| / |QH| = 1 – TC ln(V2 / V1) / TH ln(V4 / V3) (Eq. 1)1 and 2: “cold”3 and 4: “hot”Vp4312Adiabatic change Q = 0 TiVi-1 = TfVf-1
18 2 3: 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 pV diagram) is too small to overcome friction & other losses in a real engine.e = 1 - TC ln(V1 / V2) / TH ln(V4 / V3)(Eq. 1)Efficiency of Carnot engine - max possible efficiency for a heat engine operating between TC and TH
19 TC = 25 oCThe strength & hardness of metalsdecreases rapidly above 750 oC“Re-author” Y&F Example 20.3
20 Diathermal wall: A highly thermally conducting wall.
21 CAR ENGINEThe four–stroke OTTO cycle of a conventional petrol engine
22 The four–stroke OTTO cycle of a conventional petrol engine intake compression ignition power exhauststroke stroke stroke strokeintake stroke: isobaric expansioncompression stroke: adiabatic compressionignition: isochoric heating of gaspower stroke: adiabatic expansion of gasexhaust stroke: isochoric cooling of gas / isobaric compression
24 cooling of exhaust gases Otto CyclePIGNITIONfuelcombustion3adiabaticisothermalsQHpower stroke2compressionstrokeW4cooling of exhaust gasesQC51released toPsurroundingsoVintake strokeexhaust strokeV2V1 = r V2r = compression ratio
25 POtto Cycle3adiabaticisothermalsQH24QCreleased toP51surroundingsoVVV215 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 POtto Cycle3adiabaticisothermalsQH24QCreleased toP51surroundingsoVVV211 2: compression stroke inlet valve closes piston moves up compressing the air/fuel mixture adiabatically.
28 POtto Cycle3adiabaticisothermalsQH24QCP51released tosurroundingsoVVV213 4: expansion or power stroke – heated gas expands adiabatically as the piston is pushed down doing work(Vmax = r Vmin) Compression ratio, r
29 POtto Cycle3adiabaticisothermalsQH24QCreleased toP51surroundingsoVVV214 1 start of Exhaust stroke – outlet valve opens and mixture expelled at constant volume then 1 5
30 POtto Cycle3adiabaticisothermalsQH24QCreleased toP51surroundingsoVVV211 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 the thermodynamic processes in a typical car engine.For compression ratio, r ~ 8 and = 1.4 (air)TH (peak) ~ 1800 °C TC (base) ~ 50 °Ce ~ 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.
32 Comparison of theoretical and actual pV diagrams for the four-stroke Otto Cycle engine.
33 P Q Q P V V V Diesel Cycle 5 to 1: intake stroke isobaric expansion 1 to 2: compression strokeabiabatic compression2 to 3: ignitionisobaric heating3 to 4: power strokeadiabatic expansion4 to 1: exhaust stroke (start)isochoric cooling1 to 5: exhaust stroke (finish)isobaric compressionPQadiabaticisothermalsH234released toQsurroundingsP51CoVVV21Rudolf Diesel
34 P Diesel Cycle Q Q P V V V adiabatic isothermals 2 3 4 5 1 H C o 2 1 fuelignitionadiabaticisothermalsH23power strokecompressionstroke4Qcooling of exhaust gasesCreleased toP51surroundingsoVVV21
36 DIESEL CYCLENo 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 TC (base) ~ 50 °Ce ~ 68% (ideal engine) real efficiency ~ 40 %Petrol engine ~ 56% (ideal engine) real efficiency ~ 35 %
37 Diesel enginesHeavier (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 compressionThey need no carburettor or ignition system, but the fuel-injection systemrequires expensive high-precision machining.Use cheaper fuels less refined heavy oils – fuel does need to bevaporized in carburettor, fuel less volatile hence safer from fire orexplosion.Diesel cycle – can control amount of injected fuel per cycle – less fuel used atlow speeds.
38 These engines were designed primarily for very large container ships. These engines were designed primarily for very large container ships.
39 ProblemFor the theoretical Otto cycle, calculate:(a) max cycle temperature(b) work per kilogram of fuel(c) Efficiency(d) Max efficiency Carnot engine working between same temperaturesEngine characteristics: cp = kJ.kg-1.K-1 cV = kJ.kg-1.K-1compression ratio = 8:1Inlet conditions p = 97.5 kPa and T = 50 oCHeat supplied = 950 kJ.kg-1
40 Qp = m cp T QV = m cV T W = |QH| - |QC| SolutionIdentify / Setupcp = kJ.kg-1.K-1cV = kJ.kg-1.K-1V1/ V2 = 8p1 = 97.5103 PaT1 = 50 oC = 323 KQH = 950 kJ.kg-1TH = T3 = ? KW = ? kJ.kg-1e = ? eCarnot = ?Adiabatic change Q = 0 T V -1 = constant = cp / cV = / = 1.4Qp = m cp T QV = m cV T W = |QH| - |QC|e = W / |QH| eCarnot = 1 – TC / THPOtto Cycle3adiabaticisothermalsQH24QCreleased toP51surroundingsoVVV21
42 Diesel CycleCalculate the above quantities for the diesel cycle with a compression ratio = 20PQadiabaticisothermalsH234released toQsurroundingsP51CoVVV21
43 Adiabatic compression V1 / V2 = 20 T2 / T1 = (V1/V2) T2 = T1(V1 / V2)-1 = (323)(20)0.4 K = 1071 KIsobaric heatingQH = m cp (T3-T2) QH/m = cp (T3 –T2)T3 = T2 + (1/cp)(QH/m) = (1/1.005)(950) K = 2016 KIsobaric heating / Adiabatic expansionV2 / T2 = V3 / T3 V3 / V2 = T3 / T2 V4 / V2 = 20 V2 = V4 / 20V3 / V4 = (1/20)(T3/T2) = (1/20)(2016/1071) =T3V3-1 = T4V4-1 T4 = T3 (V3/V4)-1 = (2016)(0.0941)0.4 = 783 KIsochoric coolingQC = m cv (T4 – T1) QC/m = cv (T4 – T1) = (0.718)(783 – 323) K = kJ.kg-1W = QH – QC = (950 – 330) kJ = 620 kJ.kg-1e = W / QH = 620 / 950 = e = 1- QC/QH = 1 – 330/950 = 0.65The work output and efficiency are considerably higher than for Otto Cycle
44 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?
46 S(total) = - |QH|/ TH + |QC|/ TC Engine 1: S = ( ) J.K-1 = J.K-1 > 0 Second Law validatedEngine 2: S = ( ) J.K-1 = J.K-1 < 0 Second Law not validatedEngine 1 is the working engineefficiency, e = (work out / energy input) 100= (200 / 1000)(100) = 20 %
47 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.
48 SolutionIdentify / SetupefficiencySecond Law of Thermodynamics100% of heat can not be transformed into mechanical energye < 1Execute(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
50 cold evaporator gas absorbs heat from interior of frig. |QH| = |Qc| + |W|coldcompressorheats gas by compressionhotexpansion valuerapid expansion: liquid to gas, sudden large drop in temp (~35 oC)condensergas to liquid (high pressure)
51 The compressor compresses the gas (e. g. ammonia) 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.
52 K = what we want / what we pay for Refrigeration CycleHeat 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 forK = extraction of max heat from cold reservoir / least amount of work
57 PHYS Exam Question 10In the figure above, cylinder A is separated from cylinder B by an adiabatic piston which is freely movable. In cylinder A there is mole of an ideal monatomic gas with an initial temperature 300 K and a volume of 1.0x10-4 m3. In cylinder B, there is 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 the thermodynamic process of adiabatic compression. Heat flows into cylinder A until finally the gas in cylinder B is compressed to a volume of 0.5x10-5 m3. Assume that CV = J.mol-1.K-1 and the ratio of heat capacities is = (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?