Transcript

Trains and Boats and PlanesTrains and Boats and Planes

John D. BarrowJohn D. Barrow

(….and Cars Too)(….and Cars Too)

Simple Quantities We Will NeedSimple Quantities We Will Need

Distance = Speed Time: d = V t

Energy = Power Time: E = P tJoules = Watts seconds

Kinetic energy of motion = ½ M V2

Power usage units: kilowatt hours per day per person Average Briton consumes 125 kWh/day per personAverage Briton consumes 125 kWh/day per person

The Odd Units of The Odd Units of Fuel ConsumptionFuel Consumption

Miles per gallon (or litres) is length/volume = 1/areaMiles per gallon (or litres) is length/volume = 1/areaGallons per mile is volume/length = areaGallons per mile is volume/length = areaCar fuel use of 6 litres per 100km = 6 per sq mmCar fuel use of 6 litres per 100km = 6 per sq mmis 41 miles per gallon is 41 miles per gallon 1/16.7 per sq mm 1/16.7 per sq mm

What is this area ?What is this area ?

100 km

Area

Imagine fuel spurting from a tube of this area atImagine fuel spurting from a tube of this area atthe speed of your carthe speed of your car

1 litre per 100km = 1/100 sq mm

CarsCars

Kinetic Energy of MotionKinetic Energy of Motion½ mv½ mv22 = ½(m/1000kg) = ½(m/1000kg)(v/100km per hr)(v/100km per hr)22

= 4= 4101055J = 0.1KWhrJ = 0.1KWhr

Energy LossesEnergy Losses

BrakingBrakingAir resistance and swirlingAir resistance and swirling

Rolling friction, noise, vibrationRolling friction, noise, vibration

Braking EffectsBraking Effects

d d

Speed V StopStopAll KEAll KElostlost

Speed V StopStopAll KEAll KElostlost

Rate of energy lost in brakes = KE/(time between stops) Rate of energy lost in brakes = KE/(time between stops) = ½ mV= ½ mV22 ÷ ÷ (d/V) (d/V)

= ½ mV= ½ mV33/d/d

Time between stops = d/VTime between stops = d/V

Note the VNote the V33 factor factor

Air SwirlAir Swirl

Area, A(car)

Mass of tube of air swept in time, t = Mass of tube of air swept in time, t = A A V V t = m t = maa

= air density A = c = air density A = c A(car) is the effective car area A(car) is the effective car area

c = drag factorc = drag factor

Kinetic energy of swirling air = ½ mKinetic energy of swirling air = ½ maaVV2 2 = ½ = ½ A V A V33 t t

Rate of KE loss to air swirl = ½ mRate of KE loss to air swirl = ½ maaVV22 /t = /t = ½ ½ A V A V33

Speed V

What is this biggest energy loss? What is this biggest energy loss?

Ratio: (Loss to brakes/Loss to air) = m/d Ratio: (Loss to brakes/Loss to air) = m/d 1/A 1/A

So brake losses dominate ifSo brake losses dominate ifm(car) > A m(car) > A d = mass of air swept out d = mass of air swept out

between stopsbetween stops

Heavy cars: losses dominated by brakingHeavy cars: losses dominated by braking

Light cars: losses dominated by air swirlLight cars: losses dominated by air swirl

Critical distance between stops is d = D Critical distance between stops is d = D m(car)/A m(car)/A

A(car) = 1.5 m x 2 m = 3 mA(car) = 1.5 m x 2 m = 3 m22

DragDrag factor: c = 1/3factor: c = 1/3Effective A = c Effective A = c A(car) =1m A(car) =1m22

Mass: m = 1000 kgMass: m = 1000 kgAir: Air: = 1.3 kg/m = 1.3 kg/m33

D = 750 metresD = 750 metres

Braking losses dominate for d < D -- typical in for town drivingBraking losses dominate for d < D -- typical in for town drivingAir swirl losses dominate for d > D -- on the motorwayAir swirl losses dominate for d > D -- on the motorway

Both losses are Both losses are (speed) (speed)3 3 ! !

Short distance driving: reduce V, m(car), increase d, reuse brake energyShort distance driving: reduce V, m(car), increase d, reuse brake energy

Driving Without Short StopsDriving Without Short Stops M(car) not very important losses are M(car) not very important losses are AVAV33

Reduce A(car) and drag, c: Reduce A(car) and drag, c: A = A(car) A = A(car) c c Reduce Reduce VV Move slower, move less and use long, thin, streamlined Move slower, move less and use long, thin, streamlined

vehiclesvehicles

Petrol engines 25% efficient so engine power Petrol engines 25% efficient so engine power for motorway driving (v = 70mph = 31 msfor motorway driving (v = 70mph = 31 ms-2-2 and and

A = 1 mA = 1 m22 is approx is approx

4 4 ½ ½ AVAV33 = 2 = 2 1.3 1.3 1 1 31 km m 31 km m33ss-3-3 = 80 kilo Watts = 80 kilo WattsIf you drive at ½ If you drive at ½ 70 = 35 mph for 2 hrs you use only 20kW – V-cubed effect70 = 35 mph for 2 hrs you use only 20kW – V-cubed effect

Different Car ParametersDifferent Car Parameters Drag factor. We used c = Drag factor. We used c =

0.330.33 c(Honda) = 0.25c(Honda) = 0.25 c(Sierra) = 0.34c(Sierra) = 0.34 c(Citroen 2CV) = 0.51c(Citroen 2CV) = 0.51 c(bike) = 0.9c(bike) = 0.9 c(coach) = 0.42c(coach) = 0.42Effective areasEffective areas A(Discovery) = 1.6A(Discovery) = 1.6 A(typical car) = 0.8A(typical car) = 0.8 A(Honda Insight) = 0.47A(Honda Insight) = 0.47

Bikes and Cars ComparedBikes and Cars Compared

Energy/dist = 4 Energy/dist = 4 ½ ½ AVAV33 ÷÷ V = 2 V = 2 AAVV22 (E/d)(E/d)bikebike/ (E/d)/ (E/d)carcar = = [c(bk)/c(car)][c(bk)/c(car)][(A(bk)/A(car)][(A(bk)/A(car)][V[V22(bk)/V(bk)/V22(car)](car)]

= 1/0.33 = 1/0.33 ¼ ¼ (1/5) (1/5)2 2 0.03 0.03 ((V(bk)= 13mph)V(bk)= 13mph)

Bike is about 33 times more fuel efficientBike is about 33 times more fuel efficient than the motorway carthan the motorway car

Power vs speedPower vs speed

Power Power (speed) (speed)33

Energy per unit distance Energy per unit distance (speed) (speed)22

If engine efficiency stays the same thenIf engine efficiency stays the same thenHalve speed Halve speed reduce gallons per mile by 4 reduce gallons per mile by 4

D. MacKay

Braking DistancesBraking DistancesVV22 = U = U22 – 2as – 2as

If you stop (V = 0) after travelling at speed U with deceleration a then your braking distance is s

s = Us = U22/2a /2a (speed) (speed)22

Eg two cars: 65 and 60 km/hr drivers have 1.5 sec reaction timeThey travel 27.1m and 25m during the reaction time, then

s = 16.3 m and s = 13.9

Total distances travelled are 43.4m and 38.9 mTotal distances travelled are 43.4m and 38.9 m

Eg child 40 m away is not hit by the 60 km/h car but the 65 km/h hits theChild at a speed = (U2 – 2ad)1/2 = 30 km/hr = 8.2 m/s as d = 40 – 27.1 mBody thickness 20cm. Impact lasts 0.025s. Deceleration is 320ms-2.

If child weighs 59 kg, impact force ( U2) is 16,000 N (decel more than 26g).

Energy Lost in Rolling WheelsEnergy Lost in Rolling Wheels The third energy loss is in tyres, friction in bearings, axles, noise, The third energy loss is in tyres, friction in bearings, axles, noise,

heating rubber, vibrationheating rubber, vibration Energy lost = C Energy lost = C mg mg car weight car weight C = 0.002 (train), 0.005 (bike), 0.007 (truck), 0.010 (car)C = 0.002 (train), 0.005 (bike), 0.007 (truck), 0.010 (car) So rolling resistance about 100 N per ton at any speedSo rolling resistance about 100 N per ton at any speed Power reqd to overcome rolling is Power reqd to overcome rolling is

Force Force speed = 100N speed = 100N 31 m/s = 3.1 kW 31 m/s = 3.1 kW for 1 ton truck at 70mphfor 1 ton truck at 70mph

Engine 25% efficient so12 kW to engine, but 80kW needed to beat Engine 25% efficient so12 kW to engine, but 80kW needed to beat air swirl on motorway and 15% to overcome rolling resistance.air swirl on motorway and 15% to overcome rolling resistance.

Rolling > air resistance Rolling > air resistance when when

V < (2Cmg/cAV < (2Cmg/cA))1/21/2 = 7m/s = 16 mph = 7m/s = 16 mph

bikecar

Rolling losses only dominate at low speedsRolling losses only dominate at low speeds

TrainsTrains8-carriage train: m= 400,000 kg8-carriage train: m= 400,000 kgA = 11 mA = 11 m22 Air resistance > rolling when V > 75 mph = 33 m/sAir resistance > rolling when V > 75 mph = 33 m/s

1-carriage train: m = 50,000 kg1-carriage train: m = 50,000 kgAir resistance dominates at V > 26 mph = 12 m/sAir resistance dominates at V > 26 mph = 12 m/s

Kings X-Cambridge trainKings X-Cambridge train275 tonnes, 585 passengers, max V = 100 mph = 161 km/hr, Power = 1.5 MWFull speed: Energy use about 1.6 kWh per 100 passenger-km1.6 kWh per 100 passenger-kmLondon Underground trainLondon Underground trainLoaded 228 tons, max V = 45 mph (average = 30 mph), 350-620 passengersEnergy use is 4.4 kWh per 100 passenger-km4.4 kWh per 100 passenger-kmIntercity 125Intercity 125500 passengers, max V = 125 mph, Power = 2.6 MWEnergy use is 9 kWh per 100 passenger-km9 kWh per 100 passenger-km

Rowing in NumbersRowing in NumbersHow does the speed of the boat depend on the number of rowers?

Drag on boat V2 wetted surface area of boat V2 L2

Volume of boat L3 N – the number of crew

Drag Drag V V22 N N2/32/3

Crew Power overcoming drag = N P = V Drag V3 N2/3 P is the constant power exerted by each (identical rower)

V V N N1/91/9

With a cox With a cox V V N N1/31/3/[N + 1/3 ]/[N + 1/3 ]2/9 2/9 if cox is third the weight of a rower if cox is third the weight of a rower

Results of 1980 Olympic Results of 1980 Olympic coxless races (N = 1, 2, 4) + coxed N = 8coxless races (N = 1, 2, 4) + coxed N = 8

Distance is 2000 metresDistance is 2000 metres

NN Time, TTime, T

(sec)(sec)

11 429.6429.6

22 408.0408.0

44 368.2368.2

88 349.1349.1

T = (2000) / V T = (2000) / V N N-0.11-0.11

V V N N0.11 0.11 N N1/91/9

Compare with a coxCompare with a cox N = 2, T = 422.5 s N = 2, T = 422.5 s N = 4 ,T = 374.5 s N = 4 ,T = 374.5 s (predicted if cox =0.3 of a rower mass)(predicted if cox =0.3 of a rower mass)

Better without one!Better without one!

AeroplanesAeroplanesPush air down and Newton’s 3Push air down and Newton’s 3rdrd law gives law gives

(equal & opposite) upward lift(equal & opposite) upward lift

Mass of air tube = air density Mass of air tube = air density volume = volume = A V t A V t

Downward accn created in time t = momentum of plane’s weight in time tm(tube) U = VAU t = mg t

Downward speed of air: U = mg/(Downward speed of air: U = mg/(VA)VA)See how U as V as plane meets less air per sec

Why we use flaps on landing to deflect more air mass

Optimal EnergeticsOptimal EnergeticsRate of energy use to push down at speed U = KE of air tube ÷ t

P(lift) = ½ (mg)P(lift) = ½ (mg)22/ / ((VA)VA)

Total Power = P(lift) + P(drag) = ½ (mg)2/ (VA) + ½ cV3 Ap

Ap : front area of plane X-section

Fuel efficiency: P(total) / V = thrustP(total) / V = thrust

Jet engine efficiency about 30%, m = 363,000 kgAp = 180 m2, 64.5 m wingspan A, c = 0.03

Optimal when P(lift) = P(drag)Optimal when P(lift) = P(drag) VV22(opt) = mg/[(opt) = mg/[(cA(cAppA)A)1/21/2]] (540 mph) (540 mph)22 for 747 at for 747 at

30,000 ft where 30,000 ft where (air) = 1/3 (air) = 1/3 (surface) =0.13 kg/m(surface) =0.13 kg/m33

assumesubsonic< 330m/s

540 mph 220 m/s

Flight RangeFlight Range

Range = V(opt) Range = V(opt) energy/power energy/power efficiency efficiency C/g = 1/3 C/g = 1/3 (calorific value of fuel)/g (calorific value of fuel)/g

Independent of size, mass, speed etc (applies to birds too!)Independent of size, mass, speed etc (applies to birds too!)

Jet fuel (hydrocarbons): C = 40 MJ/kg Jet fuel (hydrocarbons): C = 40 MJ/kg Range Range 3.3 3.3 C/g = 3.3 C/g = 3.3 4000 km4000 km

Range = 13,300 km ie 16.7 hrs at V(opt)Range = 13,300 km ie 16.7 hrs at V(opt)Non-stop 747 range record is 16560 km (11,000 km for birds)

Note V(opt) decreases as plane loses fuel mass but can maintain speed by going to higher altitude (31000 to 39000 ft) where V(opt) 1/ is higher becauseof lower air density

Transport EfficiencyTransport EfficiencyEfficiency = No. of passenger-km per litre of fuel Efficiency = No. of passenger-km per litre of fuel

For 747 = No. For 747 = No. 1/3 1/3 1.38 MJ per litre/200,000N 1.38 MJ per litre/200,000N 25 25

Full 747 efficiency = 25 passenger-km per litre of fuelFull 747 efficiency = 25 passenger-km per litre of fuel

Ordinary car with 1 passenger has 12 passenger-km per litre of fuelOrdinary car with 1 passenger has 12 passenger-km per litre of fuel

Ordinary car with 4 passenger has 48 passenger-km per litre of fuelOrdinary car with 4 passenger has 48 passenger-km per litre of fuel

D. MacKay

Energy consumed in KWh per ton-kilometre of freight movedEnergy consumed in KWh per ton-kilometre of freight moved

WindmillsWindmills

Or …Or …

Or even…Or even…

Windmill EfficiencyWindmill EfficiencyWind power thro’ area A = ½ Wind power thro’ area A = ½ AVAV33

UIncoming

V

Average wind speed at sails = ½ (U + V)Air mass per unit time through sails is F = A ½ (U + V)

Power generated = difference in the rate of change of kinetic energy of the air before and after passing the sails.

P = ½ FUP = ½ FU22 - ½ FV - ½ FV2 2 = ¼ DA(U= ¼ DA(U22 – V – V22 )(U + V) )(U + V)

If the windmill was not there the power in the wind is PP00 = ½ DAU = ½ DAU33

P/PP/P00 = ½ {1 – (V/U) = ½ {1 – (V/U)22} } {1+ (V/U)} {1+ (V/U)}

P/PP/P00 is a maximum when V/U =1/3 is a maximum when V/U =1/3

Max P/PMax P/P00 = 16/27 = 59.26% = 16/27 = 59.26% Albert Betz’s (1919) LawAlbert Betz’s (1919) Law

Outgoing wind speed V < U

Actual Energy OutputActual Energy Output

V/U = 1/3, so PV/U = 1/3, so Pmaxmax = (8/27) = (8/27) A A U U33

Only about 20% efficient (new offshore rotors up to 50%)Only about 20% efficient (new offshore rotors up to 50%) If the diameter of the rotor is d then the area is A= If the diameter of the rotor is d then the area is A= dd22/4 and /4 and

the windmill is 20 % efficient then the power output is about the windmill is 20 % efficient then the power output is about P = 2/5 P = 2/5 1m 1m (d/2m) (d/2m)22 (U/1ms (U/1ms-1-1))33 watts. watts. But you can’t pack them too close (> 5d apart)But you can’t pack them too close (> 5d apart) [Power per mill]/[Land area per mill] = ([Power per mill]/[Land area per mill] = (/400)/400)UU33 = 2.2 W per sq metre if U = 6 m/sec or 0.7 W/m= 2.2 W per sq metre if U = 6 m/sec or 0.7 W/m2 2 if U = 4m/sif U = 4m/s A little one on your roof will give 0.2 kWh per dayA little one on your roof will give 0.2 kWh per day

Blowin’ In The WindBlowin’ In The Wind

1 person per 4000 sq m in the UK. Pack the whole country onshore with windmills1 person per 4000 sq m in the UK. Pack the whole country onshore with windmills2 W per m2 W per m22 4000 m 4000 m22 per person = 8 kW per person per person = 8 kW per person

If we covered 10% of country we get 20 kWh/day per person If we covered 10% of country we get 20 kWh/day per person This gives only 50% of power to drive average petrol car 50 km per dayThis gives only 50% of power to drive average petrol car 50 km per day


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