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Trains and Boats and Planes

Trains and Boats and Planes. (….and Cars Too). John D. Barrow. Simple Quantities We Will Need. Distance = Speed  Time: d = V  t Energy = Power  Time: E = P  t Joules = Watts  seconds Kinetic energy of motion = ½ M V 2

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Trains and Boats and Planes

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  1. Trains and Boats and Planes (….and Cars Too) John D. Barrow

  2. Simple Quantities We Will Need Distance = Speed  Time: d = V  t Energy = Power  Time: E = P  t Joules = 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 person

  3. The Odd Units of Fuel Consumption Miles per gallon (or litres) is length/volume = 1/area Gallons per mile is volume/length = area Car fuel use of 6 litres per 100km = 6 per sq mm is 41 miles per gallon  1/16.7 per sq mm What is this area ? Area 1 litre per 100km = 1/100 sq mm 100 km Imagine fuel spurting from a tube of this area at the speed of your car

  4. Cars Kinetic Energy of Motion ½ mv2 = ½(m/1000kg)(v/100km per hr)2 = 4105J = 0.1KWhr

  5. Energy Losses Braking Air resistance and swirling Rolling friction, noise, vibration

  6. Braking Effects d d Speed V Stop All KE lost Speed V Stop All KE lost Time between stops = d/V Rate of energy lost in brakes = KE/(time between stops) = ½ mV2÷ (d/V) = ½ mV3/d Note the V3 factor

  7. Air Swirl  Speed V Area, A(car) • Mass of tube of air swept in time, t =   A V  t = ma • = air density A = c  A(car) is the effective car area c = drag factor Kinetic energy of swirling air = ½ maV2 = ½  A V3 t Rate of KE loss to air swirl = ½ maV2 /t =½  A V3

  8. What is this biggest energy loss? Ratio: (Loss to brakes/Loss to air) = m/d  1/A So brake losses dominate if m(car) > A    d = mass of air swept out between stops Heavy cars: losses dominated by braking Light cars: losses dominated by air swirl

  9. Critical distance between stops is d = D  m(car)/A A(car) = 1.5 m x 2 m = 3 m2 Dragfactor: c = 1/3 Effective A = c  A(car) =1m2 Mass: m = 1000 kg Air:  = 1.3 kg/m3 D = 750 metres Braking losses dominate for d < D -- typical in for town driving Air swirl losses dominate for d > D -- on the motorway Both losses are  (speed)3 ! Short distance driving: reduce V, m(car), increase d, reuse brake energy

  10. Driving Without Short Stops • M(car) not very important losses are AV3 • Reduce A(car) and drag, c: A = A(car)  c • Reduce V • Move slower, move less and use long, thin, streamlined vehicles Petrol engines 25% efficient so engine power for motorway driving (v = 70mph = 31 ms-2 and A = 1 m2 is approx 4  ½AV3 = 2 1.3 1 31 km m3s-3 = 80 kilo Watts If you drive at ½ 70 = 35 mph for 2 hrs you use only 20kW – V-cubed effect

  11. Different Car Parameters • Drag factor. We used c = 0.33 • c(Honda) = 0.25 • c(Sierra) = 0.34 • c(Citroen 2CV) = 0.51 • c(bike) = 0.9 • c(coach) = 0.42 Effective areas • A(Discovery) = 1.6 • A(typical car) = 0.8 • A(Honda Insight) = 0.47

  12. Bikes and Cars Compared Energy/dist = 4  ½AV3÷ V = 2AV2 (E/d)bike/ (E/d)car = [c(bk)/c(car)][(A(bk)/A(car)][V2(bk)/V2(car)] = 1/0.33  ¼  (1/5)2  0.03(V(bk)= 13mph) Bike is about 33 times more fuel efficient than the motorway car

  13. Power vs speed Power  (speed)3 Energy per unit distance  (speed)2 If engine efficiency stays the same then Halve speed  reduce gallons per mile by 4 D. MacKay

  14. Braking Distances V2 = U2 – 2as If you stop (V = 0) after travelling at speed U with deceleration a then your braking distance is s s = U2/2a  (speed)2 Eg two cars: 65 and 60 km/hr drivers have 1.5 sec reaction time They 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 m Eg child 40 m away is not hit by the 60 km/h car but the 65 km/h hits the Child at a speed = (U2 – 2ad)1/2 = 30 km/hr = 8.2 m/s as d = 40 – 27.1 m Body 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).

  15. Energy Lost in Rolling Wheels • The third energy loss is in tyres, friction in bearings, axles, noise, heating rubber, vibration • Energy lost = C  mg  car weight • C = 0.002 (train), 0.005 (bike), 0.007 (truck), 0.010 (car) • So rolling resistance about 100 N per ton at any speed • Power reqd to overcome rolling is Force  speed = 100N  31 m/s = 3.1 kW for 1 ton truck at 70mph • Engine 25% efficient so12 kW to engine, but 80kW needed to beat air swirl on motorway and 15% to overcome rolling resistance.

  16. Rolling > air resistance when V < (2Cmg/cA)1/2 = 7m/s = 16 mph car bike Rolling losses only dominate at low speeds

  17. Trains 8-carriage train: m= 400,000 kg A = 11 m2 Air resistance > rolling when V > 75 mph = 33 m/s 1-carriage train: m = 50,000 kg Air resistance dominates at V > 26 mph = 12 m/s Kings X-Cambridge train 275 tonnes, 585 passengers, max V = 100 mph = 161 km/hr, Power = 1.5 MW Full speed: Energy use about 1.6 kWh per 100 passenger-km London Underground train Loaded 228 tons, max V = 45 mph (average = 30 mph), 350-620 passengers Energy use is 4.4 kWh per 100 passenger-km Intercity 125 500 passengers, max V = 125 mph, Power = 2.6 MW Energy use is 9 kWh per 100 passenger-km

  18. Rowing in Numbers How 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  V2 N2/3 Crew Power overcoming drag = N  P = V  Drag V3 N2/3 P is the constant power exerted by each (identical rower) V  N1/9 With a cox V  N1/3/[N + 1/3 ]2/9 if cox is third the weight of a rower

  19. Results of 1980 Olympic coxless races (N = 1, 2, 4) + coxed N = 8 Distance is 2000 metres Compare with a cox N = 2, T = 422.5 s N = 4 ,T = 374.5 s (predicted if cox =0.3 of a rower mass) Better without one! T = (2000) / V  N-0.11 V  N0.11  N1/9

  20. Aeroplanes Push air down and Newton’s 3rd law gives (equal & opposite) upward lift  Mass of air tube = air density  volume =   A V t Downward accn created in time t = momentum of plane’s weight in time t m(tube)  U = VAU  t = mg  t Downward speed of air: U = mg/(VA) See how U  as V  as plane meets less air per sec Why we use flaps on landing to deflect more air mass

  21. Optimal Energetics Rate of energy use to push down at speed U = KE of air tube ÷ t P(lift) = ½ (mg)2/ (VA) assume subsonic< 330m/s Total Power = P(lift) + P(drag) = ½ (mg)2/ (VA) + ½ cV3 Ap Ap : front area of plane X-section Fuel efficiency: P(total) / V = thrust Jet engine efficiency about 30%, m = 363,000 kg Ap = 180 m2, 64.5 m wingspan  A, c = 0.03 Optimal when P(lift) = P(drag) V2(opt) = mg/[(cApA)1/2] (540 mph)2 for 747 at 30,000 ft where (air) = 1/3 (surface) =0.13 kg/m3 540 mph  220 m/s

  22. Flight Range Range = V(opt)  energy/power  efficiency  C/g = 1/3  (calorific value of fuel)/g Independent of size, mass, speed etc (applies to birds too!) Jet fuel (hydrocarbons): C = 40 MJ/kg Range  3.3 C/g = 3.3 4000 km 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 because of lower air density

  23. Transport Efficiency Efficiency = No. of passenger-km per litre of fuel For 747 = No.  1/3  1.38 MJ per litre/200,000N  25 Full 747 efficiency = 25 passenger-km per litre of fuel Ordinary car with 1 passenger has 12 passenger-km per litre of fuel Ordinary car with 4 passenger has 48 passenger-km per litre of fuel

  24. Energy consumed in KWh per ton-kilometre of freight moved D. MacKay

  25. Windmills

  26. Or …

  27. Or even…

  28. Windmill Efficiency Wind power thro’ area A = ½ AV3 Outgoing wind speed V < U U Incoming 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 FU2 - ½ FV2 = ¼ DA(U2 – V2 )(U + V) If the windmill was not there the power in the wind is P0 = ½ DAU3 P/P0 = ½ {1 – (V/U)2}  {1+ (V/U)} P/P0 is a maximum when V/U =1/3 Max P/P0 = 16/27 = 59.26%Albert Betz’s (1919) Law

  29. Actual Energy Output • V/U = 1/3, so Pmax = (8/27)  A  U3 • Only about 20% efficient (new offshore rotors up to 50%) • If the diameter of the rotor is d then the area is A= d2/4 and the windmill is 20 % efficient then the power output is about • P = 2/5  1m  (d/2m)2 (U/1ms-1)3 watts. • But you can’t pack them too close (> 5d apart) • [Power per mill]/[Land area per mill] = (/400)U3 • = 2.2 W per sq metre if U = 6 m/sec or 0.7 W/m2 if U = 4m/s • A little one on your roof will give 0.2 kWh per day

  30. Blowin’ In The Wind 1 person per 4000 sq m in the UK. Pack the whole country onshore with windmills 2 W per m2 4000 m2 per person = 8 kW 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 day

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