Trains and Boats and Planes and Cars Too

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 Power usage units: kilowatt hours per day person Average Briton consumes 125 k. Wh/day person

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 100 km = 6 per sq mm is 41 miles per gallon 1/16. 7 per sq mm What is this area ? Area 100 km Imagine fuel spurting from a tube of this area at the speed of your car 1 litre per 100 km = 1/100 sq mm

Cars Kinetic Energy of Motion ½ mv 2 = ½(m/1000 kg) (v/100 km per hr)2 = 4 105 J = 0. 1 KWhr

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

Braking Effects d Speed V d 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 m. V 2 ÷ (d/V) = ½ m. V 3/d Note the V 3 factor

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 ma. V 2 = ½ A V 3 t Rate of KE loss to air swirl = ½ ma. V 2 /t = ½ A V 3

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

Critical distance between stops is d = D m(car)/A A(car) = 1. 5 m x 2 m = 3 m 2 Drag factor: c = 1/3 Effective A = c A(car) =1 m 2 Mass: m = 1000 kg Air: = 1. 3 kg/m 3 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

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

Different Car Parameters Drag factor. We used c = 0. 33 Ø c(Honda) = 0. 25 Ø c(Sierra) = 0. 34 Ø c(Citroen 2 CV) = 0. 51 Ø c(bike) = 0. 9 Ø c(coach) = 0. 42 Effective areas Ø A(Discovery) = 1. 6 Ø A(typical car) = 0. 8 Ø A(Honda Insight) = 0. 47 Ø

Bikes and Cars Compared 3 Energy/dist = 4 ½ AV ÷ V = 2 AV 2 (E/d)bike/ (E/d)car = [c(bk)/c(car)] [(A(bk)/A(car)] [V 2(bk)/V 2(car)] = 1/0. 33 ¼ (1/5)2 0. 03 (V(bk)= 13 mph) Bike is about 33 times more fuel efficient than the motorway car

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. Mac. Kay

Braking Distances V 2 = U 2 – 2 as If you stop (V = 0) after travelling at speed U with deceleration a then your braking distance is s s = U 2/2 a (speed)2 Eg two cars: 65 and 60 km/hr drivers have 1. 5 sec reaction time They travel 27. 1 m and 25 m during the reaction time, then s = 16. 3 m and s = 13. 9 Total distances travelled are 43. 4 m 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 = (U 2 – 2 ad)1/2 = 30 km/hr = 8. 2 m/s as d = 40 – 27. 1 m Body thickness 20 cm. Impact lasts 0. 025 s. Deceleration is 320 ms-2. If child weighs 59 kg, impact force ( U 2) is 16, 000 N (decel more than 26 g).

Energy Lost in Rolling Wheels Ø Ø Ø 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 = 100 N 31 m/s = 3. 1 k. W for 1 ton truck at 70 mph Engine 25% efficient so 12 k. W to engine, but 80 k. W needed to beat air swirl on motorway and 15% to overcome rolling resistance.

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

Trains 8 -carriage train: m= 400, 000 kg A = 11 m 2 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 k. Wh 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 k. Wh per 100 passenger-km Intercity 125 500 passengers, max V = 125 mph, Power = 2. 6 MW Energy use is 9 k. Wh per 100 passenger-km

Rowing in Numbers How does the speed of the boat depend on the number of rowers? Drag on boat V 2 wetted surface area of boat V 2 L 2 Volume of boat L 3 N – the number of crew Drag V 2 N 2/3 Crew Power overcoming drag = N P = V Drag V 3 N 2/3 P is the constant power exerted by each (identical rower) V N 1/9 With a cox V N 1/3/[N + 1/3 ]2/9 if cox is third the weight of a rower

Results of 1980 Olympic coxless races (N = 1, 2, 4) + coxed N = 8 Distance is 2000 metres N 1 2 4 8 Time, T (sec) 429. 6 408. 0 368. 2 349. 1 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) T = (2000) / V N-0. 11 V N 0. 11 N 1/9 Better without one!

Aeroplanes Push air down and Newton’s 3 rd 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

Optimal Energetics Rate of energy use to push down at speed U = KE of air tube ÷ t P(lift) = ½ (mg)2/ ( VA) Total Power = P(lift) + P(drag) = ½ (mg)2/ ( VA) + ½ c V 3 Ap assume subsonic < 330 m/s Ap : front area of plane X-section Fuel efficiency: P(total) / V = thrust Jet engine efficiency about 30%, m = 363, 000 kg Ap = 180 m 2, 64. 5 m wingspan A, c = 0. 03 Optimal when P(lift) = P(drag) V 2(opt) = mg/[ (c. Ap. A)1/2] (540 mph)2 for 747 at 30, 000 ft where (air) = 1/3 (surface) =0. 13 kg/m 3 540 mph 220 m/s

Flight 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

Transport Efficiency = No. of passenger-km per litre of fuel For 747 = No. 1/3 1. 38 MJ per litre/200, 000 N 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

Energy consumed in KWh per ton-kilometre of freight moved D. Mac. Kay

Windmills

Or even…

Windmill Efficiency Wind power thro’ area A = ½ AV 3 V Outgoing wind speed U Incoming V<U 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 FU 2 - ½ FV 2 = ¼ DA(U 2 – V 2 )(U + V) If the windmill was not there the power in the wind is P 0 = ½ DAU 3 P/P 0 = ½ {1 – (V/U)2} {1+ (V/U)} P/P 0 is a maximum when V/U =1/3 Max P/P 0 = 16/27 = 59. 26% Albert Betz’s (1919) Law

Actual Energy Output Ø Ø Ø Ø V/U = 1/3, so Pmax = (8/27) A U 3 Only about 20% efficient (new offshore rotors up to 50%) If the diameter of the rotor is d then the area is A= d 2/4 and the windmill is 20 % efficient then the power output is about P = 2/5 1 m (d/2 m)2 (U/1 ms-1)3 watts. But you can’t pack them too close (> 5 d apart) [Power per mill]/[Land area per mill] = ( /400) U 3 = 2. 2 W per sq metre if U = 6 m/sec or 0. 7 W/m 2 if U = 4 m/s A little on your roof will give 0. 2 k. Wh per day

Blowin’ In The Wind 1 person per 4000 sq m in the UK. Pack the whole country onshore with windmills 2 W per m 2 4000 m 2 person = 8 k. W person If we covered 10% of country we get 20 k. Wh/day person This gives only 50% of power to drive average petrol car 50 km per day