1. Definition Units Problem solving
Power
Definition
Units
Problem solving

2. Definition
So it follows that 1 Watt = 1 Joule per second

3. Useful point for engineers
1 Horsepower ~ 750 Watts
Development of the steam engine meant that it was necessary to compare them with the animals they replaced.
James Watt estimated lb-force-feet/min which is the above value when converted to SI units.

4. Example
A steam engine is to lift a bucket of coal of mass 100 kg a vertical distance 100 m up a mine shaft. If the engine’s power is 1500 W, calculate the time taken to raise the coal.

5. Solution
A steam engine is to lift a bucket of coal of mass 100 kg a vertical distance 100 m up a mine shaft. If the engine’s power is 1500 W, calculate the time taken to raise the coal. P = w/t = mgh/t t = mgh/P = 100 x 100 x 9.8/1500 = 65.3 s

6. Motive Power
Consider a vehicle driving at constant speed along a horizontal road. The motive force (provided by the engine and ultimately the tyres) is equal and opposite to the drag forces.

7. Motive Power
The power required to maintain a vehicle’s speed is equal to the total resisting force multiplied by the speed.

8. Example
A car has a power of 200 horsepower and a top speed of 148 mph. Calculate the total resisting force when the car is travelling at top speed. 1 hp = 750 W 1 mile = 1609 m

9. Solution
P=Fv F = P/v But... P and v not in SI units. 200 hp = 200 x 750 W 148 mph = 148 x 1609/3600 s So F = 200 x 750/(148 x 1609/3600) = 2270 N

10. Slopes
An escalator lifts an average mass of 500 kg per minute through a height of 20 metres. In the absence of friction, calculate the power of the escalator.

11. Solution
An escalator lifts an average mass of 500 kg per minute through a height of 20 metres. The escalator is 70 m long (along its hypotenuse) In the absence of friction, calculate the power of the escalator. P=w/t =mgh/t = 500x9.8x20/60 = 1630 W

12. Continued
Imagine now that the escalator works against a frictional force of 600N. What is the total power now? 20 m 70m P(friction) = Fd/t = 600x70/60 = 700 W P(total) = = 2330 W

13. Example
A truck of mass 5000 kg moves up a slope of gradient 1 in 20. The truck works against a resisting force of 2500N and travels with steady speed of 15ms-1. What is the engine power?

14. Solution There are two components to the energy transfer here:
Power lifting against gravity (mgh/t)
2. Power overcoming friction (Fv)

15. Solution
P(against gravity) = mgh/t = 5000 x 9.8 x 15sin x =5000 x 9.8 x 15 x(1/20) = W P(against friction) = Fv = 2500 x 15 = W P(total) = = W = 99 hp.

16. Cars - example
A car has a resisting force acting on it modelled as F = v2 where F is the force in N and v is the speed in ms-1. If car’s peak power is 150 kW, show that the top speed is between 63 and 64 ms-1.

17. Solution
P = Fv F = v2 At 63 ms-1 P =( v2)v = 350v + 0.5v3 = 350 x x633 = W At 64 ms-1 P = W

18. Example
A car of mass 1500 kg moving along level ground at 20 ms-1 requires a power of 5kW. Calculate the power required when moving at the same speed up a hill of gradient 1/15.

19. Solution
The power to move at the same speed up the hill is going to be 5kW plus the power to raise the car against gravity – mgh/t mgh/t = mgvsinθ =1500 x 9.8 x 20 x 1/15 = 19.6kW. Total power = = 24.6kW Conclusion: hills severely hinder a car’s fuel economy.

20. Example- quite challenging
A car of mass 1200kg has a resisting force F acting on it equal to v2 where F is in Newtons and v is in ms-1.
Calculate the resisting force at 20ms-1.
The car has a top speed of 50ms-1 (112 mph.)
b)Calculate the maximum motive power.
c) Show that the top speed when the car is moving up a hill of gradient 1/8 is reduced to about 35ms-1.

21. Solution
F = v2 = 0.55x20x20 = 620 N
P = Fv = ( v2)v = kW (about 66hp)

22. Solution
c) Power = Gain in GPE/time + Power overcoming drag P = mgh/t + Fv P = mgvsinθ + ( v2)v P= mgvsinθ +(400v +0.55v3) = 1470v +400v v =1870v +0.55v3  v=34.9 ms-1 (78.5mph) In other words, the top speed is reduced by 30 %.