Qu´ımica del cemento

through a height of 12 m when 7.85 kJ of energy is supplied to it. Determine the efficiency of the machine. [75%] 2. Determine the output energy of an electric motor which is 60% efficient if it uses 2 kJ of electrical energy. [1.2 kJ] 3. A machine that is used for lifting a partic- ular mass is supplied with 5 kJ of energy. If the machine has an efficiency of 65% and exerts a force of 812.5 N to what height will it lift the mass? [4 m] 4. A load is hoisted 42 m and requires a force of 100 N. The efficiency of the hoist gear is 60% and that of the motor is 70%. Determine the input energy to the hoist.

[10 kJ]

14.3

Power

Power is a measure of the rate at which work is done or at which energy is converted from one form

to another.

PowerP = energy used time taken

or P = work done

time taken

The unit of power is the watt, W, where 1 watt is equal to 1 joule per second. The watt is a small unit for many purposes and a larger unit called the kilowatt, kW, is used, where 1 kW=1000 W. The power output of a motor, which does 120 kJ of work in 30 s, is thus given by

P = 120 kJ

30 s =4 kW Since work done=force×distance,

then power= work done

time taken = force×distance time taken =force× distance time taken However, distance

time taken =velocity

Hence power=force×velocity

Problem 13. The output power of a motor is 8 kW. How much work does it do in 30 s?

Power= work done time taken, from which, work done=power×time

=8000 W×30 s =2 40 000 J=240 kJ Problem 14. Calculate the power required to lift a mass through a height of 10 m in 20 s if the force required is 3924 N.

Work done=force×distance moved =3924 N×10 m=39 240 J

Power= work done time taken =

39 240 J 20 s =1962 W or 1.962 kW

Problem 15. 10 kJ of work is done by a force in moving a body uniformly through 125 m in 50 s. Determine (a) the value of the force, and (b) the power.

(a) Work done=force×distance,

hence 10 000 J=force×125 m, from which, force= 10 000 J

125 m =80 N (b) Power= work done

time taken =

10 000 J

50 s =200 W Problem 16. A car hauls a trailer at

90 km/h when exerting a steady pull of 600 N. Calculate (a) the work done in 30 minutes and (b) the power required.

(a) Work done = force× distance moved. The distance moved in 30 min, i.e. 1₂h, at 90 km/h=45 km.

Hence, work done = 600 N×45 000 m = 27 000 kJor27 MJ (b) Power required = work done time taken = 27×106 J 30×60 s =15 000 W or 15 kW

Problem 17. To what height will a mass of weight 981 N be raised in 40 s by a machine using a power of 2 kW?

Work done=force×distance. Hence, work done=981 N×height.

Power= work done time taken, from which, work done=power×time taken

=2000 W×40 s =80 000 J

Hence 80 000=981 N×height,

from which, height= 80 000 J

981 N =81.55 m Problem 18. A planing machine has a cutting stroke of 2 m and the stroke takes 4 seconds. If the constant resistance to the cutting tool is 900 N, calculate for each cutting stroke (a) the power consumed at the tool point, and (b) the power input to the system if the efficiency of the system is 75%. (a) Work done in each cutting

stroke=force×distance =900 N×2 m=1800 J Power consumed at tool point

= work done time taken =

1800 J

4 s =450 W (b) Efficiency= output energy

input energy = output power input power Hence 75 100 = 450 input power from which, input power=450× 100

75 =600 W Problem 19. An electric motor provides power to a winding machine. The input power to the motor is 2.5 kW and the overall efficiency is 60%. Calculate (a) the output power of the machine, (b) the rate at which it can raise a 300 kg load vertically upwards. (a) Efficiency, η= power output power input i.e. 60 100 = power output 2500 from which, power output= 60 100×2500 =1500 W or 1.5 kW.

(b) Power output =force×velocity, from which, velocity= power output

force .

Force acting on the 300 kg load due to gravity=300 kg×9.81 m/s2 =2943 N Hence, velocity= 1500 2943 =0.510 m/s or 510 mm/s.

Problem 20. A lorry is travelling at a constant velocity of 72 km/h. The force resisting motion is 800 N. Calculate the tractive power necessary to keep the lorry moving at this speed.

Power=force×velocity.

The force necessary to keep the lorry moving at constant speed is equal and opposite to the force resisting motion, i.e. 800 N.

Velocity=72 km/h= 72×1000 60×60 m/s =20 m/s. Hence, power=800 N×20 m/s =16 000 N m/s=16 000 J/s =16 000 W or 16 kW.

Thus the tractive power needed to keep the lorry moving at a constant speed of 72 km/h is 16 kW.

Problem 21. The variation of tractive force with distance for a vehicle which is

accelerating from rest is:

force (kN) 8.0 7.4 5.8 4.5 3.7 3.0 distance (m) 0 10 20 30 40 50 Determine the average power necessary if the time taken to travel the 50 m from rest is 25 s. 8.0 6.0 4.0 F orce (kN) 2.0 0 10 20 Distance (m) 30 40 50 y1 y2 y3 y4 y5 Figure 14.9

The force/distance diagram is shown in Figure 14.9. The work done is determined from the area under the curve. Using the mid-ordinate rule with five intervals gives:

area=

width of

interval mid-ordinatesum of =(10)[y₁+y₂+y₃+y₄+y₅] =(10)[7.8+6.6+5.1

+4.0+3.3] =(10)[26.8]=268 kN m, i.e. work done=268 kJ

Average power= work done time taken =

268000 J 25 s =10720 Wor10.72 kW.

Now try the following exercise

Exercise 70 Further problems on power