Power is defined as the rate of transfer of energy, or the rate at which work is done (which is the same thing). The unit of power is the watt (W), named after James Watt, a British engineer who applied scientific principles to the development of the steam engine.
A rate of transfer of energy of 1 joule per second is defined as one watt, 1 W = 1 J s–1.
One kilowatt = 1 kW = 1000 W.
We could therefore say that the average power used in Worked example 9.6 is 64 watts (64 W).
watt ❯
ITQ9
A weight-lifter can lift ‘weights’ of mass 300 kg through a vertical height of 1.8 m in 4 seconds. Calculate the average power he develops in lifting the weight. [Z :WLLKTZ¶ Figure 9.10
The answers to Worked example 9.7 suggest that, while the body is accelerating, the power it must develop increases to a large value, since not only must it do work to gain kinetic energy, but it must also overcome frictional resistance. Once the body reaches its cruising speed (top speed), it only has to overcome frictional resistances (though small) to maintain that speed.
A car just starting off in first gear needs a large pulling force to overcome its inertia (see chapter 6) and get it moving. The power here should be as large as possible so that the force at this time could be as large as possible for a low speed (car just starting to move). As the car accelerates and the speed rises, this pulling force gets smaller, if we assume that the power of the engine remains constant. The car is now moving faster in a higher gear and the pull of the engine is smaller than it was in first gear.
Efficiency
Efficiency is usually associated with devices that use energy in order to fulfil a purpose and tells us how well that energy is used to complete the job. There are many such devices in use. Some devices help us to transfer energy of one kind into energy of another in order to perform the task. Others may not convert the energy supplied into another form, but may make the task to be performed easier to perform. We call such devices ‘machines’. A simple example is a hammer. Since we cannot pull a nail out of a piece of wood with our bare hands, we use a hammer. How does the hammer work to do this?
We first apply a force to the hammer and as we pull backwards on the hammer handle, we do work (force moves through a distance) and transfer energy to the hammer. What the hammer is expected to do now is to transfer as much of this energy as possible to the nail (through its ‘claw’) in order to draw the nail out of the wood against the frictional resistance (called the ‘load’) of the wood and what fraction of this energy is transferred by the hammer depends on the efficiency of the hammer. The force that is applied is called the effort. The size of the effort needed, in its turn, depends on the efficiency of the hammer. So how is efficiency determined?
We would ideally like to find that all the energy we put into the hammer was used to get the job done, that is, to draw the nail. If we were so fortunate as to be able to achieve this, we would say that the process was 100% efficient,
ITQ10
I can run up a flight of stairs 3 m high (not long!) in 5 s. What is the power I generate if my mass is 65 kg?
CHAPTER 6
Note, interestingly, that the power used is proportional to the speed while the net force remains constant. It therefore increased as the speed increased.
Method 2
The average power = initial power + final power 2 = 0 + 600 W 2 = 300 W, as before.
Method 3
Rather than finding the work done by the net (accelerating) force in attaining the speed of 6 m s–1, we could find the increase in the kinetic
energy of the athlete (energy transformed) and divide that by the time taken. This gives
increase in KE = 12 × 50 × 62 – 1
2 × 50 × 02
= 900 J
So
average power developed = 900 J3 s
meaning that all the energy applied to the hammer was properly used to draw the nail and none of it was wasted. But we find in practice that we can never achieve such an ideal result, since there is no device in which some of the input energy is not lost in the course of using it through having to overcome friction (that cannot be avoided, since surfaces will move against others, involving frictional forces) or for some other undesirable, but necessary, reason.
The energy lost through friction (which we have discussed earlier) appears as heat and is deemed to be ‘wasted’ energy. This generation of heat as a device is used is very typical of mechanical machines which employ mechanical forces (see chapter 6). Electrical and electronic machines are not exempt from energy loss in the form of heat either since, apart from the fact that they may have moving parts (think of the fan, the hard-drive and the CD-drive of your computer which rotate about their axles) and this will result in frictional forces being present and therefore heat being generated. Currents flowing in conductors will invariably also produce heat.
In effect, then, how much of the energy input to a device is usefully converted by the device to perform the expected task (or ‘to do the job’), e.g. raise a load (as in a fork-lift), give a light (as in a torch), produce sound (as in a loudspeaker), light up your computer screen and provide information (as in your computer), etc., is expressed as a percentage, the formula used being:
efficiency of the device = total energy provided to the deviceenergy used to do the job × 100 (%) or, efficiency, ε = energy usefully converted
total energy provided × 100 (%)
The efficiency may also be given in terms of the power input to the machine and the power output from it. The input power is the rate at which energy is supplied to the machine to get the job done, and the output power is the rate at which the machine converts energy to do the job. It is clear that if the numerator and the denominator of the expression above are each divided by the time the machine is used, we will have
efficiency, ε = output powerinput power × 100 (%)
We can also use the law of energy conservation to describe efficiency. Since energy is conserved, then energy put into the machine = energy obtained from it to ‘do the job’ + energy used otherwise (or wasted energy)
or, using symbols, we can say that
Ein = Eout + Ewasted
where Ewasted is the total‘wasted’ energy. Dividing throughout by Ein, we have 1 = (Eout /Ein) + (Ewasted /Ein)
But Eout/Ein = the efficiency, ε, of the machine, so from the last equation, 1 = ε + fw ,
fw beingthe fraction of the input energy that is wasted. We have then ε = 1 – fw.
This shows then that the efficiency of a machine, ε, is not constant, but depends on the fraction of the input energy that is wasted. It tends to become smaller and smaller as the load (which is the magnitude of the job to be done) becomes greater.
We would all agree that you would not use a fork-lift truck to lift a carton of soft-drink. Why?
There is no device, whether natural or man- made, that is so perfect, as to have an efficiency
of 1. They all fall short of perfection in that only a fraction of the energy supplied to the device to perform the specific task is usefully used.
‘wasted’ energy ❯
CHAPTER 6
The Greek letter ‘ε’ is pronounced ‘ep-s(eye)-lon’
or ‘ep-si-lon’.
ITQ11
A small motor draws 120 J of electrical energy from the mains to lift a book of mass 1 kg through a vertical distance of 8.0 m. It takes 5 s to do this.
Calculate: (a) the power input to the motor; (b) the work done by the motor in lifting the book; (c) the fraction of the input power that is wasted.
Worked example 9.9
A certain make ‘X’ of low energy electric lamp is advertised as being six times as efficient as a certain brand ‘Y’ of incandescent bulb.
(i) What do you understand by the statement above?
The amount of light obtained from an incandescent lamp bulb for every watt of power it receives from the source is as low as 8%.
(ii) Express the statement above in another way. What is the form of energy of the remaining 92% of the energy converted in the incandescent bulb? Write an energy flow diagram for the energy conversion in the bulb. What happens to the energy produced by the bulb eventually? (iii) How many joules of light energy are obtained from the low energy
lamp when it takes 100 W of electrical energy from the mains?
Solution
(i) The statement means that: if both lamps are supplied with the same amount of electrical energy, the low energy lamp will convert six times as much of this energy to light as the incandescent lamp will.
(ii) Another way of saying this is to say that the efficiency of the incandescent lamp is 8%. The remaining 92% is converted to heat: electrical energy from the mains
p
(heat + light)The heat produced is transferred to the atmosphere, where it is lost irretrievably.
Carry out your own research to determine whether the following claim made by a certain manufacturer of low energy lamps is a reasonable one: ‘The low energy lamps we produce have an efficiency which is about 6 times that of most incandescent bulbs.’ (Remember that the purpose of the lamp is to
produce light and not heat!)
Worked example 9.8
A warehouse fork-lift takes 4 s to lift a crate of mass 1000 kg through 5 m working at a rate of 20 kW. Five per cent (5%) of the wasted energy in the forklift is used to lift moving parts of the fork-lift and the rest is lost as electrical heating and frictional heating. Calculate:
(i) the efficiency of the fork-lift at this load; (ii) the value of the total energy wasted;
(iii) the amount of heat developed in lifting the crate.
Solution
To say that the fork-lift is working at the rate of 20 kW is to say that the power drawn from the source (normally batteries) is 20 000 W.
The useful energy converted by the fork-lift is that energy which raises the crate and this energy = mgΔh = 1000 kg × 10 N kg–1 × 5 m = 50 000 J.
(i) Efficiency = useful energy converted to work in lifting the cratetotal electrical energy supplied to the fork-lift × 100%
= 20 000 W × 4 s50 000 J (since energy = power × time)
= 62.5%
(ii) If 62.5% of the energy supplied to the fork-lift is usefully used (to lift the load), then 37.5% is wasted.
Wasted energy = 37.5% of 80 000 J = 30 000 J
(iii) If 5% of this wasted energy is used to lift certain moving parts, then 95% of it is wasted as heat.
So, amount of energy lost as heat = 95% of 30 000 J
= (10095) × 30 000 J
= 28 500 J