WORK POWER AND ENERGY____________________________________________
3.1. Effect of Electric Current.
It is a matter of common experience that a conductor, when carrying current, becomes hot after some time. As explained earlier, an electric current is just a directed flow or drift of electrons through a substance. The moving electrons as the pass through molecules of atoms of that sub- stance, collide with other electrons. This electronic collision results in the production of heat. This explains why passage of current is always accompanied by generation of heat.
3.2 Joules Law of Electric Heating The amount of work required to maintain a current of I amperes through a resistance of R ohm for t second is W.D. = PRt joules
= VIt joules (": R = VI!)
= Wt joules (": W = VI)
= V t lR joules (": I = VIR)
This work is converted into heat and is dissipated away. The amount of heat produced is H = work done = W.D.
mechanical equivalent of heat J
J = 4,186 joules/kcal = 4,200joulesI kcal(approx) 2 H = I RtI4,200 kcal = VltI4,200kcal
= WtI4,200 kcal = VtI4,200 R kcal where
..
3.3 Thermal Efficiency.
It is defined as the ratio of the heat actualy utilised to the total heat J> reduced electrically. Con- sider the case of the electric kettle used for boiling water. Out of the total heat produced (i) some
goes to heat the apparatus itself i.e. kettle (ii) some is lost by radiation and convection ect. and (iii) the rest is utilised for heating the water. Out of these, the heat utilised for useful purpose is that in (iii). Hence, thermal efficiency of this electric apparatus is the ratio of the heat utilised for heating the Watertown the total heat produced. Hence. the relation between heat produced electrically and heat absorbed usefully becomes
-xll = ms(82-81) Example 3.1. The heater element of an electric kettle has a constant resistance of 100 Q and the applied voltage is 250 V. Calculate the time taken to raise the temperature of one litre of water
from J5°C to 90°C assuming that 85% of the power input to the kettle is usefully employed. If the water equivalent of the kettle is 100 g. find how long will it take to raise a second litre of water through the same temperature range immediately after the first. (Electrical Engineering, Calcutta Univ. 1980) Solution. Mass of water = lO g = I kg (": I cm 3 weight I 2. Force. Unit of force is newton (N). Law of Motion i.e. F =ma.2 If m = 1 kg; a = Im 1s , then F = 1 newton.
Hence, one newton is that force which can give an acceleration of I ml s2 to a mass of I kg. Gravitational unit of force is kilogram- weight (kg- wt). It may be defined as follows:
or It is the force which can impart an acceleration of 9.8 ml s2 to a mass of 1 kg.
It is the force which can impart an acceleration of 1 ml s2 to a mass of 9.8 kg.
Obviously, 1 kg- wt. = 9.8 N
3. Weight. It is the force with which earth pulls a body downwards. Obviously, its units are the same as for force.
(a) Unit of weight is newton (N)
(b) Gravitational unit of weight is kg- wt.*
Note. If a body has a mass of m kg, then its weight, W =mg Newton's =9.8 Newton's.
4. Work, If a force of F moves a body through a distance S in its direction of application, then Work done W = Fx
(a) Unit of work is joule (1).
If.. in the above equation, F =IN: S =1 m ; then work done =1 m.N or joule.
Hence, one joule is the work done when a force of 1 N moves a body through a distance of 1 m in the direction of its application.
(b) Gravitational unit of work is m- kg. wt or m- kg**.
If F = I kg- wt; S = 1 m; then W.D. = 1 m- kg. Wt =1 m- kg.
Hence, one m- kg is the work done by a force of one kg- wt when applied over a distance of one metre.
Obviously, 1m-kg =9.8 m- Nor J.
5. Power. It is the rate of doing work. Its units is watt (W) which represents I joule per second. I W = 1 Jls .
If a force of F newton moves a body with a velocity of v m.ls then
power = F x V watt
If the velocity v is in km ls, then
power = F x v kilowatt
If the velocity v is in then
power = F x v kilowatt
6. Kilowatt-hour (kWh) and kilocalorie(kcal) .
1 kWh = 1000 x 11. x 3600 s =36 X 105 J . s
I =4,186 J :. 1 kWh =36 x 105/4,186=860 7. Miscellaneous Units
. J (l) I watt hour (Wh) =1- x 3600s =3600 J s
(if) 1 horse power (metric) =75 m- kg/s =75 x 9.8=735.5 Jl or watt (ii) 1kilowatt(kW)=1000W and 1megawatt(MW)=106W 3.5. Calculation of Kilo- watt Power of a Hydroelectric Station Let Q =water discharge rate in cubic metres/second (m3/s), H = net water headin metre (m). g =9.81 11; overall efficiency of the hydroelectric station expressed as a fraction. Since I m 3of water weighs 1000 kg., discharge rate is 1000 Q kg/so
When this amount of water falls through a height of H metre, then energy or work available per second or available power is
Electrical Technology Its definition may be obtained from Newton's Second = 1000 QgH JLS or W =Q gH kW
* Often it is referred to as a force of I kg, the word 'wl' being omitted. To avoid confusion with mass of 1 kg, the force of 1 kg is written in engineering literature as kg f instead of kg. wt. ** Generally the work 'wl' is omitted and the unit is simply written as m- kg.
Thank you.
Written by. Abhishek Singh E.E
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