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Power

 

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While defining work, time element is not involved in it. Sometimes it is necessary to consider the rate at which work is done as well as the total work done.

The rate at which work is done is called Power.

If work is done at a uniform rate, average power P is defined as,

P = W/t

If the work is done at a variable rate, instantaneous power P is defined as,

P = dW/dt

(a) Instantaneous Power in terms of Velocity

Let a force \vec{F}, acting on a body, displace it through a displacement  \vec{s}?. Work done, W, is given by:

W = \vec{F}.\vec{s}

Thus, Instantaneous power, 

P = \frac{dW}{dt} = \frac{d}{dt}(\vec{F}.\vec{s})

For a constant force,

P = \vec{F}.\frac{d}{dt}(\vec{s}) = \vec{F}.\vec{v}

Here, ‘v’ is the velocity of the body.

Therefore, Instantaneous power is defined as the product of force and displacement.

(b) Instantaneous Power in terms of Mechanical Energy

Mechanical energy ‘E’ of body is the sum total of its kinetic energy (K) and potential energy (U).

Or,

Therefore, instantaneous power, P = dE/dt

Thus,  instantaneous power is defined as the rate at which mechanical energy is transformed.

Dimensions of Power

Power, P = Work / Time

                = [M1L2T-2] / [T1]

               = [M1L2T-3]

So, Dimensions of power are 1,2,-3 in mass length and time, respectively.

Units of Power

(a) Absolute Units:-

(i) C.G.S. System:-

Absolute units of power, in C.G.S. system, is erg s-1.

Erg/second:- Power of an agent is said to be 1 erg s-1 if it does one erg of work in one second.

(ii) S.I:-

Absolute unit of power, in S.I., is joule per second or watt (W).

Watt (Js-1):- Power of an agent is said to be one watt if it does 1 joule of work in one second.

1 W = 1 Js-1.

A bigger unit is also, often used.

1 kilowatt (kW) =1000 W

                          = 1000 Js-1

1 Megawatt = 106 Watt

Another unit of power, horse power (HP), is often used in our daily life.

1 HP = 746 W

1 kW = 1.34 HP

1 MW = 1340 HP

(b) Gravitational Units:-

(i) C.G.S. System:-

Gravitational units of power, in C.G.S. system, is gram centimeter per second (g cm s-1).

Gram centimeter/second (g cm s-1):- Power of an agent is said to be one garm centimeter per second if it does work at the rate of 980 erg per second.

(ii) S.I.:-

Gravitational unit of power, in S.I. is kilogram meter per second.

Kilogram meter per second (kg m s-1):-Power of an agent is said to be one kilogram meter per second if it does work at the rate of 9.8 joule per second.

Solved Problems

Problem 1:-

A 57-kg woman runs up a flight of stairs having a rise of 4.5 m in 3.5 s. What average power must she supply?

Concept:-

Work done W is defined as,

W = Fs

Here F is the force and s is the displacement.

Power P is defined as,

P = W/t

Here W is the work done and t is the time.

Solution:-

As the woman runs up a flight of stairs, there are two forces will act on the woman. One is the gravitational force which is in the downward direction and other one is the normal force N of the floor which balances the weight of the body (gravitational force) and the direction is just opposite to the direction of gravitational force. 

So, F = N

          = weight of the body

          = mg         (Since, weight of the body = mg)

Here m is the mass of the body and g is the acceleration due to gravity.

To obtain the force F exerted by the woman to propel herself up the stairs, substitute 57 kg for m and 9.8 m/s2 for g in the equation F = mg,

 F = mg

    = (57 kg) (9.8 m/s2)

   = (560 kg.m/s2) (1 N/1 kg.m/s2)

  = 560 N

In accordance to the reference of woman the stairs are moving down. Thus the woman will exert a force in the down ward direction.

To obtain the work done W by the woman, substitute 560 N for F and 4.5 m for s in the equation W = Fs,

W = Fs

    = (560 N) (4.5 m)

    = (2500 N.m) (1 J/1 N.m)

   = 2500 J

As the direction of force is in the direction of displacement, thus the work done is positive here.

To obtain the average power P supplied by the woman, substitute 2500 J for W and 3.5 s for t in the equation P = W/t,

P = W/t

   = 2500 J/3.5 s

   = (710 J/s) (1 W/ 1 J/s)

   = 710 W

From the above observation we conclude that, the average power P supplied by the woman would be 710 W.

Problem 2:-

The motor on a water pump is rated at 6.6 hp. From how far down a well can water be pumped up at the rate of 220 gal/min? 

Concept:-

Mass flow rate (R) of water is equal to the volume flow rate (Q) times density of the water (d).

So, R = (Q) (d)   

The power (P) of the motor on the water pump is defined as,

P = (h) (g) (R)  

Where h is the height, g is the acceleration due to gravity and R is the mass flow rate.

Solution:-

First we convert the units from non- SI system to SI system.

Power of motor, P = 6.6 hp

                             = (6.6 hp) (745.6 W/1 hp)

                            = 4920.96 W

Rounding off to three significant figure, the power will be 4920 W.

Volume Flow rate, Q = 220 gal/min

                     = (220 gal/min) (3.785 L/1 gal) (1 min/60 s)

                     = 13.9 L/s     

Mass flow rate (R) of water is equal to the volume flow rate (Q) times density of the water (d).

So, R = (Q) (d)     …… (1)

To find mass flow rate (R), substitute 13.9 L/s for the volume flow rate Q and 1.00 kg/L for the density of the water (d) in the equation R = (Q) (d),

R = (Q) (d)

   = (13.9 L/s) (1.00 kg/L)

  = 13.9 kg/s      …… (2)

The power (P) of the motor on the water pump is defined as,

P = (h) (g) (R)      …… (3)

Where h is the height, g is the acceleration due to gravity and R is the mass flow rate.

From equation (3), the height would be,

h = P/gR

To find the height (h), substitute 4920 W for power (P), 9.81 m/s2 for g and 13.9 kg/s for R in the equation h = P/gR,

h = P/gR

  = (4920 W)/( 9.81 m/s2) (13.9 kg/s)

 = 36.1 W/ (m/s2. kg/s)

 = (36.1 W/ (m/s2. kg/s)) ((kg.m2/s3) / 1W)

 = (36.1 m) (1 ft/0.3048 m)      

= 118.11 ft    …… (4)

Rounding off to two significant figures, the height would be 120 ft.

From the above observation we conclude that, the water can be pumped from a well at a height 118.11 ft.

Problem 3:-

What power is developed by a grinding machine whose wheel has a radius of 20.7 cm and runs at 2.53 rev/s when the tool to be sharpened is held against the wheel with a force of 180 N? The coefficient of friction between the tool and the wheel is 0.32.

Concept:-

Power delivered (P) to the body is equal to the force acting on the body (F) times speed of the body (v).

So, P = Fv     …… (1)

Frictional force (F) acting on the body is equal to the coefficient of friction (µk) times the normal force (N).

F = µkN     …… (2)

Solution:-

A wheel has a radius of 20.7 cm and run at 2.53 rev/s when the tool to be sharpened is held against the wheel with a force of 180 N.

The linear velocity (v) of the wheel is equal to the angular velocity of the wheel (w) times radius of the wheel (r).

v= wr     …… (3)

But angular velocity of the wheel is equal to the 2π times frequency (f) of the wheel.

w = 2πf       …… (4)

To obtain linear velocity (v) in terms of frequency f, substitute 2πf for w in the equation v= wr,

v= wr    

 =2πfr   …… (5)

To obtain linear velocity (v) of wheel, put 20.7 cm for radius of the wheel r, 2.53 rev/s for w, and 3.14 for π in the equation v =2πfr,

v =2πfr

  =2(3.14)( 2.53 rev/s)(20.7 cm)

  =2(3.14)( 2.53 rev/s)(20.7 cm) (10-2 m/1 cm)

 =3.29 m/s      …… (6)

To obtain the frictional force (F), substitute 0.32 for the coefficient of friction µk and 180 N for normal force (N) in the equation F = µkN,

F = µkN

   = 0.32 (180 N)

  = 57.6 N       …… (7)

To obtain the power (P) developed by the grinding machine, substitute 57.6 N for F and 3.29 m/s for linear velocity v in the equation P = Fv,

P = Fv

   = (57.6 N) (3.29 m/s)

   = 189.504 N. m/s

  = (189.504 N. m/s) (1 W/ N. m/s)

  = 189.504 W       …… (8)

Rounding off to two significant figure, power (P) developed by the grinding machine would be 190 W. 

Problem 4:-

A jet airplane is traveling 184 m/s. In each second the engine takes in 68.2 m3 of air having a mass of 70.2 kg. The air is used to burn 2.92 kg of fuel each second. The  energy is used to compress the products of combustion and to eject them at the rear of the engine at 497 m/s relative to the plane. Find (a) the thrust of the jet engine and (b) the delivered power (horse power).

Concept:-

Force or thrust F is equal to the rate of change of momentum (Δp/ Δt).

F = Δp/ Δt

Power P is defined as,

P = Fv

Here F is the force and v is the velocity.

Solution:-

(a)

First we have to find out the change in momentum Δp of the jet engine.

The change in momentum Δp will be,

Δp = (70.2 kg) (497 m/s-184 m/s) +(2.92 kg) (497 m/s

     = 2.34×104 kg.m/s

To obtain the thrust F of the jet engine, substitute 2.34×104 kg.m/s for Δp and 1 s for Δt in the equation F = Δp/ Δt,

F = Δp/ Δt

   = (2.34×104 kg.m/s)/(1 s)

  = (2.34×104 kg.m/s2) (1 N/1 kg.m/s2)

 =2.34×104 N

From the above observation we conclude that, the thrust F of the jet engine would be 2.34×104 N.

(b)

To obtain the delivered power P of the jet engine, substitute 2.34×104 N for thrust F and 184 m/s for the velocity v of the jet airplane in the equation P = Fv, we get,

P = Fv

   = (2.34×104 N) (184 m/s)

   = (4.31×106 N.m/s) (1 W/1 N.m/s)

   = 4.31×106 W

   = (4.31×106 W) (0.001341 hp/1 W)

  = 5780 hp

From the above observation we conclude that, the delivered power P of the jet engine would be 5780 hp.

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