DE4101 Work Power Energy

7/25/2019 DE4101 Work Power Energy

1/18

Jonathan Leaver PhD


2/18

TheoryWork When a force F moves an object of mass m through a distance s work is performed.

The amount of work is determined from the formula

Work (Joules) = Force (N) x distance (m) or W = F x s

1 Joule = 1 N-m

Power Power (Watts) = work (J)/time (s) 1

or P = W/t

= Fx

s/tP = F x v

Power = force x velocity 2


3/18

Theory (cont.)Energy A law of physics developed in the modern day by Albert Einstein

states that energy can neither be created nor destroyed it can onlybe transformed from one form to another.

Energy can take many forms:

Chemical, potential, kinetic, nuclear, heat, mechanical, electrical,

electromagnetic

Potential energy is energy possessed by an object due to its position

relative to a fixed datum or height.

Kinetic energy is energy possessed by an object due to its motion


4/18

Theory (cont.)Energy is transformed from one form to

another in hydro projects

The water behind the dam has

potential energy.

When the water travels into the

penstocks some of the potential

energy is converted into kinetic energy

When the water turns the turbine

blade some of the kinetic energy of

the water is converted to kineticenergy in the turbine.

Some of the turbine energy is lost by

heat emission from friction in the shaft

bearings.

Much of the rest of the

turbine energy is converted to

electrical energy.


5/18

Theory (cont.)Calculate Energy of Motion (Kinetic Energy)

Now F = m a ...... 3

Also = + 2

Hence a =

Consider an object initially at rest. Then u = 0 and a =

Substituting for a in equation 3 above gives:

F = m

But W = F x s

Therefore W = KE (Kinetic Energy) =1

mv2 .4


6/18

Theory (cont.)Calculate Potential Energy due to gravity

Now F = m a .. 3

Acceleration = gravity = g (9.81 m/s2)

Therefore F = mg

But W = F x s

where s is the height above ground level i.e. s = h.

Therefore W = PE (Potential Energy) = mgh 5


7/18

SpringsF = -kX where k = spring constant

Energy = 0 = 1 kX2

Source: http://www.thetrc.org/pda_content/texasphysics/e-BookData/Images/SB/58/LR/Spring%20Potential%20Energy.png


8/18

Example 1: EnergyA steel ball bearing of mass 50 grams is dropped from a height of 2 metres on

to a smooth, flat, rigid plate and rebounds to a height of 1.75 metres.

Calculate (i) Its original potential energy (ii) Its kinetic energy the instant

before it hits; (iii) Its kinetic energy the instant it starts to rebound; (iv) Itsvelocity on impact. How do you account for energy lost in the collision of ball

and ground?

1.75m 2m

mass=50 g

Solution

(i) The original potential energy of the ball, using the

plate as a datum isPE = mgh = 0.05 x 9.81 x 2 = 0.981 Joules (J)

(ii) KE at the instant it hits is the same as the PE at the

start as no energy has been lost only transformed.

KE = 0.981 J


9/18

Example 1: Energy (cont.)

1.75m 2m

mass=50 g

(iii) After impact the KE will equal the PE at the

maximum height it rises to.

PE = mgh = 0.05 x 9.81 x 1.75 = 0.858 J

(iv) The velocity on impact is determined from kinetic

energy.

KE = 0.5mv2 = 0.981 JTherefore v2 = (0.981 x 2)/0.05 = 39.24

and v = 6.26 m/s

(v) Energy is lost at the point of impact in heat and

noise by deformation of the ball.


10/18

Example 2: Work = 600

F = 400 N

Work is only done in the direction of

movement by the component of force in

that direction.

Consider an object moving 20 m under a

force of 400 kN acting at an angle of 60

degrees to the horizontal.

WD = force x distance = FcosF x s= 0.4(kN)cos60 x 20m

= 4.00 kJ


11/18

Example 3: Power

An electrically driven conveyor belt carries a 60,000 packages per hour adistance of 80 m up an incline of 1 in 12. Each package weighs 48 N and the

power absorbed by friction in the drive is 2 kW. What is the power output of the

motor?

sin = 1/12


12/18

Example 3: Power (cont.)An electrically driven conveyor belt carries a 60,000 packages perhour a distance of 80 m up an incline of 1 in 12. Each package weighs

48 N and the power absorbed by friction in the drive is 2 kW. What is

the power output of the motor?

sin = 1/12(i) To determine the power output of the

motor determine firstly how much work is

done on each package and then how long it

takes to do this work.

(ii) Then we can use the formula

Power = Work/time

(iii) Work = force x distance where the force

must act in the direction of movement.

The force in this case is gravity.

The component of weight of one

package in the direction of the

conveyor = mgsin

mg

mgsin


13/18

Example 3: Power (cont.)An electrically driven conveyor belt carries a 60,000packages per hour a distance of 80 m up an incline of 1 in

12. Each package weighs 48 N and the power absorbed

by friction in the drive is 2 kW. What is the power output

of the motor?

sin = 1/12

The time each package takes on the belt is

t = 3600/60,000 = 0.06 s

The power ignoring friction

P = WD/t = Fsin *s/t

= 48(1/12) x 80/0.06= 5.333 kW

However we are given 2 kW is lost in friction

therefore total power required is:

P = 5.333 + 2

= 7.333 kW


14/18

Example 4: Power (Dodd & Richardson Ch7 No. 6 pg 99)

A diesel-electric locomotive of mass 300 tonnes and tractive resistance 120 N/ttravels up a 1:100 slope at 60 km/h. The train pulls carriages with a mass of 700

tonnes and tractive resistance 50 N/t. Find the power developed at the driving

wheels. Ans. 2.818 MW

120N/t

300 t

50 N/t

700 t

Slope1 in 100


15/18

Solution: Power (Dodd & Richardson Ch7 No. 6 pg 99)

mg

mgsin

Power = force (up slope) * velocity (up slope)

We must find the force exerted by the wheels to

overcome gravity and tractive resistance.

Summing forces up the slope

Fengine(300+700)gsin (120*300+700*50)/1000 = 0

Fengine = 169.1 kN

Power = F * v = 169.1 kN * 60/3.6 m/s = 2.818 MW

120N/t

300 t

50 N/t

700 t

Slope1 in 100

Fengine


16/18

Problem 5 (Ivanoff 20.1)A 180 tonne train climbs an incline of 1.5o for 2 km. Its

initial velocity before the climb is 90 km/h. The tractive

effort exerted by the engine is 53.2 kN and tractive

resistance is 95 N/t. Determine the final speed after the

climb.Ans: 72 km/h


17/18

Problem 6: (Ivanoff 20.13)A stationary 18 kg block sides down a 30 degree slope a

distance of 2.5 m where it compresses a spring with

stiffness 2.7 N/mm also aligned at 30 degrees. If the

coefficient of friction between the block and the surface is

0.2 find how much the spring will be compressed.

Ans: 349 mm Hint: Use energy and work done


18/18

Self StudyCh 7 (Dodd and Richardson)

(p 99&100) 3,4,5,9,10,15,16,17

DE4101 Work Power Energy

Documents

Transcript of DE4101 Work Power Energy