Transcript Energy - Moodle

Work & Energy
If an object moves due to the action of an applied force the force is
said to have done WORK on the object.
Work is the product of force applied and the distance moved in the
direction of the force.
WORK DONE = FORCE x DISTANCE
The unit of work can be found from
Force = Newtons
Distance = metres
Work is therefore measured in Newtons x metres or Newton . metres
However:- Work is usually defined in units of Joules
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(J)
Work & Energy
The Joule is defined as:The amount of work done when a force of 1 Newton acts for a distance
of 1 metre.
1 Joule = 1 Newton metre
1 J = 1 N.m
If a graph is plotted of Force ( vertical y axis) against :
distance ( horizontal x axis) the force/distance graph is given.
This graph is known as a WORK DIAGRAM
The area beneath the graph equals the work done
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Work & Energy
The work diagram
a.
Constant force
Work done = area of graph
WD
Force
= F.s (Nm or J)
F
Distance (s)
b.
Changing force
Work done = area of graph
WD
Force
F
Distance (s)
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= ½ F.s (Nm or J)
Work & Energy
Ex1
A spring is stretched 25mm by a force of 50N. Draw a
diagram of the work done over the next 50mm of stretch and from this
find the work done in that 50mm and the average force applied
Solution:Spring stiffness = 50N/25mm
150
= 2N/mm = 2kN/m
The next 50mm needs 50x 2N/mm
Force
= 100 N
N
Work done = area of graph
50
WD = ½ (50 + 150) x 50 x10-3
= 5 Nm or 5 J
25
75
Distance (mm)
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Average force = (50 + 150)/2
= 100N
Work & Energy
Examples:
1.
Calculate the work done in lifting a mass of 0.5 tonne through a
vertical distance of 10m
Soln.
Work = force x distance
In this case the work is done against gravity – thus force = weight (mg).
When:- mass = 0.5 tonne = 500 kg, g = 9.81, distance = height 10m
Work = mass x g x height = 500 kg x 9.81 x 10 = 49050 J = 49.05 kJ
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Work & Energy
Examples:
2.
A sledge is pulled along horizontal ground at a constant speed
by a rope which makes an angle of 25° with the horizontal. If the tension
in the rope remains constant at 100N calculate the energy supplied in
moving the sledge 1 km.
Soln.
Work = force x distance
In this case the work is done in the horizontal direction. To find the force in
this direction we must resolve the tension (force) in the pulling rope.
Horizontal pull = T cos 25° = 100 cos25°
= 90.63 N
Work = force x distance = 90.63 x 1000m
25°
Horizontal pull = T cos 25 ° Work = 90.63 kJ (in the horizontal direction)
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Work & Energy
Examples:
3.
A force acts on a body of mass 15kg, which is initially at rest,
and imparts a constant acceleration of 2 ms-2 for 10 seconds. Calculate
the energy transfer (work done ) to the body.
Soln.
Work done = Energy used = force x distance
In this case the equations of motion must be used to find distance and force.
u = 0 ms-1
v= ?
a = 2 ms-2
s=?
t = 10 seconds
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From s = ut + ½ at2 distance s = ½ x 2 x 102 = 100m
From N2 F = ma, Force = 15 kg x 2 ms-2 = 30N
Work done = Energy transferred = 30N x 100m = 3000 J
Energy transfer = 3 kJ
Work & Energy
Energy is defined as the ability to do work
WORK DONE = ENERGY USED
The unit of ENERGY is as work:-
Newtons x metres or
Newton . Metres or
Energy is given up when work is done
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Joules (J)
Work & Energy
Typical forms of energy are:
Mechanical
Electrical
Heat
Chemical
Nuclear
Light
Sound
Energy may be converted from one form into another
In doing so its value remains the same
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Work & Energy
Conservation of Energy:- the principle states that:-
ENERGY CANNOT BE CREATED OR DESTROYED
Typical conversions include:Mechanical into electrical by a generator
Electrical into mechanical by a motor
Heat into mechanical by a steam turbine
Chemical to heat by combustion
Nuclear to electrical via a steam turbine & generator
Sound to electrical by a microphone
Many other conversions are regularly seen, try to discover an area
where energy is simply ‘lost’ or created from nothing.
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Work & Energy
The conversion of energy is never simple – and the conversion is
never 100 % efficient, losses will always occur.
A loss of Energy does not mean a reduction in the overall quantity but
a conversion into a form which is not required.
A simple example is the petrol engined motor car.
The fuel (chemical energy) is intended to provide motion.
In fact only 25% of the energy within the fuel provides motion, the
remaining 75% is converted to heat, vibration, sound etc.
The car is said to be 25% efficient.
Great efforts are made in engineering to try and improve the efficiency
of energy conversion.
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Work & Energy
The conversion process can be simplified with a block diagram:-
INPUT
PROCESS
OUTPUT
LOSSES
Input Energy = Output Energy + ‘Lost’ Energy
EFFICIENCY =
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OUTPUT
INPUT
x100%
Work & Energy
Using the equation:- EFFICIENCY = OUTPUT x100%
INPUT
It can be seen that output will always be < input thus efficiency will
always be < 100%
The symbol used for efficiency is η (eta) and is given as a percentage
Ex1.
An electric motor takes 5000 J from the mains supply. If the
output shaft of the motor provides 3600J find the efficiency of the
motor.
Soln.
Efficiency = Output/Input x 100%
η
= (3600 J ÷ 5000 J) x 100% = 72%
the 28% lost energy is probably converted to heat
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Work & Energy
Ex2.
A petrol engine has an efficiency of 32% and delivers 12.5
kJ/second at the crankshaft. What is the energy input to the engine.
Soln.
Engine
η = 32%
Input
Efficiency
32/100
Input
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=
=
=
12.5 kJ/s
output/input
12.5 kJ/s ÷ Input
12.5 ÷ 0.32 kJ/s
= 39.0625 kJ/s
Work & Energy
Ex3.
A motor has an efficiency of 75% and drives a gearbox with an
efficiency of 87%. If the output from the gearbox must be 20.76 kJ/s
calculate the motor input in kJ/s
Soln.
Input
motor
η = 75%
‘A’ kJ/s
gearbox
η = 87%
20.76 kJ/s
Method 1 : - interim stages
Find output from motor ‘A’ kJ/s; (this must be the required input to the
gearbox).
A = 20.76/ 0.87 = 23.862 kJ/s which must be the motor output
Motor input = 23.862 kJ/s / 0.75 = 31.82 kJ/s
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Work & Energy
Soln.
Input
motor
η = 75%
gearbox
η = 87%
Method 2 : - single stage
Motor input = 20.76 kJ/s / (0.87 x 0.75) = 31.82 kJ/s
Note: 87 % = 87/100 = 0.87
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20.76 kJ/s
Work & Energy
POTENTIAL ENERGY PE
This is the energy possessed by a body due to its position in
space
The amount of energy held will depend upon the height of the body
above a specified datum.
CoG
Height h
Above datum
datum
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Work done = Force x distance
WD = Energy = Weight x height
Energy = mg x h
Potential Energy (PE) = mgh
When m = mass kg
g = 9.81 ms-2
h = height m
Work & Energy
KINETIC ENERGY KE
This is the energy possessed by a body due to its motion in
space
The amount of energy held will depend upon the mass and the speed
of the body.
Consider mass m kg, moving at
velocity v ms-1 which is brought to
rest in a time of t seconds
For this to occur then Newton’s 1st law suggests an opposing force
must be applied. The force needed to cause the deceleration = ma
The rate of deceleration = v/t thus Force needed = m v/t
The distance moved in this period = ½ vt
As Work = Force x distance
= m v/t x vt/2 = ½ mv2
KINETIC ENERGY = ½ mv2
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Work & Energy
Examples
1.
A crane requires 29.5 kJ of energy to raise a mass of 500 kg to a
certain height. Determine the height of the load after the crane
has stopped lifting.
Soln.
The crane will do work lifting the mass against gravity (weight) and the
load will gain potential energy mgh
From PE = mgh the height gain h = PE/mg
When PE = 29.5 kJ = 29500 J
m = 500 kg
g = 9.81 ms-2
Height h = 29500/ (500 x 9.81) = 6.01 metres
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Work & Energy
Examples
2.
A motor car of mass 900 kg is travelling at 50 km/hr
Determine: a. The kinetic energy of the vehicle
b. The speed of the car after it has made a brake
application in which 45 kJ of energy is lost
Soln.
a.
KE = ½ mv2 when: m = 900 kg
v = 50 x (1000/3600) = 13.889 ms-1
KE = ½ x 900 x 13.8892 = 86806.9 J
= 86.81 kJ
b.
If the brakes take out 45 kJ of energy the car is left with:86.81 kJ - 45 kJ = 41.81 kJ this is in the form of KE ( ½ mv2)
Transpose for v = 2 KE = 2 x 41810 = 9.64 ms-1 (34.7 km/hr)
m
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Work & Energy
Examples
3.
A body of mass 800 kg is hauled up an incline plane through a
vertical distance of 20m. If the work done against friction is 12 kJ
calculate the total work done and the potential energy of the body
at its elevated position.
Soln. (tip - work out the second part first)
Work done lifting a height of 20m is equal to the potential energy gain
PE = mgh
when:- m = 800 kg
g = 9.81 ms-2
h = 20m
PE = 800 x 9.81 x 20 = 156960 J = 156.96 kJ
BUT 12 kJ of energy was used to overcome the friction on the incline
Thus total energy used to lift = 156.96 + 12 = 168.96 kJ
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Work & Energy
Examples
4.
A railway train of mass 300 tonnes is travelling at 80 km/hr
when its brakes are applied to reduce its reduce its speed to 35 km/hr
Determine the amount of energy used in the brake application.
Soln.
Initial KE = ½ mv2 when: m = 300 tonnes = 300 x 103 kg
v = 80 x (1000/3600) = 22.222 ms-1
KE = ½ x 300x103 x 22.2222 = 74072592.6 J = 74.07 MJ
Final KE = ½ mv2 when: v = 35 x (1000/3600) = 9.722 ms-1
KE = ½ x 300x103 x 9.7222 = 14177592.6 J = 14.18 MJ
The energy taken out by the brakes is the ‘loss’ in Kinetic Energy
Change in energy (∆ KE) = Initial – Final = 74.07 – 14.18 = 59.89 MJ
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