Transcript Energy, work and Power

Energy, work and Power
Energy, work and Power
Work is defined as using energy
Work and energy are both measured in
joules
Energy, work and Power
• Work done is defined as Force x distance
moved in the direction of the force
• The SI unit is the newton metre (Nm)
• 1Nm is 1Joule
Energy, work and Power
A 10 kg mass has a weight of 98.1
Newtons (weight is a force and is found by
multiplying mass x the gravitational field strength
9.81N/kg)
2metres
10kg
If the mass is lifted vertically against the
force by a distance of 2 metres then
98.1 x 2 (f x d) = 196.2 joules of energy
are needed.
196.2 joules of work has been done
Energy, work and Power
10kg
2metres
The mass has now
gained
196.2 joules of
potential energy
Power
• Power is the rate at which energy is used (or work
is done)
• Power = energy/time
• Measured in Joules/seconds
• I Watt = I Joule/sec
Power (question)
If the power of the kettle is 2400W and
there are assumed to be no losses,
calculate how much energy will be used
if the kettle is on for I minute .
Power (question) - answer
Power = energy ÷ time ( in seconds)
Energy = power x time ( in seconds)
Energy = 2400 x 60
= 144000 joules
1.44 x 105 joules
Relationship between work, power and velocity
Work done = Force x distance
Velocity = distance / time
Work done/time = ( Force x distance )/ time
Work done /time = power and distance/ time = velocity
Power = Force x velocity
Relationship between power and velocity (question)
A vehicle has a maximum speed of 50
m/s and a power output of 120 kW at this
speed. Calculate the driving force
generated by the engine.
Relationship between work, power and velocity
(question) - answer
Power = Force x velocity
Force = Power/velocity
Force = 120000 watts/50 m/s
= 2400 Newtons
Relationship between work, power and velocity
(question)
• The brakes exert a total force of 8 kN
and the car stops from 50 m/s in 200m.
Calculate the work done by the brakes.
Relationship between work, power and velocity
(question) - answer
Work done = force x distance
Work done = 8kN x 200m
1600 kJ or 1.6MJ
Specific Heat Capacity
The specific heat is the amount of
heat per unit mass required to
raise the temperature by one
degree Celsius.
Specific Heat Capacity
Q = c.m.∆T
Q = heat added C = specific heat capacity m = mass
and ∆T = change in temperature
Specific Heat Capacity (question)
• The brake discs on the car in the last question
have a total mass of 8 kg and are made from a
material with a specific heat capacity of 1180
J/kgK. Neglecting any cooling effects, what will be
the temperature rise of the brake discs during the
stopping manoeuvre?
Specific Heat Capacity (question)
Q/c.m =∆T
∆T = 1600000/ 1180 x8
169K (or 169oC)
Specific Heat Capacity (question)
The specific heat capacity of water is
4200 J/Kg oC
How much heat is required to raise the
temperature of 10 Kg of water by 20oC
Specific Heat Capacity (question) - answer
Q = c.m.∆T
Q = 4200 x 10 x 20
= 840000 joules
or
8.4 x 105 joules
Enthalpy
Enthalpy is a measure of the
total energy of a
thermodynamic system
Enthalpy (question)
• If the specific enthalpy of water at 10oC
is 43 kJ/kg and the specific enthalpy of
steam at 100oC is 2675 kJ/kg, calculate
the energy required to convert 1.5 kg of
water at 10oC to steam at 100oC.
Enthalpy (question) -answer
• Q = m(H – h) = 1.5( 2675 – 43) = 3948 kJ
Where H = enthalpy of steam, h = enthalpy of
water, Q = quantity of energy and m = mass
Gas Laws
• Volume is inversely proportional to
pressure
• V = 1/P
• Pressure x volume is constant. If
pressure increases volume decreases
Gas Laws
Pressure x volume is constant
p1 V1 = p2 V2
Where:
p1 is the starting pressure
V1 is the starting volume
p2 is the finishing pressure
V2 is the finishing volume
This is known as Boyle’s Law
Gas Laws (question)
• An air tight syringe contains a volume of 8 x10-6
m3 of air at 1 × 105 Pa. pressure. The plunger is
pushed until the volume of trapped air is 4 x 10-6
m3. If there is no change in temperature what is
the new pressure of the gas?
Gas Laws (question) - answer
p1 V1 = p2 V2
p2 = (p1 V1) ÷ V2
= (1 x 105 Pa x 8 x10-6 m3) ÷ 4 x 10-6 m3
= 2 x 105 Pa
Gas Laws
Volume is directly proportional to
temperature
• volume ÷ temperature is constant. If
temperature increases volume
increases
Gas Laws
Pressure is directly proportional to
temperature
• pressure ÷ temperature is constant.
If temperature increases pressure
increases
Gas Laws
Combining all three laws we get
(pressure x volume) ÷ volume
is constant
Gas Laws
This equates to
PV/T = constant
(P1V1)/T1 = (P2V2)/T2 = constant
Gas Laws
• This means that in any system, if any of the
quantities change then the others will
change in proportion
• Temperature has to be measured in Kelvin
• (degrees Celsius + 273)
• - 273oC = absolute zero
Gas Laws (Question)
Steam at a pressure of 20 x 105 Pa and a
temperature of 127oC is compressed to 1/2 of
its original volume, after which the temperature
recorded is 327oC. Calculate the pressure
required to achieve this.
.
Gas Laws (Question) - answer
• (P1V1)/T1 = (P2V2)/T2
• P1 = 20 x 105 Pa
P2 = ?
• V2 = 1/2 of V1 so V1/ V2 = 2/1
• T1 = 400 Kelvin (K) and T2 = 600 Kelvin(K)
Gas Laws (Question) - answer
P2 = (P1V1 T2)/(V2T1)
(20 x 105 Pa x 600 x 2)/400
(V1/ V2 = 2)
= 60 x 105 Pa or 6 x 106 Pa
Gas Laws ( Question)
Diesel fuel is injected into a cylinder at an inlet pressure
of 1.2 bar and a temperature of 24oC. The
compression ratio of the engine is 18 : 1 and the
minimum ignition temperature of diesel fuel is 400oC.
Calculate the minimum cylinder pressure in bar
required to achieve this
(1 bar is approximately atmospheric pressure)
Gas Laws (Question) - answer
P2 = (P1V1 T2)/(V2T1)
P2 = (1.2 x18 x 673)/ (1 x 297)
= 48.9 bar
Kinetic energy (movement)
The equation for calculating kinetic
energy is
½(mv2)
Where m = mass and v = velocity
Kinetic energy (question)
• A car of mass 1500 kg is moving at
20m/s. Calculate its kinetic energy.
Kinetic energy (question) - answer
kinetic energy = ½(mv2)
Mass = 1500 kg
Velocity = 20m/s
Ke = 0.5 x 1500 x202
300000 Joules (300kJ)
Relationship between Gravitational potential energy
and kinetic energy
Gravitational potential energy is the energy an object has
due to it’s perpendicular height
Gravitational potential energy = mass x gravitational field
strength x height
GPE = mgh
Relationship between Gravitational potential energy
and kinetic energy (question)
A 20kg mass is suspended 20 metres above the
ground. Calculate the gravitational energy of
the mass
Relationship between Gravitational potential energy
and kinetic energy (question) - answer
GPE = mgh
m = mass (kg),
g = gravitational field strength 9.81 (N/kg)
h = height (metres)
Relationship between Gravitational potential energy
and kinetic energy (question) - answer
GPE = mgh
GPE = 20 x 9.81 x 20
= 3924 Joules
Relationship between Gravitational potential energy
and kinetic energy
If the 20kg mass falls to the ground the
potential energy converts to kinetic energy.
When the mass hits the ground all the
energy is kinetic energy
Relationship between Gravitational potential energy
and kinetic energy (question)
Calculate the velocity of the 20kg mass as it
hits the ground
Relationship between Gravitational potential energy
and kinetic energy (question) - answer
GPE = 3924 Joules
Kinetic of the mass when it hits the ground
= 3924 Joules
Relationship between Gravitational potential energy
and kinetic energy (question) - answer
Kinetic energy = ½(mv2) = 3924 Joules
v2 = (2 x 3924)/20
v2 = 392.4
v = √392.4
velocity = 19.8m/s
50o
500g
A pendulum of length
3.0m is pulled aside to an
angle of 50o with the
vertical. Calculate
the gain in potential
energy of the 500g bob. (g
= 9.81m/s2)
A
50o
B
500g
To calculate the length AB
Cosine 50o =
adjacent/hypotenuse
= AB/3
AB = 3 x Cosine 50o
1.9 m
Increase in vertical height of
the 500g bob
= 3 – 1.9
= 1.1m
A
Increase in GPE
= 0.5kg x 9.81 x 1.1m
5.4 Joules
50o
B
500g
A
Calculate the velocity of the bob
at the bottom of its swing.
50o
B
500g
Ke = ½(mv2) = 5.4J
v2 = (2 x 5.4)/0.5
= 21.6
V = √21.6
V = 4.6 m/s