Download Note 30 Power

Note 30 Power
Power is the rate at which work is done by a source. It can also be defined as the rate at which
the kinetic energy of a target is changing. Here is its most general definition. This is the
instantaneous power.
P=
dW
dt
You can also talk about the average power which is just the difference version of the equation.
P=
W
Δt
You can talk about a single force. This is the power output from a single force.
Pforce =
dWforce
dt
You can talk about the net force. This is the power due to all forces on a target. This is also the
rate of change of the energy of the system, but this requires that the transmission of the energy
from source to target be 100%.
Ptotal =
dWtotal
dK
=
dt
dt
If the force is independent of time, then the instantaneous power is this.
! !
! dr!
! !
dW
d(F ⋅ r )
P=
=
=F⋅
= F ⋅v
dt
dt
dt
Unit of Power
The unit of power is called the watt (W). The watt is a relatively small unit so other units are used
in industrial settings. The horsepower is 746 watts.
The watt is also used to define energy because it is convenient. A kilowatt-hour (kW•h) is a unit of
energy that lets the source produce a kilowatt of power for an hour. Here are some examples of
energy storage.
AA battery
3.0 W•h
11 kJ
ipad air
32 W•h
115 kJ
40 W bulb 8 hours
320 W•h
1.2 MJ
1 kg TNT
1.2 kW•h
4.2 MJ
prius
4.4 kW•h
16 MJ
1 kg gasoline
12 kW•h
42 MJ
tesla roadster
53 kW•h
190 MJ
atom bomb
1.4 TW•h
5.0 PJ
page 1
Efficiency
The efficiency of a power source is the ratio of the energy input into the source divided by the
useful work that it produces.
The efficiency of a device is the ratio of the useful work done by the device divided by the energy
input into the device.
eff =
Wuseful
Einput
page 2
Example: Gravity Power
What is the power output of gravity from the Earth when a 1 kg object is dropped? Is it constant?
The power output of the Earth’s gravity is
! !
! dy!
! !
dWg
d(Fg ⋅ y )
ˆ ⋅ v (−j)
ˆ = mgv
P=
=
= Fg ⋅
= Fg ⋅ vy = mg(−j)
y
y
dt
dt
dt
Since the velocity increases over time, the power output increases.
What are the average powers over the first and the second second?
Over the first second, the velocity goes from 0 to 9.8 m/s. The average power is
1
mv f2 − 12 mvi2
W
1
1
2
P=
=
= mv f2 = m(9.8)2 = 48 W
Δt
Δt
2
2
Over the second second, the velocity goes from 9.8 m/s to 19.6 m/s. The average power is
1
mv f2 − 12 mvi2
W
1
1
1
1
3
2
P=
=
= mv f2 − mvi2 = m(19.6)2 − m(9.8)2 = m(9.8)2 = 144 W
Δt
Δt
2
2
2
2
2
The work is three times larger during the second second since the displacements also increase
over time. Over the first second, it is this.
1
Δy1 = − (9.8)(1 − 0)2 = −4.9 m
2
Over the second second it is this, three times larger.
1
Δy1 = −9.8(2 − 1) − (9.8)(2 − 1)2 = −9.8 − 4.9 = −14.7 m
2
page 3
Example: Cruising and Passing
A 3,000 kg car is traveling at a constant speed of 15 m/s (33 mph). If the air resistance 600 N and
the internal friction produces another constant 2,000 N of force, how much power is required to
keep the car going at 15 m/s?
At 15 m/s, the total resistive force acting on the car is 2,600 N. The engine has to supply 2600 N
to keep the car at a constant speed. The power required to do this is
⎛ 1 hp ⎞⎟
P = Fv = (2, 600)(15) = 39, 000 W ⎜⎜
⎟ = 52.3 hp
⎜⎝ 746 W ⎟⎟⎠
Let’s say this car needs to pass a slower vehicle. How much power is required to accelerate from
15 m/s to 20 m/s (44 mph) over a time of 3 seconds? Assume that the average air resistance
force is now 750 N at the average speed of 17.5 m/s.
The total resistive force is now 2,750 N. The power required to overcome this is
P = Fv = (2750)(17.5) = 48,125 W
The average power required to go from 15 to 20 m/s is
1
mv f2 − 12 mvi2
W
1
202 − 152
2
P=
=
= 3, 000
= 87,500 W
Δt
Δt
2
3
The total power needed is
⎛ 1 hp ⎞⎟
P = 48,125 W + 87,500 W = 135, 630 W ⎜⎜
⎟ = 182 hp
⎜⎝ 746 W ⎟⎟⎠
page 4
Example: Elevator
A motor accelerates an elevator 500 kg in mass with a 100 kg of people inside from rest to 2 m/s
over a time interval of 2 seconds.
What is the power output of the motor?
The work output of the motor can be found from the work-energy theorem. All of the work goes
into changing the energy of the elevator.
Wnc = ΔK + ΔU
1
Wmotor = mv f2 + mgh
2
The height at the end of the the 2 seconds is found like this. The acceleration is
Δv = aΔt
⇒
2 = a(2) ⇒ a = 1
1
1
h = at 2 = (1)(2)2 = 2
2
2
So,
1
Wmotor = (600)(2)2 + (600)(10)(2) = 31, 200 J
2
The power output of the motor is
P=
31, 200 J
= 16, 600 W
2s
This is why counter-weights are used to reduce this power requirement. If a counter-weight of 500
kg is used, what is the power output?
Let the elevator and the counter-weight be the system.
Wnc = ΔK + ΔU
1
1
Wmotor = ΔK system + ΔU system = melevatorv f2 + mcounterv f2 + melevator gh − mcounter gh
2
2
1
1
Wmotor = (600)(2)2 + (500)(2)2 + (600)(10)(2) − (500)(10)(2)
2
2
= 1200 + 1000 + 12, 000 − 10, 000 = 4, 200 J
The power output of the motor is now
P=
4, 200 J
= 2,100 W
2s
page 5
Let’s say there is also a friction force on the elevator with an effective force of 400 newtons. What
is the power output of the motor now?
The work output of the motor is
Wmotor +Wfriction = ΔK system + ΔU system
⇒ Wmotor = ΔK system + ΔU system −Wfriction
The work done by the motor is now
1
1
Wmotor = melevatorv f2 + mcounterv f2 + melevator gh − mcounter gh − −Ffrictionh
2
2
Wmotor = 4, 200 + (400)(2) = 5, 000 J
The power output of the motor is now
P=
5, 000 J
= 2,500 W
2s
page 6