Download Magnetic Mass Balance

The Electric Motor
Introduction: For 50 years the car
industry has spent $1 Billion
dollars to introduce each new
model car. Millions of cars had to
be sold to make a profit. Today, a
dozen new car companies are
spending about $5 Million to
produce, from scratch, new
electric cars capable of going from
0 – 100kph in under 5 seconds.
As these cars retail for $300,000, a
company only needs to sell tens of
cars in order to move into profit.
In this experiment, we will
examine the electric motor
underlying these electric cars.
Theory:
Battery Power: 48 kWh lithium-ion battery pack
Acceleration: 0-100kph in under 5 seconds
Range: of 320 km on a single charge
Recharge: Eight hours from a conventional mains
socket.
http://www.economist.com/blogs/babbage/2011/05/new_electric
-vehicle_producers
Magnetic Force: F = B I L sin θ
Ohm’s Law: V = I R
Resistance: R = ρ L / A (ρ is resistivity for copper of 1.7 × 10-8 Ωm,
L is wire length, and A is area of wire)
a = (v – u) / t
F=ma
Aim: To investigate the operation of an electric motor and compare theoretical
predictions with experiment.
Hypothesis: The force on the rotor of an electric motor is given by F = I B L sin θ.
Apparatus and Operational Test:
Field
Field
wire
wire
+
Battery
-
North
Construct the electric motor shown to the
left.
Warning: Don’t bend the wire rotor as it
has been balanced to minimize friction.
Run the electric motor for only a short time.
Essentially, we are short circuiting the
battery generating a current of many Amps
and it will go flat very quickly.
If possible, count the rotations per second of
the rotor after 1sec and 5sec of operation.
Magnets
South
Additional Equipment: Micrometers, rulers,
Ammeters, Voltmeter, cables, Compass,
Digital Scales
Results and Measurements:
Why does the rotor turn? On the diagram above, draw in the flow of conventional
current through each wire of the rotor. Now, locate the two regions where the current
is flowing at approximately 90o to the direction of the magnetic field. (In other places
the angle is 0o or a small angle so sin θ is approximately zero.) Use your right hand
rule to draw in a force vector for those two regions. From this, explain why the rotor
turns.
In order to quantify the operation of the electric motor, we need to determine the force.
To use F = B I L sin θ we need to measure the quantities on the right hand side.
Use a ruler to measure the length of the wire in the region where θ=90o.
L = …………..m.
(Guess: L = 2cm for each wire.)
Dismantle the motor and use the compass to measure the strength of the magnetic
field in the region occupied by the wire. Recall tan θ = Bmagnet / BEarth, so
Bmagnet = tan θ × BEarth,
Bmagnet = …………….T
(Guess: Bmagnet = 10 BEarth = 5 × 10-4T at top of rotor.)
-5
where BEarth = 5 × 10 T. (You will have to point the magnet East-West so Bmagnet is
perpendicular to BEarth.)
We now need to know the current going through the wire. Ohm’s law V = I R says if
we know the voltage of the battery and the resistance of the circuit, then we would be
able to calculate the current. Every battery has some internal resistance and the wire
has its own resistance. We need to work out both of these. The battery internal
resistance is about 0.5Ω. We can calculate the resistance of the wire using resistivity
theory which gives
R = ρ Ltotal / A
where ρ is resistivity which for copper is 1.7 × 10-8 Ωm and L is wire length, and A is
area of wire. Use a ruler to estimate the total length of a single wire in the rotor and
use the micrometer to measure the diameter of the wire. From this calculate the area
of the wire (don’t forget there are two wires in the rotor) and calculate the resistance
of the rotor.
Rrotor = ρ L / A = ……………..Ω
(Guess: R = 1.7 × 10-8Ωm × 0.2m / (2 π 0.00052) = 0.002Ω.)
The total resistance of the circuit is the sum of the internal battery resistance and the
resistance of the two wires in the rotor.
Rtotal = 0.5 + 2 Rrotor = …………….Ω
(Guess: Rtot = 0.502Ω.)
Ohm’s law then gives the current running through each rotor wire.
I = …………………..A
(Guess: I = 5.9A.)
Now calculate the force acting on each arm of the rotor wire,
F = B I L sin 90o = …………………….N
(Guess: F = 0.00006N.)
We can now use Newton’s Laws F = m a to calculate how fast this force accelerates
the rotor.
Measure the mass of the rotor.
m = ………………..kg
(Guess: m = 0.005kg.)
Just for fun, let’s use our measurements to calculate the density of copper and
compare this to accepted values. The density of the metal in the rotor is the
mass divided by the volume. The volume of the two cylindrical wires in the
rotor is
V = 2 × Area of One Wire × Length of One Wire
= 2 × π r2 × Ltotal = …………………..m3
Evaluate the density and compare to the accepted value (8.93kg m-3).
Density = m / V = ………………kg / m3
Evaluate the percentage error in your measurement and explain this.
Error = |Measured Value – Accepted Value| / Accepted Value
We now know the force F acting on half the rotor (not the total mass). What is the
acceleration of the rotor.
a = F / (m/2) = …………… ms-2
(Guess: a = 0.00006N / 0.0025kg = 0.024ms-2.)
What is the linear velocity of the rotor after it has moved one revolution, i.e. the rotor
mass (half the total mass) has moved a distance of one circumference of C = 2 π r.
v2 = u2 + 2 a s
v = …………m/s
(Guess: v2 = 2 × 0.024 × 2π 0.05: v = 0.13ms-1=13cm/s)
How long does it take to make this first revolution.
v = u + a t, so t = v/a = …………m/s
(Guess: t=5.11s)
If friction and air resistance stabilizes the speed at this velocity, how many revolutions
per second and revolutions per minute is this.
RPS = ……………. s-1
RPM = ……………../ min
(Guess: RPS = 2.6s-1, RPM=157/min)
How does this compare with your observations.
Note that as the battery becomes flat, the current delivered decreases and the rotor
slows down.
Discussion and Applications to Electric Motors for Cars:
What is the acceleration of the car pictured above.
a = (v – u) / t = ………………..m s-2
(Guess: a = 5.6ms-2)
This linear acceleration is generated from two electric motors attached to each of the
rear wheels. The total car mass is 800kg (guess) so each motor must move 400kg at
this acceleration. This means the effective mass of the rotor of each electric motor is
400kg. If the rotor radius of each electric motor is about15cm, what force must be
generated by the motor.
F=m/a
(Guess: F = 2240N)
The electric motors are designed so the magnetic fields are curved so θ is large for
almost all of the wire (so sin θ = 1), and the rotor wire has many turns, N>>1. How
would you design an electric motor to generate this force.
F = B (N I) L
A strong field is about B = 0.5T, the rotor might have N = 2000 turns, while the
length of each wire loop L = 2 π 0.15 so the required current must be
I = F / (B N L) = ……………A
(Guess: I = 2.4A)
Regenerative Braking: Electric cars use the battery to drive an electric motor to
generate kinetic energy. If the car uses disk brakes to slow the wheels via friction,
then all this energy is lost to heat as the disks become red hot. An electric car
captures most of the kinetic energy of the car by using the electric motor to recharge
the battery. To move, the car battery supplies a current to the electric motor to
generate kinetic energy. When run in reverse, the motion of the car is used to drive
the electric motor around which generates a current to recharge the battery. While
there are losses, most of the kinetic energy is captured and converted back to electric
energy to extend the range of the car.
Conclusion: We investigated the operation of an electric motor and confirmed the
validity of the law F = B I L sin θ. This law was then applied to the design on an
electric car.