MMME2104
Design and Selection of Mining Equipment
Electrical Component
Tutorial Sheet 2 Solutions
Question 1 (Problem 2-5 in the text)
A coil having 200 turns links a flux of 3mWb, produced by a permanent magnet. The
magnet is moved, and the flux linking the coil falls to 1.2 mWb in 0.2s. Calculate the
average voltage induced in the coil.
We use Faraday’s Law to calculate the voltage induced in the coil, due to the
changing flux linkage:
V =N
∆Φ
0.0012 − 0.003
dΦ
≈N
= 200 ×
= −1.8V
∆t
0 .2
dt
*Question 2 (Similar to problem 2-4 in the text)
A perfect conductor 2m long is moved at a speed of 36km/h through a magnetic field
having a flux density of 1.0 T.
a) Calculate the induced voltage
If a 1Ω resistor is connected across the ends of the conductor, calculate the
following:
b) The current in the circuit
c) The force required to move the conductor through the field
d) The mechanical power required to move the conductor through the field
e) The electrical power generated
a) We calculate the induced voltage from Faraday’s Law for a linear conductor:
V = BLv = 1T × 2m × 10m / s = 20V
V 20V
=
= 20 A
R 1Ω
c) The force required to move the conductor is the Lorentz Force:
F = BiL = 1T × 20 A × 2m = 40 N
b) With a 1Ω resistor, the current: i =
d) The mechanical power is force multiplied by velocity:
PM = Fv = 40 N × 10m / s = 400W
e) The electrical power is voltage multiplied by current:
PE = Vi = 20V × 20 A = 400W
The mechanical and electrical powers are the same, and this process demonstrates
ideal electromechanical energy conversion (such as in an electric motor).
Question 3 (Problem 9-1 in the text)
A coil has 500 turns and a reactance of 60 Ω with negligible resistance. If it is
connected to a 120V, 60Hz source calculate the following:
a) The value of the magnetizing current
b) The peak value of magnetizing current
c) The peak MMF produced by the coil
d) The peak magnetic flux produced by the coil
a) The magnetizing current is found from the source voltage and coil reactance:
V 120V
iM =
=
= 2A
X
60Ω
b) The values in a) are RMS quantities, therefore, the peak magnetizing current is:
i Mpeak = 2 × i M = 2.83 A
c) The peak MMF produced by the coil is therefore:
MMFpeak = Ni Mpeak = 500turns × 2.83 A = 1414 A
d) The peak magnetic flux is obtained from the flux linkage equation (another form of
Faraday’s Law):
V = 4.44 NfΦ peak
Therefore:
V
120V
Φ peak =
=
= 0.9mWb
4.44 Nf 4.44 × 500turns × 60 Hz
*Question 4 (Similar to Problem 9-4 in the text)
An ideal transformer has 400 turns on the primary and 200 turns on the secondary.
The primary is supplied by a voltage of 600V, and the load is a resistance of 12Ω.
Calculate the following:
a) The secondary voltage E2
b) The secondary current I2
c) The primary current I1
d) The power delivered to the primary
e) The power output from the secondary
f) The impedance seen by the supply
Firstly we calculate the turns ratio of the transformer: a =
Ep
Np
Ns
=
400
=2
200
600V
= 300V
2
a
E
300V
b) The secondary current: i s = s =
= 25 A
R
12Ω
i
25 A
= 12.5 A
c) The primary current: i p = s =
a
2
d) Power delivered to primary: Pp = E p i p = 600V × 12.5 A = 7500W
a) The secondary voltage:
Es =
=
e) Power output to secondary: Ps = E s i s = 300V × 25 A = 7500W
Answers d) and e) are the same and this demonstrates ideal electromagnetic energy
conversion in a transformer.
f) The impedance seen by the supply: Z p = a 2 Z s = a 2 R = 2 2 × 12Ω = 48Ω
*Question 5 (Similar to Problem 10-30 in the text)
During a short-circuit test on a 10MVA, 66kV/7.2kV transformer (see Figure 10.28 in
the text), the following results were obtained:
Voltage applied to primary = 2600 V
Measured primary current = 70 A
Measured electrical power = 10 kW
Calculate the following:
a) The total resistance and the total leakage reactance referred to the 66 kV
primary side
b) The nominal impedance of the transformer referred to the primary side
c) The percent impedance of the transformer
d) If the iron losses at rated voltage are 35kW, calculate the full-load efficiency
of the transformer if the power factor of the load is 85 percent
a) The results of the short-circuit test allow us to calculate the resistance and leakage
reactance referred to the primary side, as follows:
V
2600V
Z p = sc =
= 37.14Ω
i sc
70 A
P
10kW
R p = sc2 =
= 2.04Ω
(70 A)2
i sc
2
2
X p = Z p − R p = 37.08Ω
b) To calculate the nominal impedance referred to the primary side, we may first
calculate the rated primary-side current:
S
10MVA
irated = rated =
= 151.5 A
Vrated
66kV
The nominal impedance is then:
V
66kV
Z nom = rated =
= 435.6Ω
i rated 151.5 A
Alternatively, we could have calculated the nominal impedance directly from the
apparent power and voltage ratings:
2
2
Vrated
(
66kV )
Z nom =
=
= 435.6Ω
S rated
10 MVA
c) The percent impedance is then:
Zp
37.14Ω
Z p ( pu ) =
× 100 = 8.5%
=
Z nom 435.6Ω
Pout
where Pin and Pout are active powers.
Pin
We are asked to find the full-load efficiency for the transformer. At full load, the
transformer’s rated apparent power output is 10MVA. Since we are given the power
factor of the load, the output power is:
Pout = S rated × pf = 10 MVA × 0.85 = 8.5MW
d) The definition of efficiency is η =
The input power, Pin, consists of the output power plus the losses. In a transformer,
there are two kinds of active power loss: core (iron) losses and resistive (copper)
losses. We are given the core losses at rated voltage = 35kW. By definition, the
transformer operates at rated voltage when at full load. Therefore, the core losses at
full load = 35kW. By definition, the transformer operates at rated current when at full
load. Therefore, at full load the primary-side current will be 151.5A. We can
therefore calculate the resistive losses using the transformer’s resistance referred to
the primary side:
2
2
Presistive = i rated R p = (151.5 A) × 2.04Ω = 46.8kW
Therefore, the transformer’s full load efficiency is:
P
Pout
8500kW
η = out =
=
× 100 = 99.0%
Pin
Pout + Pcore + Presistive 8500kW + 35kW + 46.8kW
Question 6 (Problem 12-2 in the text)
Three single-phase transformers rated at 250kVA, 7200V/600V, 60Hz are connected
in wye-delta on a 12470V, 3-phase line. If the load is 450kVA, calculate the following
currents:
a) In the incoming and outgoing transmission lines
b) In the primary and secondary windings
Firstly, we calculate the turns ratio of each single phase transformer:
N p V p − rated 7200V
a=
=
=
= 12
600V
N x V s −rated
a) The load is given as 450kVA. In an ideal transformer, the output power equals the
input power. Therefore, the input power is also 450kVA. For a 3-phase system, we
know the total apparent power is given in terms of line voltage and current as:
S tot = 3VL i L
Therefore, the input line current will be:
S tot
450kVA
i L −in =
=
= 20.83 A
3 × V L −in
3 × 12470V
For a wye-delta connection, the output current is given as:
i L −output = 3 × a × i L −input = 433 A
b) For a wye-connection, the line current equals the branch current. Therefore, the
current in the primary winding is the input line current = 20.83A. For a delta-
connection, the line current is 3 times the branch current. Therefore the current in
the secondary winding is 433 / 3 = 250A.
*Question 7 (Similar to Problem 12-7 in the text)
Three single phase transformers rated 100kVA, 13.2 kV / 3.3kV are connected in wyedelta on a 3-phase 15kV line:
a) What is the maximum load that can be connected to the transformer bank?
b) What is the outgoing line voltage?
Firstly, we calculate the turns ratio of each single phase transformer:
N p V p − rated 13.2kV
a=
=
=
=4
3.3kV
N x V s −rated
a) The primary side is wye-connected, therefore the rated line voltage of the primary
is 13.2kV × 3 = 22.9kV. However, the primary is connected to a line voltage of
only 15kV; therefore the transformer’s apparent power will be reduced by a factor of
15/22.9 (the current rating does not change!). Therefore, the maximum permissible 3phase supply to the primary is:
15
S tot = 3 × 100kVA ×
= 196.5kVA
22.9
Since for an ideal transformer the input and output powers are equal, the maximum
load for this transformer bank will be 196.5kVA.
b) For wye-delta connection, the outgoing line voltage is given as:
VL −input
15kV
VL −output =
=
= 2.17kV
3×a
3×4