EE11001 Tutorial 6a (Induction Machine)
1) A 4-pole 50 Hz induction motor is driving a load at 1450 r/min.
a) What is the slip of the motor ?
[3.33 %]
b) What is the frequency of the rotor currents ?
[1.67 Hz]
c) What is the angular velocity of the stator field with respect to the stator ? With
respect to the rotor ?
[1500 rpm, 50 rpm]
d) What is the angular velocity of the rotor field with respect to the rotor ? With
respect to the stator ?
[50 rpm, 1500 rpm]
2) A 3-ph induction motor runs at almost 1000 r/min at no load and 950 r/min at full
load when supplied with power from a 50 Hz, 3-ph line.
a) How many poles does the motor have ?
[6 poles]
b) What is the % slip at full load ?
[5 %]
c) What is the corresponding frequency of the rotor voltages ?
[2.5 Hz]
d) What is the corresponding speed of the rotor field with respect to the rotor ? Of
the rotor field with respect to the stator ?
[50 rpm, 1000 rpm]
e) What speed would the rotor have at a slip of 10 % ?
[900 rpm]
f) What is the rotor frequency at this speed ?
[5 Hz]
g) Repeat part (d) for a slip of 10 %.
[100 rpm, 1000 rpm]
3) The power input to the rotor of 440 V, 50 Hz, 6-pole, 3-ph induction motor is 80
kW. The rotor emf is observed to make 100 complete alterations per min. Calculate,
a) the slip
[3.33 %]
b) the rotor speed
[967 rpm]
c) the mechanical power developed
[77 kW]
d) the rotor copper loss per phase,
[890 W]
e) the rotor resistance per phase if the rotor current is 65 A.
[0.21 Ω]
4) In a 3-ph induction motor, the stator reactance is equal to the equivalent rotor
standstill reactance, whilst each resistance is equal to one-fourth of this value. A
torque of 405 N-m is developed at a slip of 3 %. Find the values of the starting torque
and the maximum torques.
[806 Nm, 1712 Nm]
5) Describe the effect on the normal torque-speed characterstics of an induction motor
produced by
a) halving the applied voltage with normal frequency ?
b) halving both the applied voltage and the frequency ?
Sketch the associated torque-speed characteristics in their approximate relative
positions with respect to the normal one. Neglect the effects of stator resistance and
leakage reactance.
6) A 208V, 4-pole, 50 Hz, Y-connected, wound rotor induction motor is rated at 15HP.
Its equivalent circuit components referred to the stator side are:
R1=0.22 Ω; R2=0.127Ω; Xφ=15.0 Ω; X1=0.43 Ω; X2=0.43 Ω
Pmech=300 W; Pmisc≈ 0; Pcore=200W. For a slip of 0.05, find:
a) The line current, Il.
[46.1 A]
b) The stator copper losses Pscl and the air-gap power Pag.
[1.138 kW, 13.14kW]
c) The power converted from electrical to mechanical form, Pconv.
[12.49 kW]
d) The induced torque τind and the load torque τload.
[83.71 Nm, 81.7 Nm]
e) The overall machine efficiency.
[84.14 %]
7) A 3-ph induction motor, at rated voltage and frequency, has a starting torque of 160
% and a maximum torque of 200 % of full-load torque. Neglect stator resistance and
rotational losses and assume constant rotor resistance. Determine:
a) the slip at full load,
[0.133]
b) the slip at maximum torque
[0.5]
c) the rotor current at starting, in per unit of full-load current.
[I2st=3.468I2fl]
8) A 2000 HP, 2300V, 3-ph, Y-connected, 4-pole, 50 Hz induction motor has the
following specifications referred to the stator:
R1=0.02 Ω; R2=0.12 Ω; Rc= 451.2 Ω; Xφ=50 Ω; X1=X2=0.32 Ω
Find the full-load efficiency of the motor if the full-load slip = 0.03476.
[94.95 %]
9) A 3-ph, Y-connected, 220 V(line-to-line), 50 Hz, 6-pole induction motor has the
following constants in Ω/phase referred to the stator:
R1=0.294; R2=0.144; X1=0.42; X2=0.209; Xφ=11.04
The total friction, windage and core losses may be assumed to be constant at 403W,
independent of load.
For a slip of 2%, compute the speed, output torque and power, stator current, power
factor, and efficiency when the motor is operated at rated voltage and frequency.
Neglect the impedance of the source.
[980 rpm, 50.97 Nm, 5.23 kW, 19.74 A, 0.81 (lag), 85.8 %]
10) For the motor of Problem 8, determine:
a) the load component I2 of the stator current, the internal torque T and the internal
power P for a slip s=0.03; [23.9 A, 78.79 Nm, 8 kW]
b) the maximum internal torque and the corresponding speed; [148.35 Nm, 311 rpm]
c) the internal starting torque and the corresponding value of I2.
[113.18 Nm, 163.58 A]
11) A 3000V, 24-pole, 50 Hz, 3-ph, Y-connected induction motor has a slip-ring rotor
of resistance 0.016 Ω and stand-still reactance of 0.265 Ω per phase. Full-load torque
is obtained at a speed of 247 rpm. Neglecting stator impedance, calculate the ratio of
maximum to full-load torque.
[2.6]
12) An 8-pole, 50 Hz, 3 phase induction motor has an equivalent rotor resistance of
0.07 Ω/phase. If its stalling speed is 630 rpm, how much resistance must be included
per phase to obtain maximum torque at starting ? Ignore magnetizing current. [0.37 Ω]
13) A 230 V, 3-ph, Y-connected, 50 Hz, 4-pole squirrel cage induction motor
develops full-load internal torque at a slip of 0.04 when operated at rated voltage and
frequency. For the purpose of this problem, rotational and core losses can be
neglected. Impedance data on the motor in Ω/phase are as follows:
R1=0.36; X1=X2=0.39; Xφ=12.92
Determine the maximum internal torque at rated voltage and frequency, the slip at
maximum torque, and the internal starting torque at rated voltage and frequency.
Express the torque in N-m.
14) The rotor resistance and stand-still reactance of a 3-ph induction motor are
respectively 0.015 Ω and 0.09 Ω per phase. At normal voltage, the full-load slip is 3%.
Estimate the percentage reduction in stator voltage to develop full-load torque at onehalf of full-load speed. What is the power factor ?
[22.5 %, 0.31 (lag)]
15) When operated at rated voltage and frequency, a 3-ph squirrel cage induction
motor (of the design classification known as high-slip motor) delivers full load at a
slip of 8.5% and develops a maximum torque of 250% of full load torque at a slip of
50%. Neglect core and rotational losses, and assume that the resistances and
inductances of the motor are constant. Determine the torque and rotor current at
starting with rated voltage and frequency. Express the torque and rotor current in per
unit based on their full load values.
16) A 500 HP wound rotor induction motor, with its slip rings short-circuited has the
following properties:
Full load slip = 1.5 %
Rotor I2R at full load torque = 5.69 kW
Slip at maximum torque = 6 %
Rotor current at maximum torque=2.82I2fl where I2fl is the full load rotor current.
Torque at 20% slip = 3.95Tfl
If the rotor-circuit resistance is increased to 5Rrotor by connecting non-inductive
resistance in series with each rotor slip ring, determine:
a) the slip at which the motor will develop the same full-load torque,
b) the total rotor circuit I2R loss at full-load torque
c) the horse power output at full-load torque,
d) the slip at maximum torque,
e) the rotor current at maximum torque
f) the starting torque, and
g) the rotor current at starting. Express the torques and rotor currents in per unit
based on the full-load torque values.
EE11001 Tutorial 6b (DC MACHINES)
1. A six-pole DC machine has an armature connected in 6 parallel paths (lap winding). The armature has 48
slots with four conductors per slot. The armature is rotated at 600 rpm and the flux per pole is 30 mWb.
Calculate:
a. The rotational speed of the armature relative to the magnetic field.
b. The induced voltage
c. The coefficient Ka' that relates induced voltage with rotational speed and flux
[Result: part b, Ea = 57.6 V]
2. A 50. kW, 240 V DC shunt generator has an armature resistance of 0.10 Ω, a field circuit resistance of 120
Ω, and a total brush voltage drop of 2V. The generator delivers rated current at rated speed and rated voltage.
Calculate the following:
a) Load current
b) Field current
c) Armature current
d) Armature induced voltage
[Result: part c, Ia = 210.3 A]
3. A 220 V DC shunt motor. has armature and field winding resistances of 0.15 Ω and 110 Ω, respectively. The
motor draws a line current of 5 A while running on no load. When driving a load, the motor runs at 1100 rpm
and draws 48 A of line current. Calculate the no-load speed.
[Result: Nnl = 1133.3 rpm]
4. . A DC series motor is rated 230 V, 12 hp, and 1200 rpm. It is connected to a 230 V supply and it draws a
current of 40 A while rotating at 1200 rpm. The armature and series field winding resistances are 0.25 Ω and
0.1 Ω, respectively.
a) Determine the power and torque developed by the motor
b) Determine the speed, torque, and power if the motor draws 20 amperes
[Result: part b, Tem = 17.2 N-m]
5. A 240 V, DC shunt motor has an armature winding resistance of 0.2 Ω. The full-load armature current is 50
A. Find the value of the resistance to be connected in series with the armature circuit so that the speed is 75%
of the rated speed if full-load current flows.
6. A 220 V shunt motor delivers20 hp to a load connected to its shaft on full load at 1150 rpm. The motor fullload efficiency is 85%. The armature and field winding resistances are 0.15 Ω and 110 Ω, respectively.
Determine:
a) The starting resistance such that the starting line current does not exceed twice the full-load current
b) The starting torque with the starting resistance computed in part (a) inserted in the armature
circuit.
[Result: part b, Tem = 272.6 N-m]
7. A 12 kW, 100 V, 1000 rpm DC shunt machine having field and armature winding resistances of 80 Ω and
0.1 Ω respectively is connected to a 100 V DC supply and is operated as a motor. The brush resistance is
negligible. The motor draws 6 A while running under no load. Resistance of the brush is negligible.
(a) Find the speed, electromagnetic torque and efficiency of the motor when rated current flows in the
armature if the air gap flux remains the same as that at no load.
(b) Find the starting torque if the starting armature current is limited to 150% of its rated value. Neglect
armature reaction.
8. (a) The speed of a 500V, 1500 rpm DC shunt motor is varied from 0 to 1500 rpm (base speed) by varying the
terminal voltage from 0 to 500 V with the field current maintained constant. Determine the motor armature
current if the torque is held constant at 300 N-m upto the base speed. Assume that the terminal voltage (Vt) is
equal to the back emf (Ea).
(b) Now the same motor is run at 3000 rpm by field control keeping the terminal voltage fixed at 500 V and
the armature current constant at the value obtained in part (a). Determine the available torque. Neglect all
losses and assume that the terminal voltage (Vt) is equal to the back emf (Ea).
9. At standstill, a dc series motor draws 5 A and develops a torque of 5 N-m when connected to a 5 V DC
supply. The motor is mechanically coupled to a load. It draws 10 A when connected to a 120 V DC supply
and drives the load at 300 rpm. Neglecting saturation determine the torque developed by the motor and the
value of the external resistance required to be connected in series with the motor.
10. A 240 V DC motor with full load current of 110 A has magnetization curve as shown in the Fig.Q13). The
motor can be connected cumulatively compounded, where part of the armature with Nse = 14 turns/pole adds
to the magnetic field produced by the field winding which has NF = 2700 turns/pole. The armature and field
winding resistances are 0.19 Ω and 240 Ω, respectively. Determine: a) The no-load speed; b) The full load
speed; c) Speed regulation.
[Result: part b, Nfl = 907.4 rpm]
11. Additional Problems: Parker Smith 9th edition: Page: 257, Problems 62, 63, 64, 66, 67, 78, 79, 83, 84