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Tutorial Questions for Electrical Machine (EET306)
LECTURER
1. MUZAMIR ISA
2. IR SYAFRUDIN HASSAN
PLV
1. ROSNAZRI ALI
2. MUHD HATTA HUSSEIN
3. SALSABILA AHMAD
Problem 1
The power input to a 380 volt, 50 Hz, 6 pole, 3-phase induction motor running at 735 rpm
is 50 kW at 0,85 power factor lag. Stator losses are 1,2 kW, the friction and windage
losses are 2,5 kW. Calculate :
a). the slip
b). the rotor copper loss
c). the brake hp
d) efficiency
e) input current
f) draw it’s power flow
Solution
f = 50 Hz, P = 8, nr = 735 rpm
Pin = 50 kW, Pcus = 1.2 kW, Pag = 2.5 kW, pf = cos φ = 0.85
a) Synchronous speed
ns = 120 f / P = 120x50/8 = 750 rpm
slip
s
n s  nr
ns
s=
750  735
= 0.02 or 2 %
750
b) Rotor input = air-gap power
PAG = Pin – Pcus
= 50 – 1.2 = 48.8 kW
PAG : Pdev : Pcur = 1 : (1-s) : s
or
PAG : Pcur = 1 : s
Pcur = s PAG = 0.02 x 48.8 = 0.976 kW
c)
Pdev = PAG - Pcur = 48.8 – 0.976 = 47.824 kW
Pout = Pdev – Pag = PAG - Pcur – Pag = 47.824 – 2.5 = 45.324 kW
BHP = Pout (kW) / 0.746 = 45.324 / 0.746 = 60.75 hp ~ 61 hp
d) Efficiency

Pout
= 45.324 / 50 = 0.91 or 91%
Pin
e). Pin =
3 .V.I cos φ
I
Pin
3V cos 
=
50000
3 x380 x 0.85
= 89.4 A
f) Power flow
Pin
Pdev
PAG
50 kW
1.2 kW
48.2 kW
0.976 kW
47.824 kW
Pout
2.5 kW
45.324 kW
Problem 2
A 25-hp, 4-pole, 50 Hz, 3-phase induction motor has friction and windage losses of 2,5 %
of the output. S The full-load slip is 4 %. Compute for full-load :
a. the rotor copper loss
b. the rotor input
c. the output torque
d induction torque
e. input power
f. efficiency
g. draw it’s power flow
Solution
Pout = 35 hp = 35 x 746 = 26,110 W
Pag = 2.5 % Pout = 0.025 x 26110 = 652.75 W
s = 4 % = 0.04
a)
Pdev = Pout + Pag = 26110 + 652.75 = 26762.75 W
PAG : Pdev : Pcur = 1 : (1-s) : s
or
Pdev : Pcur = (1-s) : s
Pcur 
s
0.04
Pdev =
x 26762.75 = 1115.1 W
1 s
1  0.04
b) Rotor input = PAG = Pcur / s = 1115.1 / 0.04 = 27877.85 W
Remember PAG : Pcur = 1 : s
c) Synchronous speed
ns = 120 f / P = 120x50/4 = 1500 rpm = 25 rps
Rotor speed
nr = (1 –s) ns = (1 – 0.04) x 1500 = 1440 rpm = 24 rps
Output torque = load torque
TL 
Pout (W )
26110
=
= 173.24 N-m
2 .nr (rps ) 2x 24
d) Induction torque
Tind 
Pdev (W )
26762.75
=
= 177.6 N-m
2 (24)
2 .nr (rps )
e) Input power = PAG + Pstator losses = 27877.85 + 650 = 28527.85 W
f) Efficiency

Pout
= 26110 / 28527.85 = 0.915 or 91.5 %
Pin
g) Power flow
Pin
28527.85 W
Pdev
PAG
650 W
27877.85 W
1115.1 W
26762.75 W
Pout
652.75 W
26110 W
Problem 3
The power input to the rotor of a 440 V, 50 Hz, 6-pole, 3 phase induction motor is 22
kW, The slip is 4 %, calculate :
a). the frequency of rotor currents
b). rotor speed
c). rotor copper losses
d). mechanical power developed
e). the rotor resistance per phase if rotor current is 65 A
Solution
V = 400 V, f = 50 Hz, P = 6, s = 4 % = 0.04
P input rotor = PAG = 22 kW
a) Frequency of rotor current
fr = s f = 0.04 x 50 = 2 Hz
b) Synchronous speed
ns = 120 f / P = 120x50/6 = 1000 rpm
Rotor speed
nr = (1 –s) ns = (1 – 0.04) x 1000 = 960 rpm
c) Rotor cu loss
Pcur = s PAG
Pcur = 0.04 x 20 = 0.8 kW = 800 W
d) Pdev = (1-s) PAG = (1 – 0.04) x 20 = 19.2 kW = 19200 W
or Pdev = PAG – Pcur = 20 – 0.8 = 19.2 kW = 19200 W
e) Let R2 be the rotor resistance per phase, then
3 I22.R2 = Pcur = 800
3 (65)2 R2 = 800
R2 = 800 / (3 x 652) = 0.63 Ω
Problem 4
A 460 V, 25 hp, 60 Hz, four pole, Y-connected induction motor has the following
impedances in ohms per phase referred to the stator circuit:
R1 = 0.641 Ω
R2 = 0.332 Ω
X1 = 1.106 Ω
X2 = 0.464 Ω
XM = 26.3 Ω
The total rational losses are 1100 W and are assumed to be constant. The core loss is
lumped in with the rotational losses. For a rotor slip of 2.2 percent at the rated voltage
and rated frequency, find the motor’s
a. Speed
b. Stator current
c.
d.
e.
f.
Power factor
Pconv and Pout
Tind and Tload
Efficiency
Solution
The per-phase equivalent circuit of this motor is shown in Figure 4a and the power-flow
diagram is shown in Figure 4b.
Figure 4a
Figure 4b
Since the core losses are lumped together with the friction and windage losses and the
stray losses, they will be treated like the mechanical losses and be substracted after P conv
in the power-flow diagram.
a). The synchronous speed is
ns = 120f/p = 120x60/4 = 1800 rpm
or
ωs = 1800 x 2  / 60 = 188.5 rad/s
The rotor’s mechanical shaft speed is
nr = (1 – s) ns
= (1 – 0.022) x 1800 = 1760 rpm
or
ωr = 1760 x 2  / 60 = 184.4 rad/s
b). To find the stator current, get the equivalent impedance of the circuit. The first step is
to combine the referred rotor impedance in parallel with the magnetization branch,
and then to add the stator impedance to that combination in series.
The referred rotor impedance is
Z2 = (R2/s) + jX2
= (0.332/0.022) + j0.464
= 15.09 + j0.464 = 15.10  1.76o Ω
The combined magnetization plus rotor impedance is given by
Zf 
1
1 / jX M  1 / Z 2
Zf 
1
 j 0.038  0.0662  1.76 o
Zf 
1
 12.9431.1o
o
0.0773  31.1
Therefore, the total impedance is
Ztot = Zstat + Zf
= 0.641 + j1.106 + 12.94  31.1o
= 11.72 + j7.79 = 14.07  33.6o
The resulting stator current is
I1 = Vp / Ztot
=
2660 o
 18.88  33.6 o A
o
14.0733.6
c). The motor power factor is
PF = cos 33.6o = 0.833 lagging
d). The input power to this motor is
Pin  3VT I L cos
Pin =
3 (460V)(18.88A)(0.833) = 12,530 W
The stator copper losses in this machine is
PSCL = 3 I12.R1
= 3 (18.88A)2(0.641 Ω) = 685 W
The air-gap power is given by
PAG = Pin – PSCL
= 12,530 W – 685 W = 11,845 W
Therefore, the power converted is
Pconv = (1 – s) PAG
= (1 – 0.022) (11,845W) 11,585 W
The power Pout is given by
Pout = Pconv – Prot
= 11,585 W – 1100 W = 10,585 W
or, in horse power
Pout = (10,585 W) x 1hp/746 W = 14.1 hp
Problem 5
From the following test results, determine the voltage regulation of a 2000 V, singlephase synchronous generator delivering a current of 100 A at (i) unity pf, (ii) 0,8 leading
pf and (iii) 0,8 lagging pf. The result :
Full-load current 100 A is produced on short circuit by a field excitation 2,5 A.
An emf 500 V is produced on open-circuit by the same excitation.
Armature resistance is 0,8 Ω, and draw it phasor diagram.
Solution
Single-phase generator
V = 2200 V, Ra = 0.5 Ω , I = 100 A
E
500 V
ISC
I = 100 A
If
3A
Synchronous impedance
Zs 
E1 (occ)
= 500 / 100 = 5 Ω
I1 ( scc )
Synchronous reactance
Zs2 = Ra2 + Xs2
or
X s  Z s  Ra
2
2
=
5 2  (0.5) 2  24.75 = 4.97 Ω
Diagram phasor
IXs
E
V
IRa
V sin φ
φ
V cos φ
I
General form of voltage equation is

E  (V cos   IRa ) 2  (V sin   IX s ) 2
positive sign for inductive load
negative sign for capacitive load

1/ 2
i) Unity power factor
cos φ = 1 ,
sin φ = 0
I Ra = 100 x 0.5 = 50 V
I Xs = 100 x 4.97 = 497 V

E  (V cos   IRa ) 2  ( IX s ) 2

1/ 2
E = [(2200 + 50)2 + 4972]1/2
5309509 = 2304 V
E=
%VR 
E V
2304  2200
x100% =
x100%
V
2200
%VR = 4.72 %
ii) Pf 0.8 lead
cos φ = 0.8 ,
φ = 36.87o , sin φ = 0.6

E  (V cos   IRa ) 2  (V sin   IX s ) 2

1/ 2
E = [(2200x0.8 + 50)2 + (2200x0.6 – 497)2]1/2
E = [(1760 + 50)2 + (1320 – 497)2]1/2
E = [(1810)2 + (823)2]1/2 = (3953429)1/2 = 1988 V
%VR 
E V
1988  2200
x100% =
x100%
V
2200
%VR = - 9.7 %
iii) Pf 0.8 lag
cos φ = 0.8,
sin φ = 0.6

E  (V cos   IRa ) 2  (V sin   IX s ) 2

1/ 2
E = [(2200x0.8 + 50)2 + (2200x0.6 + 497)2]1/2
E = [(1760 + 50)2 + (1320 + 497)2]1/2
E = [(1810)2 + (1817)2]1/2 = (6577589)1/2 = 2565 V
%VR 
E V
2565  2200
x100% =
x100%
V
2200
%VR = 16.6 %
Problem 6
A 3-phase, 800 kVA, 11 kV, star connected synchronous generator has resistance of 1,5
Ω /phase and synchronizing reactance of 25 Ω/phase. Find the percentage regulation for
a load of 600kW at 0,8 leading power factor.
Solution
3 phase generator
S = 800 kVA = 800,000 VA
VLL = 11 kV = 11,000 V
Ra = 1.5 Ω, Xs = 25 Ω
Voltage/phase = 11000 /
3 x VLL x I
S=
or
3 = 6350 V
S
I
3V LL
=
800000
3 x11000
= 42 A
I Ra = 42 x 1.5 = 63 V
I Xs = 42 x 25 = 1050 V
Pf 0.8 lead
cos φ = 0.8 ,
φ = 36.87o ,
sin φ = 0.6

E  (V cos   IRa ) 2  (V sin   IX s ) 2

1/ 2
E = [(6350x0.8 + 63)2 + (6350x0.6 – 1050)2]1/2
E = [(5080 + 63)2 + (3810 – 1050)2]1/2
E = [(5143)2 + (2760)2]1/2 = (34068049)1/2 = 5837 V
%VR 
E V
5837  6350
x100% =
x100%
V
6350
%VR = - 8 %
THEORY – REVIEW QUESTIONS
1. What types of construction are used in induction motors (IMs)? What are the relative
advantages of each construction and where is each type operated?
Ans
There are two types of IM construction. They have the same kind of stator but different
rotor construction. These are elaborated below:
Squirrel cage, (SC) construction: conducting bars usually of aluminium are placed in
rotor slots. These are shorted at both ends by conducting rings. In small and medium size
cage with rings is die cast in rotor slots. Rings also have projecting fins which provide
fan action.
Slip-ring (SR) construction: the rotor is wound (3-phase) and three connections are
brought out through slipring (copper) and copper-carbon brushes. The rotor winding is
short circuited externally.
Advantages and use:
Squirrel Cage Induction Motor (SCIM)
(i) Rotor is permanently shorted and has low rotor resistance
(ii) Low starting torque and high running efficiency. Reduced voltage start required.
(iii) Low cost sturdy motor
(iv) Most commonly used motor except where starting on load is needed.
Slip-ring Induction Motor (SRIM) or Wound Rotor Induction Motor (WRIM)
(i) Resistance can be included in rotor circuit at start.
(ii) High starting torque, full voltage start, low starting current
(iii) High cost-copper winding for rotor + slip-ring brushes
(iv) Used only where full-load start is a must.
2. Why do IMs run at low power factor when lightly loaded?
Ans
At no load an IM draws large magnetizing current (30-40% of full load current) to create
flux in the air gap. This current has very low pf of the order 0.2 lag. The power is mainly
drawn to provide for core loss.
At the motor is loaded, it draws a load component of current with much higher pf.
Therefore, the pf on the net current drawn from mains increases with load. It rises to 0.80.85 at full load. At light load the pf will be slightly more than at no load. This is
illustrated by the phasor diagram of Fig. 1.
V1
IT
I2'
Io
Fig. 1
3. Describe briefly the methods of controlling speed of an IM
Ans.
Synchronous speed of an IM is given by
120 f
ns 
P
Various methods of controlling IM speed are briefly describe below:
(i) Change the stator poles---stator phase windings are reconnected to create
consequent poles. So the number of motor poles become 2 P, i.e speed is
halved. This method is used for two-speed operation.
(ii) Reducing stator voltage—The motor runs at lower speed at a given load torque.
But the torque developed decreases as square of voltage. Suitable only for small
motors with fan type blowers, slip ring ceiling fans, etc.
(iii) For slip-ring motors speed for given torque is reduced by adding resistance in
rotor circuit. Because of power loss in rotor resistance (external), the motor
efficiency is considerably reduced. The method can be used only for speed
reduction for short time period.
(iv) V/f control—As motor mains frequency is reduced or increased keeping V/f
constant, the motor speed is controlled accordingly proportional to frequency.
At the same time air-gap flux and and torque (max) is maintained at constant
value.
4. Explain why a single-phase single-winding IM produces no starting torque
Ans.
The single winding of stator of a single phase IM is distributed in space so that space
fundamental of mmf is the most dominant component of the actual mmf distribution.
When the winding carries a sinusoidal current, it produces a sinusoidally spacedistributed mmf whose peak value pulsates with time. The mmf at any time is
F = Fpeak cos θ
where θ is the angle measured from the winding axis
Now Fpeak = Fmax cos ωt
so that the mmf has both space and time distribution expressed as
F = Fmax cos θ cos ωt
F = ½ Fmax cos (θ- ωt) + ½ Fmax cos (θ+ωt)
This equation tell us that pulsating single-phase field can be considered as superposition
of two rotating field rotates at synchronous speed in opposite directions (Fig.2), i.e
the forward rotating field, Ff = ½ Fmax cos (θ- ωt)
the backward rotating field, Fb = ½ Fmax cos (θ+ωt)
ns
Ff=½ Famx
ns
Fb=½ Famx


Fig.2
5. What design features are incorporated in a split-phase motor to make it self-starting?
Ans.
The split-phase motor is provided with windings, main winding and auxiliary winding.
These two windings are excited from the same voltage (supply being single phase). The
current in the two windings can made out-of-phase by adjustment of the impedance of
auxiliary winding in relation to the main winding. As a result the mmf of main winding
Fm and mmf of auxiliary winding Fa constitute an unbalance field set with 90o
elect.space-phase relationship. The two symmetrical component now being unequal
Ff  Fb, the forward rotating field is made stronger than the backward rotating field,
which result in the net production of starting torque. Thus the two winding, with 90o
elct.phase difference, make the split phase motor self starting.
6. What are the advantage of a capacitor run over a capacitor start motor ?
Ans.
The advantage of a capacitor run over a capacitor start motor are:
(i) It has a better power factor
(ii) It has more efficiency
(iii) It has quieter and smoother operation
(iv) It used for both easy and hard to start loads