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## 2re_5cap

### Hierarchy of files Downloads Files created online (20706) HP-Prime (2086) mViewer GX Creator App (1676)

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Auteur Author: pauloreto
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5-6. If a 60-Hz synchronous motor is to be operated at 50 Hz, will its synchronous reactance be the same as at
60 Hz, or will it change? (Hint: Think about the derivation of X S .)

SOLUTION The synchronous reactance represents the effects of the armature reaction voltage Estat and the
armature self-inductance. The armature reaction voltage is caused by the armature magnetic field B S ,
and the amount of voltage is directly proportional to the speed with which the magnetic field sweeps over
the stator surface. The higher the frequency, the faster B S sweeps over the stator, and the higher the
armature reaction voltage Estat is. Therefore, the armature reaction voltage is directly proportional to
frequency. Similarly, the reactance of the armature self-inductance is directly proportional to frequency,
so the total synchronous reactance X S is directly proportional to frequency. If the frequency is changed
from 60 Hz to 50 Hz, the synchronous reactance will be decreased by a factor of 5/6.
5-7. A 208-V Y-connected synchronous motor is drawing 50 A at unity power factor from a 208-V power
system. The field current flowing under these conditions is 2.7 A. Its synchronous reactance is 1.6 .
Assume a linear open-circuit characteristic.
(a) Find V and E A for these conditions.

(b) Find the torque angle  .
(c) What is the static stability power limit under these conditions?
(d) How much field current would be required to make the motor operate at 0.80 PF leading?
(e) What is the new torque angle in part (d)?
SOLUTION
(a) The phase voltage of this motor is V = 120 V, and the armature current is I A  500 A .
Therefore, the internal generated voltage is
E A  V  RAI A  jX S I A
E A  1200 V  j 1.6   500 A 
E A  144  33.7 V
133
(b) The torque angle  of this machine is –33.7.
(c) The static stability power limit is given by
3V E A 3 120 V 144 V 
Pmax    32.4 kW
XS 1.6  
(d) A phasor diagram of the motor operating at a power factor of 0.78 leading is shown below.
P
} I A2
I A1 V

E A1
jX I
S A

E A2
} P

Since the power supplied by the motor is constant, the quantity I A cos  , which is directly proportional
to power, must be constant. Therefore,
I A2  0.8    50 A 1.00 
I A 2  62.536.87 A
The internal generated voltage required to produce this current would be
E A2  V  RAI A2  jX S I A2
E A 2  1200 V  j 1.6   62.5036.87 A 
E A 2  197  23.9 V

The internal generated voltage E A is directly proportional to the field flux, and we have assumed in this
problem that the flux is directly proportional to the field current. Therefore, the required field current is
E A2 197 V
IF 2  I F1   2.7 A   3.70 A
E A1 144 V

(e) The new torque angle  of this machine is –23.9.
5-8. A 4.12 kV, 60 Hz, 3000-hp 0.8-PF-leading, Δ-connected, three-phase synchronous motor has a
synchronous reactance of 1.1 per unit and an armature resistance of 0.1 per unit. If this motor is running
at rated voltage with a line current of 300 A at 0.85 PF leading, what is the internal generated voltage per
phase inside this motor? What is the torque angle δ?
SOLUTION
The output power of the motor is  3000 hp  746 W/hp   2238 kW . If we take this as rated power,
the ratings of this machine are
S base  P / PF   2238 kW  /  0.8   2798 kVA

VL ,base  4120 V

V ,base  4120 V

134
S base 2798 kVA
I L ,base    392 A
3VL ,base 3  4120 V 

I L ,base 392 A
I ,base    226 A
3 3
Therefore, the line current of 300 A in per-unit is
IL 300 A
I L ,pu    0.765 pu
I L ,base 392 A
and the final per-unit current is
I pu  0.76531.8 pu
The internal generated voltage in per-unit is
E A  V  RAI A  jX S I A
E A  10   0.1 0.76531.8   j 1.1 0.76531.8 
E A  1572  28.7 pu
In volts, the internal generated voltage is
E A  1.572  28.7 pu  4120 V   6477  28.7 V

And the torque angle δ is -28.7.
5-9. Figure P5-2 shows a synchronous motor phasor diagram for a motor operating at a leading power factor
with no RA . For this motor, the torque angle is given by

X S I A cos 
tan  
V  X S I A sin 

 X S I A cos  
  tan -1  
 V  X S I A sin  
Derive an equation for the torque angle of the synchronous motor if the armature resistance is included.
SOLUTION The phasor diagram with the armature resistance considered is shown below.

135
I
A
X I sin 
S A
V

}

}

X I cos 
jX I 
S A
S A

R I
 A A

}
EA
R I cos 
A A

Therefore,
X S I A cos   RA I A sin 
tan  
V  X S I A sin   RA I A cos 

 X S I A cos   RA I A sin  
  tan 1  
 V  X S I A sin   RA I A cos 

5-10. A synchronous machine has a synchronous reactance of 1.0  per phase and an armature resistance of 0.1
 per phase. If E A = 460-10 V and V = 4800 V, is this machine a motor or a generator? How
much power P is this machine consuming from or supplying to the electrical system? How much reactive
power Q is this machine consuming from or supplying to the electrical system?
SOLUTION This machine is a motor, consuming power from the power system, because E A is lagging V .
It is also consuming reactive power, because E A cos   V . The current flowing in this machine is

V  E A 4800 V  460  10 V
IA    83.9  13 A
RA  jX S 0.1  j1.0 
Therefore the real power consumed by this motor is
P  3V I A cos  3...

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