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Catégorie :Category: mViewer GX Creator App HP-Prime
Auteur Author: pauloreto
Type : Application
Page(s) : 9
Taille Size: 580.40 Ko KB
Mis en ligne Uploaded: 17/10/2020 - 22:53:15
Uploadeur Uploader: pauloreto (Profil)
Téléchargements Downloads: 1
Visibilité Visibility: Archive publique
Shortlink : http://ti-pla.net/a2648548

Description 

(b) The input power is
POUT 298 kW
PIN    331 kW
 0.90
(c) The mechanical speed is
nm  900 r/min
(d) The armature current is
PIN 331 kW
I A  IL    97.8 A
3 VT PF 3  2300 V  0.85 

I A  97.831.8 A

The phase voltage is V   2300 V  / 30  13280 V . Therefore, E A is

E A  V  RAI A  jX S I A
E A  13280 V    0.8   97.831.8 A   j 11   97.831.8 A 
E A  2063  27.6 V
(e) The magnitude of the armature current is 97.8 A.
(f) The power converted from electrical to mechanical form is given by the equation Pconv  PIN  PCU

PCU  3I A 2 RA  3  97.8 A   0.8    23 kW
2



Pconv  PIN  PCU  331 kW  23 kW  308 kW
(g) The mechanical, core, and stray losses are given by the equation
Pmech  Pcore  Pstray  Pconv  POUT  308 kW  298 kW  10 kW

5-15. The Y-connected synchronous motor whose nameplate is shown in Figure 5-21 has a per-unit
synchronous reactance of 0.70 and a per-unit resistance of 0.02.
(a) What is the rated input power of this motor?
(b) What is the magnitude of E A at rated conditions?
(c) If the input power of this motor is 12 MW, what is the maximum reactive power the motor can
simultaneously supply? Is it the armature current or the field current that limits the reactive power
output?
(d) How much power does the field circuit consume at the rated conditions?
(e) What is the efficiency of this motor at full load?
(f) What is the output torque of the motor at the rated conditions? Express the answer both in
newton-meters and in pound-feet.




143
SOLUTION The base quantities for this motor are:
VT ,base  6600 V
6600 V
V ,base   3811 V
3
I A,base  I L ,base  1404 A
S base  Prated  3 VT I L PF  3  6600 V 1404 A 1.0  16.05 MW

(a) The rated input power of this motor is
PIN  3 VT I L PF  3  6600 V 1404 A 1.0  16.05 MW

(b) At rated conditions, V  1.00 pu and I  1.00 pu , so E A is given in per-unit quantities as

E A  V  RAI A  jX S I A
E A  10    0.02 1.00   j  0.70 10 
E A  1.20  35.5 pu
The base phase voltage of this motor is V ,base  3811 V , so E A is
E A  1.20  35.5  3811 V   4573  35.5 V

(c) From the capability diagram, we know that there are two possible constraints on the maximum
reactive power—the maximum stator current and the maximum rotor current. We will have to check each
one separately, and limit the reactive power to the lesser of the two limits.
The stator apparent power limit defines a maximum safe stator current. This limit is the same as the rated
input power for this motor, since the motor is rated at unity power factor. Therefore, the stator apparent
power limit is 16.05 MVA. If the input power is 12 MW, then the maximum reactive power load that still
protects the stator current is

16.05 MVA   12 MW   10.7 MVAR
2 2
Q  S 2  P2 

Now we must determine the rotor current limit. The per-unit power supplied to the motor is 12 MW /
16.05 MW = 0.748. The maximum E A is 4573 V or 1.20 pu, so with E A set to maximum and the motor
consuming 12 MW, the torque angle (ignoring armature resistance) is

144
 XSP    0.70  0.748  
  sin 1   sin 1    8.4
 3V E   3 1.0 1.20  
  A
(The negative sign on  comes from the fact that E A lags V in a motor.)

At rated voltage and 12 MW of power supplied, the armature current will be
V  E A 10  1.20  8.4
IA    0.36646.9 pu
RA  jX S j 0.70
In actual amps, this current is
I A  1404 A  0.39646.9   55646.9 A

The reactive power supplied at the conditions of maximum E A and 12 MW power is

Q  3V I A sin   3  3811 V  556 A  sin  46.9   4.6 MVAR

Therefore, the field current limit occurs before the stator current limit for these conditions, and the
maximum reactive power that the motor can supply is 4.6 MVAR under these conditions.
(d) At rated conditions, the field circuit consumes
Pfield  VF I F  125 V  5.2 A   650 W

(e) The efficiency of this motor at full load is


POUT
 100% 
 21000 hp 746 W/hp  100%  97.6%
PIN 16.05 MW
(f) The output torque in SI and English units is

 load 
POUT

 21000 hp 746 W/hp  124,700 N  m
m 1 min   2 rad 
1200 r/min     
60 s 1r

5252 P 5252  21000 hp 
 load    91,910 lb  ft
nm 1200 r/min 
5-16. A 480-V, 500-kVA, 0.8-PF-lagging, Y-connected synchronous generator has a synchronous reactance of
0.4  and a negligible armature resistance. This generator is supplying power to a 480-V, 80-kW, 0.8-
PF-leading, Y-connected synchronous motor with a synchronous reactance of 2.0  and a negligible
armature resistance. The synchronous generator is adjusted to have a terminal voltage of 480 V when the
motor is drawing the rated power at unity power factor.
(a) Calculate the magnitudes and angles of E A for both machines.
(b) If the flux of the motor is increased by 10 percent, what happens to the terminal voltage of the
power system? What is its new value?
(c) What is the power factor of the motor after the increase in motor flux?
SOLUTION
(a) The motor is Y-connected, and is operating at rated voltage (480 V), rated power (80 kW), and unity
power factor. The current flowing in the motor is



145
P 80 kW
I A,m  I L ,m    96.2 A
3 VT PF 3  480 V 1.0

so I A,m  96.20 A . This machine is Y-connected, so the phase voltage is V = 480 / 3 = 277 V.
The internal generated voltage of the motor is
E A,m  V ,m  jX S I A,m
E A,m  2770 V  j  2.0   96.20 A 
E A,m  337  34.8 V

The current supplied to the motor comes from the generator, so the internal generated voltage of the
generator is
E A,g  V ,g  jX S I A,g
E A,g  2770 V  j  0.4   96.20 A 
E A,g  2807.9 V

IA,g IA,m
j0.4 ? j2.0 ?


+ +


+ + EA,m
EA,g - V?,g V?,m -



- -
E A,g

I jX I I V
A A
S,g A

V jX I
S,m A

E A,m


Generator Motor
(b) The power supplied by the generator to the motor will be constant as the field current of the motor ...

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3.06 Ko KB readme.txt
3.65 Ko KB lisezmoi.txt
1.08 Ko KB 3re_5cap.hpprgm
577.31 Ko KB 3re_5cap.hpappdir.zip
95 octets bytes appslist.txt

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