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Unformatted text preview: 52. A 2300—V 1000kVA 0.8—PFlagging 60Hz two~pole Y—connected synchronous generator has a
synchronous reactance of 1.1 Q and an armature resistance of 0.15 9. At 60 Hz, its friction and windage
losses are 24 kW, and its core losses are 18 kW. The ﬁeld circuit has a dc voltage of 200 V, and the
maximum I F is 10 A. The resistance of the ﬁeld circuit is adjustable over the range from 20 to 200 Q. The OCC of this generator is shown in Figure P5—1. (a) How much ﬁeld current is required to make VT equal to 2300 V when the generator is running at no
load? (27) What is the internal generated voltage of this machine at rated conditions? (c) How much ﬁeld current is required to make VT equal to 2300 V when the generator is running at rated
conditions? (d) How much power and torque must the generator’s prime mover be capable of supplying? SOLUTION (a) If the no—load terminal voltage is 2300 V, the required ﬁeld current can be read directly from the
opencircuit characteristic. It is 4.25 A. (b) This generator is Y—connected, so I L = I A. At rated conditions, the line and phase current in this
generator is P 1000 kVA I :I = :———
A L ﬁvL 6(2300v) = 251 A at an angle of —36.87° The phase voltage of this machine is V¢ z VT / J3 = 1328 V. The internal generated voltage of the machine
1s EA =V¢+RAIA+jXSIA EA 21328400 + (0.15 Q)(251£— 36.87° A)+ j(1.1 Q)(251£ — 36.87° A) EA :153717.4° V (c) The equivalent opencircuit terminal voltage corresponding to an E A of 1537 volts is V = 73(1527 v) = 2662 V T,oc From the OCC, the required ﬁeld current is 5.9 A.
(d) The input power to this generator is equal to the output power plus losses. The rated output power is POUT = (1000 kVA)(O.8) = 800 kW PCU = 31131;, = 3(251 A)2(0.15 o) = 28.4 kW
PMW = 24 kW P =18kW core P = (assumed 0) stray PIN = POUT +PCU +13%W +P +Pstray =870.4 kW core Therefore the prime mover must be capable of supplying 175 kW. Since the generator is a twopole 60 HZ
machine, to must be turning at 3600 r/min. The required torque is P 175.2 kW TAPP = #1“  2 d
60,. (3600r/min)(1613m)( ”1 :a j
S =465N‘m Assume that the ﬁeld current of the generator in Problem 5—2 is adjusted to achieve rated voltage (2300 V)
at full load conditions in each of the questions below. (a) What is the efﬁciency of the generator at rated load? (b) What is the voltage regulation of the generator if it is loaded to rated kilovoltamperes with 0.8—PF—
lagging loads? (c) What is the voltage regulation of the generator if it is loaded to rated kilovoltamperes with 0.8—PF
leading loads? I (d) What is the voltage regulation of the generator if it is loaded to rated kilovoltamperes with unityvpower
factor loads? (e) Use MATLAB to plot the terminal voltage of the generator as a function of load for all three power
factors. (a) This generator is Yconnected, so I L = I A. At rated conditions, the line and phase current in this generator is [A = 1L 2 P —M = 251 A at an angle of—36.87° «5 V, _ \/§(2300 v) The phase voltage of this machine is V¢ = VT / J3 = 1328 V. The internal generated voltage of the machine
1s EA = V¢+RAIA + jXSIA EA = 1328z0°+ (0.15 Q)(251£ 36.87° A)+ j(1.1 Q)(251A — 36.87° A) E, = 1537174" V The input power to this generator is equal to the output power plus losses. The rated output power is POUT = (1000 kVA)(O.8) 2800 kW Pcu = 31,219 = 3(251A)2(0.15 Q) = 28.4 kW
10,,ng = 24 kW P =18 kW core P = (assumed 0) stray P1N = POUT +PCU +PF&W +P +P =870.4 kW core may
77 = POUT x100% = Mx100% = 91.9%
pIN 870.4 kW (b) If the generator is loaded to rated kVA with lagging loads, the phase voltage is V4, = 1328100 V and the internal generated voltage is E A = 1537174“ V. Therefore, the phase voltage at no—load would be
VP 2 153740" V . The voltage regulation would be: _ 1537 — 1328
1328 (c) If the generator is loaded to rated kVA with leading loads, the phase voltage is V¢ = 1328100 V and VR X 100% = 15.7% the internal generated voltage is
EA = V, + RAIA + jXSIA
EA 2 132810°+(0.15 Q)(25 1136.87O A)+ j(1.1 Q)(251136.87° A)
EA =1217z11.5° V The voltage regulation would be: _12171328
1328 (d) If the generator is loaded to rated kVA at unity power factor, the phase voltage is
Vq, = 1328400 V and the internal generated voltage is VR X 100% = —8.4% EA =V¢+RAIA+jXSIA E, = 1328z0°+ (0.15 o)(251z0° A)+ j(1.1 o)(25140° A)
EA =1393411.4° V The voltage regulation would be: _ 1393— 1328
1328 VR x 100% = 4.9% (e) For this problem, we will assume that the terminal voltage is adjusted to 2300 V at no load
conditions, and see what happens to the voltage as load increases at 0.8 lagging, unity, and 0.8 leading
power factors. Note that the maximum current will be 251 A in any case. A phasor diagram representing
the situation at lagging power factor is shown below: By the Pythagorean Theorem,
EAZ = (v, + RAIA 0086+ X51, sin 602 +(XSIA cosa— RAIS sin 6)2
V¢ 2 EA2 —— (XSIA cosQ—RAIS sing)2 —RAIA cost9— XSIA sing A phasor diagram representing the situation at leading power factor is shown below: By the Pythagorean Theorem, E; = (V¢ + RAIA 0086— X31, sin 6)2 +(XSIA cos9+ RAIS sin 6)2 V¢ =\/EA2 —(XSIA cosi9+RAIS sint9)2 —RAIA cos6+XSIAsini9 A phasor diagram representing the situation at unity power factor is shown below: By the Pythagorean Theorem,
2 EA2 = v; + (351A) 2 V¢ = E; — (XSIA) 55. Terminal Voltage Versus Load B
8 Terminal Voluge (V)
75 g
0
O — 0.8 PF lagg'ng
'  1.0 PF
m" 0‘8 PF leadng ’ 1500 i ‘ I
0 so 100 150 200 250 Loud (A) Assume that the ﬁeld current of the generator in Problem 52 has been adjusted so that it supplies rated
voltage when loaded with rated current at unity power factor. (You may ignore the effects of R A when answering these questions.)
(a) What is the torque angle 6 of the generator when supplying rated current at unity power factor? (b) When this generator is running at full load with unity power factor, how close is it to the static stability
limit of the machine? SOLUTION
(a) The torque 5 angle can be found by calculating E A :
EA 2V¢+RAIA+jXSIA
EA = 132810°+(0.15 Q)(25110° A)+ j(1.1 Q)(25110° A) EA =1393411.4° V
Thus the torque angle 5: 11.4°. (b) The static stability limit occurs at 5 = 90°. This generator is a very long way from that limit. If we
ignore the internal resistance of the generator, the output power will be given by
3V E
P 2 ¢ A sin 5
XS and the output power is proportional to sin§ . Since sin 11.4° 20.198, and sin 90° = 1.00, the static
stability limit is about 5 times the current output power of the generator. 57. A 13.8—kV 10MVA 0.8PFlagging 60Hz twopole Yconnected steamturbine generator has a 1
synchronous reactance of 12 9 per phase and an armature resistance of 1.5 (2 per phase. This generator is ‘
operating in parallel with a large power system (inﬁnite bus). 1 l
1
1 (a) What is the magnitude of E A at rated conditions?
([9) What is the torque angle of the generator at rated conditions? I (c) If the ﬁeld current is constant, what is the maximum power possible out of this generator? How much ;
reserve power or torque does this generator have at full load? ‘ (d) At the absolute maximum power possible, how much reactive power will this generator be supplying or
consuming? Sketch the corresponding phasor diagram. (Assume I F is still unchanged.) SOLUTION (a) The phase voltage of this generator at rated conditions is V _ 13,800 V
¢ J3
The armature current per phase at rated conditions is s _ 10,000,000 VA J5 V, _ £03,800 v) Therefore, the internal generated voltage at rated conditions is 1
E, = V¢ + RAIA + jXSIA :
EA = 7967z0° + (1.5 Q)(4184 — 36.87° A) + j(12.0 Q)(4184 — 36.87° A) :
EA =12,040117.6° V E = 7967 V =418A IA: The magnitude of E A is 12,040 V. (b) The torque angle of the generator at rated conditions is 5: 17.6°. (c) Ignoring R A, the maximum output power of the generator is given by P = 3 V¢ EA 2 3(7967 V)(12,040 v) = 24.0 MW
MAX XS 12 £2 The power at maximum load is 8 MW, so the maximum output power is three times the full load output
power. (d) The phasor diagram at these conditions is shown below: EA J XS IA
RAIA
"1
Under these conditions, the armature current is
E  V 12,040190o V  7967A °
_ A ¢ _ _____0 V 2119414060 A A _ RA + jXS _ 1.5+ 112.0 Q The reactive power produced by the generator at this point is
Q = 3 V, 1, sin 6: 3(7967 v)(1194 A) sin(0°—40.6°) = —18.6 MVAR i The generator is actually consuming reactive power at this time. ll A 2300V 1000hp 0.8PF leading 60Hz twopole Y—connected synchronous motor has a synchronous
reactance of 2.8 Q and an armature resistance of 0.4 9. At 60 Hz, its friction and windage losses are 24
kW, and its core losses are 18 kW. The ﬁeld circuit has a dc voltage of 200 V, and the maximum I F is 10 A. The opencircuit characteristic of this motor is shown in Figure P61. Answer the following questions
about the motor, assuming that it is being supplied by an inﬁnite bus. (a) How much ﬁeld current would be required to make this machine operate at unity power factor when
supplying full load? (b) What is the motor’s efﬁciency at full load and unity power factor? (c) If the ﬁeld current were increased by 5 percent, what would the new value of the armature current be?
What would the new power factor be? How much reactive power is being consumed or supplied by the motor? (d) What is the maximum torque this machine is theoretically capable of supplying at unity p0wer factor?
At 0.8 PF leading? SOLUTION (a) At full load, the input power to the motor is Pm: OUT+P +P +PCU mesh core We can’t know the copper losses until the armature current is known, so we will ﬁnd the input power and
armature current ignoring that term, and then correct the input power after we know it. P1N = (1000 hp)(746 W/hp)+ 24 kW + 18 kW = 788 kW Therefore, the line and phase current at unity power factor is kW
1A=1 — P L2198A L _ [Ti/:13? : @9300 v)(1.0)
The copper losses due to a current of 198 A are
PCU = 31,212, = 3(198 A)2(0.4 o) = 47.0 kW
Therefore, a better estimate of the input power at full load is
PIN = (1000 hp)(746 W/hp)+24 kW + 18 kW + 47 kW 2 835 kW and a better estimate of the line and phase current at unity power factor is P 835 kW 6 V, PF : \/§(2300 v)(1.0) : 210 A IAZIL: The phasor diagram of this motor operating a unity power factor is shown below: IA V¢
“ JXSIA EA R I A A The phase voltage of this motor is 2300 / J3 = 1328 V. The required internal generated voltage is
EA =V¢—RAIA—jXSIA
EA =132820° v(0.4 r2)(21020° A)— j(2.8 o)(21020° A)
EA =1376z—25.3° V This internal generated yoltage corresponds to a terminal voltage of «[3 (1376) = 2383 V. This voltage
would require a ﬁeld current of 4.6 A. (b) The motor’s efﬁciency at full load and unity power factor is POUT X100% 2 746 kW
835 kW 7]: x100% 289.3% IN mmmvmwn’aﬁwl‘wnﬁmvr :r, i
E
E mw.,rwm....~...mww,u=w (c) To solve this problem, we will temporarily ignore the effects of the armature resistance R A. If R A is
ignored, then E A sin 5 is directly proportional to the power supplied by the motor. Since the power
supplied by the motor does not change when I F is changed, this quantity will be a constant. If the ﬁeld current is increased by 5%, then the new ﬁeld current will be 4.83 A, and the new value of
the opencircuit terminal voltage will be 2450 V. The new value of E A will be 2450 V/ J8 = 1415 V. Therefore, the new torque angle 5 will be V
E“ siné‘1 = sin 1 1376
EA, 1415 v 62 = sin‘1 sin( 253°) = —24.6° Therefore, the new armature current will be V _E o _ _ o
IA= , A 2132840 v 14152 25.3 V=214.523.5°A RA + jXS 0.4 + j2.8 o The new current is about the same as before, but the phase angle has become positive. The new power
factor is cos 35" = 0.998 leading, and the reactive power supplied by the motor is Q = J3 V, IL sin 19: \/§ (2300 V)(214.5 A)sin(3.5°) = 52.2 kVAR (d) The maximum torque possible at unity power factor (ignoring the effects of R A) is: T _3V¢EA 3(1328V)(1376V) 519” m
ind,max — — =  = ‘
“’m XS (3600mm) 1‘5““ 2’: rad (2.8 Q)
S 1' If we are ignoring the resistance of the motor, then the input power would be 788 kW (note that copper
losses are ignored!) At a power factor of 0. 8 leading, the current ﬂow will be
P 788 kW
I A = I — = — 247 A L \/§ V, PF J5 (2300 V)(0.8) so I A = 247436.87° A . The internal generated voltage at 0.8 PF leading (ignoring copper losses) is
EA =V¢—RAIA —jXSIA
E, = 132810" V— j(2.8 Q)(247z36.87° A)
EA =1829z ——17.6° V Therefore, the maximum torque at a power factor of 0.8 leading is 3V E 31328V 1829V
“u='—’(—'.L—)*—~=6093N~m . 1 mm 27: rad
3600 /
( r mm) 60 s 1 r (2.8 Q) Tind,max = (0m XS 3
t
3 610. A 480V 375kVA 0.8PFlagging Y—connected synchronous generator has a synchronous reactance of 0.4
Q and a negligible armature resistance. This generator is supplying power to a 480—V 80—kW 0.8PF
leading Yconnected synchronous motor with a synchronous reactance of 1.1 Q 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. (1;) 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 operating at rated power and unity power factor, so the current ﬂowing in the motor is P 80 kW I L"“ = J3 VT PF 2 J3 (480 v)(1.0) =I =96.2A A,m so I Am 2 96210" A. This machine is Y—connected, so the phase voltage is V¢ = 480/ J5 = 277 V. The internal generated voltage of the motor is EA,m : V¢,m _ jXSIAm
Em = 27740° v — j(1.1 o)(96.240° A)
Em = 29M — 20.9° v This same current comes from the generator, so the internal generated voltage of the generator is EA,g = v‘,‘g + jXSIAlg
E” = 277z0° v + j(0.4 Q)(96.2£0° A)
EA‘g = 28047.9o v Generator Motor (b) The power supplied by the generator to the motor will be constant as the ﬁeld current of the motor is
varied. The 10% increase in ﬂux will raise the internal generated voltage of the motor to (1.1)(297 V) = 327 V. To make finding the new conditions easier, we will make the angle of the phasor E A_ g the reference during the following calculations. The resulting phasor diagram is shown below.
I E
A A,g Then by Kirchhoff’ s Voltage Law,
EM 2 EM + j(XS,g +XS‘m)IA EA,g — EA,m or IA =.———
J(Xs_g +ng,,,) Note that this combined phasor diagram looks just like the diagram of a synchronous motor, so we can
apply the power equation for synchronous motors to this system. 3E E
P = ———A’g A‘m sin y
X S, g + X SM
where 7 = 68 + 5m . From this equation,
X +XmP 1.59 80kW
7: 51111 ng—S) = sin“ L_)(_.__) = 25.90
313,813” 3(2sov)(327 v)
Therefore,
E _ E o __ _ 0
IA = ‘ Ag A," ___ 28010 V .3271 25.9 V =95.745.7°A
J(X5,g+Xs,m) jljg The phase voltage of the system would be
V¢ = EM — jXS’g A = 280z0° V — j(0.4 Q)(95.745.7° A) = 2861 —7.6° v If we make V¢ the reference (as we usually do), these voltages and currents become: EA = 28047.6o V
V, = 286400 v EM = 3274—18.? v
IA = 95.7z13.3° A The new terminal voltage is VT = J37 (286 V) = 495 V , so the system voltage has increased. (c) The power factor of the motor is now PF = cos(—13.3°) = 0.973 leading, since a current angle of 18.3° implies an impedance angle of 18.3”. Note: The reactive power in the motor is now QM = 3V¢IA sine = 3(286 v)(95.7 A)sin(—13.3°) = —18.9 kVAR The motor is now supplying 18.9 kVAR to the system. Note that an increase in machine ﬂux has
increased the reactive power supplied by the motor and also raised the terminal voltage of the system
This is consistent with what we learned about reactive power sharing in Chapter 5. 615. A 100~hp 440V 0.8PF—1eading Aconnected synchronous motor has an armature resistance of 0.22 9 and
a synchronous reactance of 3.0 (2. Its efficiency at full load is 89 percent. (a) What is the input power to the motor at rated conditions? (b) What is the line current of the motor at rated conditions? What is the phase current of the motor at
rated conditions? (c) What is the reactive power consumed by or supplied by the motor at rated conditions? ((1) What is the internal generated voltage E A of this motor at rated conditions? (e) What are the stator copper losses in the motor at rated conditions? (f) What is P at rated conditions? CODV (g) If E A is decreased by 10 percent, how much reactive power will be consumed by or supplied by the 1
motor? I SOLUTION (a) The input power to the motor at rated conditions is p =E=W=g3gkw
‘” 7; 0.89 ‘ (b) The line current to the motor at rated conditions is
I _ P _ 83.8 kW L «0 V, PF J3 (440 v)(os) The phase current to the motor at rated conditions is 1 =I_L=lﬂ=79.4A ¢J§J§ (c) The reactive power supplied by this motor to the power system at rated conditions is QMd = 3V,,IA sine = 3(440 v)(79.4 A)sin36.87° = 62.9 kVAR =137A g
i
E
E
g
% (d) The internal generated voltage at rated conditions is
EA = V¢—RAIA — jXSIA
EA = 44020° V —(0.22 Q)(79.4z36.87° A)— j(3.0 Q)(79.4z36.87° A)
EA = 603.4 —19.5° V (e) The stator copper losses at rated conditions are PCU = 313R, = 3(79.4 A)2 (0.22 (2) = 4.16 kW
(f) Pconv at rated conditions is P 2 PIN —PCU =83.8 kW—4.16kW=79.6 kW COHV (g) If E A is decreased by 10%, the new value if E A = (0.9)(603 V) = 543 V. To simplify this part of the
problem, we will ignore R A. Then the quantity E A sin6 will be constant as E A changes. Therefore, V
6, =sin“ Elana, =sin‘ 603 sin (—19.5°) =—21.8°
EA2 V
Therefore,
V _E o _ _ 0
IA : ¢l A 2440.40 V .5431 21.8 =70.5417.7°A
jXS 13.0 and the reactive power supplied by the motor to the power system will be Q = 3V¢IA sine = 3(440 v)(70.5 A)sin(17.7°) = 28.3 kVAR 616. Answer the following questions about the machine of Problem 615.
(a) If E A = 43041350 V and Va = 440100 V, is this machine consuming real power from or supplying real power to the power system? Is it consuming reactive power from or supplying reactive power to
the power system? (b) Calculate the real power P and reactive power Q supplied or consumed by the machine under the
conditions in part (a ). Is the machine operating within its ratings under these circumstances? (0) If E A = 470412o V and V9 = 440100 V, is this machine consuming real power from or supplying real power to the power system? Is it consuming reactive power from or supplying reactive power to
the power system? (d) Calculate the real power P and reactive power Q supplied or consumed by the machine under the
conditions in part (c). Is the machine operating within its ratings under these circumstances? SOLUTION
(a) This machine is a generator supplying real power to the power system, because E A is ahead of V». It is consuming reactive power because E A cos 6 < V¢. (b) This machine is acting as a generator, and the current ﬂow in these conditions is I _ EA—V¢ _430£l3.5°—44040°V A _ _—_.— _ . = 34.21165O A
RA +jXS 0.22+}3.0 The real power supplied by this machine is P = 3V¢IA cosl9= 3(440 v)(34.2 A)cos(—16.5°) = 43.3 kW The reactive power supplied by this machine is Q = 3V¢IA sine = 3(440 v)(34.2 A)sin (~16.5°) = —12.8 kVAR (c) This machine is a motor consuming real power from the power system, because E A is behind V¢. It is supplying reactive power because E A cos6 > V¢. (d) This machine is acting as a motor, and the current ﬂow in these conditions is _ V¢—EA _ 44040" V4704—12" _ _ . =33.1115.6° A
RA +jXS 0.22+_]3.0 IA
The real power consumed by this machine is
P = 3V¢IA cos6 = 3(440 V)(33.1A)cos(15.6°)= 42.1 kW The reactive power supplied by this machine is Q = 3V¢IA sin6= 3(440 V)(33.1A)sin(15.6°)= +11.7 kVAR wmwwnmmumemmmmmw ...
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 Spring '05
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