For a four-conductor circuit, we need three wattmeters if the three-phase circuit is unbalanced, but only two wattmeters if it is balanced, because in the latter case there is no current
Trang 1416 Balanced Three-Phase Circuits
Applying this general observation, we can see that for a three-conductor circuit, whether balanced or not, we need only two wattmeters
to measure the total power For a four-conductor circuit, we need three wattmeters if the three-phase circuit is unbalanced, but only two wattmeters if it is balanced, because in the latter case there is no current in the neutral line Thus, only two wattmeters are needed to measure the total average power in any balanced three-phase system
The two-wattmeter method reduces to determining the magnitude and algebraic sign of the average power indicated by each wattmeter We can describe the basic problem in terms of the circuit shown in Fig 11.20, where the two wattmeters are indicated by the shaded boxes and labeled
W\ and W 2 The coil notations cc and pc stand for current coil and
poten-tial coil, respectively We have elected to insert the current coils of the wattmeters in lines aA and cC Thus, line bB is the reference line for the two potential coils The load is connected as a wye, and the per-phase load
impedance is designated as Z^ = \Z\ /jh This is a general representation,
as any A-connected load can be represented by its Y equivalent;
further-more, for the balanced case, the impedance angle 0 is unaffected by the
A-to-Y transformation
We now develop general equations for the readings of the two wattmeters We assume that the current drawn by the potential coil of the wattmeter is negligible compared with the line current measured by the cur-rent coil We further assume that the loads can be modeled by passive circuit
elements so that the phase angle of the load impedance (0 in Fig 11.20) lies
between -90° (pure capacitance) and +90" (pure inductance) Finally, we assume a positive phase sequence
From our introductory discussion of the average deflection of the wattmeter, we can see that wattmeter 1 will respond to the product of
|V A JJ|, |IaA^ a nd the cosine of the angle between V A B a nd IaA- M w e denote
this wattmeter reading as W h we can write
Wi - I V A B I P U C O S * ,
It follows that
W2 = |VCBI|ICC|COS02
In Eq 11.54,0] is the phase angle between VAB and Ia A, and in Eq 11.55,
0 2 is the phase angle between VCB and Ic(>
To calculate Wj and W 2 , we express 0] and 0 2 in terms of the imped-ance angle 0, which is also the same as the phase angle between the phase voltage and current For a positive phase sequence,
The derivation of Eqs 11.56 and 11.57 is left as an exercise (see Problem 11.34) When we substitute Eqs 11.56 and 11.57 into Eqs 11.54 and 11.55, respectively, we get
h t
+
v i
h
+
v 2
/3
+
v 3
m 1
• 1
• Z
*
General network
Figure 11.19 • A general circuit whose power is
supplied by n conductors
IaA
cc
A
Wi pc4 Z^ = |Z|Z0 VAN
b«-'CN
c+
-•— -¾
c(
N
Figure 11.20 • A circuit used to analyze the
two-wattmeter method of measuring average power
delivered to a balanced load
Trang 2(11.60)
To find the total power, we add W ] and W 2 ; thus
p T = W ] + W 2 = 2 V L / L C O S ^ C O S 3 0 "
= V 3 V L / L C O S ^ , which is the expression for the total power in a three-phase circuit
Therefore we have confirmed that the sum of the two wattmeter readings
yields the total average power
A closer look at Eqs 11.58 and 11.59 reveals the following about the
readings of the two wattmeters:
1 If the power factor is greater than 0.5, both wattmeters read positive
2 If the power factor equals 0.5, one wattmeter reads zero
3 If the power factor is less than 0.5, one wattmeter reads negative
4 Reversing the phase sequence will interchange the readings on the
two wattmeters
These observations are illustrated in the following example and in
Problems 11.41-11.51
| Computing Wattmeter Readi
Calculate the reading of each wattmeter in
circuit in Fig 11.20 if the phase voltage at
load is 120 V and (a) Z+ = 8• + ;
(b) Z0 = 8 - ;6 ft; (c) Z^ = 5 + ;5 V3 12;
(d) Z s = 10 / - 7 5 ° ft (e) Verify for (a)-(d)
the sum of the wattmeter readings equals the
power delivered to the load
Solution
a) Z (b = 10 /36.87° ft, V L = 120 V 3 V , and
/L = 120/10 = 12 A
W ] = (120V3~)(12) cos (36.87° + 30°)
= 979.75 W,
W 2 = (120 V3)(12) cos (36.87° - 30°)
= 2476.25 W
b) Z & = 10 / - 3 6 8 7 ° ft, V L = 120V^ V, and
/L= 120/10 = 12 A
Wi = (120V3)(12) cos (-36.87° + 30°)
= 2476.25 W,
W 2 = (120V3)(12) cos (-36.87° - 30°)
= 979.75 W
ngs in
the the
6 ft;
and that total
Three-Phase Circuits
c) Z^ = 5(1 + jV3) = 10 / 6 0 ° O, V L = 120V3 V,
a n d /L = 12 A
Wi = (120V5)(12) cos (60° + 30°) = 0,
W 2 = (120 V3)(12) cos (60° - 30°)
= 2160 W
d) Z 4> = 10 / - 7 5 ° ft, V L = 12QV3" V, and
/L = 12 A
Wi = (120V3)(12)cos(-75° + 30°) = 1763.63 W,
W 2 = (120V5)(12) cos (-75° - 30°) = -645.53 W
e) Pr(a) = 3(12)2(8) - 3456 W,
W } +W 2 = 979.75 + 2476.25
= 3456 W,
Pr(b) = Pr(a) = 3456 W,
Wi + W 2 = 2476.25 + 979.75
= 3456 W, /V(c) = 3(12)2(5) = 2160 W,
W t + W 2 = 0 + 2160
= 2160W,
Pr(d) = 3(12)2(2.5882) = 1118.10 W,
Wi + W 2 = 1763.63 - 645.53
= 1118.10W
NOTE: Assess your understanding of the two-wattmeter method by trying Chapter Problems 11.43 and 11.44
Trang 3418 Balanced Three-Phase Circuits
3-0 line -)
Generator
Plant
4
*)
/
Sub Station
Figure 11.21 A A substation connected to a power
plant via a three-phase line
0.6 O
- A W - /4.8 n
^ Y Y Y V
A
v.„ 13,800 /r\0 y ^3 *~ MW 1.2 MVAR 1.2
n N
Figure 11.22 • A single phase equivalent circuit for
the system in Fig 11.21
Practical Perspective
Transmission and Distribution of Electric Power
At the start of this chapter we pointed out the obligation utilities have to maintain the rms voltage level at their customer's premises Although the acceptable deviation from a nominal level may vary among different utilities
we will assume for purposes of discussion that an allowable tolerance is
± 5.8% Thus a nominal rms voltage of 120 V could range from 113 to
127 V We also pointed out that capacitors strategically located on the sys-tem could be used to support voltage levels
The circuit shown in Fig 11.21 represents a substation on a Midwestern municipal system We will assume the system is balanced, the line-to-line voltage at the substation is 13.8 k V , the phase impedance of the distribu-tion line is 0.6 + /4.811, and the load at the substadistribu-tion at 3 PM on a hot, humid day in July is 3.6 M W and 3.6 magnetizing M V A R
Using the line-to-neutral voltage at the substation as a reference, the single-phase equivalent circuit for the system in Fig 11.21 is shown in Fig 11.22 The line current can be calculated from the expression for the complex power at the substation Thus,
13,800
V3 I*A = (1.2 + /1.2)10*
It follows that
or
I*A = 150.61 + /150.61 A
I a A = 150.61 - /150.61 A The line-to-neutral voltage at the generating plant is
13,800
V3 '0° + (0.6 + /4.8)(150.61 - /150.61)
= 8780.74 + /632.58
= 8803.50/4.12° V
Therefore the magnitude of the line voltage at the generating plant is
|V ab | = V3(8803.50) = 15,248.11V
We are assuming the utility is required to keep the voltage level within
± 5.8% of the nominal value This means the magnitude of the line-to-line
voltage at the power plant should not exceed 14.6 k V nor be less than
13 k V Therefore, the magnitude of the line voltage at the generating plant could cause problems for customers
When the magnetizing vars are supplied by a capacitor bank connected
to the substation bus, the line current I a A becomes
I = 150.61 + /0 A
Trang 4Therefore the voltage at the generating plant necessary to maintain a
line-to-line voltage of 13,800 V at the substation is
13,800
Van = - ^ - /O: + (0-6 + /4.8)(150.61 + /0)
= 8057.80 + /722.94
= 8090.17/5.13° V
Hence
|Vab| = V3(8090.17) = 14,012.58 V
This voltage level falls within the allowable range of 13 kV to 14.6 kV
NOTE: Assess your understanding of this Practical Perspective by trying Chapter
Problems 11.52(a)-(b) and 11.53, 11.56, and 11.57
Summary
• When analyzing balanced three-phase circuits, the first
step is to transform any A connections into Y connections,
so that the overall circuit is of the Y-Y configuration (See
page 402.)
• A single-phase equivalent circuit is used to calculate the
line current and the phase voltage in one phase of the
Y-Y structure The a-phase is normally chosen for this
purpose (See page 404.)
• Once we know the line current and phase voltage in the
a-phase equivalent circuit, we can take analytical
short-cuts to find any current or voltage in a balanced
three-phase circuit, based on the following facts:
• The b- and c-phase currents and voltages are
identi-cal to the a-phase current and voltage except for a
120° shift in phase In a positive-sequence circuit, the
b-phase quantity lags the a-phase quantity by 120°,
and the c-phase quantity leads the a-phase quantity
by 120° For a negative sequence circuit, phases b and
c are interchanged with respect to phase a
• The set of line voltages is out of phase with the set of
phase voltages by ±30° The plus or minus sign
corre-sponds to positive and negative sequence, respectively
• In a Y-Y circuit the magnitude of a line voltage is
V3 times the magnitude of a phase voltage
• The set of line currents is out of phase with the set of phase currents in A-connected sources and loads by T30° The minus or plus sign corresponds to positive and negative sequence, respectively
• The magnitude of a line current is V 3 times the mag-nitude of a phase current in a A-connected source
or load, (See pages 404-405 and 407-408.)
• The techniques for calculating per-phase average power, reactive power, and complex power are identical
to those introduced in Chapter 10 (See page 410.)
• The total real, reactive, and complex power can be deter-mined either by multiplying the corresponding per phase quantity by 3 or by using the expressions based on line current and line voltage, as given by Eqs 11.36,11.38, and 11.41 (See pages 410 and 411.)
• The total instantaneous power in a balanced three-phase circuit is constant and equals 1.5 times the average power per phase (See page 412.)
• A wattmeter measures the average power delivered to a load by using a current coil connected in series with the load and a potential coil connected in parallel with the load (See page 415.)
• The total average power in a balanced three-phase cir-cuit can be measured by summing the readings of two wattmeters connected in two different phases of the circuit (See page 415.)
Trang 5420 Balanced Three-Phase Circuits
Problems
AH phasor voltages in the following Problems are stated in
terms of the rms value
Section 11.1
11.1 Verify that Eq 11.3 is true for either Eq 11.1 or
Eq.11.2
11.2 What is the phase sequence of each of the following
sets of voltages?
a) ya = 208 cos (OJ? + 27°) V,
^ = 208 cos (cot + 147°) V,
v c = 208 cos (cot - 93°) V
b) v a = 4160 cos (cot - 18°) V,
v b = 4160 cos (cot - 138°) V,
?;c = 4160 cos (cot + 102°) V
11.3 For each set of voltages, state whether or not the
volt-PSPICE ages form a balanced three-phase set If the set is
bal-anced, state whether the phase sequence is positive or
negative If the set is not balanced, explain why
a) v a = 139 cos 377? V,
v b = 139 cos (377? + 120°) V,
v c = 139 cos (377? - 120°) V
b) v d = 381 cos 377? V,
v b = 381 cos (377? + 240°) V,
v c = 381 cos (377? + 120°) V
c) v a = 2771 sin (377? - 30°) V,
v b = 2771 cos 377? V,
v c = 2771 sin (377? + 210°) V
d) v A = 170 sin (cot + 30°) V,
v b = - 1 7 0 cos cot V,
v c = 170 cos (cot + 60°) V
e) v a = 339 cos cot V,
% = 339 cos (cot + 120) V,
v c = 393 cos (cot - 120°) V
f) v A = 3983 sin (cot + 50°) V,
v b = 3983 cos (cot - 160°) V,
v c = 3983 cos (cot + 70°) V
Section 11.2 11.4 Refer to the circuit in Fig 11.5(b) Assume that there
are no external connections to the terminals a,b,c Assume further that the three windings are from a balanced three-phase generator How much current will circulate in the A-connected generator?
Section 11.3 11.5 A balanced three-phase circuit has the following
characteristics:
• Y-Y connected;
• The line voltage at the source, Vab, is
240 V3~/90° V;
• The phase sequence is negative;
• The line impedance is 4 -I- /5 £l/<f>',
• The load impedance is 76 + /55 ft/cf)
a) Draw the single phase equivalent circuit for the a-phase
b) Calculated the line current in the a-phase c) Calculated the line voltage at the load in the a-phase
11.6 Find the rms value of I„ in the unbalanced
three-phase circuit seen in Fig PI 1.6
Figure PI 1.6
0.1 n /0.8ft a 0.4 ft /3.2 ft A 59.5 ft /76 ft
->WV »****"¥"> 0 /yy*^ ( V Y Y > 0 ' W W
0.1ft /0.8 ft b 0.4 ft /3.2 ft B 39.5 ft /26 ft
- • "VS/V r r v Y > 9 V W —
M ^ 240/-240/-120° V
0.1ft /0.8 ft c 0.4 ft /3.2 ft C 19.5 ft / l i f t
- • /vW «^w> » VW rvw>_
N
Trang 611.7 The time-domain expressions for three line-to-neutral
voltages at the terminals of a Y-connected load are
v AN = 7620 cos (W + 30°) V,
v m = 7620 cos (a)t + 150°) V,
'»CN = 7620 cos (cot - 90") V
What are the time-domain expressions for the three
line-to-line voltages v Aii , I>BO and VQA?
11.8 a) Is the circuit in Fig PI 1.8 a balanced or
unbal-PSPICE anced three-phase system? Explain
MULTISIM
b) Find I„
Figure PI 1.8
^2()0 /0!V
11.9 The magnitude of the line voltage at the terminals
of a balanced Y-connected load is 6600 V The load
impedance is 240 - /70 0/<£ The load is fed from a
line that has an impedance of 0.5 + /4 0/(£
a) What is the magnitude of the line current?
b) What is the magnitude of the line voltage at
the source?
11.10 a) Find I„ in the circuit in Fig PI 1.10
c) FindVA B
d) Is the circuit a balanced or unbalanced
three-phase system?
11.11 The magnitude of the phase voltage of an ideal
balanced three-phase Y-connected source is
125 V The source is connected to a balanced Y-connected load by a distribution line that has an
impedance of 0.1 + /0.8 Cl/<j) The load impedance
is 19.9 + /14.2 Cl/4> The phase sequence of the
source is acb Use the a-phase voltage of the source as the reference Specify the magnitude and phase angle of the following quantities: (a) the three line currents, (b) the three line voltages at the source, (c) the three phase voltages at the load, and (d) the three line voltages at the load
Section 11.4 11.12 A balanced, three-phase circuit is characterized as
follows:
• Y-A connected;
• Source voltage in the c-phase is 2 0 / - 9 0 ° V;
• Source phase sequence is abc;
• Line impedance is 1 + / 3 CL/<f>;
• Load impedance is 117 — /99 12/(/)
a) Draw the single phase equivalent for the a-phase b) Calculate the a-phase line current
c) Calculate the a-phase line voltage for the three-phase load
11.13 An acb sequence balanced three-phase Y-connected
source supplies power to a balanced, three-phase
A-connected load with an impedance of 12 + /9 fl/4>
The source voltage in the b-phase is 2 4 0 / - 5 0 ° V
The line impedance is 1 + /1 fl/4> Draw the single
phase equivalent circuit for the a-phase and use it
to find the current in the a-phase of the load
11.14 A balanced A-connected load has an impedance of
864 — /252 Q,/(f) The load is fed through a line hav-ing an impedance of 0.5 4- /4 Q/<f) The phase
volt-age at the terminals of the load is 69 kV The phase sequence is positive Use VA B as the reference a) Calculate the three phase currents of the load b) Calculate the three line currents
c) Calculate the three line voltages at the sending end of the line
Figure P11.10
0.4 ft /212 a 1.6 n
- v w - /4 ft - A l V i r 7811
/54 a
O Till (FV
/1.6 b 2.6 0 /2.4 ft B 77 fl /56 O
— • W V o o o r > 1 w ^
0.2 ft /1.2 ft c 0.8 ft /3.8 ft C 79 ft /55 ft
N
Trang 7422 Balanced Three-Phase Circuits
11.15 A balanced Y-connected load having an
imped-ance of 72 + /21 Cl/(f) is connected in parallel with
a balanced A-connected load having an
imped-ance of 150/0° Q,/<f> The paralleled loads are fed
from a line having an impedance of /1 0/4> The
magnitude of the line-to-neutral voltage of the
Y-load is 7650 V
a) Calculate the magnitude of the current in the
line feeding the loads
b) Calculate the magnitude of the phase current in
the A-connected load
c) Calculate the magnitude of the phase current in
the Y-connected load
d) Calculate the magnitude of the line voltage at
the sending end of the line
11.16 In a balanced three-phase system, the source is a
bal-anced Y with an abc phase sequence and a line voltage
Va b = 208/50° V The load is a balanced Y in
paral-lel with a balanced A.The phase impedance of the Y is
4 + / 3 il/cf) and the phase impedance of the A is
3 - /9 0./4) The line impedance is 1.4 -I- /0.8 Oj4>
Draw the single phase equivalent circuit and use it to
calculate the line voltage at the load in the a-phase
11.17 The impedance Z in the balanced three-phase
cir-cuit in Fig PI 1.17 is 100 - /75 O Find
a) IAB> I B O and I C A ,
b) IaA* \ ^ and IcC,
c) Iba, Icb, and Iac
Figure P11.17
13.2/-120° kV
11.18 For the circuit shown in Fig PI 1.18, find
PSPKE a) the phase currents IA B, IRC, and IC A
b) the line currents IaA, IbB, and FcC
when Z, = 2.4 - /0.7 ft, Z2 = 8 + /6 Q, and
Z3 = 20 + / 0 ft
Figure P11.18
a A
480/-120° V
11.19 A three-phase A-connected generator has an
inter-nal impedance of 9 -(- /90 mft/<jf> When the load is removed from the generator, the magnitude of the terminal voltage is 13,800 V The generator feeds a A-connected load through a transmission line with
an impedance of 20 + /180 m f t / $ The per-phase impedance of the load is 7.056 + /3.417 ft
a) Construct a single-phase equivalent circuit b) Calculate the magnitude of the line current c) Calculate the magnitude of the line voltage at the terminals of the load
d) Calculate the magnitude of the line voltage at the terminals of the source
e) Calculate the magnitude of the phase current in the load
f) Calculate the magnitude of the phase current in the source
11.20 A balanced three-phase A-connected source is
shown in Fig PI 1.20
a) Find the Y-connected equivalent circuit
b) Show that the Y-connected equivalent circuit delivers the same open-circuit voltage as the original A-connected source
c) Apply an external short circuit to the terminals
A, B, and C Use the A-connected source to find the three line currents Ia A, IbB, and IcC
d) Repeat (c) but use the Y-equivalent source to find the three line currents
Figure PI 1.20
• A
2.7 O
/13.511
4156/-120° V
• C
11.21 The A-connected source of Problem 11.20 is
con-nected to a Y-concon-nected load by means of a bal-anced three-phase distribution line The load
Trang 8impedance is 1910-/636 fl/<f) and the line
imped-ance is 9.1 + /71.5(2/0
a) Construct a single-phase equivalent circuit of
the system
b) Determine the magnitude of the line voltage at
the terminals of the load
c) Determine the magnitude of the phase current
in the A-source
d) Determine the magnitude of the line voltage at
the terminals of the source
11.27 The three pieces of computer equipment described
below are installed as part of a computation center Each piece of equipment is a balanced three-phase load rated at 208 V Calculate (a) the magnitude of the line current supplying these three devices and (b) the power factor of the combined load
• Hard Drive: 4.864 kW at 0.79 pf lag
• CD/DVD drive: 17.636 kVA at 0.96 pf lag CPU: line current 73.8 A, 13.853 kVAR
11.28 Calculate the complex power in each phase of the
unbalanced load in Problem 11.18
Section 11.5
11.22 In a balanced three-phase system, the source
has an abc sequence, is Y-connected, and
Van = 120/20° V The source feeds two loads, both
of which are Y-connected The impedance of load 1
is 8 + /6 il/cf) The complex power for the a-phase
of load 2 is 600/36° VA Find the total complex
power supplied by the source
11.23 A balanced three-phase source is supplying 60 kVA
at 0.6 lagging to two balanced Y-connected parallel
loads The distribution line connecting the source to
the load has negligible impedance Load 1 is purely
resistive and absorbs 30 kW Find the per-phase
impedance of Load 2 if the line voltage is 120 V5 V
and the impedance components are in series
11.24 A three-phase positive sequence Y-connected
source supplies 14 kVA with a power factor of 0.75
lagging to a parallel combination of a Y-connected
load and a A-connected load The Y-connected load
uses 9 kVA at a power factor of 0.6 lagging and has
an a-phase current of 1 0 / - 3 0 ° A
a) Find the complex power per phase of the
A-connected load
b) Find the magnitude of the line voltage
11.29 A balanced three-phase distribution line has an
impedance of 1 + /8 fl/</> This line is used to sup-ply three balanced three-phase loads that are con-nected in parallel The three loads are L! = 120 kVA at 0.96 pf lead, L2 = 180 kVA at 0.80 pf lag, and L3 = 100.8 kW and 15.6 kVAR (magnetizing) The magnitude of the line voltage at the terminals of the loads is 2400 V3 V
a) What is the magnitude of the line voltage at the sending end of the line?
b) What is the percent efficiency of the distribution line with respect to average power?
11.30 The line-to-neutral voltage at the terminals of the
balanced three-phase load in the circuit shown in Fig PI 1.30 is 1200 V At this voltage, the load is absorbing 500 kVA at 0.96 pf lag
a) Use VA N as the reference and express Ina in polar form
b) Calculate the complex power associated with the ideal three-phase source
c) Check that the total average power delivered equals the total average power absorbed
d) Check that the total magnetizing reactive power delivered equals the total magnetizing reactive power absorbed
11.25 The total apparent power supplied in a balanced,
three-phase Y-A system is 4800 VA The line
volt-age is 240 V If the line impedance is negligible and
the power factor angle of the load is - 5 0 ° ,
deter-mine the impedance of the load
11.26 Show that the total instantaneous power in a
bal-anced three-phase circuit is constant and equal to
1,5V m I m cos 0$, where V m and I m represent the
maximum amplitudes of the phase voltage and
phase current, respectively
Figure PI 1.30
0.18 fi J iA4n
# Vs/V ^VY-r>_
:-/180 a
0.18 0 /1.44«
— / y y V e-v-Y-v>_
0.18 a /1-440
* W V i^r>nr\_
A
500 kVA
0.% pf lag
-•C
Trang 94 2 4 Balanced Three-Phase Circuits
PSPICE
MULTISIM
11.31 a) Find the rms magnitude and the phase angle of
IC A in the circuit shown in Fig PI 1.31
b) What percent of the average power delivered by
the phase source is dissipated in the
three-phase load?
Figure PI 1.31
/2 ft a 1.5 ft
-•—'vw-f^J 1365/0!
6
565/0° V
/2 0 b i.5r>
1365/-120° V
1365/120° V
/2 ft
:85.5 0 /114 a
B 85.5 XI
!yii4n
:85.5 ft
;/ii4 n
c 1.5 ft
11.32 At full load, a commercially available 100 hp,
three-phase induction motor operates at an efficiency of
97% and a power factor of 0.88 lag The motor is
sup-plied from a three-phase outlet with a line-voltage
rating of 208 V
a) What is the magnitude of the line current drawn
from the 208 V outlet? (1 hp = 746 W.)
b) Calculate the reactive power supplied to
the motor
11.33 Three balanced three-phase loads are connected in
parallel Load 1 is Y-connected with an impedance
of 400 + /300 ft/0; load 2 is A-connected with an
impedance of 2400 - /1800 ft/<£; and load 3 is
172.8 + /2203.2 kVA The loads are fed from a
dis-tribution line with an impedance of 2 + /16 fl/cj)
The magnitude of the line-to-neutral voltage at the
load end of the line is 24 V 3 kV
a) Calculate the total complex power at the
send-ing end of the line
b) What percentage of the average power at the
sending end of the line is delivered to the loads?
11.34 A three-phase line has an impedance of
0.1 + /0.8 Cl/(f) The line feeds two balanced
three-phase loads connected in parallel The first
load is absorbing a total of 630 kW and absorbing
840 kVAR magnetizing vars The second load
is Y-connected and has an impedance of
15.36 - /4.48 0,/4> The line-to-neutral voltage at
the load end of the line is 4000 V What is the
magnitude of the line voltage at the source end of
the line?
11.35 A balanced three-phase load absorbs 96 kVA at a
lagging power factor of 0.8 when the line voltage at the terminals of the load is 480 V Find four equiva-lent circuits that can be used to model this load
11.36 The total power delivered to a balanced
three-phase load when operating at a line voltage of
2400 V3 V is 720 kW at a lagging power factor of 0.8 The impedance of the distribution line
sup-plying the load is 0.8 + /6.4 Q,/<fi Under these
operating conditions, the drop in the magnitude
of the line voltage between the sending end and the load end of the line is excessive To compen-sate, a bank of A-connected capacitors is placed
in parallel with the load The capacitor bank is designed to furnish 576 kVAR of magnetizing reactive power when operated at a line voltage
of 2400 V3 V
a) What is the magnitude of the voltage at the sending end of the line when the load is operat-ing at a line voltage of 2400 V3 V and the capac-itor bank is disconnected?
b) Repeat (a) with the capacitor bank connected c) What is the average power efficiency of the line
in (a)?
d) What is the average power efficiency in (b)? e) If the system is operating at a frequency of 60 Hz, what is the size of each capacitor in microfarads?
11.37 A balanced bank of delta-connected capacitors is
connected in parallel with the load described in Assessment Problem 11.9 The effect is to place a capacitor in parallel with the load in each phase The line voltage at the terminals of the load thus remains at 2450 V The circuit is operating at a fre-quency of 60 Hz.The capacitors are adjusted so that the magnitude of the line current feeding the paral-lel combination of the load and capacitor bank is at its minimum
a) What is the size of each capacitor in microfarads? b) Repeat (a) for wye-connected capacitors
c) What is the magnitude of the line current?
11.38 A balanced three-phase source is supplying 150 kVA
at 0.8 pf lead to two balanced Y-connected parallel loads The distribution line connecting the source to the load has negligible impedance The power associ-ated with load 1 is 30 + /30 kVA
a) Determine the types of components and their impedances in each phase of load 2 if the line voltage is 2500 V 3 V and the impedance compo-nents are in series
b) Repeat (a) with the impedance components in parallel
Trang 1011.39 The output of the balanced positive-sequence
three-phase source in Fig PI 1.39 is 41.6 kVA at a
lagging power factor of 0.707 The line voltage at
the source is 240 V
a) Find the magnitude of the line voltage at the load
b) Find the total complex power at the terminals of
the load
Figure P11.39
Balanced
three-phase'
source
0.04 n / ° -0 3 n
W V /-Y-YV^_
0.04 a / ° - 0 3 n
—Vs/V /"YVY-N—
0.04 0 /0-°3 U
—VvV o r - o r ^
Balanced
"•three-phase load
11.43 The two wattmeters in Fig 11.20 can be used to
compute the total reactive power of the load
a) Prove this statement by showing that
V3(W 2 - Wj) = V 3 VL/Ls i i i V b) Compute the total reactive power from the wattmeter readings for each of the loads in Example 11.6 Check your computations by cal-culating the total reactive power directly from the given voltage and impedance
11.44 The two-wattmeter method is used to measure the
power at the load end of the line in Example 11.1 Calculate the reading of each wattmeter
11.45 The wattmeters in the circuit in Fig 11.20 read as
follows: W x = 40,823.09 W, and W 2 = 103,176.91 W
The magnitude of the line voltage is 2400 V 5 V The phase sequence is positive Find Z^
Section 11.6
11.40 Derive Eqs 11.56 and 11.57
11.41 In the balanced three-phase circuit shown in
Fig PI 1.41, the current coil of the wattmeter is
con-nected in line aA, and the potential coil of the
wattmeter is connected across lines b and c Show
that the wattmeter reading multiplied by V 3 equals
the total reactive power associated with the load
The phase sequence is positive
Figure PI 1.41
11.46 a) Calculate the reading of each wattmeter in the
circuit shown in Fig PI 1.46 The value of Z^ is
4 0 / - 3 0 ° a
b) Verify that the sum of the wattmeter readings equals the total average power delivered to the A-connected load
Figure PI 1.46
24Q/W V |
a •
K m
D •
r
-C'
±
cc
[
[
W
1
1
pcj
A
B
C
Z<b
z*
z*
N
b l ±
Wi
-1-) »_| /-YVV^
240/120o~V~
4 ^ 240/-120° V
-•
W,
+ ™ ~ l-\c
11.42 The line-to-neutral voltage in the circuit in
Fig PI 1.41 is 680 V, the phase sequence is positive,
and the load impedance is 16 - j 12 0 / $
a) Calculate the wattmeter reading
b) Calculate the total reactive power associated
with the load
11.47 The two-wattmeter method is used to measure
the power delivered to the unbalanced load in Problem 11.18 The current coil of wattmeter 1 is placed in line aA and that of wattmeter 2 is placed in line bB
a) Calculate the reading of wattmeter 1
b) Calculate the reading of wattmeter 2
c) Show that the sum of the two wattmeter read-ings equals the total power delivered to the unbalanced load