08/10/2012
CHAPTER 3
Engr. Dr. Kok Boon Boon Ching 2012@JEK/FKEE
Introduction o a e an es
Definition of Voltage Drop
Cable Impedances
Transformer ransforme r Voltage Drop Dro p
Voltage Drop Due to Motor Starting
Symmetrical and Asymmetrical Fault Currents
Equivalent System Impedance
Short Circuit Analysis in Three-phase Systems
Short Circuit Analysis in Single-phase Systems
3 r t p a h C – 3 0 8 2 4 X E B
2
1
08/10/2012
Voltage drop on electrical power distribution , transformers, and motors.
Voltage drop happens when load current (Ib) flows through a conductor or transformer having a finite impedance. impedance.
Severe voltage drop will result in motor failures, dimming of lamps, and CPU shutdown.
Voltage drop calculation is important to system designer for maintaining nominal voltage at servicing sides.
3 r t p a h C – 3 0 8 2 4 X E B
3
According to 17th Edition of IEE Wiring Regulations between the origin of an installation and any load point should not be greater than the values in Table 12A expressed with respect to the value of the nominal voltage of the installation. Max. 100m only. Increase 0.005% per meter if beyond 100m. a e
– o tage rop
Lighting
Other uses
i. Low voltag voltage e instal installat lation ion suppli supplied ed directly from a public low voltage distribution system
3%
5%
ii. Low voltage voltage installati installation on supplied supplied from private LV supply
6%
8%
3 r t p a h C – 3 0 8 2 4 X E B
4
2
08/10/2012
Voltage drop on electrical power distribution , transformers, and motors.
Voltage drop happens when load current (Ib) flows through a conductor or transformer having a finite impedance. impedance.
Severe voltage drop will result in motor failures, dimming of lamps, and CPU shutdown.
Voltage drop calculation is important to system designer for maintaining nominal voltage at servicing sides.
3 r t p a h C – 3 0 8 2 4 X E B
3
According to 17th Edition of IEE Wiring Regulations between the origin of an installation and any load point should not be greater than the values in Table 12A expressed with respect to the value of the nominal voltage of the installation. Max. 100m only. Increase 0.005% per meter if beyond 100m. a e
– o tage rop
Lighting
Other uses
i. Low voltag voltage e instal installat lation ion suppli supplied ed directly from a public low voltage distribution system
3%
5%
ii. Low voltage voltage installati installation on supplied supplied from private LV supply
6%
8%
3 r t p a h C – 3 0 8 2 4 X E B
4
2
08/10/2012
3 r
V VD(X)
VSEND
θ
VREC
IA
IARA
jXSIA
t p a h C – 3 0 8 2 4 X E B
5
Approximation method: method: Vdrop
≈
I b
×
R L × cosθ − X L × sinθ
RL = circuit resistance in Ohms XL = circuit reactance in Ohms Ib = design current/ line current
θ = phase angle of line current
If VA = system voltage, %Vdrop
=
Vdrop VA
3 r t p a h C – 3 0 8 2 4 X E B
× 100% 6
3
08/10/2012
Dete Determ rmin ine e the the perc percen enta tage ge volt voltag age e drop drop alon along g , , , consisting of one 400 THW (Thermoplastic Heat and Water Resistant Insulated Wire) copper conductor per phase. The current is 350A at 0.85 PF lagging. Assume steel conduit.
3 r t p a h C – 3 0 8 2 4 X E B
7
From the table of 600V cables, resistance = . , = . . R L
=
X L
=
0.035 1000 0.049 1000
× 85 = 0.002975Ω × 85 = 0.004165Ω
θ = − cos −1 (0.85) = −31.79° V
= 350 A × 0.002975Ω × cos −31.79° − 0.004165Ω × sin −31.79° = 1.65V
%V drop
=
1.65V 239.6V
× 100% = 0.69%
3 r t p a h C – 3 0 8 2 4 X E B
8
4
08/10/2012
Tabulated mV/A/m values: rop
3 r
×l
1000
Taking account of load power factor (for AC circuits using conductors of 16mm2 or less), Vdrop
=
tabulated (mV/A/m)z × I
=
tabulated (mV/A/m) z × I b × l × cos θ 1000
volts
For AC circuits using conductors of 25mm2 or , Vdrop
=
[tabulated (mV/A/m)r × cos θ + tabulated (mV/A/m)x × sin θ ] × I b × l 1000
volts
t p a h C – 3 0 8 2 4 X E B
Note: Refer to Tabulated Table of Voltage Drop (17 th IEE Regulations) 9
A 415V three-phase AC circuit is wired in a XLPE insulation and aluminum conductors of 35mm2 cross-sectional area. If Ib = 120A, length = 27m, and (mV/A/m)z = 1.95, what is the percentage voltage drop?
3 r t p a h C – 3 0 8 2 4 X E B
10
5
08/10/2012
V drop
=
1.95 × 120 × 27
%V drop
=
1000 . 415 / 3
V = 6.32V
×100% = 2.64%
3 r t p a h C – 3 0 8 2 4 X E B
11
Three-phase voltage drop calculations is “ ” . The ohmic cable impedances: R L
=
XL
=
Resistance in Ohms/1000f t 1000 Reactance in Ohms/1000f t 1000
× (Cable length in ft) × (Cable length in ft)
3 r t p a h C – 3 0 8 2 4 X E B
12
6
08/10/2012
Single-phase voltage drop considers the load and back to the source. Thus, the ohmic cable impedances are calculated as: R L
= 2×
XL
= 2×
Resistance in Ohms/1000f t 1000 Reactance in Ohms/1000f t 1000
× (Cable length in ft) × (Cable length in ft)
3 r t p a h C – 3 0 8 2 4 X E B
13
Determine the voltage drop along a 240V, , consisting of #10 AWG THW copper conductor. The load current is 13∠-25°A. Assume PVC conduit.
3 r t p a h C – 3 0 8 2 4 X E B
14
7
08/10/2012
From the table of 600V cables, resistance = . , = . . 1.2
R L
= 2×
X L
= 2×
V drop
× 135 = 0.324Ω
1000 0.050 1000
× 135 = 0.0135Ω
= 13 A × [0.324Ω × cos(−25°) − 0.0135Ω × sin(−25°)] = 3.89V
%V dro
3.89V
=
V
× 100% = 1.62%
3 r t p a h C – 3 0 8 2 4 X E B
15
The voltage drop through the transformer , Vdrop
≈
I LS
× [R TR × cosθ − X TR × sinθ ]
In percentage, drop
=
Vdrop VLS
×
3 r t p a h C – 3 0 8 2 4 X E B
16
8
08/10/2012
Three-phase
transformer -
impedances ,
1 ⎡ (%R)(Secondary line voltage)2 ⎤ ⎢ ⎥ R TR = 100 ⎢ Transformer voltampere rating ⎥ ⎣ ⎦ 2 1 ⎡ (%X)(Secondary line voltage) ⎤ ⎢ ⎥ X TR = 100 ⎢ Transformer voltampere rating ⎥ ⎣ ⎦
ven n
TR
an
ra o,
-1
θ = tan (X/R)
3 r t p a h C – 3 0 8 2 4 X E B
%R TR = %ZTR × cosθ %X TR = %ZTR × sinθ 17
Determine the voltage drop through a , , transformer having an impedance of 4%, and an X/R ratio of 1.8. The transformer is operating at full load, 0.82 lagging power factor.
3 r t p a h C – 3 0 8 2 4 X E B
18
9
08/10/2012
The impedance angle, = . = . ° Transformer %RTR and %XTR,
%RTR = 4% x cos(60.95°) = 1.94% %XTR = 4% x sin(60.95°) = 3.50% RTR and XTR,
⎡ (1.94%)(415V ) 2 ⎤ RTR = ⎢ 115kVA ⎥ = 0.02905Ω 100 ⎣ ⎦ 1 ⎡ (3.50%)(415V ) 2 ⎤ X TR = ⎢ ⎥ = 0.05242Ω 100 ⎣ 115kVA ⎦ 1
3 r t p a h C – 3 0 8 2 4 X E B
19
Full-load current of the transformer, I LS
=
3 × 415V
∠ − cos −1 (0.82) = 160∠ − 34.92° A
Voltage drop, V drop = 160 A × [0.02905Ω × cos(−34.92°) − 0.05242Ω × sin(−34.92°)] = 8.61V %V drop
=
8.61V
×100% = 2.07%
3 r t p a h C – 3 0 8 2 4 X E B
20
10
08/10/2012
Voltage drop or voltage dips occurs due to .
The voltage drop is large when starting large motors applied to systems having a relatively high source impedance.
2 common methods to determine voltage drop due to the motor starting:
Constant Impedance Constant Current
3 r t p a h C – 3 0 8 2 4 X E B
21
A 50HP, 415V, code letter G induction motor from a 415V/240V system whose equivalent impedance is 0.01+j0.02 ohms/phase. Assume a locked-rotor power factor of 35% lagging. Calculate the percentage voltage drop during starting using (a) the constant mpe ance an e cons an curren representations.
3 r t p a h C – 3 0 8 2 4 X E B
22
11
08/10/2012
(a) The locked-rotor kVA/HP is 6.3 (take the worst case . The locked-rotor kVA durin starting, kVALR = 6.3 kVA/HP x 50HP = 315 kVA The locked-rotor current, I LR
315kVA
=
3 × 415V
= 438.23 A
T e active an reactive power uring starting, P = (315kVA) x (0.35) = 110.25 kW Q = (315kVA) x [sin(cos-1(0.35))] = 295.1kVAr
3 r t p a h C – 3 0 8 2 4 X E B
23
Locked-rotor R and X, . ⎟ = 0.1913Ω × ⎜⎜ 3 ⎝ 438.232 ⎠⎟ 1 ⎛ 295.1kVAr ⎞ ⎟ = 0.5122Ω X = × ⎜⎜ 3 ⎝ 438.232 ⎠⎟
R =
Voltage at motor terminal, V M
⎛ 0.1913 + j 0.5122 ⎟⎟ = 230.62∠0.24°V = 240∠0°⎜⎜ ⎝ 0.1913 + j 0.5122 + 0.01 + j 0.02 ⎠
3 r t p a h C – 3 0 8 2 4 X E B
Voltage drop, %V drop
=
240V − 230.62V 240V
× 100% = 3.91% 24
12
08/10/2012
(b) Constant current, = . - . ° Voltage drop (using approximation method), M
V drop
= 438.23 A × [0.01Ω × cos(−69.51°) − 0.02Ω × sin( −69.51°)] = 9.74V
%V drop
=
9.74V 240V
× 100% = 4.05%
3 r t p a h C – 3 0 8 2 4 X E B
25
In some cases, the locked-rotor power factor .
In this case, it is possible to have an approximation value of voltage drop due to motor starting by assuming that the voltage drop is in-phase with the source voltage.
For previous Example 5, the Vdro ,
Vdrop = I x Z = 438.23A x | 0.01+j0.02| = 9.8V. The %Vdrop,
3 r t p a h C – 3 0 8 2 4 X E B
%Vdrop = (9.8V/240V) x 100% = 4.08% 26
13
08/10/2012
Determine the voltage drop at the service . 8kVA@0.85 lagging power factor. 415-240V 30kVA R=1.8% X=1.5%
Service #3/0 AWG aluminum con u 120ft
3 r t p a h C – 3 0 8 2 4 X E B
40A
Service entrance panel 8kVA, 0.85 PF lagging
Step 1 – Determine all system impedances rans ormer: RTR X TR
⎡ (1.8%)(240V ) ⎤ ⎢ ⎥ = 0.03456Ω 100 ⎣ 30kVA ⎦ 2 1 ⎡ (1.5%)(240V ) ⎤ = ⎢ ⎥ = 0.0288Ω 100 ⎣ 30kVA ⎦
=
2
1
Cable/Wire: 0.13
R L
= 2×
X L
= 2×
× (120 ft ) = 0.0312Ω
1000 0.042 1000
27
3 r t p a h C – 3 0 8 2 4 X E B
× (120 ft ) = 0.0101Ω 28
14
08/10/2012
Step 2 – Determine load supplied at the end . The loading is 8kVA@0.85 PF lagging.
Step 3 – Determine the load current magnitude and phase angle. I =
240V
= 33.33 A∠ − cos −1 (0.85)° = 33.33∠ − 31.79° A
3 r t p a h C – 3 0 8 2 4 X E B
29
Step 4 – Calculate the %Vdrop along each , . Transformer: V drop
≈ 33.33 A × [0.03456Ω × cos( −31.79°) − 0.0288Ω × sin(−31.79°)] = 1.48V
%V drop
=
1.48V 240V
a e V drop
× 100% = 0.62%
re:
≈ 33.33 A × [0.0312Ω × cos( −31.79°) − 0.0101Ω × sin( −31.79°)] = 1.06V
%V drop
=
1.06V 240V
3 r t p a h C – 3 0 8 2 4 X E B
×100% = 0.44% 30
15
08/10/2012
Step 5 – Add the %Vdrop along each segment, interest. The total voltage drop at the panel is: Transformer Cable/Wire
: 0.62% : 0.44% .
3 r t p a h C – 3 0 8 2 4 X E B
31
Determine the %Vdrop at the Main Distribution Panel (SP) for the system shown below.
3 r t p a h C – 3 0 8 2 4 X E B
32
16
08/10/2012
Transformer TR1 800kVA 6600 – 1000V R = 1.5%, X = 5% 30ft two 400 kcmil copper/phase, steel conduit 2000A MDP 600kVA, 0.9 lagging PF
1500A 400A 250A 35ft #8 AWG aluminum, steel conduit
Transformer TR2 40kVA 1000 – 415V R = 3.5%, X = 4% 12ft #12 AWG copper, aluminum conduit
3 r t p a h C – 3 0 8 2 4 X E B
150A 100A
SP 20kVA, 0.85 lagging PF
33
Step 1 – Determine all system impedances rans ormer : 1 ⎡ (1.5%)(1000V ) 2 ⎤ RTR1 = ⎢ 800kVA ⎥ = 0.01875Ω 100 ⎣ ⎦ 1 ⎡ (5%)(1000V ) 2 ⎤ X TR1 = ⎢ ⎥ = 0.0625Ω 100 ⎣ 800kVA ⎦ Cable (400 kcmil): R L
=
X L
=
0.035
× (30 ft ) = 0.00105Ω
1000 0.049 1000
3 r t p a h C – 3 0 8 2 4 X E B
× (30 ft ) = 0.00147Ω 34
17
08/10/2012
Cable (#8 AWG): R L
=
X L
=
.
× (35 ft ) = 0.0455Ω
1000 0.065 1000
× (35 ft ) = 0.002275Ω
Transformer (TR2):
⎡ (3.5%)(415V ) 2 ⎤ RTR 2 = ⎢ ⎥ = 0.1507Ω 100 ⎣ 40kVA ⎦ 2 1 ⎡ (4%)(415V ) ⎤ X TR 2 = ⎢ 40kVA ⎥ = 0.1722Ω 100 ⎣ ⎦ 1
3 r t p a h C – 3 0 8 2 4 X E B
35
Cable (#12 AWG):
R L
=
X L
=
.
× (12 ft ) = 0.024Ω
1000 0.054 1000
× (12 ft ) = 0.000648Ω
Step 2 – Determine load supplied at the end o eac se men or por on o e sys em. Loading until MDP is 600kVA@0.9 lagging PF.
3 r t p a h C – 3 0 8 2 4 X E B
Loading from MDP until SP is 20kVA@0.85 lagging PF. 36
18
08/10/2012
Step 3 – Determine the load current . Through Transformer TR1, I =
600kVA 3 ×1000V
= 346.41∠ − cos −1 (0.90)° A = 346.41∠ − 25.84° A
Through Cable (#8 AWG), I =
3 ×1000V
= 11.55∠ − cos −1 (0.85)° A = 11.55∠ − 31.79° A
3 r t p a h C – 3 0 8 2 4 X E B
Through Transformer TR2, I =
20kVA 3 × 415V
= 27.82∠ − cos −1 (0.85)° A = 27.82∠ − 31.79° A 37
Summary of load currents:
3 r
Component
Load Current (A)
Transformer (TR1)
346.41∠ − 25.84° A
Cable (400 kcmil)
173.21∠ − 25.84° A / conductor
Cable (#8 AWG)
11.55∠ − 31.79° A
Transformer (TR2)
27.82∠ − 31.79° A
a e
.
−
.
t p a h C – 3 0 8 2 4 X E B
38
19
08/10/2012
Step 4 – Calculate the %Vdrop along each , . Transformer (TR1): ≈ 346.41 A × [0.01875Ω × cos(−25.84°) − 0.0625Ω × sin(−25.84°)] = 15.28V
V drop
%V drop
=
15.28V 1000V
×100% = 1.53%
a e V drop
cm :
≈ 173.21 A × [0.00105Ω × cos(−25.84°) − 0.00147Ω × sin(−25.84°)] = 0.27V
%V drop
=
0.27V 577.35V
3 r t p a h C – 3 0 8 2 4 X E B
×100% = 0.05%
39
Cable (#8 AWG): V drop
≈ 11.55 A × [0.0455Ω × cos(−31.79°) − 0.002275Ω × sin(−31.79°)] = 0.46V
%V drop
=
0.46V 577.35V
×100% = 0.08%
Transformer (TR2): V drop
≈ 27.82 A × [0.1507Ω × cos(−31.79°) − 0.1722Ω × sin(−31.79°)] = 6.09V
%V drop
=
6.09V 415V
×100% = 1.47%
3 r t p a h C – 3 0 8 2 4 X E B
40
20
08/10/2012
Cable (#12 AWG): V drop
3 r
≈ 27.82 A × [0.024Ω × cos(−31.79°) − 0.000648Ω × sin(−31.79°)] = 0.58V
%V drop
=
0.58V 239.6V
×100% = 0.24%
t p a h C – 3 0 8 2 4 X E B
41
Step 5 – Add the %Vdrop along each segment, . The total voltage drop at service panel is: Transformer (TR1) Cable (400 kcmil) a e
: 1.53% : 0.05% : .
Transformer (TR2)
: 1.47%
Cable (#12 AWG)
: 0.24%
Total
: 3.37%
3 r t p a h C – 3 0 8 2 4 X E B
42
21
08/10/2012
3 r
Types of current
Normal current
Overload current
Short-circuit current
Ground-fault current
t p a h C – 3 0 8 2 4 X E B
43
Normal, or load, current may be defined as the load under normal operating conditions.
Normal motor current varies from low values (under light loading) to medium values (under medium loading) to maximum values (under maximum loading).
Normal current flows only in the normal circuit path. The normal circuit path includes the phase and neutral conductors. It does not include equipment grounding conductors.
3 r t p a h C – 3 0 8 2 4 X E B
44
22
08/10/2012
Overload current is greater in magnitude than full load current and flows onl in the normal circuit path. It is commonly caused by overloaded equipment, single-phasing, or low line voltage, and thus is considered to be an abnormal current. Some overload currents such as motor startin currents (or locked-rotor current), are only temporary. Overload current is greater in magnitude than full-load amperes but less than locked-rotor amperes.
3 r t p a h C – 3 0 8 2 4 X E B
45
Short-circuit current is greater than lockedthousands of amperes.
The maximum value is limited by the maximum short-circuit current available on the system at the fault point.
Short-circuit current may be further classified as bolted or arcing. Large amounts of shortcircuit current will flow into a bolted fault than the arcing fault.
3 r t p a h C – 3 0 8 2 4 X E B
46
23
08/10/2012
Ground-fault current consists of any current .
Ground-fault current flow in the equipment grounding conductor for low-voltage systems.
In medium- and high-voltage systems, groundfault current may return to the source through the earth.
Ground-fault current on low-voltage systems may be classified as leakage, bolted, or arcing.
3 r t p a h C – 3 0 8 2 4 X E B
47
Synchronous generators - when a short-circuit occurs downstream of a s nchronous enerator it may continue to produce output voltage and current. Synchronous motors - delivers short-circuit current into the fault until the motor completely stops Induction motors - short-circuit current deca s very quickly. Supply transformers - Transformer impedances will also limit the amount of short-circuit current from the utility generators.
3 r t p a h C – 3 0 8 2 4 X E B
48
24
08/10/2012
3 r t p a h C – 3 0 8 2 4
Synchronous Generator
Induction Motor
X E B
Synchronous Motor 49
3 r
Totally Symmetrical Current
t p a h C – 3 0 8 2 4 X E B
Totally Asymmetrical Current
Partially Asymmetrical Current
50
25
08/10/2012
“Symmetrical" and “Asymmetrical” are terms circuit current waveform around the zero axis.
If a short-circuit occurs in an inductive reactive circuit at the peak of the voltage waveform, the resulting short-circuit current will be totally symmetrical. If a short-circuit, in the same circuit, occurs at the zero of the voltage waveform, the resulting short-circuit current will be totally asymmetrical.
3 r t p a h C – 3 0 8 2 4 X E B
51
The symmetrical short circuit current consists only waveform. It is applicable only for balanced threephase power system and can be calculated as the total line-to-neutral voltage over the total impedances on the power system.
The asymmetrical short circuit current is the actual current that flows durin a fault condition. It consists of DC and AC components that contribute to a certain amount of ‘DC offset’ in the waveform immediately after the initiation of the fault. The amount of ‘DC offset’ or asymmetry depends on the point when the fault occurs.
3 r t p a h C – 3 0 8 2 4 X E B
52
26
08/10/2012
The instantaneous peak short circuit current is on the asymmetrical short circuit current waveform. It is a function of X/R of the system. Instantaneous peak short cir cuit cu rrent Asymmetr ic al s ho rt circ ui t c ur ren t Symmetrical short circuit current
3 r t p a h C – 3 0 8 2 4 X E B
53
jXL
R
3 r
i(t)
t p a h C – 3 0 8 2 4
t=0s
+ V m sin(ω t + θ )
Fault
Line-to-Neutral Equivalent Circuit
i (t ) = I rms
=
2 ⋅ I rms sin(ω t − θ Z ) + sin(θ Z ) ⋅ e −(ω R / X ) t V m
2 ⋅ Z S
θ Z
=
tan
−1
⎛ X ⎞ ⎜ ⎟ ⎝ R ⎠
Z S
X E B
=
R 2
+ X 2 54
27
08/10/2012
First half-cycle asymmetrical fault current: I rms ,1 2
- cycle factor) × I rms = (rms half
The rms half-cycle factor: 1 First half - cycle rms multiplying factor =
T
∫ i (t )dt 2
T 0
rms S mmetrical short circuit current
3 r t p a h C – 3 0 8 2 4 X E B
55
The source impedance at a 12.47kV distribution . . . Calculate (a) the rms fault current, (b) the maximum peak instantaneous value of fault current, and (c) the rms value of the half-cycle fault current if a balanced three-phase fault occurs.
3 r t p a h C – 3 0 8 2 4 X E B
56
28
08/10/2012
(a) The line-to-neutral voltage: V LN
=
. 3
= 7.2kV
The rms symmetrical fault current: I rms
=
7.2kV (0.4 2 + 1.52 )
= 4638 A
3 r t p a h C – 3 0 8 2 4 X E B
57
(b) The system X/R ratio = 1.5/0.4 = 3.75 From table, the instantaneous peak factor is determine by interpolation: = (2.0892 – 1.9495)(3.75 – 3.0) + 1.9495 = 2.0543 The maximum peak instantaneous value of fault current is Ip = (2.0543)(4638A) = 9528A
3 r t p a h C – 3 0 8 2 4 X E B
58
29
08/10/2012
(c) The rms half-cycle multiplying factor is , = (1.191 – 1.115)(3.75 – 3.0) + 1.115 = 1.172 The rms half-cycle asymmetrical fault , Irms,1/2 = (1.172)(4638A) = 5436A
3 r t p a h C – 3 0 8 2 4 X E B
59
To determine short circuit current, the total must be established.
Common system impedances – system, transformers, cables, etc.
equivalent
All impedances placed before transformer need to be reflected to its low voltage side.
3 r t p a h C – 3 0 8 2 4 X E B
60
30
08/10/2012
Since the three-phase fault condition results in , short circuit current at a particular fault point is calculated as: I rms
=
Line - to - neutral voltage Z total
The X/R ratio is used to determine the instantaneous peak factor and half-cycle factor.
3 r t p a h C – 3 0 8 2 4 X E B
Half-cycle factor is used to calculate the asymmetrical fault current. 61
Determine
the
RMS
symmetrical,
RMS
magnitudes for a three-phase fault occurring at (a) F1 and (b) F2 for the power system shown in Figure below.
3 r t p a h C – 3 0 8 2 4 X E B
62
31
08/10/2012
Equivalent system 3-phase SC MVA = 65MVA@3.3kV, X/R = 3 Transformer TR1 750kVA 3300 – 1100V Z = 5.75%, X/R = 5 50ft three 400 kcmil copper/phase, steel conduit 1000A F1
1000A 400A 250A 10ft #4/0 AWG copper, steel conduit
Transformer TR2 75kVA 1100 – 415V Z = 1.8%, X/R = 1.5 10ft 250 kcmil copper, steel conduit
3 r t p a h C – 3 0 8 2 4 X E B
400A 250A
F2
63
Equivalent system impedance, Z sys
=
(3300V ) 2 65 MVA
= 0.1675Ω
Impedance angle, θ = tan −1 (3) = 71.57°
= 0.1675 × cos( 71.57°) = 0.05295Ω X sys = 0.1675 × sin( 71.57°) = 0.1589Ω
Rsys
3 r t p a h C – 3 0 8 2 4 X E B
64
32
08/10/2012
The equivalent system impedance referred to , 2
1100V ⎞ = 0.05295 × ⎛ ⎜ ⎟ = 0.005883Ω ⎝ 3300V ⎠ 2 1100V ⎞ ⎛ X sys ( LS ) = 0.1589 × ⎜ ⎟ = 0.01766Ω ⎝ 3300V ⎠
Rsys ( LS )
an
o
,
3 r t p a h C – 3 0 8 2 4 X E B
−1
θ = tan (5) = 78.69°
= 5.75% × cos(78.69°) = 1.13% % X TR1 = 5.75% × sin(78.69°) = 5.64% % RTR1
65
RTR1 and XTR1 of TR1,
3 r
(1.13%)(1100V ) 2
t p a h C – 3 0 8 2 4
⎢ ⎥ = 0.01823Ω 100 ⎣ 750kVA ⎦ 2 1 ⎡ (5.64%)(1100V ) ⎤ X TR1 = ⎢ ⎥ = 0.09099Ω 100 ⎣ 750kVA ⎦
RTR1
=
1
Rc and Xc of 400 kcmil cable (From Table), ⎛ 1 ⎞ ⎡ 0.035 ⎤ × 50 ft = 0.00058Ω ⎟ ⎝ 3 ⎠ ⎢⎣ 1000 ⎥⎦ ⎛ 1 ⎞ ⎡ 0.049 ⎤ × 50 ft = 0.00082Ω Xc = ⎜ ⎟ ⎢ ⎝ 3 ⎠ ⎣ 1000 ⎥⎦
Rc = ⎜
R = 0.035Ω / 1000 ft
X E B
X = 0.049Ω / 1000 ft
66
33
08/10/2012
(a) The total impedance to fault at F1, . . Transformer (TR1)
: 0.01832 + j0.09099
Cable (400 kcmil)
: 0.00058 + j0.00082
Total Z sys
: 0.024783 + j0.10947
= 0 .1122Ω
[ X / R]sys = 4.42
3 r t p a h C – 3 0 8 2 4 X E B
67
The RMS symmetrical short circuit current at , I RMS =
635.09V 0.1122Ω
= 5,660 A
Instantaneous peak factor (interpolation), = (2.1924 – 2.0892) x (4.42 – 4.0) + 2.0892 = .
3 r t p a h C – 3 0 8 2 4 X E B
Peak instantaneous current, Ip = (2.1325) x (5,660A) = 12,070A 68
34
08/10/2012
Half-cycle factor, = . – . = 1.2212
.
– .
.
Half-cycle RMS asymmetrical current, IRMS/1/2 = (1.2212) x (5,660A) = 6,912A
3 r t p a h C – 3 0 8 2 4 X E B
69
(b) Rc and Xc of #4/0 AWG cable (from table), Rc = ⎢
.
× 10 ft = 0.00063Ω ⎣ 1000 ⎥⎦ ⎡ 0.051⎤ × 10 ft = 0.00051Ω Xc = ⎢ ⎣ 1000 ⎥⎦
= 0.063Ω / 1000 ft X = 0.051Ω / 1000 ft
R
Rc and Xc (#4/0 AWG) referred to low voltage s eo , ⎛ 415V ⎞
Rc = 0.00063 × ⎜
2
⎟ = 0.00009Ω ⎝ 1100V ⎠ 2 415V ⎞ ⎛ Xc = 0.00051 × ⎜ ⎟ = 0.00007Ω ⎝ 1100V ⎠
3 r t p a h C – 3 0 8 2 4 X E B
70
35
08/10/2012
%R and %X of TR2, =
3 r
an −1 .
= . % RTR 2 = 1.8% × cos(56.31°) = 1.00% % X TR 2 = 1.8% × sin(56.31°) = 1.50%
t p a h C – 3 0 8 2 4
⎡ (1.00%)(415V ) 2 ⎤ ⎢ ⎥ = 0.02296Ω 100 ⎣ 75kVA ⎦ 1 ⎡ 1.50% 415V 2 ⎤ = = . TR 2 100 ⎣ 75kVA ⎦
RTR 2
=
1
X E B
71
The Rc and Xc of 250 kcmil cable, Rc = ⎢
.
× 10 ft = 0.00054Ω ⎣ 1000 ⎥⎦ ⎡ 0.052 ⎤ × 10 ft = 0.00052Ω Xc = ⎢ ⎣ 1000 ⎥⎦
= 0.054Ω / 1000 ft X = 0.052Ω / 1000 ft
R
R and X up to point F1 reflected to the low vo a e s e o , 2
415V ⎞ = 0.024783 × ⎛ ⎜ ⎟ = 0.00353Ω ⎝ 1100V ⎠ 2 415V ⎞ ⎛ X sys ,TR 2 = 0.10947 × ⎜ ⎟ = 0.01558Ω ⎝ 1100V ⎠
Rsys ,TR 2
3 r t p a h C – 3 0 8 2 4 X E B
72
36
08/10/2012
The total impedance to F2, .
.
Cable #4/0 AWG
: 0.00009 + j0.00007
Transformer (TR2)
: 0.02296 + j0.03445
Cable 250 kcmil Total
: 0.00054 + j0.00052 : 0.02712 + j0.05062
Z sys
= 0 .05743Ω
3 r t p a h C – 3 0 8 2 4 X E B
[ X / R]sys = 1.87
73
The RMS symmetrical short circuit current at , I RMS =
239.6V 0.05743Ω
= 4,172 A
Instantaneous peak factor (interpolation), = (1.7560 – 1.5122) x (1.87 – 1.0) + 1.5122 = .
3 r t p a h C – 3 0 8 2 4 X E B
Peak instantaneous current, Ip = (1.7243) x (4,172A) = 7,194A 74
37
08/10/2012
Half-cycle factor, = . – . = 1.0368
.
– .
.
Half-cycle RMS asymmetrical current, IRMS/1/2 = (1.0368) x (4,172A) = 4,326A
3 r t p a h C – 3 0 8 2 4 X E B
75
In single-phase system, the only possible fault - . The short circuit current is calculated as: I rms =
Line - to - neutral voltage Z total
=
240V Z total
3 r t p a h C – 3 0 8 2 4 X E B
76
38
08/10/2012
Determine
the rms symmetrical, rms , magnitudes for a single-phase, line-to-ground fault occurring at point F1 for the power system shown below.
3 r t p a h C – 3 0 8 2 4 X E B
77
Equivalent system Three-phase: MVA = 65MVA@6.6kV, X/R = 3 Single-phase: ILG = 2kA, X/R = 2.4 50kVA 6600 – 415V R = 1.9% X = 2.2% 240V
Service 100ft #4/0 AWG AL, Steel conduit
240V
3 r t p a h C – 3 0 8 2 4 X E B
50ft #12 AWG copper, steel conduit F1 78
39
08/10/2012
The equivalent system impedance, Z sys,1φ
=
2000 A
3 r t p a h C – 3 0 8 2 4
= 1.9053Ω
θ = tan −1 (2.4) = 67.4° Rsys,1φ = 1.9053 × cos(67.4°) = 0.7322Ω X sys,1φ = 1.9053 × sin(67.4°) = 1.7590Ω
Re ecting system R an X to t e 240V, ⎛ 240V ⎝ 6600V / ⎛ 240V ' X sys ,1φ = 1.7590Ω × ⎜ ⎝ 6600V /
' Rsys ,1φ = 0.7322Ω × ⎜
X E B
2
⎞ ⎟ = 0.002905Ω 3 ⎠ 2 ⎞ ⎟ = 0.006978Ω 3 ⎠
79
Transformer R and X referred to low voltage , ⎡ (1.9%)(415V ) 2 ⎤ RTR = ⎢ ⎥ = 0.06544Ω 100 ⎣ 50kVA ⎦ 1 ⎡ (2.2%)(415V ) 2 ⎤ X TR = ⎢ ⎥ = 0.07578Ω 100 ⎣ 50kVA ⎦ 1
rans ormer an or single-phase) condition, RTR,1φ = (1 / 3)(0.06544Ω) = 0.02181Ω X TR ,1φ = (1 / 3)(0.07578Ω) = 0.02526Ω
e
a -w n n
or
3 r t p a h C – 3 0 8 2 4 X E B
80
40
08/10/2012
The Rc and Xc of the #4/0 AWG AL cable, Rc = 2 × Xc = 2 ×
.
×100 ft = 0.02Ω
1000 0.051 1000
×100 ft = 0.0102Ω
The Rc and Xc of the #12 AWG copper cable, Rc = 2 ×
2.0
× 50 ft = 0.2Ω 1000 0.0541 Xc = 2 × × 50 ft = 0.0054Ω 1000
3 r t p a h C – 3 0 8 2 4 X E B
81
Total impedances to F1, System impedance
: 0.002905 + j0.006978
Transformer
: 0.02181 + j0.02526
Cable #4/0
: 0.02 + j0.0102
Cable #12
: 0.2 + j0.0054
Total
: 0.244715 + j0.047838
= 0.2493 X / R ratio = 0.2 Z total
3 r t p a h C – 3 0 8 2 4 X E B
82
41
