Comentarios y Análisis
O) Certificados bursátiles-
Permissible voltage dip is 3% of 230 volt = 6.9 volt Maximum starting current is
0.2707 6.9
= ≈ 25.489 amp
Assuming a motor efficiency of 80% and power factor of 0.8 lag, the maximum rating of a direct on-line motor is:
5
Current Starting ×
3× Volts × Efficiency × Power Factor (kW) = 5 25.489 × 1.732 × 0.4 × 0.8 × 0.8 kW = kW Or, 0.746 2.26 = ≈ 3.03 H.P.
3.2 Example 2 - Three Phase 400 volt Motor
a) Fault level at primary 11 kV busbars - 75 MVA
b) From the primary substation to the distribution substation there are 3 miles of 100 mm2 and one mile of 50 mm2 Al overhead lines.
c) At the distribution substation a 50 kVA transformer
d) From the distribution transformer to the point of common coupling there is 400 yards of 50 mm2 Al 400 volt overhead line.
Table Section of System
Equivalent Resistance
Equivalent Resistance
Ohms Ohms
J Source to 11 kV busbars at the primary substation 0 0.00320 K 100 mm2 11kV line (3 miles) 0.002115 0.002106
K 50 mm2 11kV line (1 miles) 0.001416 0.00075
M 50 KVA transformer 0.0755 0.1165 A 50 mm2 400 volt line (400 yds) 0.243409 0.108409
TOTALS => 0.32244 0.230965
≈ 0.3224 ≈ 0.2310 Assuming a starting current of five times full load current at a power factor of 0.3 lag, The volt drop per ampere of starting current is
R Cosφ + X Sinφ
= (0.3224 × 0.3) + (0.2310 × 0.9539) = volt
Case 1 - Frequent Start
Permissible voltage dip is 1% of 230 volt = 2.3 volt Maximum starting current is
0.317 2.3
= amp
Assuming a motor efficiency of 80% and power factor of 0.8 lag, the maximum rating of direct on-line motor is:
5
Current Starting ×
= 5 7.255 × 1.732 × 0.4 × 0.8 × 0.8 kW = kW Or, 0.746 0.643 = ≈ 0.86 H.P.
Case 2 - Infrequent Start
Permissible voltage dip is 3% of 230 volt = 6.9 volt Maximum starting current is
0.317 6.9
= ≈ 21.766 amp
Assuming a motor efficiency of 80% and power factor of 0.8 lag, the maximum rating of a direct on-line motor is:
5
Current Starting ×
3× Volts × Efficiency × Power Factor (kW) = 5 21.766 × 1.732 × 0.4 × 0.8 × 0.8 kW = kW Or, 0.746 1.93 = ≈ 2.59 H.P.
3.3 Example 3:- Single Phase 230 Volt Motor (from three phase supply)
(a) Fault level at primary 11kV basbars (33/11kV substation) -100 MVA.
(b) From the primary substation to the distribution substation there are 3 miles of 100 mm2 and one mile of 50 mm2 Al overhead line.
(c) At the distribution substation a 50 kVA three phase transformer.
(d) From the distribution transformer to the point of common coupling there is 400 yards of 50 mm2 Al. distribution line.
In this example it is necessary to use voltage factor to convert the equivalent impedance per phase in a 400 volt system to equivalent impedance per phase in a 230 volt system.
The voltage factor is
2 400 230 =
The return path of the single phase load current must also be taken into account.
SECTION OF SYSTEM
BASIC EQUIV.
RESISTANCE BASIC EQUIV. REACTANCE VOLTAGE FACTOR
EQUIVALENT
RESISTANCE EQUIVALENT REACTANCE
Ohms Ohms Ohms Ohms
Source to 11 kV Busbar Table - J 0 0.00160 × 0.33 0.000528 11kV O/H Line Table – K 0.002115 0.002106 × 0.33 0.00071 0.000695 0.002115 0.002106 × 0.33 0.00071 0.000695 0.001416 0.00075 × 0.33 0.000467 0.000248 0.001416 0.00075 × 0.33 0.000467 0.000248 TRANSFORMER Table – M 0.0755 0.1165 × 0.33 0.024915 0.038445 0.0755 0.1165 × 0.33 0.024915 0.038445 230 Volt Line Table - A 0.24341 0.10841 0.24341 0.10841 TOTAL 0.539004≈ 0.5390 0.296124≈ 0.2961
Assuming a starting current of 5 times full load current at a power factor of 0.3 lag, The volt drop per ampere of starting current is
R Cosφ + X Sinφ
= (0.5390 × 0.3) + (0.2961 × 0.9539) = volt
Case 1 - Frequent Start
Permissible voltage dip is 1% of 230 volt = 2.3 volt Maximum starting current is
0.444 2.3
= amp
Assuming a motor efficiency of 75% and power factor of 0.8 lag, the maximum size of direct on-line motor is:
5
Current Starting ×
Volts × Efficiency × Power Factor (kW) = 5 5.1802 × 0.23 × 0.75 × 0.8 kW = kW Or, 0.746 0.143 = ≈ 0.192 H.P.
Case 2 - Infrequent Start
Permissible voltage dip is 3% of 230 volt = 6.9 volt Maximum starting current is
0.444 6.9
= amp
Assuming a motor efficiency of 75% and power factor of 0.8 lag, the maximum size of direct on-line motor is:
5
Current Starting
× Volts × Efficiency × Power Factor (kW) = 5 15.5405 × 0.23 × 0.75 × 0.8 kW = ≈ 0.429 kW Or, 0.746 0.429 = ≈ 0.575 H.P.
3.4 Example 4: Single phase motor (from Single Phase transformer)
As example 3 but with a 15 kVA single phase distribution transformer,
The equivalent resistances and reactance remain the example 3 except for the transformer. From Table – M the transformer resistance is 0.11 and the reactance is 0.13. Note that these values are not multiplied by two in this case.
Total equivalent resistance is now 0.5992
Total equivalent reactance is now 0.3492 The volt drop per ampere of starting current is
R Cosφ + X Sinφ
= (0.5992 × 0.3) + (0.3492 × 0.9539) = ≈ 0.513 volt
Case 1 - Frequent Start
Maximum starting current is
0.513 2.3
= amp
Assuming a motor efficiency of 75% and power factor of 0.8 lag, the maximum size of direct on-line motor is:
5
Current Starting ×
Volts × Efficiency × Power Factor (kW) = 5 4.483 × 0.23 × 0.75 × 0.8 kW = kW Or, 0.746 0.124 = ≈ 0.166 H.P.
Case 2 - Infrequent Start
Permissible voltage dip is 3% of 230 volt = 6.9 volt Maximum starting current is
0.513 6.9
= amp
Assuming a motor efficiency of 75% and power factor of 0.8 lag, the maximum size of direct on-line motor is:
5
Current Starting ×
Volts × Efficiency × Power Factor (kW) = 5 13.4503 × 0.23 × 0.75 × 0.8 kW = ≈ 0.371 kW Or, 0.746 0.3712 = ≈ 0.498 H.P.
3.5 Example 5: Three Phase 400 volt Motor direct from transformer terminals
a) Fault level at primary 11kV busbars - 100 MVA
b) From the primary substation to the distribution transformer there are 3 miles of 100 mm2 and one mile of 50 mm2 Al overhead line.
c) At the distribution substation there is a 50 kVA 3-Phase transformer.
TABLE SECTION OF SYSTEM
EQUIVALENT RESISTANCE Ohms EQUIVALENT REACTANCE Ohms J Source to 11 kV Busbars 0 0.0016
K 100 mm2 Al. 11kV line (3 miles) 0.002115 0.002106 K 50 mm2 Al. 11kV line (1 mile) 0.001416 0.00075
M 50 KVA Transformer 0.0755 0.1165
TOTAL 0.079031 0.120956
≈ 0.0790 ≈ 0.1210
Assuming a starting current of 5 times full load current at a power factor of 0.3 lag, The volt drop per ampere of starting current is
R Cosφ + X Sinφ
= (0.079 × 0.3) + (0.121 × 0.9539) = volt
Case 1 - Frequent Start
Maximum starting current is
0.139 2.3
= ≈ 16.547 amp
Assuming a motor efficiency of 80% and power factor of 0.8 lag, the maximum size of direct on-line motor is:
5
Current Starting ×
3× Volts × Efficiency × Power Factor (kW) = 5 16.547× 1.732 × 0.4 × 0.8 × 0.8 kW = kW Or, 0.746 1.467 = ≈ 1.97 H.P.
Case 2 - Infrequent Start
Permissible voltage dip is 3% of 230 volt = 6.9 volt Maximum starting current is
0.139 6.9
= ≈ 49.64 amp
Assuming a motor efficiency of 87% and power factor of 0.87 lag, the maximum rating of a direct on-line motor is:
5
Current Starting ×
3× Volts × Efficiency × Power Factor (kW)
= 5 49.64× 1.732 × 0.4 × 0.87 × 0.87 kW = kW Or, 0.746 5.206 = ≈ 6.98 H.P.
3.6 Example 6: Single Phase 230 volt Motor direct from 3-phase transformer terminals
a) Fault level at primary 11kV busbars - 100 MVA.
b) From the primary substation to the distribution substation there are 3 miles of 100 mm2 and one mile of 50 mm2 Al overhead 11kV line.
c) At the distribution substation there is a 25 kVA, 3-Phase transformer.
SECTION OF SYSTEM BASIC EQUIV. RESISTANCE BASIC EQUIV.
REACTANCE VOLTAGEFACTOR
EQUIVALENT RESISTANCE
EQUIVALENT REACTANCE
Ohms Ohms Ohms Ohms
Source to 11 kV Busbar Table - J 0 0.00160 × 0.33 0 0.000528 11kV O/H Line Table – K 0.002115 0.002106 × 0.33 0.00071 0.000695 0.002115 0.002106 × 0.33 0.00071 0.000695 0.001416 0.00075 × 0.33 0.000467 0.000248 0.001416 0.00075 × 0.33 0.000467 0.000248 TRANSFORMER Table – M 0.17 0.21 × 0.33 0.0561 0.0693 0.17 0.21 × 0.33 0.0561 0.0693 TOTAL 0.114554≈ 0.1146 0.141014≈ 0.1410
Assuming a starting current of 5 times full load current at a power factor of 0.3 lag, The volt drop per ampere of starting current is
R Cosφ + X Sinφ
= volt
Case 1 - Frequent Start
Permissible voltage dip is 1% of 230 volt = 2.3 volt Maximum starting current is
0.1689 2.3
= ≈ 13.62 amp
Assuming a motor efficiency of 75% and power factor of 0.8 lag, the maximum size of direct on-line motor is:
5
Current Starting ×
Volts × Efficiency × Power Factor (kW) = 5 13.62 × 0.23 × 0.75 × 0.8 kW = kW Or, 0.746 0.38 = ≈ 0.50 H.P.
Case 2 - Infrequent Start
Permissible voltage dip is 3% of 230 volt = 6.9 volt Maximum starting current is
0.1689 6.9
= ≈ 40.853 amp
Assuming a motor efficiency of 75% and power factor of 0.8 lag, the maximum size of direct on-line motor is:
5
Current Starting ×
Volts × Efficiency × Power Factor (kW) = 5 40.853 × 0.23 × 0.75 × 0.8 kW = ≈ 1.13 kW Or, 0.746 1.13 = ≈ 1.51 H.P.
3.7 Example 7: High Voltage Consumer (11000 volts) 3-Phase Motor
a) Fault level at primary 11kV busbars (at 33/11kV substation) is 100 MVA
b) From the primary substation to the point of common coupling on the 11000 volts distribution system there are 800 yards of 185 mm2 Aluminium cable and 2400 yards of
WAPS three phase line.
TABLE SECTION OF SYSTEM
EQUIVALENT RESISTANCE
EQUIVALENT REACTANCE
OHMS OHMS
J Source to 11 kV Bus bars at the
primary substation 0 0.001600
L 185 mm2 11kV cable (800 yards) 0.000196 0.000088
K 100 mm2 11kV line (2400 yards) 0.000961 0.000957
TOTAL 0.001157 0.002645
Assuming a motor starting current power factor of 0.3 lag, The volt drop per ampere of starting current will be:
R Cosφ + X Sinφ
= (0.001157 × 0.3) + (0.002645 × 0.9539) = volt per ampere of starting current
Assuming the direct-on-line starting current to be 5 times the full load current
Case 1 - Frequent Start
Permissible voltage dip is 1% of 6350 volt = 0.0635 kV Maximum starting current is
0.00287 0.0635
= ≈ 22.13 amp (at 11kV)
Assuming a motor efficiency of 90% and power factor of 0.9 lag, the maximum rating of a direct on-line motor is:
5
Current Starting ×
3× Volts × Efficiency × Power Factor (kW) = 5 22.13× 1.732 × 11 × 0.9 × 0.9 kW = kW Or, 0.746 68.303 = ≈ 91.56 H.P.