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I've an old handbook from 2005... but I don't think this has changed.
In the Annex, it provides an example for 'ranges of unequal rating' (Example D6B)
However, this doesn't specify whether the service is 3ph or 1ph. I'm assuming 1ph, as if you refer to Example D5A, which is for 208Y/120 3ph, it specifically addresses the provision in 220.55 "Where two or more single-phase ranges are supplied by 3-phase, 4-wire feeder or service, the total load shall be calculated on the basis of twice the maximum number connected between any two phases.
My question specifically is related to extending the example in D6B to a 3ph service. Is my procedure below sound?
For 3ph service
30 Ranges
Maximum between any two phase legs = 30 ÷ 3 = 10
10 x 2 = 20
Table 220.55 Column C for 20 ranges = 35kW
408 kW ÷ 30 = 13.6 kW average
13.6 kW exceeds 12 kW by 1.6 kW --> use 2 kW
5% x 2 = 10%
35 kW x 10% = 3.5 kW increase
35 kW + 3.5 kW = 38.5 kW
Per phase demand = 38.5 kW ÷ 2 = 19.25 kW
Equivalent 3Φ load = 19.25 x 3 = 57.75 kW (value to be used in selection of feeders)
Is there any error in the above procedure?
Computing currents to see if the numbers seem reasonable...
if 1ph at 240v, the feeder current would be 49.5kW ÷ 240 = 206.26A
for my above computation of 3ph, the feeder current would be 57.75 kW ÷ (208 * sqrt(3)) = 160.4 A
Thx for any guidance/clarification.
In the Annex, it provides an example for 'ranges of unequal rating' (Example D6B)
However, this doesn't specify whether the service is 3ph or 1ph. I'm assuming 1ph, as if you refer to Example D5A, which is for 208Y/120 3ph, it specifically addresses the provision in 220.55 "Where two or more single-phase ranges are supplied by 3-phase, 4-wire feeder or service, the total load shall be calculated on the basis of twice the maximum number connected between any two phases.
My question specifically is related to extending the example in D6B to a 3ph service. Is my procedure below sound?
For 3ph service
30 Ranges
Maximum between any two phase legs = 30 ÷ 3 = 10
10 x 2 = 20
Table 220.55 Column C for 20 ranges = 35kW
408 kW ÷ 30 = 13.6 kW average
13.6 kW exceeds 12 kW by 1.6 kW --> use 2 kW
5% x 2 = 10%
35 kW x 10% = 3.5 kW increase
35 kW + 3.5 kW = 38.5 kW
Per phase demand = 38.5 kW ÷ 2 = 19.25 kW
Equivalent 3Φ load = 19.25 x 3 = 57.75 kW (value to be used in selection of feeders)
Is there any error in the above procedure?
Computing currents to see if the numbers seem reasonable...
if 1ph at 240v, the feeder current would be 49.5kW ÷ 240 = 206.26A
for my above computation of 3ph, the feeder current would be 57.75 kW ÷ (208 * sqrt(3)) = 160.4 A
Thx for any guidance/clarification.