I've been reading similar post here, but just wanted to run my understanding of the sizing a main service given a combination of single and three phase loads. I've put together some sample loads and a example main service below.
Example Main Service
3 Phase, 200A, 120/240V High Leg Delta --> (240V)(200A)(1.73) = 83040VA = 83.040kVA available
Loads
Single Phase
3) 240V, 1.6A, pf=0.9 --> (240V)(1.6A)(1.73)/.9 = 738.1VA
4) 240V, 4A, pf=0.85 --> (240V)(4)(1.730)/.85 = 1953.9VA
Now I sum all single and three phase VA.
Total Used kVA: 6.37
My final kVA is way smaller than my avaible, so can I safely say (in this bad example), that my service is adequate?
Assuming all my loads are balanced the total max amperage I could see across any phase would be (1000)(6.37kVA)/((1.73)(240V)) = 15.3A. So I could theoretically have a 3ph 20A 120/240V service be adequate for the sample loads listed.
Is my rough understanding of the process correct?
I'm obviously not an electrician, just interested in understanding the process more. Thank you for taking the time to read through it all.
Example Main Service
3 Phase, 200A, 120/240V High Leg Delta --> (240V)(200A)(1.73) = 83040VA = 83.040kVA available
Loads
Single Phase
- 120V, 15 A, pf=0.9 --> (120V)(15A)/0.9 = 2000VA
- 240V, 7A, pf=1 --> (240V)(7A)/1 = 1680kVA
3) 240V, 1.6A, pf=0.9 --> (240V)(1.6A)(1.73)/.9 = 738.1VA
4) 240V, 4A, pf=0.85 --> (240V)(4)(1.730)/.85 = 1953.9VA
Now I sum all single and three phase VA.
Total Used kVA: 6.37
My final kVA is way smaller than my avaible, so can I safely say (in this bad example), that my service is adequate?
Assuming all my loads are balanced the total max amperage I could see across any phase would be (1000)(6.37kVA)/((1.73)(240V)) = 15.3A. So I could theoretically have a 3ph 20A 120/240V service be adequate for the sample loads listed.
Is my rough understanding of the process correct?
I'm obviously not an electrician, just interested in understanding the process more. Thank you for taking the time to read through it all.
