MDP caculation

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Dbronx

Senior Member
Location
Colorado
ten 50 amp subpanels needed averaging 30-40 amp loads. How would the MDP load be caculated by the SDP circuit breakers, or the sum total amps or watts of sub panels.
 

topgone

Senior Member
ten 50 amp subpanels needed averaging 30-40 amp loads. How would the MDP load be caculated by the SDP circuit breakers, or the sum total amps or watts of sub panels.

Let me imagine what you got there!
You should know your loads first. Which loads are considered as 'continues' and which are not. You need to classify them in order to compute for the correct size of the wires (125% of continues load amps plus the rest of the load in the circuit). With the size of wires done, you decide which breaker size protects the wires used, keeping in mind that you will only load up the breakers up to 80% of its rating (if not indicated as 100%-rated breakers).

The purpose of the main circuit breaker is to protect the panelboard bus, the transformer, and the wiring between the transformer and the panelboard. Keep in mind that the main breaker will be the rated higher than the sub-panel breaker sizes, else, your main will trip ahead of the sub-panel breakers. The MDP breaker size is found by multiplying the largest single phase amps on any leg of the breaker by 125%. ( I am assuming you cannot balance the loads especially if there are single-phase loads).
 

Dbronx

Senior Member
Location
Colorado
Caculated MDP

Caculated MDP

The SOW specified 50 amp services for housing units. So a 3 phase 50 amp service at 80% would be 14,410.66 watts X 10 = 144,106.63 watts divided by 360 = 400.29619 X 1.25 = 500.37024
A 500 amp service with two cables in parallel 4C 250 with 2/0 ground.The 50 amp services will need #8 with #8 ground. Is that correct ?
 

Dbronx

Senior Member
Location
Colorado
Split unit Air conitioners caculation

Split unit Air conitioners caculation

Fifty 9K Air-conditioners will be installed at a large K-span building. THe 9K air-conditioner is rated at 6.9 amps 115V specified a 15A CB, 6.9 X 115 = 793.5 watts The largest load is heating a resistive load no 125% for largest motor caculated. So 793.5 X 50 = 39,677 watts divided by 360 for 3 phase load = 110.208 X 125% = 137.760.
A 150 AMP service is needed a 4C 1/0 with #6 ground is needed correct.
 

jumper

Senior Member
Fifty 9K Air-conditioners will be installed at a large K-span building. THe 9K air-conditioner is rated at 6.9 amps 115V specified a 15A CB, 6.9 X 115 = 793.5 watts The largest load is heating a resistive load no 125% for largest motor caculated. So 793.5 X 50 = 39,677 watts divided by 360 for 3 phase load = 110.208 X 125% = 137.760.
A 150 AMP service is needed a 4C 1/0 with #6 ground is needed correct.

Lights, fridges, stoves, WHs, etc???

What is the total speced load and sq.ft. per unit?

These are dwelling units,correct?
 

Dbronx

Senior Member
Location
Colorado
HVAC designed BY Engineer

HVAC designed BY Engineer

These units have been specified, and a dedicated SDP has been specified for air-conditioners
 
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Dbronx

Senior Member
Location
Colorado
This is a additional SDP for a large building, it is only for air conditioners loads. I want to caculate the total air condition load, then service, then feeders for service.
 

topgone

Senior Member
This is a additional SDP for a large building, it is only for air conditioners loads. I want to caculate the total air condition load, then service, then feeders for service.

For us to be able to chip-in, can you tabulate your loads per feeder. It would be a lot easier that way than wording it all out.:?:?
 

Dbronx

Senior Member
Location
Colorado
quote_icon.png
Originally Posted by Dbronx
Fifty 9K Air-conditioners will be installed at a large K-span building. THe 9K air-conditioner is rated at 6.9 amps 115V specified a 15A CB, 6.9 X 115 = 793.5 watts The largest load is heating a resistive load no 125% for largest motor caculated. So 793.5 X 50 = 39,677 watts divided by 360 for 3 phase load = 110.208 X 125% = 137.760.
A 150 AMP service is needed a 4C 1/0 with #6 ground is needed correct.

This is a additional SDP for a large building, it is only for air conditioners loads. I want to caculate the service, then feeders .
 

topgone

Senior Member
quote_icon.png
Originally Posted by Dbronx
Fifty 9K Air-conditioners will be installed at a large K-span building. THe 9K air-conditioner is rated at 6.9 amps 115V specified a 15A CB, 6.9 X 115 = 793.5 watts The largest load is heating a resistive load no 125% for largest motor caculated. So 793.5 X 50 = 39,677 watts divided by 360 for 3 phase load = 110.208 X 125% = 137.760.
A 150 AMP service is needed a 4C 1/0 with #6 ground is needed correct.

This is a additional SDP for a large building, it is only for air conditioners loads. I want to caculate the service, then feeders .

Lemme try.

If you supply the load with 3-phase, it will be impossible for you to split your loads equally on the three phases. You will be forced to load 16 units on each of the two phases and the remaining 18 units loaded on the last phase.

Phase currents: 110.4 amps (16 x 6.9); 110.4 amps and 124.20 amps (18 x 6.9).

That would result into line currents of:
Ia =203.7 amps,
Ib = 191.22 amps, and
Ic = 203.7 amps.
So you take the largest current and design around that value. A 4/0 copper will do if the distance is just within 100 feet or a bigger wire if the distance is quite far.

203.7 amps x 1.25 = 254.625 amps. say 250 AT breaker.
Others, please check my math, been rusty for a while.
 

Dbronx

Senior Member
Location
Colorado
For us to be able to chip-in, can you tabulate your loads per feeder. It would be a lot easier that way than wording it all out.:?:?

THe 9K air-conditioner is rated at 6.9 amps 115V specified a 15A CB, 6.9 X 115 = 793.5 watts The largest load is heating a resistive load no 125% for largest motor caculated. So 793.5 X 50 = 39,677 watts divided by 360 for 3 phase load = 110.208 X 125% = 137.760.
A 150 AMP service is needed a 4C 1/0 with #6 ground is needed correct.
 
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