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Distribution Transformer Sizing

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jmlporter

Electrical
Jan 10, 2005
1
Can you determine the transformer size (kVA) given the estimated consumption (kWh) of that customer? If yes with what formulas?
 
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I am afraid, not.

If you can estimate kwh, why can't you estimate maximum kW?

kVA=kW/pf. Assume pf of 0.8 if not known. Have some margin like 25% for growth or not stressing the system.

 
Some utilities have criteria for sizing residential transformers based on the number and size of residences served.
 
Think about this. Two houses have identical load. One has no insulation and the other is fully insulated. The demand would be about the same but the KWH for the two homes would be vastly different.
 
It's not quite a "no" - if you know the power factor (and its variability) of the loads, and the length of time (and the variation with time) of the energy consumption, you can. The transformer size is a power rating. The consumption is an energy rating. The factor that ties them together is time (and pf to get from kW to kVA).

The engineering depts of most utilities work with the kVA (or kW) ratings. It's only the bean counters and revenue guys who hassle about the kWhrs :)


Bung
Life is non-linear...
 
With wareagle's assumption of identical load, I think he is mostly right. The a/c would be on some of the time in both homes, and if they have the same size a/c unit, the demand would be the same (for the a/c). The a/c would stay on longer in the uninsulated house, increasing the kWh. The uninsulated house probably would have a little more demand on the average because there is a greater chance of having other loads on at the same time as the a/c if the a/c runs longer. The same logic would apply for electric heat in the winter.

Although engineering depts work with kVA or kW ratings, kWh is usually the only thing measured for individual residential customers. Many utilities have developed relationships between kWh usage and kW demand. There is no universal formula because it depends a lot on local conditions. The REA (now RUS) used to use a historic relationship for estimating feeder loads and published a series of tables. These were the basis for the "a" and "b" factors that you see in some distribution feeder analysis programs that can use kWh as input.

 
We calculate a 'load factor' to approximate a relationship between kWHr and kW (or kVA).

The load factor is defined as the ratio of the average kW (or kVA) for a service for all hours of a period, divided by the peak kW (or kVA).

LF = kWhr / (hours * kW(peak) )

Many industries or commerical services follow fairly consistant patterns. For example, a 24/7/365 automated warehouse will tend to have a laod factor of 75-85%. A 'mom'n'pop' restauraunt will have 30-45%, etc.

In my areas, feeders as a whole tend to have a monthly load factor of 70% (average/peak) and an annual load factor of 50-55%.

So long as you accept that these are averages and individiual services may vary, this can be a useful tool.
 
david das

To solve sizing transformer,when three trans. are to supply a 225kva load-with connections wye to wye-can you assume supply at 480.
 
You might try looking in Westinghouse's "Electric Utility Engineering Reference Book", vol. 3. Yeah, it's old, but I think it's still more or less quite useful.
 
For padmount transformers you need to be a little more careful because anything over 51 degrees Celsius or 124 degrees Fahrenheit will instantly cause a first or second degree burn when touched. According to General Electric's direction for installing dry transformers ( under 600 volts ) the case will have a surface temperature of 90 degree Celsius at full load which is hot enough to cause 2nd degree burns.

Also, when you heat up a transformer you are paying for electrical generation and not getting to bill it. The practice of deliberately overloading transformers was to save on no load core losses. Since modern transformers have something like 10% of what the core loss was 100 years ago, some formulas are apt to be out of date.

Also, wound cores with very thin laminations have been able acheive losses comparable to stacked cores that use amorphous iron. The problem with stacked cores is that you have air gaps ( which increase magnetic density in adjacent laminations ) at the corners and at the corners the flux is 45 to 90 degrees against the grain orientation. Wound cores avoid that problem and Acme has found a way to economically produce wound cores.

Deliberate overloading of oil filled distribution transformers is also less acceptable if the customer has a heat pump or some other form of both air conditioning and electric heat. I have installed York's relay box for heat pumps tht are added onto a gas or oil furnace. This relay box uses an outside thermostat with form C contacts to determine when it is time to switch to fossil fuel. Add on heat pumps are apt to become more popular as gas prices go up. Here in northern Ohio most of the heating season is mild enough for heat pumps and the technology is better due in part to heftier motors.
 
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