how to calculate apparent power ratings

[mari]

Hello,

Anybody can help me with the following problem:
say you have a building that needs 208Y/120. The utility primary distribution is12.47KV. On the problem they give you the single phase load 30 W at .95 lagging and also they give you three phase load 150 at .98 lagging.
Why on the solution they add 30W + 150W to get the total active power. Also they add the reactive power by single phase and 3 phase. Then they divide it by 3 to get rating for each transformer.
The formula that i dont get is that they add single phase power and three phase power together, never seen that before..
Anybody know why?

Thanks in advance

 

[Ingenieur]

Because it is all in watts
I assume you meant kw?

S va = P w + jQ var
where:
P = cos ang x S
Q = sin ang x S
cos ang = pf

total P = 180 kw

Find S kva for each
3 ph: S = 150/0.98 = 153.1 kva
1 ph: S = 30/0.95 = 31.6 kva
total S 184.7 kva

Find ang for each
ang = arccos pf
3 ph 11.5 deg
1 ph 18.2 deg

Find Q var for each
Q var = sin ang x S
3 ph 30.5 kvar
1 ph 9.9 kva
total Q 40.4 kvar

Total
184.7 kva = 180 kw + j40.4 kvar = 184.7/12.65 deg

 

[Smart $]

Welcome... :thumbsup:



Who are they? :roll:

Should we assume the single phase load is balanced (as it appears, Ingenieur already has).


Personally, all I would have done is this...

...
Find S kva for each
3 ph: S = 150/0.98 = 153.1 kva
1 ph: S = 30/0.95 = 31.6 kva
total S 184.7 kva

...and divided by 3.

 

[Ingenieur]

Welcome... :thumbsup:



Who are they? :roll:

Should we assume the single phase load is balanced (as it appears, Ingenieur already has).


Personally, all I would have done is this...

...and divided by 3.

I wanted him to see the concept and relationship of S, P and Q
makes it easier to derive any qty once the overall relationship is grasped

 

[Sahib]

Because it is all in watts
I assume you meant kw?

S va = P w + jQ var
where:
P = cos ang x S
Q = sin ang x S
cos ang = pf

total P = 180 kw

Find S kva for each
3 ph: S = 150/0.98 = 153.1 kva
1 ph: S = 30/0.95 = 31.6 kva
total S 184.7 kva

Find ang for each
ang = arccos pf
3 ph 11.5 deg
1 ph 18.2 deg

Find Q var for each
Q var = sin ang x S
3 ph 30.5 kvar
1 ph 9.9 kva
total Q 40.4 kvar

Total
184.7 kva = 180 kw + j40.4 kvar = 184.7/12.65 deg

Wont the 184.7 kva transformer size undersized for any future expansion
because per phase 3 phase load=153.1/3 and single phase load=31.6kva and so total per phase load={153.1/3)+31.6. So total three phase load=3*[{153.1/3)+31.6]=248kva?

 

[Smart $]

Wont the 184.7 kva transformer size undersized for any future expansion
because per phase 3 phase load=153.1/3 and single phase load=31.6kva and so total per phase load={153.1/3)+31.6. So total three phase load=3*[{153.1/3)+31.6]=248kva?

Assume the OP means total single phase load (30kW) on all three phases (10kW each).

I already asked about unbalanced single phase load with no reply.

...
Should we assume the single phase load is balanced (as it appears, Ingenieur already has).
...

 

[Ingenieur]

Wont the 184.7 kva transformer size undersized for any future expansion
because per phase 3 phase load=153.1/3 and single phase load=31.6kva and so total per phase load={153.1/3)+31.6. So total three phase load=3*[{153.1/3)+31.6]=248kva?

I did not size the transformer
only determind the load

 

[Sahib]

I did not size the transformer
only determind the load

I think OP means load for 3 single phase transformers.