220.44 question

[patsfan51]

Here is a question i seem to be having a little trouble with. If anyone could help pls..

How many 120v 20a 2-wire branch circuits are required for an office area that measures 43'x80' with an unknown receptacle outlet load? Receptacles and lighting are not permitted onthe same branch circuit for this installation.
a) 6
b) 7
c) 8
d) 9

My math..
43x80x3.5=12040 t.220.44=> 10000+(2040x.50)=11020
120vx20a=2400 11020/2400=4.59 x 2 = 9.18==> 10 circuits

Do I not round down... I was sure that we round up on number of circuits.....
or do i not mutiply by 2 and find 4 mandatory outlets in art. 210.

thanks

 

[Volta]

Here is a question i seem to be having a little trouble with. If anyone could help pls..

How many 120v 20a 2-wire branch circuits are required for an office area that measures 43'x80' with an unknown receptacle outlet load? Receptacles and lighting are not permitted onthe same branch circuit for this installation.
a) 6
b) 7
c) 8
d) 9

My math..
43x80x3.5=12040 t.220.44=> 10000+(2040x.50)=11020
120vx20a=2400 11020/2400=4.59 x 2 = 9.18==> 10 circuits

Do I not round down... I was sure that we round up on number of circuits.....
or do i not mutiply by 2 and find 4 mandatory outlets in art. 210.

thanks

Given the text of the question, should you be looking at 220.44 at all?

Maybe start it again at 220.12 and 220.14.

 

[jumper]

I think the key to this problem is find your lighting load and unknown rec. load. I believe Volta is correct, go back to 220.14.

 

[patsfan51]

the way that i read 220.14 (I) would be 180va per outlet. that would be for known outlet loads.220.14 (K) at 1va/sqft. One va per square foot throws my math off. And if the 3.5va for offices is for general lighting where would I begin....

not sure what you mean by finding the lighting load to get the receptacle load when question says not on same circuit.

 

[jumper]

the way that i read 220.14 (I) would be 180va per outlet. that would be for known outlet loads.220.14 (K) at 1va/sqft. One va per square foot throws my math off. And if the 3.5va for offices is for general lighting where would I begin....

not sure what you mean by finding the lighting load to get the receptacle load when question says not on same circuit.

Your lighting load is 12040 and your rec. is based on 1va/ sqft. Forget about 180va per rec. you do not know how many recs. Now apply T 220.42 to lights and T220.44 to recs.

 

[patsfan51]

total load minus lighting is recept. load?
lighting:
12040-3000=9040x.35=3164+3000=6164va
recep.:
12040-6164=5878va

??????

 

[jumper]

Okay, here is what I come up with:
Lights
12040va@ 100%=12040va
12040va/120v=100.1666667a
100.16666667a=100a 220.5(B)
100a=5 circuits

Recs
43*80=3440
3440* 1va=3440va @ 100%
3440va/120v=28.66667a or 29a
29a=2 circuits

7 circuits if I am correct.

 

[patsfan51]

ah ha!
but office lighting should be at 125% no?

12040x1.25=15050
15050/120=125.4
125/20=6.25
7 circuits for lighting
2 circuits for recept.
9 total?

 

[patsfan51]

Thank you volta and jumper. i appreciate your help with this one.

 

[jumper]

ah ha!
but office lighting should be at 125% no?

12040x1.25=15050
15050/120=125.4
125/20=6.25
7 circuits for lighting
2 circuits for recept.
9 total?

Are using 210.19(A)(1) for that 125%, because I believe that should apply to the conductors. #12 is good for 25 amps. 25*.8=20 or 20*1.25=25. I am back to 7 circuits, unless I missed something else

 

[jumper]

ah ha!
but office lighting should be at 125% no?

12040x1.25=15050
15050/120=125.4
125/20=6.25
7 circuits for lighting
2 circuits for recept.
9 total?

Correct I think

Are using 210.19(A)(1) for that 125%, because I believe that should apply to the conductors. #12 is good for 25 amps. 25*.8=20 or 20*1.25=25. I am back to 7 circuits, unless I missed something else

I was wrong. I missed 210.20, thanks Volta, 9 is the right answer, I think.