Continouous Load - study question

[starbright28]

Hi

I've had this on my test and I am not sure on how to find the answer.

You have a 120/240 volt feeder that is protected by a 150 ampere break that supplies 40 amps of continuous load. How much continuous load my be added to his feeder?

This one truly has me stumped.

Its that studying time before the exam again. I want to list things that i struggle with - but not sure if to combine in one post or not.

Thanks!

 

[iwire]

You may add up to 110 amps of continuous load.

Standard breakers and fuses may be loaded to 100% non-continuous and 80% continuous.

Obviously not a great design but that brings us to 90.1(B)s 'may not be adequate for good service or future expansion' etc.

 

[gregorsc]

Good luck

Good luck

The multiplier for contiuous lodas is 125%. So I think you would take the 150 amp breaker and divide it by 1.25(same as 125%) = 120amps.If you have 40 amps contiuous subtract that from 120 and that leaves 80 amps. So you can add 80 amps cont. load to the 40 and the 150 amp breaker will be ok. I believe one of the code sections that states this is 215.2a1.

 

[charlie b]

80% of 150 is 120. You have 40; you can add 80 more.

But the trick part of this question is this, how did you (meaning the test question's author) come to know that the existing load is 40? If that is based on a load calculation that accounts for all connected load, then my answer above stands.

But if it is based on a "30-day load measurement," and if that represents the maximum demand measured during that period, then I would first add 25% to the measured value of 40. That leaves room for only 70 more amps. That, at least, is how my local authority would make me do the math.

 

[charlie b]

Bob, please note that the original question discussed only continuous loads.

 

[iwire]

I apologize for being so far off, please ignore my posts above.

Thanks for setting me straight Charlie and gregorsc.

 

[starbright28]

Thanks for the replies. I was looking in section 215 but sometimes that key word or answer doesn't jump off the page at you.

my poor code book is highlighted so much from studying and classes - i will be ready for a new code book. Im not purchasing one until i take this test - since the test is still on the '05 code.

for clarificition of the problem
i basically take the total amp of the breaker, divide it by 125% to get total continuous watts. From those continuous watts total I subtract load that was stated to find out how much more can be added, correct?

seems way too simple for a problem that just looks complex! :smile:

 

[Gustav]

Well looking to the overall dimensioning of main disconnecting means I would suggest, the following
1. disconnect is not stated for 100% rating, i.e. 80% rated
this results in 120A max. cont. load.
2. assuming that the 40A is the highest load at all
==>> 40A x 1.25 = 50A
120A - 50A = 70A
So you can add another 70A if the 40A is max.

==>> for your explanation:
I suggest that you have to take the highest load multiplying with 1.25 and adding all other loads mulitplied with 1.0

 

[starbright28]

stumped again

stumped again

what is the correct way to solve this porblem, so i can write it in for my notes?

do I take 150amps and divide by 1.25%??

or do I take 150amp and times by 1.25%???

just wondering.

didn't get the clarification.

 

[don_resqcapt19]

Amanda,
Divide by 1.25 or multilpy by .8
don

 

[Dennis Alwon]

what is the correct way to solve this porblem, so i can write it in for my notes?

do I take 150amps and divide by 1.25%??

or do I take 150amp and times by 1.25%???

just wondering.

didn't get the clarification.

Amanda-- there are 2 ways to go about this you may take 150 and divide by 1.25 not 1.25%--- 125%=1.25

I find it easier to take the 125 and multiply it by 80%. Since 80%=.80 you would then say 150 X .80. Either scenario will give you 120 amps.

 

[charlie b]

40A x 1.25 = 50A

120A - 50A = 70A
So you can add another 70A

Sorry, Gustav, but the answer is 80, not 70. Your error is that you ?double-counted? the 80% factor. You took 80% of the 150 to get the 120. Then you added 25% to the 40 to get 50. That second step was incorrect. You can use every bit of the 120 as continuous loads. You already have 40 continuous amps; you don?t add 25% to that because you have room for 120 continuous amps. Therefore, you can add 80 more before you exceed 120.

 

[steveng]

please help

please help

what is the correct way to solve this porblem, so i can write it in for my notes?

is it allowed to write the formula's in the blank section pages of the code book, in the back, before you go in to take the exam jmen?:-?

 

[starbright28]

What is the correct answer to the problem?

What is the correct equation?

I'm seeing varances again.

 

[Dennis Alwon]

What is the correct answer to the problem?

What is the correct equation?

I'm seeing varances again.

Amanda I think it has been made clear Charlie and I have stated the correct method. What are you having trouble with.

150amp X .8= 120

Since you used 40 amps already it means you have 80 left.

 

[starbright28]

SORRY i get confused sometimes with the go between others but it doesn't seem to be clear.

thanks for clearing it up. no problem. just wanted to know.

 

[pgordon]

80 more here