I need help with these calculations.

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kefox81

Member
Location
Rainelle WV USA
I am currently studying for my Masters test in West Virginia. This math is giving me trouble. I have a practice test but I keep getting the wrong answers even though I think I am right. And some of it though I cant figure out what to do. Should I post them individually or can I put them all in this thread and someone show me step by step how to do them? Any help would be greatly appreciated.
 
junkhound

junkhound

Senior Member
Location
Renton, WA
Occupation
EE, power electronics specialty
And hopefully folks can help you see the WHY and HOW the math can work, rather than just plugging numbers into a standard set of rules.
 
K

kefox81

Member
Location
Rainelle WV USA
Here is problem one.

Here is problem one.

First let me start out by saying I appreciate the responses. Here is question number 1.
For the purpose of calculating neutral demand, a 15 kW range would add:
a. 15 kW
b. 9.2kW
c. 7.36 kW
d. 6.44 kW

The book says the correct answer is D. But I come up with B, 9.2 kW.
Here is how I am doing it. I go to chart 220.55 in my NEC book and read note 1. I then counted up from 12 to 15 which gave me a total of 3. I then added 5% to each one to get 15%. I then take and go to column C of chart 220.55 and find that I use the number 8. I then take and do 8kW x 15% which gives me 1.2 kW. I then add the 1.2 kW to the 8 and get 9.2 kW. Am I right or is the book right? If I am wrong can someone please show me what I did wrong? Thanks again.
 
Ponchik

Ponchik

Senior Member
Location
CA
Occupation
Electronologist
You take your calculation that gave 9.2 KW then take 70% of it in order to size the neutral. The question is asking about the neutral.

This number 6.44 is added to other neutral calculation to come up with a TOTAL neutral load inorder to size the neutral conductor for the service.
 
T

topgone

Senior Member
The catch is "neutral demand"! NEC allows computing the neutral load equal to 70% of dryers, counter-mounted cooking units, electric ranges and ovens. Please read 220.61.

The value would then be: 9.2 kW X 0.70 = 6.44 kW!
 
Smart $

Smart $

Esteemed Member
Location
Ohio
kefox81 said:
... Here is question number 1.
...
Click to expand...
I've not taken the WV master electrician test but I have heard it is quite tough to pass. My understanding it is intentional because master electricians serve as the AHJ representive... i.e. there are no fire marshall inspections.

At one time I was prepared to take the master or journeyman test, but the contractor wanted me on the job working and they already had a master-E on site. Test schedule was months out. I had to settle for getting an apprentice license even though I had an out-of-state IBEW journeyman ticket, a letter from my BA to verify passing my local's journeyman test, and several Ohio-city journeyman licenses. The only Ohio license WV reciprocates is a contractor license.
 
K

kefox81

Member
Location
Rainelle WV USA
Thanks for the help. Here is problem 2.

Thanks for the help. Here is problem 2.

I really appreciate all of the help. I see what I was doing wrong now. Here is problem number 2.
What would be the minimum general lighting load for a 12' x 60' mobile home?
a. 2.16 KW
b. 2.7 KW
c. 2160 KW
d. 2700 KW

I turned to Article 550.18 (A) Then I take 12 x 60 x 3 which give me 2160. But the book says the answer is A.
Am I supposed to take and move the decimal to go from VA to KW?

Again I appreciate all of the help.
 
jumper

jumper

Senior Member
Location
3 Hr 2 Min from Winged Horses
kefox81 said:
I really appreciate all of the help. I see what I was doing wrong now. Here is problem number 2.
What would be the minimum general lighting load for a 12' x 60' mobile home?
a. 2.16 KW
b. 2.7 KW
c. 2160 KW
d. 2700 KW

I turned to Article 550.18 (A) Then I take 12 x 60 x 3 which give me 2160. But the book says the answer is A.
Am I supposed to take and move the decimal to go from VA to KW?

Again I appreciate all of the help.
Click to expand...

Then I take 12 x 60 x 3 which give me 2160 Watts = 2.16 KW. KW is kilowatts = 1000 watts.

FYI. the true answer is actually 2160 VA or 2.16 KVA.
 
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G

ggunn

PE (Electrical), NABCEP certified
Location
Austin, TX, USA
Occupation
Consulting Electrical Engineer - Photovoltaic Systems
kefox81 said:
I really appreciate all of the help. I see what I was doing wrong now. Here is problem number 2.
What would be the minimum general lighting load for a 12' x 60' mobile home?
a. 2.16 KW
b. 2.7 KW
c. 2160 KW
d. 2700 KW

I turned to Article 550.18 (A) Then I take 12 x 60 x 3 which give me 2160. But the book says the answer is A.
Am I supposed to take and move the decimal to go from VA to KW?

Again I appreciate all of the help.
Click to expand...

You can immediately throw out c and d for being ridiculously too large (over two megawatts). A big part of a successful test taking strategy is to eliminate unreasonable answers.
 
K

kefox81

Member
Location
Rainelle WV USA
I really appreciate all of the help. I'm learning a lot.

I really appreciate all of the help. I'm learning a lot.

Thanks everyone. I am learning a lot. I really appreciate all of the help. Here is the next question.
A restaurant has the following:
One 16 kW cook top plus one 8 kW dishwasher plus one 5 kW water heater plus one 3 kW dough roller plus one 2 kW exhaust fan.
Using the optional calculation method for kitchen equipment, the minimum service demand for these loads would be:
a. 32 k
b. 25.6 k
c. 25.6 kW
d. 22.4 kW

I turn to article 220.56 and use chart 220.56. I add 16kW for the cooktop, 8 kW for the dishwasher, 5 kW for the water heater and 3 kW for the dough roller. This gives me 32 kW. I use table 220.56 and find out that the demand factor is 80%. I then take 32 x .8 = 25.6 kW. This is the same answer the book gives. Am I correct?
 
Smart $

Smart $

Esteemed Member
Location
Ohio
kefox81 said:
Thanks everyone. I am learning a lot. I really appreciate all of the help. Here is the next question.
A restaurant has the following:
One 16 kW cook top plus one 8 kW dishwasher plus one 5 kW water heater plus one 3 kW dough roller plus one 2 kW exhaust fan.
Using the optional calculation method for kitchen equipment, the minimum service demand for these loads would be:
a. 32 k
b. 25.6 k
c. 25.6 kW
d. 22.4 kW

I turn to article 220.56 and use chart 220.56. I add 16kW for the cooktop, 8 kW for the dishwasher, 5 kW for the water heater and 3 kW for the dough roller. This gives me 32 kW. I use table 220.56 and find out that the demand factor is 80%. I then take 32 x .8 = 25.6 kW. This is the same answer the book gives. Am I correct?
Click to expand...
The question says to use optional calculation method. See 220.88.
 
G

ggunn

PE (Electrical), NABCEP certified
Location
Austin, TX, USA
Occupation
Consulting Electrical Engineer - Photovoltaic Systems
kefox81 said:
Thanks everyone. I am learning a lot. I really appreciate all of the help. Here is the next question.
A restaurant has the following:
One 16 kW cook top plus one 8 kW dishwasher plus one 5 kW water heater plus one 3 kW dough roller plus one 2 kW exhaust fan.
Using the optional calculation method for kitchen equipment, the minimum service demand for these loads would be:
a. 32 k
b. 25.6 k
c. 25.6 kW
d. 22.4 kW

I turn to article 220.56 and use chart 220.56. I add 16kW for the cooktop, 8 kW for the dishwasher, 5 kW for the water heater and 3 kW for the dough roller. This gives me 32 kW. I use table 220.56 and find out that the demand factor is 80%. I then take 32 x .8 = 25.6 kW. This is the same answer the book gives. Am I correct?
Click to expand...
Once again, two answers (a and b) can be eliminated out of hand; the units expression is wrong.
 
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