I am abut t take y journeyman and I need help....ASAP

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tbr

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I have some quetions that are on my pre test study guide that I need some brushing up.. The first question is A 240 volt 2 horse power , single phase AC motor requires a branch circuit capable of carrying a minimum of A. 30 amps B. 25 amps C. 20 amps D. 15 amps The nxt question is The minimum neutral demand on an apartment service for 11 8kw ranges would be.... A 88kw B 32kw C 20.8 kw D 18.2 kw the next question is For the purpose of computing neutral demand, a 15 kw range would add A 15kw B 9.2kw C 7.36kw D 6.44 kw The next questin is A 230 volt 1 horse power singl phase AC motor is connected to a branch circuit supplied by #14 THW conductors. The MAXIMUM inverse time delay breaker permitted to protect this circuit would be a A20 amp breaker B 25 amp breaker C 30 amp breaker D 40 amp breaker. The next question is a 230 volt 1 horse power single phase AC motor is connected to a branch circuit supplied by #14 THW conductors. The overload protection for this circuit would be a A 11.5 amps B 12.5 amps C 15.0 amps D 20.0 amps The next question is a 100' run of rigid nonmetallic conduit is installed on a day when the temprature is 700 F. If the temprature increases to 1000 , this conduit would be expected to increase how many inches in length? A o.25 B 1.22 C 1.42 D 2.31 If anyone can please take the time to give me the answers and explain how the answer is found or the steps it will be VERY appreciated..... I have taken electrical classes in the past but it has been awhile since then.... Thanks
 
I'll give some pointers for #1.

Step 1: Go to Table 430.248 - to find your 'full load current' for this motor.
Step: Apply the requirements of 430.22(A).

This is simple arithmetic and will give you the answer you are looking for.
 
tbr said:
I have some quetions that are on my pre test study guide that I need some brushing up..

The first question is :
A 240 volt 2 horse power , single phase AC motor requires a branch circuit capable of carrying a minimum of
A. 30 amps
B. 25 amps
C. 20 amps
D. 15 amps

The next question is:
The minimum neutral demand on an apartment service for 11 8kw ranges would be....
A 88kw
B 32kw
C 20.8 kw
D 18.2 kw

The next question is:
For the purpose of computing neutral demand, a 15 kw range would add
A 15kw
B 9.2kw
C 7.36kw
D 6.44 kw

The next questin is:
A 230 volt 1 horse power singl phase AC motor is connected to a branch circuit supplied by #14 THW conductors. The MAXIMUM inverse time delay breaker permitted to protect this circuit would be a
A 20 amp breaker
B 25 amp breaker
C 30 amp breaker
D 40 amp breaker.

The next question is:
A 230 volt 1 horse power single phase AC motor is connected to a branch circuit supplied by #14 THW conductors. The overload protection for this circuit would be a
A 11.5 amps
B 12.5 amps
C 15.0 amps
D 20.0 amps

The next question is a 100' run of rigid nonmetallic conduit is installed on a day when the temprature is 700 F. If the temprature increases to 1000 , this conduit would be expected to increase how many inches in length?
A o.25
B 1.22
C 1.42
D 2.31

If anyone can please take the time to give me the answers and explain how the answer is found or the steps it will be VERY appreciated..... I have taken electrical classes in the past but it has been awhile since then.... Thanks
Click to expand...

I dont do tests, but I did neaten up your post. :)
 
TBR, the standard procedure in these situations is for the original poster to answer the questions himself and seek verification. We will not do your homework for you, that doesn't help you. Please use Stickboy's neater version. :)
 
ok tbr. i'll take guestion # 4 for 500.00 dollars please ..an say the answer is (c)30 amps .....:D the key word is maximum first look in your code book an look up single phase motors 430-248 230 volts single phase = 8 amps now look at 430-52 but the key word is maximum.now look at 430.52 exception#2 (c)up to 400% so 8 amps x 400% = 32 amps but you shall not exceed 400% so drop down to a 30 amp cb....
 
I was wondering about question #5.

I did the calcs per 430.32(A)(1) & (C) - none of the answers matched. [I used the 'all other motors at 115%...made the 'assumption' that the nameplate matched the tables...then ignored all this and used the max values...still no match].

...then I took a look at 430.32(D)(2)(a), x - is that code section applicable? I can't otherwise figure a way for one of the answers to match...even if that code section is relevant, which overload rating would you use?
 
I noticed several answers don't match also

Calc for the 11 8kw ranges...

700 degree day (what death valley?) A 300 degree diference would make it grow over a foot!

How can you calc OL without nameplate info? fla,Temp rise, etc
 
Q1. 430.22(A) requires a minimum conductor ampacity (MCA) of 125% of motor FLC as determined by 430.6(A)(1): Table 430.248 yields FLC = 12 amps. MCA = 1.25 x 12 = 15.0.
[Answer D].

Q2. 220.61(B)(1) permits the neutral to be sized at 70% of a feeder supplying household electric ranges where the maximum unbalanced load has been determined in accordance with Table 220.55. For 11 ranges, Column C of the Table yields 26 kW, which, when multiplied by .70 = 18.2 kW. [Answer D].

Q3. As in Q2, 220.61(B)(1) permits the neutral to be sized at 70% of a feeder supplying household electric ranges where the maximum unbalanced load has been determined in accordance with Table 220.55. Note 1 to the Table requires the feeder to a single 15 kW range to be sized at [8 kW + (15% x 8 kW)] = [8 kW + 1.2 kW] = 9.2 kW. The neutral for the feeder is 0.7 x 9.2 = 6.44. [Answer D].

Q4. As in Q1, 430.22(A) requires a minimum conductor ampacity (MCA) of 125% of motor FLC as determined by 430.6(A)(1): Table 430.248 yields FLC = 8 amps. The circuit protection is determined by 430.52(A), (B), and (C), which normally leads to Table 430.52:
250 % of 8.0 = 20 amps. However, the question asks for the MAXIMUM rating, so one must look at 430.52(C)(1) Ex. No. 2(c) which allows the rating of an inverse time circuit breaker to be increased to a maximum of 400% for this particular motor: 400% x 8.0 = 32 amps. There is no provision here to round up to the next standard size breaker, so one must round down to 30 amps. [Answer C].

Q5. The answer is determined as shown in the first half of Q4, above: 20 amps. [Answer D].

Q6. It is reasonable to assume that 700F really means 70 deg F and that 1000F means 100 deg F. That being the case, 352.44 requires the use of expansion fittings when Type PVC expands or contracts more than 1/4 inch in accordance with Table 352.44 under the
conditions stated. For a temperature change of 30 deg F (100 - 700 = 30), the Length Change of PVC Conduit is 1.22 in. [Answer B].
 
I am not in disagreement with You, however my job status does not reflect my talent nor the ability to calculate electrical calculations. I am 24 years old and for the matter of fact you telling Me Im in need of more classes only determines Me to study harder. I didnt ask for anybody to do my home work for me....I am human and only human... I asked for help and guidance...... For those who did reply in a positive way I appreciate it greatly....
 
Wow, I'm impressed this took this long to get to page 2.

TBR, I think what is being said (in general) is that most of us like to help those who are willing to help themselves, and don't take kindly to, seeming, demands for hand-outs.

Being that this is a semi-open forum, I hope you are further encouraged to provide in a well formatted (for sticky-boy) response including your thoughts regarding your purposed answers to the previously posted questions.

I further believe that once you have followed the suggestion in the previous paragraph that you will provoke responses from those here that will be far more beneficial than ones limited to a A, B, C or D.

Good Luck,
Doug S.

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If in doubt pick B! or C and sometimes D but never A. :D
 
Kevin,

Kudos!

That long analysis is certainly the most Encouraging response
I have heard from an electrician this month!

Well, maybe I just had a bad week,
but I liked the way you worked everything out.

Glen.
 
how can you put #14 wire on a 30 amp cb. I believe the answer you are looking for is 20amp. Look at table 310-16
 
the key word in q #4 is size 14 thw. You can't put a conductor on a cb that is greater than the ampacity of the conductor. I believe the answer you are looking for is 20amps. Refer to Table 310-16.
 
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