load calculations exam question

[calwat]

If anyone could help me with an exam question, I would be grateful. It was a load calculation with many motors, The problem i had was getting the input watts correct to convert it to amps. All they gave me on the exam was a 50 hp at 480v three phase. It is my understanding that i need pf + eff to get there is there another way. thanks for any response.

 

[cadpoint]

1 hp = 746 watts

1 hp = 746 watts

use I = VA / V VA = watts ; single pole
use I = VA /V(1.732) ; Three pole

 

[cadpoint]

1hp = 746

1hp = 746

VA= wattage not a mutplier in fitst equation (space didn't Keep)
I= VA / V
; for 1 pole

I= VA/ V(1.732)
; for 3 pole

 

[Charles R. Miller]

Look up 220.50 and also read the first sentence in 430.6(A)(1).

Or, wait for the June issue of Electrical Contractor magazine. Although it covers motors on branch circuits, read ?Code In Focus? in the December 2006 issue of Electrical Contractor magazine.

Charles

 

[kingpb]

If anyone could help me with an exam question, I would be grateful. It was a load calculation with many motors, The problem i had was getting the input watts correct to convert it to amps. All they gave me on the exam was a 50 hp at 480v three phase. It is my understanding that i need pf + eff to get there is there another way. thanks for any response.

You are correct in thinking the pf and eff need to be known. There are some rule of thumb assumptions you can make for these values but it won't get you any closer then the following:

Simply use the approximation of 1KVA = 1 Hp, e.g. 50Hp would be 50KVA. Then calculate current as you normally would.

How this works:

If you assume a power factor 0f 0.85, and an efficiency of say 94%, you find that the the value obtained is very near to 1 Hp = 1 KVA. Since your approximating anyway for pf and eff, simplify the approach and this will keep you on the conservative side.

 

[lloyd B]

Here you go

Here you go

Step 1. F.L.C (full load current) T 430.248 -1 phase-T430.250 3 phase.

Step 2. Motor running overload protection or heaters 430.32a1 min,430.32c max.(Use motor name plate only for heaters if given)

Step 3.Branch Circuit Wire Size.430.22 FLCx125%=required ampacity
Table 310.16 size wire ampacity to insulation.

Step 4. Branch circiut over current protection(fuse or cb) shall be selected from table 430.52.First select the type of motor (1 phase or 3 phase AC or DC wound rotor,code letter) next select type of protection(non-time delay,dual element,inverse time CB) now select the percentage from the propper colunm and multiply it times the FLC of the motor.Use 240.6 to select the standard size,code permits the next size up per 430.52 ex. 1

Step 5 Feeder conductor size 430.24. Multiply the largest rated motor in FLC by 125% and add the FLC of all the other motors connected to the (SAME feeder) conductor for required ampacity T 310.16.

Step 6 Feeder over current protection 430.62.Select the largest branch circuit (over current device) and add all the other motor FLC connected on the (SAME feeder) to select feeder fuse or .Code does not permit going up must go down . (Rember on three phase motors its per phase so draw it out on paper dont add up all the motors.)

 

[wirenut1980]

While kingpb's rule of thumb is a decent approximation for the real world, I agree with Lloyd's response if this is for a test question. They will expect you to use the table in Article 430 to get Full load current from HP.