hi all,

need a rough relationship between a given kva feeding transformer and the maximum hp motor that would run good with dol connection

 

the summary (the one liner) "maximum hp motor for dol starting with given kva transformer" as its written, is a mistake.

kindly read it as "maximum hp dol connected motor with a given kva transformer"

 

As you did not specify a transformer type, voltage class, kVA or hp, I will give you an example:

Typically you would use a 2,000 kVA power transformer to start and run a 1,500 hp motor across-the-line (dol)

This is for a 4,160 volt 3-phase 4-wire feed to the drive motor in an air compressor.

I hope this helps,
William Hinton

 

William... I would be very interested in the motor's kVA/Hp ratio?

Regards, Phil

 

William... I meant NEMA Locked-Rotor Code.

Phil

 

dear phil corso and william hinton,
thanks

the basic idea i started out with is, the maximum hp (dol) starting depends on the capacity of the power system, its ability to provide current and tolerate voltage drop.

if i increase the kva of the transformer, i can increase the hp of the motor?
i was looking for a rough estimate, how much hp change for some kva change(general, a rough graph), if its there.

and please correct me if i am wrong somewhere

 

Engineer... since the V-Drop experienced by the motor is proportional to the load carried by the xfmr, there is an easier way!

If the Transformer Capacity (kVAc) and Percent Impedance Voltage (Zt) are known, as well as the Starting kVA, then use the following solution:

1) Calculate the short-circuit duty at the transformer secondary:

kVAsc = (kVAc)/(%Zt/100)

2) Calculate the Starting kVA of the motor.

kVAsm = (kVA/Hp)x Hp

3) Calculate the V-Drop to Motor.

V-Drop, % = 100x(kVAsm/kVAsc)

4) Example:
Xfmr = 1,500kVA,Z = 5.0%;
Mtr = 200Hp, Start kVA/Hp = 6.0

a) kVAsc = 1,500/(5.0/100) = 30,000kVA

b) kVAsm = 6.0 x 300 = 1,800kVA

c) V-D,% = 100 x 1,800/30,000 = 6.0%

The above should be adequate for the solution to your problem.

Regards,
Phil

 

Engineer... I committed three (3) 'OOPs' (the most feared word in an operating-room):

OOPs 1) I neglected to mention my solution was loosely based on a transformer's Regulation Factor. That is Reg (%) = 100 x (VNL-VFL)/VFL.

OOPs 2) Motor Starting-current is assumed to be inductive, thus lagging.

OOPs 3) The final V-Drop, then equals, VDm'= V-Drop x Zt. So for the example, VDm' = 0.05 x 5% = 0.3%.

Phil Corso