# Starting Torque- DC



## clemente (Mar 28, 2008)

Heres a question:

A DC motor is rated 900v, 200a, 225hp, 1500 rpm. the armature circuit resistance, including series winding is .001 ohm. if the starting current is limited to 6 times the rated current what is the starting torque.

Thanks for help in advance. I'm having problems with how to utilize the starting current, to get starting torque.


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## Dark Knight (Mar 28, 2008)

clemente said:


> Heres a question:
> A DC motor is rated 900v, 200a, 225hp, 1500 rpm. the armature circuit resistance, including series winding is .001 ohm. if the starting current is limited to 6 times the rated current what is the starting torque.
> 
> Thanks for help in advance. I'm having problems with how to utilize the starting current, to get starting torque.


Clememte,

Don't have my books with me right now and I have no way to check if my comment is correct or how to really help you. But I think that the starting torque it is not a function of the starting current.

I am shooting from the top of my head but check first if you can get something using the horse power and the rpm(omega). I am going to look into it as soon as I have a chance.


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## mudpuppy (Mar 29, 2008)

Motors are not my area of expertise, but I'll take a stab at it. Someone let me know if anything looks wrong here.

From the problem statement, it _sounds_ like this is a series-wired DC motor. If we can assume it is, in fact, series wound then see the EERM section 39.9. Specifically, see equation 39.16:

T1/T2 = (Ia,1/Ia,2)2. In words: the torque is directly proportional to the square of the current.

Let's assign T1 as the starting torque and T2 as the rated torque. We are given Ia,2 (rated current) is 200, so Ia,1 is 1200. So (Ia,1/Ia,2)2 is (1200/200)2 = 36. It follows from 39.16 that T1 = 36*T2. All we need to solve for T1 is to know T2.

The EERM equation 39.1 (and 40.5) gives torque as a function of horsepower and speed. Plugging in the given rated horsepower and speed gives:

T2 = 5252*225/1500 = 787.8 ft-lb. Solving for the T1 above gives 28361 ft-lb. (Did I get it right?)

Also see EERM section 39.14 for a little more explanation on starting torque vs. starting current. And Figure 39.9 for a figure of the torque vs. current for the different DC motor types. Note that if this is not, in fact, a series-wired motor (e.g. if it is compound), then my answer is wrong and I'd have to do a lot more digging to solve it.


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## clemente (Mar 29, 2008)

great help. you got the correct answer, 28,368 ft-lb.

your assumption of a series wired dc machine was correct. I overlooked the speed, torque, and current equation 39.16.

Now its very clear. Thanks again.


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## mudpuppy (Mar 30, 2008)

Glad I could help. I learned something on this too. Like BIO, I didn't realize starting torque was directly related to current, but it makes sense--a higher current will give a higher flux, which should result in higher torque.


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## Dark Knight (Mar 30, 2008)

mudpuppy said:


> Glad I could help. I learned something on this too. Like BIO, I didn't realize starting torque was directly related to current, but it makes sense--a higher current will give a higher flux, which should result in higher torque.


The day you stop learning that is it!!!!!

Thanks MP and Clemente.


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