Expressing motor loads in polar form

dumbkiwi
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Homework Statement


I am having difficulty with a question on my Electrical course and the expressing a motor load voltage and current in polar form. The motor is single phase 2.8 KW efficiency 70% and 80% of full load runs with a power factor of 0.8 lagging. Voltage 400v.

Express the motor loads as impedances in polar and complex number forms assuming the motors are operating at 400v. This "express" has me baffled and have spent 4 hours trying to get past this bit of the problem. My notes are not clear on the form this should take and I am now stuck as it were.



Homework Equations





The Attempt at a Solution


I have worked out that 80% of 2.8 Kw is 2.24kw which with a voltage of 400v gives a current of 7amps. With a power factor of 0.8 I get a Cos of 36.86 degrees. The current is lagging. If current is expressed as 7xCos38.86 is voltage then 400xCos38.86? My notes are not very clear on these two expressions and can't progress until I have this information.

Many thanks, Dumbkiwi.
 
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Hi dumbkiwi! Welcome to PF! :smile:
dumbkiwi said:
Express the motor loads as impedances in polar and complex number forms assuming the motors are operating at 400v. This "express" has me baffled and have spent 4 hours trying to get past this bit of the problem. My notes are not clear on the form this should take and I am now stuck as it were.

The impedance of a load (a combination of components) in an AC current is a complex number Z = R + jX where R is the resistance of the load and X is its reactance.

For a load for which the voltage leads the current by a phase angle φ:

Z = |Z|cosφ + j|Z|sinφ.​

Polar form is the form re, or in this case Z = |Z|e :wink:
 
Hi Tiny Tim,

Thanks for your reply, I think it will help but I am too tired to try and work through it just now but will post again when I have had time to look at.

Regards Dumbkiwi.
 
hello dumbkiwi,

Is that you Calum If it is I am struggling on this assignment too! fancy meeting up and putting our heads together?

salv
 
Hey Salv,

Yeah could do that. I'll send you a pm with my phone number.

See you soon mate.

Regards Callum.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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