Graffeo: Ind. Motor

Professional Engineer & PE Exam Forum

Help Support Professional Engineer & PE Exam Forum:

This site may earn a commission from merchant affiliate links, including eBay, Amazon, and others.
Joined
Dec 6, 2018
Messages
65
Reaction score
2
Can someone please help how Pmax derived here?

image.png.9b565a62491545141de5d9d368ff9d78.png


 
Work through chapter 15 of Wildi's book. He goes through the equivalent circuit step by step.

Much more thorough than what can be explained on a message board.

If you don't have the book - get it and read it! I read it cover to cover and was glad I did. I picked up an copy on eBay for ~$20. 

 
Sorry for the late response. Thank you  @ellen3720 for your kind suggestion. I do have Wildi book. I just want to know if Graffeo Pmax came from same logic as mentioned in Wildi. Yes definitely i will follow Wildi. :)

 
I tried to solve Graffeo Ex38 by Wilid method but i got different answers than one in the Graffeo.

I think i am missing something with my calculation.

Please let me know who have done this problem with Wildi method.

For those who don't know the question:

"A 120V, 6OHz, 6 pole, delta connected, three phase induction motor has the following characteristics. At standstill: Z stator = 0.2+j 0.25 ohm; Z rotor = 0.3+j 0.35 ohm. Find Pmax. The slip at which Pmax occurs and the Torque N- m at that point."

Book ans: Pmax= 16.88 KW; Slip=0.278; Torque=186.05 N-m

My ans: Pmax=8.687 KW; Slip=0.476; Torque=69 N-m

 
I tried to solve Graffeo Ex38 by Wilid method but i got different answers than one in the Graffeo.

I think i am missing something with my calculation.

Please let me know who have done this problem with Wildi method.

For those who don't know the question:

"A 120V, 6OHz, 6 pole, delta connected, three phase induction motor has the following characteristics. At standstill: Z stator = 0.2+j 0.25 ohm; Z rotor = 0.3+j 0.35 ohm. Find Pmax. The slip at which Pmax occurs and the Torque N- m at that point."

Book ans: Pmax= 16.88 KW; Slip=0.278; Torque=186.05 N-m

My ans: Pmax=8.687 KW; Slip=0.476; Torque=69 N-m
Still waiting for someone to response on this.

 
What formula are you using for slip?
From Wildi equation 15.8:

For max power/torque; R2/S=IZ1I

where R2= 0.3

Z1= r1+jx = 0.2+j(0.25+0.35) = 0.2+j0.6 = 0.632<71.6 deg

Therfore, S=R2/IZ1I = 0.3/0.632 = 0.475

 
Page 103 if Geaffeo explains that max power occurs when the load resistance is the same as stand still impedance.

for this condition, slip = r_rotor divided by

(r_rotor + Ze)

Ze is the square root of re ^2 + xe^2

where re is r_rotor plus r_ stator

and xe is x_rotor plus x_ stator

 
Finally torque is pmax/w_m

pmax calculated above is 16875 watts

w_m= (1-s_p) *w_s

s_p is defined as slip for pmax, 0.278

w_s = 4pi f/ p

this is the same as 120f/p multiplied by 2pi/ 60 to convert to rad per sec

w_s=4pi* 60/6 poles = 125.6

w_m= (1-.278)* 125.6= 90.6 rad per second

torque=16875/90.6 = 186.258Nm

 

Latest posts

Back
Top