# Graffeo: Ind. Motor



## Sthabik PE (Jan 12, 2019)

Can someone please help how *Pmax* derived here?


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## Sthabik PE (Jan 28, 2019)

Is there any one who can help me with this one?


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## ellen3720 (Feb 2, 2019)

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.


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## Sthabik PE (Mar 1, 2019)

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.


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## Sthabik PE (Mar 1, 2019)

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


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## Sthabik PE (Mar 5, 2019)

Sdhabik said:


> 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.
> 
> ...


Still waiting for someone to response on this.


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## Wow_PE! (Jul 9, 2019)

Sdhabik said:


> Still waiting for someone to response on this.


What formula are you using for slip?


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## Sthabik PE (Jul 16, 2019)

Wow! said:


> 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&lt;71.6 deg

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


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## Wow_PE! (Jul 16, 2019)

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


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## Wow_PE! (Jul 16, 2019)

So max slip equals

.3/(.3+.78)=.278


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## Wow_PE! (Jul 16, 2019)

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


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## Sthabik PE (Jul 18, 2019)

can you try with wildi method as well plz.


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