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Alexander Rajan
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Join Date: Feb 2008
Posts: 3

Motor & Compressor

09/01/2008 8:25 AM

Hi buddy,

Please clarify my doubt. We have coupled one motor and compressor by using gear drive.Gear Ratio is 2.7.The compressor RPM is 655rpm and the motor rpm is 1800rpm.The torque at compressor end is 482.5Nm.We just want to know whether the delivering torque of the motor shaft and the compressor receving end (after the speed reduction by using the Gear) torque same or not?
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Rica
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Join Date: Sep 2008
Posts: 3
#1

Re: Motor & Compressor

09/01/2008 8:49 AM

I suppose yes.
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kik0kik0
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Join Date: Sep 2006
Posts: 31
#2

Re: Motor & Compressor

09/01/2008 11:02 AM

I do not get it, how comes that the torques are equal after you had reduced the speed?
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JohnDG
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#3

Re: Motor & Compressor

09/01/2008 11:15 AM

For an ideal (lossless) gearbox, the torque is increased or reduced by the gear ratio. 2:1 reduction gives ≈2 x torque, and so on.

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user-deleted-1104
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#4
In reply to #3

Re: Motor & Compressor

09/02/2008 1:20 AM

Well done JohnDG!

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sayed sarhan
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Join Date: Apr 2008
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#5

Re: Motor & Compressor

09/02/2008 5:22 AM

Hi Alexander,

power=torque*omega

where omega is in rev. per sec.

therefore,T1*omega 1=T2*omega 2

i.e. T1=T2*omega2/omega1

If motor speed is 1800 rpm,gear ratio is 2.7 and T2 is482.5 Nm

then,T1 will be 1302.88Nm

note:

Compressor speed in this case is 666.6 rpm,not 655 rpm.(1800/2.7=666.6)

sayed sarhan
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Anonymous Poster
#6
In reply to #5

Re: Motor & Compressor

09/02/2008 5:54 AM

Hi.....Thank for yhe reply.....But yu answer not suppose to be correct..bcoz the deliverin torque is different before gear and after gear.......>>>We to multiply the compressor torque by the 1/X(Wr X stands for Gear Ration)...then we will get the motor torque....Actually the delivering torque is different ...the compressor neees high torque....and the motor delivering torque is lesss with keeping of Power remain same.....
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Anonymous Poster
#7
In reply to #5

Re: Motor & Compressor

09/02/2008 5:55 AM

Hi.....Thank for yhe reply.....But yu answer not suppose to be correct..bcoz the deliverin torque is different before gear and after gear.......>>>We to multiply the compressor torque by the 1/X(Wr X stands for Gear Ration)...then we will get the motor torque....Actually the delivering torque is different ...the compressor neees high torque....and the motor delivering torque is lesss with keeping of Power remain same.....
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JohnDG
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#8
In reply to #7

Re: Motor & Compressor

09/02/2008 6:24 AM

Motor speed = 1800 rpm

Gear ratio = 2.7:1

Compressor speed = 1800 / 2.7 = 667 rpm

Compressor torque = 482.5 Nm

Motor torque ≈ 482.5 / 2.7 ≈ 179 Nm

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Anonymous Poster (2); JohnDG (2); kik0kik0 (1); Rica (1); sayed sarhan (1); user-deleted-1104 (1)

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