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Posted
Gentleman are plant has been running for 4 month know, and the avergare running hours per start of our large motor (6.6kV and more than 1000KW) is more less 42 hours, this number seems very small for me on a plant that supposed to be 24 hour operation.

I would expect a least 700 hours.

CAn some one help me on this one, maybe operation are starting and stoping equipment to often,.....

Attach file with critical motor and with all motor

Excel SpreadsheetStatistics.xls (832 Kb, 19 downloads)
 
Posts: 5 | Location: Rass Laffan, Qatar | Registered: 14 August 2007Reply With QuoteEdit or Delete MessageReport This Post
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Daniel,
If a motor has to accelerate large inertia's like heavy duty fans with run-up times of 5 - 10 seconds or more then one should realize that the lifetime of the rotor is predominantly determined by the number of starts (or accumulated total [I^2 * t]), not running hours. Please refer to various recent postings on rotor bar problems.
Regards, Arie Mol, NL
 
Posts: 124 | Location: Wierden, Netherlands | Registered: 06 March 2006Reply With QuoteEdit or Delete MessageReport This Post
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In order to determine the heating effects on a rotor, you need to determine two things. One is the acceleration time and the other relates to the losses by the rotor. The effect is heating of the rotor bars and core which, when excessive, will cause the bars to become brittle.

The acceleration time is calculated as follows:

Accel time in seconds = [(WK^2)(rpmf-rpmi)]/[(308)(accel T, lb-ft)]

Where WK^2 is the inertia of the system, rpmf is the final speed being considered, rpmi is the initial speed being considered, 308 is a constant and T is the torque at that point.

For the Torque, you cannot just take the average torque as the formula requires a fairly constant torque across the speed being evaluated. This means that you must run this formula for segments of the torque curve up to full speed.

Then you need to determine the rotor loss in kw-sec (kilowatt-seconds) for each of the above time periods:

[(0.00136)(WK^2)(synch rpm)^2(initial slip^2-final slip^2)]/5910

You then figure the amount of time for each segment which will give you the total heat, in kW) generated by the rotor.

The amount of running time after start actually provides time for the rotor to cool as most of the heat is generated as the motor accelerates due to the high rotor slip frequency. Once up to speed, the I^2R losses are significantly lower.

Howard


Howard W Penrose, Ph.D., CMRP
President, SUCCESS by DESIGN Reliability Services
Author: "Physical Asset Management for the Executive (Caution: Don't Read this on an Airplane)" and;
"Electrical Motor Diagnostics: 2nd Edition"
 
Posts: 803 | Location: Connecticut | Registered: 12 April 2005Reply With QuoteEdit or Delete MessageReport This Post
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