I NEED help to finding heat output of 15 hp electric motor and then use that data to calculate heat radiation?
hi
i m working on my corn mill and my corn mill need new electric or diesel motor with 15 hp. i need to calculate the heat output of a 15 HP motor and then use that data to determine the heat radiated at(btw grinding unit and motor at different distances);
1 inch away
2 inches away
3 inches away
…uptil 6 inches away.
the motor info is :
Baldor JPM2333T 15HP 1755RPM 3 PHASE 60HZ 254JP 0748M TEFC F MOTOR
Catalog Number: JPM2333T
Specification Number: 07H301X798
Horsepower: 15
Voltage: 208-230/460
Hertz: 60
Phase: 3
Full Load Amps: 41-37.8/18.9
Usable at 208 Volts: N/A
RPM: 1755
Frame Size: 254JP
Service Factor: 1.15
Rating: 40C AMB-CONT
Locked Rotor Code: J
NEMA Design Code: A
Insulation Class: F
Full Load Efficiency: 89.5
Power Factor: 82
Enclosure: TEFC
Baldor Type: 0748M
DE Bearing: 6309
ODE Bearing: 6307
Electrical Specification Number: 07WGX798
Mechanical Specification Number: 07H301
Base: RG
Mounting: F1
i need help like how to calculate and which eqn i have to use it.
really need help please please.
thanks
The rated mechanical output of the motor is 15 Hp or 15 X 746 = 11190 watts.
The nameplate efficiency is 89.5% or 0.895.
Using losses = output power X (1-efficiency) / efficiency), the losses are 11190 X (1-.895) / 0.895 = 1313 watts.
The motor’s nameplate rated ambient temperature is 40C. The performance information on the Baldor web site indicates that the typical temperature rise of the windings at full load is 80C. The motor’s surface temperature would be considerably less that the winding temperature, but assume that the surface temperature is 120C or 393K.
The radiated heat in watts can be calculated using q(W) = rT^4Ae, where r is the Steffan-Boltzman constant = 5.67E-8, T is the Kelvin temperature, A is the surface area in m^2 and e is the emissivity. The surface area can be estimated at 0.6 m^2 from the drawing provided on the Baldor web site. The emissivity of gray painted steel can be estimated to be 0.9. The maximum radiated heat can then be calculated to be 730 watts. The remaining 583 watts can be presumed to be transferred to the air by direct conduction. The radiated energy would also be absorbed by the air if not transferred to a fairly close object.
Obviously the heat dissipated by a diesel engine would be considerably more since the efficiency is only about 30%. However, quite a bit of the heat could be carried out of the area by piping the cooling water to a remote radiator.
Revision:
A small diesel engine is going to be a lot less efficient than 30%. I think 15% may be more likely.
September 27th, 2009 at 6:43 pm
This is not an easy question; I could calculate this for you but, really, I won’t do that for free. You really need to hire a professional for this.
References :
September 27th, 2009 at 7:11 pm
The rated mechanical output of the motor is 15 Hp or 15 X 746 = 11190 watts.
The nameplate efficiency is 89.5% or 0.895.
Using losses = output power X (1-efficiency) / efficiency), the losses are 11190 X (1-.895) / 0.895 = 1313 watts.
The motor’s nameplate rated ambient temperature is 40C. The performance information on the Baldor web site indicates that the typical temperature rise of the windings at full load is 80C. The motor’s surface temperature would be considerably less that the winding temperature, but assume that the surface temperature is 120C or 393K.
The radiated heat in watts can be calculated using q(W) = rT^4Ae, where r is the Steffan-Boltzman constant = 5.67E-8, T is the Kelvin temperature, A is the surface area in m^2 and e is the emissivity. The surface area can be estimated at 0.6 m^2 from the drawing provided on the Baldor web site. The emissivity of gray painted steel can be estimated to be 0.9. The maximum radiated heat can then be calculated to be 730 watts. The remaining 583 watts can be presumed to be transferred to the air by direct conduction. The radiated energy would also be absorbed by the air if not transferred to a fairly close object.
Obviously the heat dissipated by a diesel engine would be considerably more since the efficiency is only about 30%. However, quite a bit of the heat could be carried out of the area by piping the cooling water to a remote radiator.
Revision:
A small diesel engine is going to be a lot less efficient than 30%. I think 15% may be more likely.
References :
September 27th, 2009 at 7:24 pm
There is no solution unless you can actually measure the surface temperature. Since this is a TEFC motor (totally enclosed fan cooled) most of the heat energy from efficiency loss is going to be removed by convection through the fan and never show up as a radiation loss.
References :