hi i m working on my corn mill and my corn mill need new electric ordiesel motor

Question

hi
i m working on my corn mill and my corn mill need new electric ordiesel motor with 15 hp. i need to calculate the heat output of a15 HP motor and then use that data to determine the heat radiatedat(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 FMOTOR

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 useit.
really need help please please.
thanks

Explanation / Answer

Assuming that your motor is running at full load, that means it isputting out 15hp: The output power in watts is P_out = (15hp)/(745.7W/hp)=11185Watts The full load efficiency is 89.5%. Efficiency = P_out/P_in. Thatmeans your input power is: P_in=P_out/efficiency= 11185W/0.895 =12497 Watts All of your power losses are from friction, windage, and windingresistances, etc. All of these losses produce heat. Your powerlosses are P_loss = P_in - P_out = 1312 Watts of heat Now, I don't know the dimensions of your motor, but we can make arough approximation. Let's assume it is a sphere-shape with a radius of 0.25 meters. Ifwe assume an even distribution of heat loss, then the heat willflow radially from the sphere. Then the heat loss per unit area atany given point will be: heat/area = P_loss/(4r^2) , where ris the distance from the center of the sphere to the point inquestion. For example, at the surface of our spherical motor, heat/area = 1312 / (4*0.25^2) = 1670 Watts/meter^2 If you prefer English units: 1Watt*second = 1 Joule; 1 BTU =1055.06 J: then 1 Watt = (1/1055.06) BTU/second 1 meter = 39.37 inches 1 meter^2 = 1550 inches^2 So on the surface of our motor, we have: heat/area = (1760)*(1/1055.06 BTU/second) / (1550 inches^2) =0.0010762 BTU/sec/in^2 At 1 inch away, we have r = 0.25m*(39.37 in/m) + 1 in = 10.843inches P_loss = 1312/1055.06=1.2435 BTU Then our heat per unit area is 1.2435 / (4*10.843^2) =0.00084168 BTU/sec/in^2 To find the heat at farther distances, just keep adding to yourradius. Remember that this spherical approximation is not exact, sothe hot spots on your real motor are likely to have higher heatflow. Also, you can adjust the radius used in these calculationsbase on the size of your actual motor. The simplest way is to findthe distance from the middle of your motor to the point you areinterested in, and use that distance for 'r' in the abovecalculations.

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