An electrical cabinet is cooled by a 18000 BTU air conditioner. The heat that is transfered from inside the cabinet to the outside of the cabinet PLUS the heat generated by the A/C system componets = to the total power consumed by the A/C system ?
Look at the electric meter outside your homeyou can find out from that. If not you have to measure the currentflow or go with manugactures spec.
No, it most certainly does not. Some, hopefully most, of the electrical energy taken by the motor will be converted to mechanical energy. That's what a motor does. But, in the conversion process some of the energy, hopefully only a small amount, will be lost as heat.
Short answer: take the BTU rating and add about 1/6 more to account for the power consumed by the motor to get the total maximum heat rate dumped out the dump If the A/C motor consumes 6 amps of 120 volts, then that 720 watts that needs to be dissapated somewhere, somehow. About 15% will be wasted in the motor for a single-phase motor and 85% will be do useful work (and ultimately end up as heat. 0.720 kilowatts x 3413 BTU/hr per kw = 2460 BTU/hour Refridgeration, depending on the coolant used and the temperature range (inside versus outside the cabinet) have efficiencies of 4 to 11 or so. Let's try 7.5. So 2460 BTU/hour of electrical power consumed x 7.5 = 18,000 BTU/hour of heat moved from inside to outside. Add the 2460 (motor) + 18,000 (cooling) and there's your total heat dumped out the back of the cabinet. Note that A/C systems often have duty cycles and all of the above is about peak rates. When the cabinet contents generated a lot of heat and its a hot day, the A/C will run for a long time. If the contents are unpowered, the duty cycle of the A/C might 10-15% of the time.
