Tonnage of HVAC

OT: I'd suggest you don't use dear that way in your posts. It is acceptable to say something like "Dear Sir", "Dear Madam", "Dear Ikram", or "Dear Mr. Shehzador", but I don't recommend any of those in this forum either.

To just say "dear" makes many people uncomfortable.

On to the subject of your questions:

You've been told that one ton of air conditioning is 12,000 BTU/hour.

It is hard to convert that directly to kw or kwhr--well, you can convert it to kwhr, but the problem is, reading between the lines, I suspect your question is really (or partially) how much electrical energy does it take to create to remove 12,000 BTU/hr of heat energy from, let's say, a typical room.

The standard way to do that is by using a mechanical refrigeration unit (air conditioner) which typically includes a compressor (for the refrigerant in the system, along with other important components, including an evaporator coil, a condenser coil, an expansion valve, and other stuff) and an electric motor driving the compressor. All of those parts have different efficiencies (including the refrigerant), so if you want to know the amount of electrical energy used for a certain amount of cooling, you must consider those efficiencies.

Further, a mechanical refrigeration unit is a heat pump. To make an area cold, it pumps heat from that area to somewhere else. (In cooling a room, the heat is pumped outside the room.) (BTW, the mechanical refrigeration unit could be "reversed" to heat the room by pumping heat from outside the room into the room.)

The thing is, !2,000 BTU / hr. represents a certain amount of heat energy. But, the mechanical refrigeration system doesn't create that heat, it simply pumps it from one location to another. It actually (usually) takes less energy to pump that much heat than it would to take to create that much heat. So, simply finding the electrical energy equivalent of 12,000 BTU / hr. doesn't tell you how much electrical energy it takes to pump that much heat.

BTW, I won't try to go into it, but air conditioners (heat pumps for removing heat from a room) are tested to determine their efficiency, and the efficiency rating is then advertised--in the US it is known as the SEER (Seasonal Energy Efficiency Rating), and, in general terms, it also takes into account the temperature outside the room. The higher that outside temperature, the more energy it takes to pump heat there. (Just like pumping water up a higher hill takes more energy than pumping it up a lower hill.)

(When a heat pump is used to pump heat into the room, a different efficiency rating is used, COP (Coefficient of Performance). A kwhr of electrical energy is about 3400 BTU/hr. If you used an electric resistance heater to heat a room, you'd get 3400 BTU of heat for every kwhr of electricity used by the resistance heater.

As mentioned above, it (usually) takes less electricity to pump heat--with a good heat pump (and when it is not too cold outside), you might pump 3 times as much heat per kwhr. than you would get by simply converting the electric power to heat in a resistance heater. Thus, the COP at this particular operating condition would be 3, and you would get about 10,200 BTU of heat into the room for each kwhr of electricity used by the heat pump.

As I tried to imply above, how well the heat pump works (how efficient it is) depends on how much heat is available outside the room to pump into the room. The colder it is outside, the less heat is available to be pumped. When it is below a certain temperature, the COP will fall off to below 1, and when it does, it is more efficient to use a resistance heater (if your only energy source for heating is electricity).

Finally, your most recent question was about 1 phase vs. 3 phase power. As people have stated, it doesn't matter. You will use (about) the same amount of electrical power whether you have a 1 phase motor or a 3 phase motor. (I think 3 phase motors can be slightly more efficient than 1 phase motors, but let's ignore that--it is insignificant.)

What we've been looking at is how much electrical energy is needed to pump heat. We've talked about it in terms of kwhrs. (We might switch units now--an electric motor is typically (well, in the US) rated in HP, which is analogous to kw. A 1 Hp motor could also be rated at 0.746 kw.--if you operate that motor at full load for 1 hour, it will use 0.746 kwhrs (at 100% efficiency).

Either a 1 phase or a 3 phase motor can deliver a required HP. For example, you can get a 1 phase motor rated at 1 Hp., or you can get a 3 phase motor rated at 1 Hp. They both use the same amount of electrical power (for the same mechanical load): 0.746kws, and, in one hour, both will use the same amount of electrical energy: 0.746 kwhrs.

The differences are in the construction of the motor and the power supply to the motor, and, thence the voltage and current supplied to the motor.

A 1 phase (single phase) motor might be designed for a voltage of 240 volts, and then need something like 4 amps of current (considering efficiency of the motor and power factor, which I won't even try to talk about here). Two wires will carry the electrical current to and from the motor.

A 3 phase motor might also be designed for a voltage of 240 volts. 3 wires will be used to carry the electrical current to and from the motor. Because the current is split up over 3 wires instead of two, it will take less current to carry the same amount of electrical energy to the motor.

Because the electric waves in those three wires are not in phase with each other the current required in each of those three phases (for 1 Hp) will be about 57% of the current in the single phase motor (for 1 Hp). (So, about 2.28 amps per phase for the 3 phase motor, vs. 4 amps for the 1 phase of the 1 amp motor. But both deliver 1 hp / 0.746 kw of power, and use 0.746 kwhrs of electrical energy in 1 hour.)

There are other advantages of 3 phase motors, for example, it is easier to start the rotation--single phase motors need special design features to start the rotation.

I have not tried to explain power flow, but you should know that it is not the same as current flow. To make a motor run, the same amount of current must be carried away from the motor as is carried to it. The power at the motor is developed because the current going to the motor has a certain amount of potential energy as measured by the voltage. At the motor, the potential energy carried in the current is reduced--the power transferred to the motor (in the single phase case, which is easiest to understand) is the amount of current (in amps) times the voltage drop at the motor (in volts).

I can tell you all this stuff, and I think I mostly have it right, but for you to really understand it, and learn to make use of it, you need to:

• learn a lot more, which I haven't tried to tell you
• practice using these (and other) facts and laws so that you develop some intuition about what they mean

That's why you're in school. I hope you will apply yourself in school and become a productiv engineer and citizen (of the earth and universe).