Electrical Engineering

1 hp is generally around 750 watts, so you can expect a 7.5hp motor at full load to be ~5.6kW.

If lightly loaded, the motor will draw less current, thus less wattage. Also, you cannot calculate current directly from the kW rating and voltage (so full load current will not be 5600/230 = 24.3Amps), as motors are not 100% efficient, and do not have 1.0 power factor. I'd expect the full load amperes to be greater than the kw/voltage value by 20-30%.

I looked up a Baldor motor stated to be 7.5hp, and it was ~7kVA, 5.6kW at full load.

Another Answer

You don't specify whether you want to know the machine's output power or its input power.

The rated power of a motor is always its output power. In North America, this is expressed in horsepower whereas, in the rest of the world, this is expressed in watts (in practice, kilowatts). As 746 W is equivalent to 1 horsepower, then a 7.5 horsepower motor is equivalent to 5.6 kW.

If you wish to determine the rated input power to a machine, which is always larger than its output power, then you must know its efficiency at its rated output power. Divide the output power by the efficienty (expressed as a per unit value).

You really can't really "control" the speed of a the commonly available home-use 120 VAC fans.

The problem is that most of the motors on the "regular" fans are synchronous motors, and they want to sync with the line by virtue of frequency. Fans with different speeds have different windings in them that are specifically set up for the different speeds. A switch allows the different windings to be selected. Simply reducing the line voltage will only reduce the torque the motor will supply, but will not directly affect its speed. The lower torques will cause the motor to slow down, and that's true, but this is not necessarily a good thing as the fan was not designed to run like this.

Sophisticated electronic variable frequency drives (VFDs) are designed to provide AC power but at a frequency you can control. Your ability to control the frequency gives you the ability to control the speed of the fan directly. But these units cost. For the smaller non-commercial units, it is smarter (and cheaper) to just get another fan with multiple speeds than to try to fool around and control the speed by some "gizmo" or something else.

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I beg to differ. Most home fans are inexpensive single phase AC motors, either with a capacitor (PSC) or shaded pole. These are not synchronous motors. A simple light switch dimmer (aka phase controller or triac controller) can be used to control the fan. Caution: be careful to avoid fan stalling when setting at low speeds. The dimmer controls the speed by reducing the effective voltage to the fan. At low voltages (and resulting speeds) the fan torque is also reduced. Thus the likelihood of stalling is higher. If the fan stalls with the power on, the motor will overheat and possibly catch on fire.

VFDs are definitely not the right choice for a simple fan. These require motors (almost always 3 phase) designed for use with VFDs (aka, inverters) or a special output filter for the controller. Otherwise the bearings will pit within months to a year. Besides, VFDs with a single phase output is very rare.

The length of the wire causes resistive load to be distributed along line. This accumulated resistance causes the voltage drop.

The voltage drop along a high-voltage transmission line isn't large in relation to the operating voltage of the line. One of the advantages (there are many) of using high voltages is that, for a given load, the line current is relatively low -so the voltage drop along the line is also relatively low.

The current is automatically stepped up or down when the voltge is stepped down or up by a transformer.

Most transformers are designed for a constant voltage, and the current varies depending on the load. Current transformers are also made to detect the current in a power cable so that it can be measured.

Another Answer

It's a misconception that transformers step current up or down (with the exception of current transformers, that is!). Transformers step voltages up or down. The secondary current is determined by the load. If there is no load, then there is no secondary current; if there is a heavy load, then there will be a heavy secondary current.

The secondary current then determines the primary current, because the primary power must match the secondary power (ignoring losses), so if the secondary current increases, so must the primary current. The current itself isn't stepped up or down.

<<>> Normal transformers work with constant-voltage supplies and are designed to step the voltage up or down, and the current is incidentally stepped down or up in the same ratio as the voltage.

Current transformers are designed to step current up or down (usually down), and are used most commonly to sample the current in power conductors.

The simple explanation as to why a change in the coil causes a change to the electric current is that the electric field and the magnetic field are the same field and they exist in a four dimensional quaternion space.

Change in a quaternion space is four dimensional and when you take the derivative there is a real derivative d/dr=d/cdt and a vector derivative Del

X= d/dr + Del = d/cdt + Id/dx + Jd/dy + Kd/dz.

The answer to this question is the Quaternion four space derivative of the vector Electric Field, Ev.

(d/dr + Del)Ev= -Del.Ev + dEv/dr + DelxEv = -Del.Ev + dBv/dt + DelxEv

The Quaternion Derivative is X=d/dr + Del = d/cdt + Del

The vector part of the derivative dB/dt + DelxEv is the answer to the question this can be transformed by introducing the coil with dA as the area and dC as the circumference:

dBv/dt.dA + DelxEv.dA = AdB/dt + Integral loop Ev.dC