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).
7.5 horsepower is the machine's output power; without knowing its efficiency, you cannot determine its input (true) power.
To determine its apparent power, you will need to know its input power and its power factor.
The kV.A (not 'kva') rating is the total apparent power of the machine. So a 75 kV.A machine is 25 kV.A per phase.
200 and 100
For a single-phase transformer, divide the ratedapparent power (expressed in volt amperes) by the voltage rating (expressed in volts) of the primary winding; this will give you the rated primary current (expressed in amperes) of the primary winding.
HP/.00134= Watts Then Watts divided by Volts = AMPS For expample. a .75 HP electric motor running on 220VAC uses 2.544 amps .75 / .00134 = 559.7015 Watts then 559.7015 / 220 = 2.544
For single phase, KVA = (line to ground) * (phase current). A 75kVA 480 to 208Y/120 volt transformer is a fairly common transformer. I assume this is the type of transformer you are referring to. 75k / 120 = 625 Amps. As an FYI, the 208Y voltage is the line to line voltage, which is equal to (phase 1) - (phase 2), where the phases are separated by 120 degrees, thus (phase 1) * 1.732 For three phase, kVA = (line to line voltage) * (phase current) *(sqrt 3), 75k / 208 / 1.732 = 208 Amps.
The kV.A (not 'kva') rating is the total apparent power of the machine. So a 75 kV.A machine is 25 kV.A per phase.
The formula you are looking for is , A = kva x 1000/Volts.
12HP is approximately 10.8 KVA. You would want to use a 15KVA transformer to supply this motor. KW = HP * .75 KVA = KW * 1.2 (These formulas are approximate)
speculation is that it consumes about .75 ltr of diesel in an hour
Volts per hour is an invalid statement. You may have meant Watts per Hour.
starting current of 3 phase 75 KW induction motor
200 and 100
For a single-phase transformer, divide the ratedapparent power (expressed in volt amperes) by the voltage rating (expressed in volts) of the primary winding; this will give you the rated primary current (expressed in amperes) of the primary winding.
Multiply by Amps.
Each phase supplies 15 kVA. The primary has a line-to-neutral voltage of 277 v so the line current is 15,000 / 277 or 54 amps. The secondary has a line-to-neutral voltage of 120v so the current is 15,000/120 or 125 amps.
HP/.00134= Watts Then Watts divided by Volts = AMPS For expample. a .75 HP electric motor running on 220VAC uses 2.544 amps .75 / .00134 = 559.7015 Watts then 559.7015 / 220 = 2.544
For single phase, KVA = (line to ground) * (phase current). A 75kVA 480 to 208Y/120 volt transformer is a fairly common transformer. I assume this is the type of transformer you are referring to. 75k / 120 = 625 Amps. As an FYI, the 208Y voltage is the line to line voltage, which is equal to (phase 1) - (phase 2), where the phases are separated by 120 degrees, thus (phase 1) * 1.732 For three phase, kVA = (line to line voltage) * (phase current) *(sqrt 3), 75k / 208 / 1.732 = 208 Amps.