I'm understanding this question to mean that you're trying to run a three-phase motor off a single-phase line. If you're trying to run the motor directly off the AC line, the phase angles involved will make it difficult to get the motor started, which is part of the reason three-phase exists in the first place. Single-phase motors usually have motor start caps for just that reason. That sounds like what you're describing.
The simple answer to your question is that to get three-phase AC from single-phase AC, you need to rectify the single-phase AC line into DC, then run the DC back through an inverter to get controlled three-phase AC. There are other electronic approaches, but they're less common in my (limited) experience. There are also mechanical approaches, which may be more convenient if you have the parts.
I'd suggest using a drive to operate your three-phase motor. Typical variable-frequency three-phase drives are exactly what I described above: a rectifier, followed by an inverter. I can't speak as to what's on the market in a given power class, but larger three-phase drives typically have terminals for the three-phase AC line input, the DC bus, and the three-phase motor output. If you have those terminals, you have two options.
One is to run single-phase AC through the three-phase input of the drive. If the voltages are correct, the drive should operate fine. The caveat is that you'll have to derate the drive somewhat. The input diodes are spec'd assuming that the drive's constant-power load will be distributed among three legs of the rectifier. If you distribute that same load over just two legs, those diodes will get hotter. The internal bus capacitors will also get hotter, because they'll see more ripple current without the third phase. Check with the drive manufacturer for the derating info.
If your drive has DC bus terminals, your other option is to skip it's internal rectifier and use an external one. Rectify the single-phase AC, then use that DC as the input to the drive. This will let you avoid derating the drive. My company makes something exactly for that purpose, though its power range may be larger than is cost-effective for your application. You'd have to price both options out to find out for sure. Read this for more details.
As far as I know the magnetic force is responsible for the motor rotation. So, horsepower of the motor is mostly related to magnetic flux on the coils (windings). Also magnetic flux is directly related to inductance (Number of turns) and the current flow on the coil. Am I right so far?
That is not quite right. Energy is force X distance. Power is the rate of energy transfer or conversion, force x distance / time. The power delivered by a motor is torque X speed. Torque is force multiplied by the radius at which the force is applied.
1- What is magnetizing reactance of the winding?
The magnetizing reactance is the part of the stator reactance that produces useful magnetic flux. Due to limitations of the geometry of the motor structure, less than 100% of the winding flux actually contributes to producing torque and transferring energy to the rotor. Therefore the winding is shown in the equivalent circuit as two components. X1 is just an impedance to the flow of current that does not directly contribute to the function of the motor. The flux in Xm is the flux that produces torque.
2- Why do we want high power factor?
High power factor means the minimum current to deliver a given power to the load.
3- How can I calculate the horsepower and the efficiency of my motor?
You completely analyze the equivalent circuit. The mechanical power developed in the rotor is the power in R2x(1-s)/s. In the rotor circuit, R2/s is comprised of two parts, R2, the rotor resistance, and R2x(1-s)/s the representation of the mechanism of conversion of electrical power to mechanical power.
You also need to consider that some of the power developed in the rotor is lost to friction and windage rather than delivered to an external load.
Note that you have asked at least four questions. A complete answer would require a chapter in a text book assuming that you have mastered the prerequisites for the course.
Best Answer
The connection does not result in good performance, but the best that can be achieved without a 3-phase power source. The motor should be able to provide about 70% of rated power. Starting torque can be expected to be 20-30% of the motor's rated starting torque, less that that for a 2-pole motor. A 2-pole motor may not be suitable for such use at all.
With the optimum capacitor value, the capacitor current will be equal to the rated motor current.
The capacitor value can be approximated by:
C = 50 x Hp x (220/V)^2 x 50/f where:
C is in microfarads
Hp is the motor's rated horsepower
V is the motor's rated voltage
f is the motor's rated frequency
Unfortunately I copied the references I have some time ago without making a note of their origin.
Addendum 1:
The capacitor value should be optimized based on the actual motor load.
The formula came from a PDF on engineering.com clicking the Google search link downloads the PDF. I don't know how to access any related context on the site.
Single Phase Induction Motor, Charles Proteus Steinmetz, Meeting of The American Institute of Electrical Engineers, New York, February 23d, 1898
Addendum 2:
A method for optimizing the capacitor value is to adjust the capacitance such that the current in the capacitor is equal to the rated current of the motor for the delta connection.
There are variations of the Steinmetz connection for capacitor-start, capacitor-start with capacitor-run and for the wye (star) connection.