# Practical Work of Three-Phased Induction Motor Modeling in a Classical Spreadsheet Software
## 5 Practice
### Exercise 5
#### Solutions  
All the solutions of this exercise can be computed thanks to the the excel file given in the supplementary material and filled along the reading of this paper. The Figure 10 summarises all the results, each of which is detailed hereafter.  
1. Equation (6) states that \(R_{I}=3\cdot V^{2}/P_{0}\). Then, \(R_{I}=3\cdot V^{2}/P_{0}\) and \(R_{I}=3.230^{2}/300=533.3\ \Omega\). In addition, \(X_{m}=3\cdot V^{2}/Q_{0}\), as presented in Equation (7). Thanks to the Fresnel triangle, one can obtain the value of \(Q_{0}\), where \(Q_{0}=\sqrt{S_{0}^{2}-P_{0}^{2}}\) and \(Q_{0}=\sqrt{(230\cdot 15)^{2}-100^{2}}=3448.55\ \text{VAR}\). Then, \(X_{m}=230^{2}/3448.55=15.34\ \Omega\).
2. The determination of the values of \(R\) and \(l\) was directly similar to the one presented in the last question. Thanks to Equation (13), the calculation of \(R\) answers to the following calculation: \(R=P_{1}/(3\cdot I_{1}^{2})\), \(R=700/100^{2}=0.07\ \Omega\). In addition, the reactive power at a blocked rotor test can be calculated, by phase of the motor, thanks to the following formula: \(Q_{1}=\sqrt{S_{1}^{2}-P_{1}^{2}}=\sqrt{(90/\sqrt{3}\cdot 100)^{2}-700^{2}}=5148.8\ \text{VAR}\).  
Figure 10: Spreadsheet model of the induction motor applied on Exercise 5.  
Then, \(l\omega=Q_{1}/(3\cdot I_{1}^{2})\), \(l\omega=5148.8/100^{2}=0.515\)\(\Omega\) by motor phase, and \(l=0.515/(2\cdot\pi\cdot 50)=1.64\) mH.
3. \(I_{2}\) represents the current transmitted to the rotor. Thanks to Figure 6 and Equation (6), the equation giving the value of \(I_{2}\) is \(I_{2}=V/\sqrt{(R/s)^{2}+l^{2}\cdot\omega^{2}}\). \(I_{2}=230/\sqrt{(0.07/0.08)^{2}+0.515^{2}}=226.53\) A.
4. The power transmitted to the rotor \(P_{tr}\) is calculated thanks to the formula presented in Equation (12): \(P_{tr}=3\cdot(R/s)\cdot I_{2}^{2}=3\cdot(0.07/0.08)\cdot 226.53^{2}=134.7\) kW.
5. The rotor Joule effect losses were calculated thanks to the following formula: \(P_{JR}=s\cdot P_{tr}=0.08\cdot 134.7=10.78\) kW.
6. According to the scheme presented in Figure 4, the total mechanical power \(P_{M}\) is the difference between the electromagnetic power transmitted to the rotor \(P_{tr}\) and the power losses in the rotor \(P_{JR}\). Then, \(P_{M}=P_{tr}-P_{JR}\) and \(P_{M}=134.7-10.78=124\) kW.
7. The rotor rotational speed \(\Omega_{R}\) is easily deduced from the synchronous speed \(\Omega_{S}\) and slip \(s\) thanks to the formula presented in Equation (2). \(\Omega_{R}=(1-s)\cdot\Omega_{S}\), and \(\Omega_{R}=(1-0.08)\cdot 1500=0.92\cdot 1500=1380\) rpm.
8. The first way to calculate the electromagnetic torque \(T_{em}\) is to deduce it from transmitted mechanical power thanks to the following division: \(T_{em}=P_{M}/\Omega_{R}\). With the data of this problem, it provides the following: \(T_{em}=124\),\(000\cdot 60/(2\cdot\pi\cdot 1380)=858\) N\(\cdot\)m. As can be seen, the rotor rotational speed should be expressed in rad/s. Otherwise, the units equation is not respected.  
The second way to obtain the same result is to calculate the electromagnetic torque \(T_{em}\) with Equation (18). \(T_{em}=(3p/l)\cdot(V/\omega)^{2}/(R/sl\omega+sl\omega/R)\).