# Practical Work of Three-Phased Induction Motor Modeling in a Classical Spreadsheet Software
## 5 Practice
### Exercise 4  
The tests on a six-pole induction motor gave the following results:  
* No-load test (\(s=0\)): \(U_{0}=240\) V; \(f=50\) Hz; \(I_{0}=15\) A; and \(P_{0}=534\) W.  
The measure, a rated DC current, of the resistance between two stator terminals gave \(R=0.10\)\(\Omega\).
* In-load operation (\(s=1\)): \(U=240\) V (between phases); \(I=40\) A; \(P_{abs}=15.3\) kW; and \(n^{\prime}=960\) rpm; From these measures, it is possible to determine the following: 1. The slip \(s\).
2. The motor loaded power factor cos\(\phi\).
3. The stator iron losses \(P_{Is}\) and the mechanical power losses \(P_{M}\), if they are assumed to be equal.
4. The stator and rotor Joule effect losses \(P_{Js}\) and \(P_{lr}\).
5. The useful Power \(P_{u}\) and the efficiency \(\eta\) of the motor.
6. The electromagnetic \(T_{em}\) and the useful mechanical rotor \(T_{u}\) torques. _Solutions:_  
1. The slip \(s\) is calculated using Equation (2). Thus, \(s=\frac{\Omega_{s}-\Omega_{s}}{\Omega_{s}}=(1000-960)/1000=0.04=4\%\).
2. The motor loaded power factor cos\(\phi\) can be calculated using Equation (3). Thus, \(\cos\varphi=\frac{P_{ls}}{\sqrt{3}\cdot 1\cdot I}\), and the in-load motor power factor is \(\cos\varphi=15{,}300/(\sqrt{3}\cdot 400\cdot 40)=0.92\).
3. At no-load test, the power balance scheme presented in Figure 4 gives the following equation: \(P_{abs}=P_{Is}+P_{Js}+P_{M}\). Since stator iron losses \(P_{Is}\) and mechanical power losses \(P_{M}\) are assumed to be equal, calculating the Joule effect losses in the stator gives the answer to the question. At no-load test, \(P_{Js}=\frac{3}{2}\cdot R\cdot I_{0}^{2}=\frac{3}{2}\cdot 0.1\cdot 15^{2}=33.75\) W. Then, \(P_{Is}+P_{M}=P_{abs}-P_{Is}=534-33.75=500\) W, and \(P_{Is}=P_{M}=250\) W.
4. In the load test, stator Joule effect losses are calculated in exactly the same way as in the last question, with the in-load current. Thus, \(P_{Js}=\frac{3}{2}\cdot R\cdot I^{2}=\frac{3}{2}\cdot 0.1\cdot 40^{2}=240\) W. Then, rotor Joule effect losses are calculated using the power balance (Figure 4). Attention must be paid to the lack of mechanical power in the power balance presented hereafter because it is taken into account just after the rotor Joule effect losses. Thus, the rotor Joule effect losses are equal to the following: \(P_{lr}=s\cdot P_{lr}=s\cdot(P_{abs}-P_{Is}+P_{Js})=0.04\cdot(15{,}300-250-240)= 0.04\cdot 14{,}810=592.4\) W. It can be noted that rotor Joule effect losses are significantly higher than all others as the load increases.
5. The useful power \(P_{u}\) transmitted to the rotor is equal to \(P_{u}=P_{lr}+P_{lr}+P_{M}=14{,}810-592.4-250=13{,}968\)W. The resulting efficiency \(\eta\) is calculated according to Equation (4) as \(\eta=P_{u}/P_{abs}=13{,}968/15{,}300=0.91\).
6. According to the power balance presented in Figure 4, the electromagnetic torque \(T_{em}\) can be calculated as \(T_{em}=P_{lr}/\Omega_{s}=14{,}810\cdot 60/(2\cdot\pi\cdot 1000)=141.42\) N\(\cdot\)m. Similarly, the useful mechanical rotor torque \(T_{u}\) can be calculated as \(T_{u}=P_{u}/\Omega_{r}=13{,}968\cdot 60/(2\cdot\pi\cdot 960)=138{,}94\) N\(\cdot\)m.