# Online Efficiency Optimization and Speed Sensorless Control of Single-Phase Induction Motors
## 4 Energy Efficiency Optimization  
### Optimal Energy Efficiency Condition  
When the mechanical load is decreased away from its nominal value, the motor efficiency will drop. This stems from the fact that since the reactive part of the current does not decline with the load decrease, the load decrease hence is not proportional to the current and copper loss decrease. Further, the core loss which is proportional to the applied voltage will not change at all. Therefore, the motor loss in light loads would not change considerably. Given the output decrease and the rather constant loss, the efficiency will drop. On the other hand, by decreasing the input voltage the core loss will drop substantially with square of the voltage, while the copper loss by an increase in the active part of the current begins to increase slightly. This voltage decrease should be in a controlled manner to a voltage point where the total loss is minimum. It is apparent that decreasing the voltage only down to a certain value causes the total loss to decrease and efficiency to increase, after which the motor loss will start to increase [21]. The motor-developed torque is equal to the load torque plus windage and mechanical losses, which are all constant at state conditions. As such, the developed torque is also constant and its derivate is zero [21]:  
\[\frac{\partial T_{e}}{\partial s}=0 \tag{22}\]  
According to (7), the following equation will be obtained:  
\[2A_{1}I_{m}\frac{\partial I_{m}}{\partial s}+I_{m}^{2}\frac{\partial A_{1}}{ \partial s}=0 \tag{23}\]  
The optimum slip (\(s_{opt}\)) is the slip of the operating point at which the loss is at minimum. Assuming that the machine torque and supply frequency are constant (motor is operating at steady-state condition), the optimum slip can be calculated by setting the derivative of electrical losses with respect to slip to zero:  
\[\frac{\partial P_{el}}{\partial s}\bigg{|}_{T_{e},\omega_{e}}=0 \tag{24}\]  
Using (16), condition (24) is satisfied when  
\[2A_{2}I_{m}\frac{\partial I_{m}}{\partial s}+I_{m}^{2}\frac{\partial A_{2}}{ \partial s}=0 \tag{25}\]From (23), (25), by eliminating \(I_{m}\) yields:  
\[\frac{1}{A_{1}}\frac{\partial A_{1}}{\partial s}-\frac{1}{A_{2}}\frac{\partial A _{2}}{\partial s}=0 \tag{26}\]  
Since at every operating point, the supply frequency is constant, \(A_{1}\) and \(A_{2}\) are only dependent upon slip variations. Hence, from the above equation the optimum slip is attainable. Substituting \(s_{opt}\) into (1) the optimum STCR can be introduced as follows:  
\[\left(\frac{I_{m}}{I_{a}}\right)_{opt}=\left|Z_{t}\right|_{s=sopt}=K_{s} \tag{27}\]  
where, according to (1), \(K_{s}\) is only a function of frequency and motor parameters. In case of neglecting motor parameters slight variations, the optimum STCR (\(K_{s}\)) would only be dependent upon the supply frequency. \[K_{s}=h(\omega_{e}) \tag{28}\]  
To obtain \(K_{s}\), the motor is supplied with a variable frequency and amplitude AC source. Keeping the frequency and mechanical load constant, the voltage amplitude is varied between 20% to 100% of the nominal value and then the input power, stator main and auxiliary winding currents are measured. Further, STCR will be determined by dividing the two windings' currents. After drawing a diagram of input power (\(P_{in}\)) versus STCR, the minimum value of input power and its corresponding STCR value will be determined. As such, the optimum STCR (\(K_{s}\)) is determined. Repeating the same procedure for different frequencies, the \(K_{s}\) for each frequency will be obtained. Further the frequency and its corresponding \(K_{s}\) pairs will be saved in a table as a lookup table in the controller's memory and will be used as the reference value of \(K_{s}\). Hence, the optimum condition will be achieved for constant pre-known motor parameters. Nevertheless, factors such as saturation and temperature variations will cause changes in some of the motor parameters such as resistances and inductances.