Question 6.AE.8: A three-phase, squirrel-cage induction motor is fed by a vol......

The induction motor is operated at 60 Hz (natural torque-speed characteristic) [46].

a)

\begin{aligned} & n_{m s}=1800 \mathrm{rpm}, \omega_{m s}=188.49 \mathrm{rad} / \mathrm{s}, \omega_m=180.1 \mathrm{rad} / \mathrm{s}, \\ & s_{r a t}=0.044,\left|\widetilde{I}_s(t)\right|=49.82 \mathrm{~A},\left|\widetilde{I}_r^{\prime}(t)\right|=47.66 \mathrm{~A}, \\ & |\widetilde{E}|=221.75 \mathrm{~V}, \text { and } \mathrm{T}_{\mathrm{rat}}=164.31 \mathrm{Nm} .\quad (E6.8-1) \end{aligned}

b) Due to constant flux control, the (ω_m–T) or the (s–T) characteristic consists of a family of parallel lines as indicated in Fig. E6.8.2.
The maximum torque T_{max} and the slip s_m where the maximum torque occurs are [46]

T_{\max }=391.32 \mathrm{Nm}, s_m=0.20            (E6.8-2)

From the relation (see Fig. E6.8.2)

\frac{T_{\text {rat }}}{T_{\max \_ \text {fict }}}=\frac{s_{\text {rat }}}{1}            (E6.8-3)

one obtains for the maximum fictitious torque

T_{\max \_ \text {fict }}=T_{\mathrm{rat} / s_{\mathrm{rat}}}=3734.3 \mathrm{Nm}          (E6.8-4)

c) Equation of motion (or fundamental torque equation) is

T-T_L-J \frac{d \omega_m}{d t}=0            (E6.8-5)

where T_L and T are the load and the motor torques, respectively. J is the total axial moment of inertia J=J_m+J_{load}. Dividing this equation by T_{max\_fict} one gets

\frac{T}{T_{\max \_ \text {fict }}}-\frac{T_L}{T_{\max \_ \text {fict }}}-\frac{J}{T_{\max \_ \text {fict }}} \cdot \frac{d \omega_m}{d t}=0                (E6.8-6)

or with \frac{T}{T_{\max \_ \text {fict }}}=s follows

s-\frac{T_L}{T_{\max _{-} \text {fict }}}-\frac{J}{T_{\max _{-} \text {fict }}} \cdot \frac{d \omega\_m}{d t}=0 .          (E6.8-7)

with ω_m=ω_{ms}(1 – s) the differential equation becomes

\tau_m \frac{d s}{d t}+s=\frac{T_L}{T_{\max \_ \text {fict }}}              (E6.8-8)

where

\tau_m=\left(J \cdot \omega_{m s}\right) / T_{\text {max fict }}=(5 \cdot 188.49) / 3734.3=0.252 \mathrm{~s}            (E6.8-9)

is the mechanical time constant.
The general solution of this differential equation is

s(t)=\text { const } \cdot e^{\left(-\frac{t}{\tau_m}\right)}+\frac{T_L}{T_{\mathrm{max}_{-} \text {fict }}}              (E6.8-10)

For the initial condition where at t=0 one assigns s=s(0)

s(t)=\left(s(0)-\frac{T_L(t)}{T_{\max \_ \text {fict }}}\right) \cdot e^{\left(-\frac{t}{\tau_m}\right)}+\frac{T_L(t)}{T_{\max \_ \text {fict }}}                (E6.8-11)

d) Application of s(t) to eight different regions of Fig. E6.8.1.

Region I. It is assumed that the motor at time t¼0 s (at the start of the load cycle) has the steady-state slip of s(0). This slip is obtained from

\frac{T(0)}{T_{\mathrm{rat}}}=\frac{s(0)}{s_{\mathrm{rat}}}              (E6.8-12)

and with Figure E6.8.1

\begin{aligned} & T(0)=0.2 \cdot T_{\mathrm{rat}}=32.86 \mathrm{Nm} \\ & s(0)=32.86 \cdot 0.044 /(164.31)=0.0088 \end{aligned}

Therefore, the slip at the start of region I is

s\left(t^{(I)}=0\right)=0.0088

It is best to introduce for each region a new coordinate system for the time axis. Thus for region \mathrm{I}\left(0 \leq t^{(I)} \leq 0.4 \mathrm{~s}\right) one obtains from Fig. E6.8.1 the load torque

T_L^{(I)}=2.8 \cdot T_{\mathrm{rat}}=460.07 \mathrm{Nm}

With Eq. E6.8-11

\begin{aligned} s\left(t^{(\mathrm{I})}\right)= & \left(s\left(t^{(\mathrm{I})}=0\right)-\frac{T_L^{(\mathrm{I})}\left(t^{(\mathrm{I})}\right)}{T_{\mathrm{max}_{-} \mathrm{fict}}}\right) \cdot e^{\left(-\frac{t^{(\mathrm{I})}}{\tau_m}\right)}+\frac{T_L^{(\mathrm{I})}\left(t^{(\mathrm{I})}\right)}{T_{\mathrm{max}_{-} \mathrm{fict}}} \\ = & \left(0.0088-\frac{460.07}{3734.3}\right) e\left(-\frac{t^{(\mathrm{I})}}{\tau_m}\right)+\frac{460.07}{3734.3} \\ & s\left(t^{(\mathrm{I})}\right)=-0.1144 e \\ & s\left(t^{(\mathrm{I})}=0.4 \mathrm{~s}\right)=0.0998=s\left(t^{(\mathrm{II})}=0\right) \text { and } \\ & s\left(t^{(\mathrm{I})}=0.2 \mathrm{~s}\right)=0.07147 . \end{aligned}

Region II. For region \text { II }\left(0 \leq t^{(I I)} \leq 0.4 \mathrm{~s}\right) one obtains from Fig. E6.8.1 the load torque

T_L^{(\mathrm{II})}=0.2 \cdot T_{\mathrm{rat}}=32.86 \mathrm{Nm}

With Eq. E6.8-11

\begin{aligned} s\left(t^{(\mathrm{II})}\right) & =\left(s\left(t^{(\mathrm{II})}=0\right)-\frac{T_L^{(\mathrm{II})}\left(t^{(\mathrm{II})}\right)}{T_{\max \_\mathrm{fict}}}\right) \cdot e\left(-\frac{t^{(\mathrm{II})}}{\tau_m}\right)+\frac{T_L^{(\mathrm{II})}\left(t^{(\mathrm{II})}\right)}{T_{\max \_ \text {fict }}} \\ & =\left(0.0998-\frac{32.86}{3734.3}\right) e\left(-\frac{t^{(\mathrm{II})}}{\tau_m}\right)+\frac{32.86}{3734.3} \end{aligned} \\ \begin{aligned} & s\left(t^{(\mathrm{II})}\right)=0.091 e^{\left(-\frac{t^{(\mathrm{II})}}{\tau_m}\right)}+0.0088 \\ & s\left(t^{(\mathrm{II})}=0.4 \mathrm{~s}\right)=0.0274=s\left(t^{(\mathrm{III})}=0\right) \text { and } \\ & s\left(t^{(\mathrm{II})}=0.2 \mathrm{~s}\right)=0.04994 . \\ & \end{aligned}

Region III. For region \text { III }\left(0 \leq t^{\text {(III) }} \leq 0.6 \mathrm{~s}\right) one obtains from Fig. E6.8.1 the load torque

T_L^{(\mathrm{III})}=2 \cdot T_{\mathrm{rat}}=328.62 \mathrm{Nm}

With Eq. E6.8-11

\begin{aligned} s\left(t^{(\mathrm{III})}\right) & =\left(s\left(t^{(\mathrm{III})}=0\right)-\frac{T_L^{(\mathrm{III})}\left(t^{(\mathrm{III})}\right)}{T_{\max \_ \text {fict }}}\right) \cdot e^{\left(-\frac{t^{(\mathrm{III})}}{\tau_m}\right)}+\frac{T_L^{(\mathrm{III})}\left(t^{(\mathrm{III})}\right)}{T_{\max \_ \text {fict }}} \\ & =\left(0.0274-\frac{328.62}{3734.3}\right) e\left(-\frac{t^{(\mathrm{III})}}{\tau_m}\right)+\frac{328.62}{3734.3} \\ s\left(t^{(\mathrm{III})}\right) & =-0.060 e\left(-\frac{t^{(\mathrm{III})}}{\tau_m}\right)+0.088 \\ s\left(t^{(\mathrm{III})}\right. & =0.6 \mathrm{~s})=0.0824=s\left(t^{(\mathrm{IV})}=0\right) \text { and } \\ s\left(t^{(\mathrm{III})}\right. & =0.3 \mathrm{~s})=0.06957 . \end{aligned}

Region IV. For region \text { IV }\left(0 \leq t^{(I V)} \leq 0.4 \mathrm{~s}\right) one obtains from Fig. E6.8.1 the load torque

T_L^{(\mathrm{IV})}=0.2 \cdot T_{\mathrm{rat}}=32.86 \mathrm{Nm}

With Eq. E6.8-11

\begin{aligned} & s\left(t^{(\mathrm{IV})}\right)=\left(s\left(t^{(\mathrm{IV})}=0\right)-\frac{T_L^{(\mathrm{IV})}\left(t^{(\mathrm{IV})}\right)}{T_{\max \_ \text {fict }}}\right) \cdot e^{\left(-\frac{t^{(\mathrm{IV})}}{\tau_m}\right)}+\frac{T_L^{(\mathrm{IV})}\left(t^{(\mathrm{IV})}\right)}{T_{\max \_ \text {fict }}} \\ & =\left(0.0824-\frac{32.86}{3734.3}\right) e^{\left(-\frac{t^{(\mathrm{IV})}}{\tau_m}\right)}+\frac{32.86}{3734.3}, \\ & s\left(t^{(\mathrm{IV})}\right)=0.0736 e^{\left(-\frac{t^{(\mathrm{IV})}}{\tau_m}\right)}+0.0088, \\ & s\left(t^{(\mathrm{IV})}=0.4 \mathrm{~s}\right)=0.02385=s\left(t^{(\mathrm{V})}=0\right) \text { and } \\ & s\left(t^{(\mathrm{IV})}=0.2 \mathrm{~s}\right)=0.04208 \\ & \end{aligned}

Region V. For region \mathrm{V}\left(0 \leq t^{(V)} \leq 0.8 \mathrm{~s}\right) one obtains from Fig. E6.8.1 the load torque

T_L^{(\mathrm{V})}=T_{\text {rat }}=164.3 \mathrm{Nm}

With Eq. E6.8-11

\begin{aligned} & s\left(t^{(\mathrm{V})}\right)=\left(s\left(t^{(\mathrm{v})}=0\right)-\frac{T_L^{(\mathrm{V})}\left(t^{(\mathrm{V})}\right)}{T_{\max \_ \text {fict }}}\right) \cdot e\left(-\frac{t^{(\mathrm{v})}}{\tau_m}\right)+\frac{T_L^{(\mathrm{V})}\left(t^{(\mathrm{v})}\right)}{T_{\max \_\mathrm{fict}}} \\ &=\left(0.02384-\frac{164.31}{3734.3}\right) e\left(-\frac{t^{(\mathrm{v})}}{\tau_m}\right)+\frac{164.31}{3734.3} \\ & s\left(t^{(\mathrm{v})}\right)=-0.0202 e\left(-\frac{t^{(\mathrm{v})}}{\tau_m}\right)+0.044 \\ & s\left(t^{(\mathrm{V})}=0.8 \mathrm{~s}\right)=0.0432=s\left(t^{(\mathrm{VI})}=0\right) \text { and } \\ & s\left(t^{(\mathrm{V})}=0.4 \mathrm{~s}\right)=0.03987 \end{aligned}

Region VI. For region \mathrm{VI}\left(0 \leq t^{(\mathrm{VI})} \leq 0.4 \mathrm{~s}\right) one obtains from Fig. E6.8.1 the load torque

T_L^{(\mathrm{VI})}=0.2 \cdot T_{\mathrm{rat}}=32.86 \mathrm{Nm}

With Eq. E6.8-11

\begin{aligned} s\left(t^{(\mathrm{VI})}\right)= & \left(s\left(t^{(\mathrm{VI})}=0\right)-\frac{T_L^{(\mathrm{VI})}\left(t^{(\mathrm{VI})}\right)}{T_{\max \_ \text {fict }}}\right) \cdot e^{\left(-\frac{t^{(\mathrm{VI})}}{\tau_m}\right)}+\frac{T_L^{(\mathrm{VI})}\left(t^{(\mathrm{VI})}\right)}{T_{\text {max } \_ \text {fict }}} \\ = & \left(0.0432-\frac{32.86}{3734.3}\right) e\left(-\frac{t^{(\mathrm{VI})}}{\tau_m}\right)+\frac{32.86}{3734.3} \\ & s\left(t^{(\mathrm{VI})}\right)=0.0344 e \\ & \left(-\frac{t^{(\mathrm{VI})}}{\tau_m}\right)+0.0088 \\ & s\left(t^{(\mathrm{VI})}=0.4 \mathrm{~s}\right)=0.01583=s\left(t^{(\mathrm{VII})}=0.2 \mathrm{~s}\right)=0.0244 . \end{aligned}

Region VII. For region \text { VII }\left(0 \leq t^{(V I I)} \leq 1 \mathrm{~s}\right) one obtains from Fig. E6.8.1 the load torque

T_L^{(\mathrm{VII})}=0.5 \cdot T_{\mathrm{rat}}=82.155 \mathrm{Nm}

With Eq. E6.8-11

\begin{gathered} s\left(t^{(\mathrm{VII})}\right)=\left(s\left(t^{(\mathrm{VII})}=0\right)-\frac{T_L^{(\mathrm{VII})}\left(t^{(\mathrm{VII})}\right)}{T_{\max \_ \text {fict }}}\right) \cdot e^{\left(-\frac{t^{(\mathrm{VII})}}{\tau_m}\right)}+\frac{T_L^{(\mathrm{VII})}\left(t^{(\mathrm{VII})}\right)}{T_{\max \_ \text {fict }}} \\ =\left(0.01583-\frac{82.155}{3734.3}\right) e^{\left(-\frac{t^{(\mathrm{VII})}}{\tau_m}\right)}+\frac{82.155}{3734.3} \\ s\left(t^{(\mathrm{VII})}\right)=-0.00617 e\left(-\frac{t^{(\mathrm{VII})}}{\tau_m}\right)+0.022 \\ s\left(t^{(\mathrm{VII})}=1 \mathrm{~s}\right)=0.0219=s\left(t^{(\mathrm{VIII})}=0\right) \text { and } \\ s\left(t^{(\mathrm{VII})}=0.5 \mathrm{~s}\right)=0.02115 . \end{gathered}

Region VIII. For region \text { VIII }\left(0 \leq t^{(V I I I)} \leq 1 \mathrm{~s}\right) one obtains from Fig. E6.8.1 the load torque

T_L^{(\text {VIII) }}=0.2 \cdot T_{\mathrm{rat}}=32.86 \mathrm{Nm}

With Eq. E6.8-11

\begin{aligned} s\left(t^{(\mathrm{VIII})}\right) & =\left(s\left(t^{(\mathrm{VIII})}=0\right)-\frac{T_L^{(\mathrm{VIII})}\left(t^{(\mathrm{VIII})}\right)}{T_{\max \_ \text {fict }}}\right) \cdot e\left(-\frac{t^{(\mathrm{VIII})}}{\tau_m}\right)+\frac{T_L^{(\mathrm{VIII})}\left(t^{(\mathrm{VIII})}\right)}{T_{\max _{-} \text {fict }}} \\ & =\left(0.0219-\frac{32.86}{3734.3}\right) e\left(-\frac{t^{(\mathrm{VIII})}}{\tau_m}\right)+\frac{32.86}{3734.3} \\ s\left(t^{(\mathrm{VIII})}\right) & =0.0131 e\left(\frac{t^{(\mathrm{VIII})}}{\tau_m}\right)+0.0088 \end{aligned}.

s\left(t^{(\mathrm{VIII})}=1 \mathrm{~s}\right)=0.0090 \approx s\left(t^{(\mathrm{I})}=0\right)=0.0088. That is, the periodicity of the slip s(t) is about satisfied. s\left(t^{(\mathrm{VIII})}=0.5 \mathrm{~s}\right)=0.0106. The plot of s(t) is given in Fig. E6.8.3.

e) Calculation of the torque T(t) and its plot from t=0 to t=5 s. From \frac{T}{T_{\max \_ \text {fict }}}=\frac{s}{1} follows

T=s \cdot T_{max \_ fict}          (E6.8-13)

Region I

T\left(t^{(\mathrm{I})}\right)=s\left(t^{(\mathrm{I})}\right) \cdot T_{\max _{-} \text {fit }}              (E6.8-14)

\begin{aligned} & T\left(t^{(\mathrm{I})}=0\right)=0.0088 .3734 .3=32.86 \mathrm{Nm} \equiv 0.2 \mathrm{pu}^* \\ & T\left(t^{(\mathrm{I})}=0.2 \mathrm{~s}\right)=266.9 \mathrm{Nm} \equiv 1.62 \mathrm{pu} \\ & T\left(t^{(\mathrm{I})}=0.4 \mathrm{~s}\right)=372.7 \mathrm{Nm} \equiv 2.27 \mathrm{pu} \end{aligned}

Region II

\begin{aligned} & T\left(t^{(\mathrm{II})}=0 \mathrm{~s}\right)=372.7 \mathrm{Nm} \equiv 2.27 \mathrm{pu} \\ & T\left(t^{(\mathrm{II})}=0.2 \mathrm{~s}\right)=186.5 \mathrm{Nm} \equiv 1.14 \mathrm{pu} \\ & T\left(t^{(\mathrm{II})}=0.4 \mathrm{~s}\right)=102.3 \mathrm{Nm} \equiv 0.62 \mathrm{pu} \end{aligned}

Region III

\begin{aligned} & T\left(t^{(\mathrm{III})}=0 \mathrm{~s}\right)=102.3 \mathrm{Nm} \equiv 0.62 \mathrm{pu} \\ & T\left(t^{(\mathrm{III})}=0.3 \mathrm{~s}\right)=259.8 \mathrm{Nm} \equiv 1.58 \mathrm{pu} \\ & T\left(t^{(\mathrm{III})}=0.6 \mathrm{~s}\right)=307.7 \mathrm{Nm} \equiv 1.87 \mathrm{pu} \end{aligned}

Region IV

\begin{aligned} & T\left(t^{(\mathrm{IV})}=0 \mathrm{~s}\right)=307.7 \mathrm{Nm} \equiv 1.87 \mathrm{pu} \\ & T\left(t^{(\mathrm{IV})}=0.2 \mathrm{~s}\right)=157.1 \mathrm{Nm} \equiv 0.96 \mathrm{pu} \\ & T\left(t^{(\mathrm{IV})}=0.4 \mathrm{~s}\right)=89.1 \mathrm{Nm} \equiv 0.54 \mathrm{pu} \end{aligned}

Region V

\begin{aligned} & T\left(t^{(\mathrm{V})}=0 \mathrm{~s}\right)=89.1 \mathrm{Nm} \equiv 0.54 \\ & T\left(t^{(\mathrm{V})}=0.4 \mathrm{~s}\right)=148.9 \mathrm{Nm} \equiv 0.91 \mathrm{pu} \\ & T\left(t^{(\mathrm{V})}=0.8 \mathrm{~s}\right)=161.3 \mathrm{Nm} \equiv 0.98 \mathrm{pu} \end{aligned}

Region VI

\begin{aligned} & T\left(t^{(\mathrm{NI})}=0 \mathrm{~s}\right)=161.3 \mathrm{Nm} \equiv 0.982 \mathrm{pu} \\ & T\left(t^{\mathrm({VI})}=0.2 \mathrm{~s}\right)=9.91 \mathrm{Nm} \equiv 0.55 \mathrm{pu} \\ & T\left(t^{(\mathrm{VI})}=0.4 \mathrm{~s}\right)=59.1 \mathrm{Nm} \equiv 0.36 \mathrm{pu} \end{aligned}

Region VII

\begin{aligned} & T\left(t^{(\mathrm{VII})}=0 \mathrm{~s}\right)=59.1 \mathrm{Nm} \equiv 0.36 \mathrm{pu}, \\ & T\left(t^{(\mathrm{VII})}=0.5 \mathrm{~s}\right)=79 \mathrm{Nm} \equiv 0.48 \mathrm{pu}, \\ & T\left(t^{(\mathrm{VII})}=1 \mathrm{~s}\right)=81.8 \mathrm{Nm} \equiv 0.50 \mathrm{pu} \end{aligned}

Region VIII

\begin{aligned} & T\left(t^{(\mathrm{VIII})}=0 \mathrm{~s}\right)=81.8 \mathrm{Nm} \equiv 0.50 \mathrm{pu} \\ & T\left(t^{(\mathrm{VIII})}=0.5 \mathrm{~s}\right)=39.6 \mathrm{Nm} \equiv 0.24 \mathrm{pu} \\ & T\left(t^{(\mathrm{VIII})}=1 \mathrm{~s}\right)=33.6 \mathrm{Nm} \equiv 0.204 \mathrm{pu} \end{aligned}

f) Point-wise calculation of the stator current \left|\widetilde{I}_s(t)\right| for the eight different regions. For the calculation of the stator current \left|\widetilde{I}_s(t)\right| the equivalent circuit of Fig. E6.8.5 is employed.
For s\left(t^{(\mathrm{I})}=0\right)=0.0088 one gets

\frac{R_r^{\prime}}{s}=\frac{0.2}{0.0088}=22.73 \Omega

* referred to rated torque T_{rat}=164.31 Nm. Note that the motor torque is at any time less than T_{max}=391.32 Nm. The motor torque T(t) is proportional to the slip s(t) and is depicted in Fig. E6.8.4.

With \widetilde{E}=221.75 \mathrm{~V} \angle 0^{\circ} the rotor current becomes

\widetilde{I}_{\mathrm{r}}^{\prime}=\frac{\widetilde{E}}{\left(\frac{R_r^{\prime}}{s}+j X_r^{\prime}\right)}=\frac{221.75 \mathrm{~V} \angle 0^{\circ}}{(22.73+j)}=(9.737-j 0.428) \mathrm{A} .

The magnetizing current is

\widetilde{I}_m=\frac{\widetilde{E}}{j X_m}=\frac{221.75 \angle 0^{\circ}}{j 30 \Omega}=-j 7.392 \mathrm{~A}

and the stator current

\widetilde{I}_s=\widetilde{I}_m+\widetilde{I}_{\mathrm{r}}^{\prime}=(9.737-\mathrm{j} 7.82) \mathrm{A}

where \tilde{E} is the reference phasor.
Table E6.8.1 summarizes the point-wise calculation of the rms values of \left|\widetilde{I}_s(t)\right| over a load cycle.

g) Figure E6.8.6 shows the plot of \left|\widetilde{I}_s(t)\right| as a function of time and Fig. E6.8.7 depicts \left|\widetilde{I}_s(t)\right|^2.

h) The rms current value of the stator current I_{s\_rms\_actual} for the load cycle of Fig. E6.8.1 can be calculated from Fig. E6.8.7 using

I_{s\_rms\_actual}=\sqrt{\frac{1}{T} \int_t^{t+T}\left|\widetilde{I}_s(t)\right|^2 \mathrm{~d} t}

The evaluation of this integral can be done mathematically or graphically. The graphic evaluation leads to I_{s\_rms\_actual}=47.7A.

i) Comparison of rated I_{s\_rms\_rat} and actual current I_{s\_rms\_actual}. Is the motor over- or underdesigned?

The losses of an induction motor are about proportional to the square of the stator current I_{s\_rms}. The ratio \left(\frac{I_{s \_rms \text { actual }}}{I_{s\_rms\_ rat }}\right)^2=\left(\frac{47.7}{49.82}\right)^2=0.91 indicates that the motor rating can be reduced by 9%, that is, P_{out\_reduced}=0.91 P_{out\_rat}=0.91 29.594 kW= 26.93 kW or 36.1 hp.

Let us assume that the stator current is larger than the rated current, say, I_{s\_rms\_high}=52A; then the motor will experience a lifetime reduction – if the service factor of the motor is 1.0 – based on the ratio \left(\frac{I_{s_{-}r m s \_\_ \text {high }}}{I_{s_{-}r m s_{-} \mathrm{rat}}}\right)^2=\left(\frac{52}{49.82}\right)^2=1.09, that is, the losses are about △P_{loss}=9% larger than the rated losses. This leads for a steadystate rated temperature of the hottest spot of T_2=85 °C≡273+85=358 kelvin at an ambient temperature of T_{amb}=40 °C to an additional temperature rise of △T=9% or △T=4.05 °C. With an activation energy of E=0.74 eV and a rated lifetime of 40 years the reduced lifetime of the three-phase induction motor is with △T=4.05 °C.

t_1=40 e^{-8.58 \cdot 10^3}\left(\frac{\Delta T}{358(358+\Delta T)}\right)=30.6 \text { years. }

Most induction motors are designed with a service factor of 1.1–1.15, that is, the motor can be overloaded by 10 to 15%. If this motor were designed with a service factor of 1.10 then no lifetime reduction would occur for a stator current of I_{s\_rms\_high}=52 A.