This section describes various equations that express voltage, current, and flux of stator and rotor of induction motor.$$ V_{S} = R_{S} I_{S} + Plambda_{S} , $$
(1)
$$ 0 = R_{r} I_{r} + Plambda_{S} , $$
(2)
where$$ V_{S} = [begin{array}{*{20}l} {V_{as1} } & {V_{as2} } & {V_{bs} } & {V_{cs} } \ end{array} ]^{T} , $$
(3)
$$ I_{S} = [begin{array}{*{20}l} {I_{as} } & {(I_{as} – I_{f} )} & {I_{bs} } & {I_{cs} } \ end{array} ]^{T} , $$
(4)
$$ I_{r} = [begin{array}{*{20}l} {I_{ar} } & {I_{br} } & {I_{cr} } \ end{array} ]^{T} , $$
(5)
$$ lambda_{S} = left[ {begin{array}{*{20}l} {lambda_{as1} } & {lambda_{as2} } & {lambda_{bs} } & {lambda_{cs} } \ end{array} } right]^{T} . $$
(6)
V represents the voltage, I shows current, flux is represented as ({lambda }_{i}), here, “s” and “r” represent stator and rotor respectively, a, b, c denote the three-phase system. as1, as2 denote unfaulty and faulty parts of the stator, respectively. Here P is the Laplace operator, the derivative operator (frac{d}{dt}) is replaced by P.$${V}_{as2}= beta {R}_{s}left({I}_{as}-{I}_{f}+ rho {uplambda }_{as2}right)= {R}_{f}{I}_{f}, $$
(7)
$${uplambda }_{mathrm{s}}= {mathrm{L}}_{mathrm{s}} {mathrm{I}}_{mathrm{s}}+ {mathrm{L}}_{mathrm{sr}} {mathrm{I}}_{mathrm{r}},$$
(8)
$${uplambda }_{r}=[{L}_{sr}{]}^{T}{I}_{s}+{L}_{r} {mathrm{I}}_{r}.$$
(9)
These equations show shorted part of stator winding voltage. β denotes shorted turn.The resistance matrix is shown as$${R}_{s}={R}_{s} diagleft[begin{array}{cc}left(1-beta right) & beta end{array} begin{array}{cc}1& 1end{array}right],$$
(10)
$${R}_{r}= {R}_{r} [I{]}_{3times 3}.$$
(11)
Here, the equations represent mutual inductance and self-inductance of stator winding12,13,14.$${L}_{s}= {L}_{ls} ; diag ; left[begin{array}{cc}left(1-beta right) & beta end{array} begin{array}{cc}1& 1end{array}right]left[begin{array}{cccc}(1-beta {)}^{2}& beta *(1-beta )& -frac{(1-beta )}{2}& -frac{(1-beta )}{2}\ beta *(1-beta )& (beta {)}^{2}& -frac{beta }{2}& -frac{beta }{2}\ -frac{(1-beta )}{2}& -frac{beta }{2}& 1& -frac{1}{2}\ -frac{(1-beta )}{2}& -frac{beta }{2}& -frac{1}{2}& 1end{array}right],$$
(12)
$${L}_{sr}={L}_{ms}left[begin{array}{ccc}left(1-beta right)*mathrm{cos}({theta }_{r})& left(1-turnright)*mathrm{cos}({theta }_{r}+2pi/ 3)& left(1-beta right)*mathrm{cos}({theta }_{r}-2pi/3)\ beta *mathrm{cos}({theta }_{r})& beta *mathrm{cos}({theta }_{r}+2pi/3)& beta *mathrm{cos}({theta }_{r}-2pi/3)\ mathrm{cos}({theta }_{r}-2pi/3)& mathrm{cos}({theta }_{r})& mathrm{cos}({theta }_{r}+2pi/3)\ mathrm{cos}({theta }_{r}+2pi/3)& mathrm{cos}({theta }_{r}-2pi/3)& mathrm{cos}({theta }_{r})end{array}right],$$
(13)
$${L}_{r}=left[begin{array}{ccc}{L}_{lr}+{L}_{RM}& -frac{{L}_{RM}}{2}& -frac{{L}_{RM}}{2}\ -frac{{L}_{RM}}{2}& {L}_{lr}+{L}_{RM}& -frac{{L}_{RM}}{2}\ -frac{{L}_{RM}}{2}& -frac{{L}_{RM}}{2}& {L}_{lr}+{L}_{RM}end{array}right].$$
(14)
β represents number of turns in phase a, ({theta }_{r}) represents rotor position, Ls shows self-inductance, Lr shows rotor self-inductance, and Lsr represents stator to rotor mutual inductance. Water is pumped out from a constant level water tank, and the pumping system consists of water tank, an asynchronous three-phase induction motor, and other parts. The tank receives liquid with input flow (mathrm{represented ; by} {q}_{{v}_{1}}) Output flow of the control valve is represented by ({q}_{{v}_{2}}). With the help of fluid mechanics and fundamental physics laws, plant dynamics analysis has been done, and a mathematical model has been developed21. This mathematical model includes the mathematical models of centrifugal pump and tank. The counterpart of Newton’s law of force is that angular acceleration is proportional to the torque on the axis. So, the equations show the motion for the motor and pump set.$$Jfrac{domega }{dt}={M}_{a}-{M}_{p}={M}_{MT}-left({M}_{p}+right).$$
(15)
J shows the moment of inertia. Here moment of inertia is the constant of proportionality in specific case. Active torque of asynchronous motor is shown by ({M}_{MT} ; mathrm{and ; accleration ; torque ; is ; shown ; by} ; {M}_{a}), passive or resistive torque of pump is shown by (M_{p}) and viscous torque is (M_{zeta })22. Network frequency is shown by f, and it is assumed that stator pole pair number is one. Here following equation shows the torque of the asynchronous motor.$${M}_{MT}={k}_{MT}{U}^{2}left(2pi f-omega right).$$
(16)
Viscous torque and passive torque can be represented by$$ M_{zeta } = k_{zeta } omega , $$
(17)
$$ M_{p} = frac{{rho gQ_{v2} H}}{{eta_{p} omega }}. $$
(18)
Equation 18 shows the basic parameters of the centrifugal pump, and the pump flow rate is shown by Q, H shows the pump head, and the angular velocity is shown by ω. Peripheral cross-section of the impeller channels and meridian component of velocity express the pump flow. Head value is proportional to angular velocity, as the flow rate is proportional to angular velocity23.In the last equation, the pump efficiency coefficient is denoted by which is constant, and in different modes, it changes to some extent, reflecting to the other parameters.The total operating system (H_{Total}) can be defined as$${H}_{Total}={H}_{S}+{H}_{D}+left({P}_{RT}-{P}_{RES}right).$$
(19)
Here the static head is represented by ({H}_{S}), the dynamic head is shown by ({H}_{D}), the pressure on the surface of the water in the receiving tank is shown by ({P}_{RT}), and the pressure on the surface of the water in the reservoir tank is represented by ({P}_{RES})24.Based on pump height pressure changes and it is considered negligible value. But atmospheric pressure changes with the height. Equation shows the change in pressure and elevation difference between the reservoir and receiving tank. But this is not so significant and considered as negligible.$${P}_{RT}-{P}_{RES}approx 0.$$
(20)
So the equation will be$${H}_{Total}={H}_{S}+{H}_{D}.$$
(21)
The difference between the point of discharge and the surface of the reservoir into the receiving tank is the static head which is shown by ({H}_{S}).The system’s static head will vary between maximum and minimum head values because the reservoir’s water level also varies.$${H}_{{S}_{min}}=discharge ; level-reservoir ; TWL,$$
(22)
$${H}_{{S}_{max}}=discharge ; level-reservoir ; BWL.$$
(23)
Here top water level is TWL, and the bottom water level is BWL.Within the system, as a result of dynamic friction head is generated. Basic Darcy Weisbach equation helps to calculate the dynamic head$${H}_{D}=frac{K{v}^{2}}{2g}.$$
(24)
Here the loss coefficient is shown by K, velocity in the pipe is shown by and acceleration is (g).Now velocity is shown as$$v=frac{Q}{A }.$$
(25)
Here flow rate is shown by Q through the pipe, and the cross-sectional area is shown by A.Area A is shown as$$A=frac{pi {D}^{2}}{4}.$$
(26)
The loss coefficient K is a form of two elements:$$K={K}_{fittings}+{K}_{pipe}.$$
(27)
({K}_{fittings}) is shown as pumping the water from the reservoir to receiver tank fittings used for the pipeworks of the system.({K}_{pipe}) is associated with the length of the pipe, friction, and the diameter of the pipe.$$ k_{pipe} = frac{FL}{D}. $$
(28)
Here F shows the friction factor, L shows the pipe length, and D is the pipe diameter. By the modified version of the Colebrook White equation, the friction coefficient f can be found.$$ F = frac{0.25}{{left[ {log left{ {frac{k}{3.7 times D} + left. {frac{5.74}{{{text{Re}}^{0.9} }}} right}} right.} right]^{2} }}. $$
(29)
Here roughness factor is k, and the Reynolds number is Re. The roughness factor k is a standard fixed value collected from standard tables and depends on the pipe’s material and pipe condition. For any flow in the pipe, the following formula is used for the calculation of the Reynolds number25:$$Re=frac{vD}{vartheta }.$$
(30)
(vartheta ) is the kinematic viscosity. Operation of the pumping system is based on affinity law. First affinity law is shown in the equation where flow Q is proportional to shaft speed N.$$frac{{Q}_{1}}{{Q}_{2}}=frac{{N}_{1}}{{N}_{2}}.$$
(31)
As per, the second affinity law, the head is proportional to the square of the shaft speed.$$frac{{H}_{1}}{{H}_{2}}=frac{{left({N}_{1}right)}^{2}}{{left({N}_{2}right)}^{2}}.$$
(32)
The power of the pump can be calculated as$$P=frac{Qtimes Htimes gtimes rho }{Pump ; Efficiency}.$$
(33)
Here P is the power requirement for the pump, H is the head, (g) acceleration gravity, and water’s density.
Source: https://www.nature.com/articles/s41598-022-16987-6
