Application of machine learning for inter-turn fault detection in pumping system

This section describes various equations that express the voltage, current, and flux of the stator and rotor of the induction motor.

$$ V_{S} = R_{S} I_{S} + Plambda_{S} , $$


$$ 0 = R_{r} I_{r} + Plambda_{S} , $$



$$ V_{S} = [begin{array}{*{20}l} {V_{as1} } & {V_{as2} } & {V_{bs} } & {V_{cs} } end{array} ]^{T} , $$


$$ I_{S} = [begin{array}{*{20}l} {I_{as} } & {(I_{as} – I_{f} )} & {I_{bs} } & {I_{cs} } end{array} ]^{T} , $$


$$ i_{r} = [begin{array}{*{20}l} {I_{ar} } & {I_{br} } & {I_{cr} } end{array} ]^{T} , $$


$$ lambda_{S} = left[ {begin{array}{*{20}l} {lambda_{as1} } & {lambda_{as2} } & {lambda_{bs} } & {lambda_{cs} } end{array} } right]^{T} . $$


V represents the voltage, I shows the current, the flux is represented as ({lambda}_{i})here, “s” and “r” represent respectively the stator and the rotor, a, b, vs designate the three-phase system. as1, as2 refer to the non-defective and defective parts of the stator respectively. here P is the Laplace operator, the derived operator (frac{d}{dt}) is replaced by P.

$${V}_{as2}= beta {R}_{s}left({I}_{as}-{I}_{f}+ rho {uplambda }_{as2}right )= {R}_{f}{I}_{f}, $$


$${uplambda }_{mathrm{s}}= {mathrm{L}}_{mathrm{s}} {mathrm{I}}_{mathrm{s}}+ {mathrm{ L}}_{mathrm{sr}} {mathrm{I}}_{mathrm{r}},$$


$${uplambda}_{r}=[{L}_{sr}{]}^{T}{I}_{s}+{L}_{r} {mathrm{I}}_{r}.$$


These equations show a shorted portion of the stator winding voltage. β denotes a short turn.

The resistance matrix is ​​represented by

$${R}_{s}={R}_{s} diagleft[begin{array}{cc}left(1-beta right) & beta end{array} begin{array}{cc}1& 1end{array}right],$$


$${R}_{r}= {R}_{r} [I{]}_{3times 3}.$$


Here the equations represent the mutual inductance and self-inductance of the stator winding12,13,14.

$${L}_{s}= {L}_{ls}; diag; left[begin{array}{cc}left(1-beta right) & beta end{array} begin{array}{cc}1& 1end{array}right]left[begin{array}{cccc}(1-beta {)}^{2}& beta *(1-beta )& -frac{(1-beta )}{2}& -frac{(1-beta )}{2} beta *(1-beta )& (beta {)}^{2}& -frac{beta }{2}& -frac{beta }{2} -frac{(1-beta )}{2}& -frac{beta }{2}& 1& -frac{1}{2} -frac{(1-beta )}{2}& -frac{beta }{2}& -frac{1}{2}& 1end{array}right],$$


$${L}_{sr}={L}_{ms}left[begin{array}{ccc}left(1-beta right)*mathrm{cos}({theta }_{r})& left(1-turnright)*mathrm{cos}({theta }_{r}+2pi/ 3)& left(1-beta right)*mathrm{cos}({theta }_{r}-2pi/3) beta *mathrm{cos}({theta }_{r})& beta *mathrm{cos}({theta }_{r}+2pi/3)& beta *mathrm{cos}({theta }_{r}-2pi/3) mathrm{cos}({theta }_{r}-2pi/3)& mathrm{cos}({theta }_{r})& mathrm{cos}({theta }_{r}+2pi/3) mathrm{cos}({theta }_{r}+2pi/3)& mathrm{cos}({theta }_{r}-2pi/3)& mathrm{cos}({theta }_{r})end{array}right],$$


$${L}_{r}=left[begin{array}{ccc}{L}_{lr}+{L}_{RM}& -frac{{L}_{RM}}{2}& -frac{{L}_{RM}}{2} -frac{{L}_{RM}}{2}& {L}_{lr}+{L}_{RM}& -frac{{L}_{RM}}{2} -frac{{L}_{RM}}{2}& -frac{{L}_{RM}}{2}& {L}_{lr}+{L}_{RM}end{array}right].$$


β represents the number of phase turns a, ({theta}_{r}) represents the position of the rotor, Ls shows self-inductance, Lr shows the self-inductance of the rotor, and Lsr represents the mutual inductance from the stator to the rotor. Water is pumped from a constant level water tank, and the pumping system consists of a water tank, a three-phase asynchronous induction motor and other parts. The tank receives the liquid with an inlet flow (mathrm{represented ; by} {q}_{{v}_{1}}) The output flow of the control valve is represented by ({q}_{{v}_{2}}). Using fluid mechanics and fundamental laws of physics, an analysis of plant dynamics was performed and a mathematical model was developed.21. This mathematical model includes the mathematical models of the centrifugal pump and the tank. The counterpart of Newton’s law of force is that angular acceleration is proportional to torque on the axis. So the equations show the motion of the motor and the pump.

$$Jfrac{domega }{dt}={M}_{a}-{M}_{p}={M}_{MT}-left({M}_{p}+ right).$$


J indicates the moment of inertia. Here, the moment of inertia is the constant of proportionality in a specific case. The active torque of the asynchronous motor is indicated by ({M}_{MT} ; mathrm{et ; acceleration ; torque ; is ; shown ; by} ; {M}_{a})the passive or resistive torque of the pump is represented by (M_{p}) and the viscous couple is (M_{zeta })22. The network frequency is indicated by F, and the number of pole pairs of the stator is assumed to be one. The following equation shows the torque of the asynchronous motor.

$${M}_{MT}={k}_{MT}{U}^{2}left(2pi f-omega right).$$


Viscous torque and passive torque can be represented by

$$ M_{zeta } = k_{zeta } omega , $$


$$ M_{p} = frac{{rho gQ_{v2} H}}{{eta_{p} omega }}. $$


Equation 18 shows the basic parameters of the centrifugal pump, and the pump flow rate is represented by Q, H shows the pump head, and the angular velocity is indicated by ω. The peripheral cross-section of the impeller channels and the meridian component of the speed express the flow rate of the pump. The value of the load is proportional to the angular velocity, because the flow is proportional to the angular velocity23.

In the last equation, the pump efficiency coefficient is denoted by what is constant, and in different modes it changes to some extent, reflecting the other parameters.

The whole operating system (H_{Total}) can be defined as



Here, the static charge is represented by ({H}_{S})the dynamic load is represented by ({HD})the pressure at the surface of the water in the receiving tank is indicated by ({P}_{RT})and the pressure at the surface of the water in the reservoir is represented by ({CLOSE})24.

Based on the pressure changes of the height of the pump and it is considered a negligible value. But atmospheric pressure changes with altitude. The equation shows the pressure change and elevation difference between the reservoir and the receiving reservoir. But it is not so important and considered negligible.

$${P}_{RT}-{P}_{RES}about 0.$$


So the equation will be



The difference between the point of discharge and the surface of the tank in the receiving tank is the static head which is represented by ({H}_{S})The static head of the system will vary between the maximum and minimum head values ​​because the water level in the tank also varies.

$${H}_{{S}_{min}}=discharge; tank-level ; TWL,$$


$${H}_{{S}_{max}}=discharge; tank-level ; BWL.$$


Here the upper water level is TWL and the lower water level is BWL.

In the system, as a result of dynamic friction, a head is generated. Darcy Weisbach’s Basic Equation Helps Calculate Dynamic Load



Here, the loss coefficient is represented by Kthe velocity in the pipe is indicated by and the acceleration is (g).

Now the speed is represented by

Here, the flow is represented by Q through the pipe, and the cross section is indicated by A.

Area A is shown as



The loss coefficient K is a form of two elements:



({K}_{fittings}) is shown as the pumping of water from the reservoir to the receiving reservoir fittings used for system piping.

({K}_{pipe}) is associated with pipe length, friction and pipe diameter.

$$ k_{pipe} = frac{FL}{D}. $$


here F shows the friction factor, L indicates the length of the pipe, and D is the diameter of the pipe. By the modified version of Colebrook White’s equation, the coefficient of friction F can be found.

$$ F = frac{0.25}{{left[ {log left{ {frac{k}{3.7 times D} + left. {frac{5.74}{{{text{Re}}^{0.9} }}} right}} right.} right]^{2} }}. $$


Here, the 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 pipe material and pipe condition. For any flow in the pipe, the following formula is used for the calculation of the Reynolds number25:

$$Re=frac{vD}{vartheta }.$$


(vartheta) is the kinematic viscosity. The operation of the pumping system is based on the law of affinity. The first law of affinity is shown in the equation where the flux Q is proportional to the speed of the shaft NOT.



According to the second law of affinity, the head is proportional to the square of the speed of the shaft.

$$frac{{H}_{1}}{{H}_{2}}=frac{{left({N}_{1}right)}^{2}}{{left ({N}_{2}right)}^{2}}.$$


The power of the pump can be calculated as

$$P=frac{Qtimes Htimes gtimes rho }{Pump ; Efficiency}.$$


here P is the power required by the pump, H is the head, (g) acceleration, gravity and water density.

Sherry J. Basler