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Mohamed Dewidar 2013
Electric Submersible Pumps Chapter 3
Submersible Motor Table of Content Section 1 2
Content
Page 3 3
2.1 2.2 2.3 2.4 2.5
General Motor construction Stator Rotor Rotor bearing Motor thrust bearing Pothead
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3.1 3.2 3.3 3.4 3.5 3.6
Electromagnetism Magnetic field Magnetic flux and flux density Magnetic field due to current in a solenoid Changing polarity Induced voltage Electromagnetic attraction Start coil arrangement
15
Power supply Start Time 1 Time 2 360 degree rotation
15
3
4 5 5.1 5.2 5.3 5.4 6
Mathematical analysis of rotating magnetic field due to 3 phase current
19
7
Slip
24
8
Rotor current frequency
24
9
Magneto-motive force and magnetic field Strength
25
Force in current carrying conductor in magnetic field
26
Torque on a current carrying coil in magnetic field
27
12
Theory of operation
28
13
Motor configurations
30
14
Motor current
32
15
Motor rating
32
16
Motor protection
35
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Application of ESP motor
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10 11
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Fundamentals of electricity Equivalent circuit of 19.1 Effective circuit of Standstill 19.2 Effective circuit of Operating conditions 19.3 Power relations
induction motor induction motor at
39 63
induction motor under (rotor is shorted)
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Determination of motor parameters
70
21
NEMA standard for squirrel cage IM
75
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Torque of squirrel cage IM
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Submersible Motor 3.1. General Motor is an electric machine energy into mechanical energy.
which
converts
electric
Three phase induction motors are the most frequently encountered in industry. They are simple, rugged, low priced, and easy to maintain. They run at essentially constant speed from zero to full load. The speed is frequency dependent; however, variable speed electronic drives are being used more and more to control the speed of the motors. ESP motor Classified as 3 phases, induction, Alternating current motor.
squirrel
cage,
2
pole
The position of the motor in ESP integrity is just below the protector (seal).
3.2. Motor Construction The induction motor is a three phase, squirrel cage, two pole induction design, consists of: 1. Stator, which supports windings which receive energy from the mains circuit. 2. Rotor, which carries windings in which working current is induced. 3. Shaft, which transfers the mechanical energy to the pump. 4. Bearings 5. Housing 6. Insulated Magnet Wire 7. Thrust bearing
Fig (3.1) Most of the motor construction
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3.2.1. Stator The stator is the stationary electrical part of the motor. The stator core of a National Electrical Manufacturers Association (NEMA) motor is made up of several hundred thin laminations.
Fig (3.2) stator laminations
Stator laminations are stacked together forming a hollow cylinder (fig 3.2). Coils of insulated wire are inserted into slots of the stator core (fig 3.3).
Fig (3.3) stator core
Electromagnetism is the principle behind motor operation. Each grouping of coils, together with the steel core it surrounds, form an electromagnet. The stator windings are connected directly to the power source. The stator winding consists of three individual windings which overlap one another and are offset by an electrical angle of 120° (fig 3.4). When it is connected to the power supply, the incoming current will first magnetize the stator. This magnetizing current generates a rotary field which turns with synchronous speed Ns.
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Fig (3.4) stator windings
When the alternating current passes through a coil group, a magnetic field of fixed shape and sinusoidally varying amplitude will result. A magnetic pole is formed at the center of this coil group. The internal stator winding connections determine the number of poles, the voltage applied to individual windings and the direction of rotation. In a three phase induction motor, rotating magnetic field is obtained by three separate single phases with currents that differ in phase by 120 degrees. Three phases reach their maximum and minimum in a rapid succession sequence. As currents change, the effect is to rotate the magnetic fields. The magnetic field rotates continuously at a constant speed determined by the line frequency and number of poles. The laminations wound with three very big loops of wire one for each phase. When current is flowing through a phase, magnetic flux is induced as shown in fig (3.5).
Fig (3.5) induced magnetic flux due to current flow
Because of this configuration, the inside of the stator holds a strong magnetic field. The strength of the field will depend on the amount of current flowing through the wire loop (i.e. the phase winding). 5
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Electric Submersible Pumps Chapter 3
The more copper that is in the stator, the more the winding losses are reduced, making the motor more efficient. The winding is "two pole" because two magnetic poles are created (one North and one South). Motors can be wound differently to create more than two poles such as a four pole motor. Remember that the direction of the magnetic field in the stator depends on the direction of current flowing in the wire. With AC, or Alternating Current, the direction of current flow is changing 60 times every second for 60 Hz power (or 50 times per second for 50 Hz power).
3.2.2. Rotor The rotor circuit.
is
the
rotating
part
of
the
electromagnetic
The most common type of rotor is the “squirrel cage” rotor.
Fig (3.6) squirrel cage rotor
The rotor consists of a stack of steel laminations. The squirrel cage rotor consists of copper or aluminum bars accommodated in slots of rotor core (fig 3.7).
Fig (3.7) rotor core
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Electric Submersible Pumps Chapter 3
At each end the bard are connected to heavy conducting end rings which serve the purpose of short circuit bars. In the absence of the end rings, emfs would be induced in the rotor bars but no current would flow through them and no torque would be produced. The wound rotor has a 3 phase winding placed in the slots of the rotor core. The rotor is wound for the same number of poles as the stator. The terminals of the rotor winding are brought out of three slip rings mounted on the machine shaft. During running condition the slip ring are short circuited so as to close the rotor circuit. The air gap or more accurately, the clearance between the stator and rotor, should be as small as possible in order to the primary and secondary leakage fluxes to minimum.
3.2.3. Rotor bearing Rotor Bearings are one of the most vital parts of the motor. The Bearing Material is Babbitt-lined steel and machined after processing. There are fluid holes to insure oil circulation and wide angle oil grooves on the OD to distribute lubrication evenly over the entire length of the bearing surface.
Fig (3.8) rotor bearing
The bearing sleeve is a bronze material for the sleeve construction of the bearing. This part is keyed to the shaft and the hole on the sleeve is aligned with the hole on the shaft to insure proper cooling and lubrication.
3.2.4. Motor thrust bearing The motor thrust bearing is installed at the top of the rotor string. It is designed to hold the weight of the entire rotor string.
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Electric Submersible Pumps Chapter 3
Fig (3.9) motor thrust bearing
Currently three types of motor thrust bearings are used. Babbitt Glacier KMC Bronze pads The thrust bearing limits on the system will indicate the type of load required for the selected bearing material.
3.2.5. Pothead Pothead is the place where the motor lead extension cable is connected to the motor three phase windings (fig 3.10). There are two types of pothead, they are: 1. Tape in type, where tape wrapped around individual connector leads inside motor. 2. Plug in type, where mating block mounted in motor.
Fig (3.10) motor pothead
3.3. Electromagnetism 3.3.1. Magnetic field When an electric current is passed through a conductor, a magnetic field is set up around the conductor. The direction 8
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Electric Submersible Pumps Chapter 3
of the magnetic field can be found by using right hand rule or the right hand screw. The right hand rule states “Grasp the wire in the right hand, with the thumb pointing in the direction of the current. The fingers will curl around the wire in the direction of the magnetic field” The right hand screw is explained in this way, as a wood screw is turned clockwise it progresses into the wood. The horizontal direction of screw is analogous to the direction of current in a conductor. The circular motion of the screw shows the direction of the magnetic flux around the conductor. (fig 3.11).
Fig (3.11) Magnetic field around the conductor carrying current
In fig 3.12 (a) the dot inside the circle is the standard symbol used to show that the direction of current flow is out of the page. Then, by right hand rule, the magnetic field is counterclockwise. In fig 3.12 (b) the cross is the standard symbol used to show that the current is entering the page. The magnetic field is clockwise. The strength of the magnetic field is proportional to the current, i.e. if the current is doubled, the magnetic field will be doubled.
(a) Current coming out of the page
(b) Current entering the page
Fig (3.12) Magnetic field surrounding the conductor
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Since a current carrying conductor has a magnetic field around it, when two current carrying conductors are brought close together, there will be interactive between the fields. When the currents in the two conductors are in opposite direction, the fields are as shown in fig 3.13 (a) and the force of repulsion is experienced. When the currents are in the same direction, the field are as shown in fig 3.13 (b) and a fore of attractive is experienced.
Fig (3.13) force between parallel current carrying conductors Consider a single turn coil carrying current. As shown in figure 3.5 the hole of the magnetic flux generated by electric current passed through the centre of the coil. Therefore the coil acts like a little magnet and has a magnetic field with identifiable N and S poles. The coil may also have more than one turn. The flux generated by each of the individual turns of the coil tends to link up and pass out of one end of coil and back into the other end. Such an arrangement is known as solenoid (figure 3.6) and has a magnetic field pattern very similar to that of bar magnet. The right hand rule for determining the direction of flux from solenoid states “Grasp the solenoid in the right hand such that the fingers point in the direction of current flow in the coil. The thumb will point towards N pole of field”. As discussed above, a current flow in the conductors produces a magnetic field. The converse is also possible; that is, a magnetic field can produce a current in a conductor. This is known as the phenomenon of electromagnetic induction (this will be discussed later).
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3.3.2. Magnetic flux and flux density The total lines of force in a magnetic field are called magnetic flux. Flux density is the flux per unit area of cross section. Weber (Wb) is the SI unit of magnetic flux. Tesla (T) is the SI unit of flux density and represents Wb/m2. Flux density is also known as magnetic induction. From its definition:
B
A
Where B is the flux density in teslas, φ is the total flux in webers and A is the cross sectional area in m2.
3.3.3. Magnetic field due to current in a solenoid When an electric current passed through a solenoid, the resultant magnetic flux is very similar to that of a bar magnet. The magnetic flux lines make complete circuit inside and outside the coil; each line is a closed path. The side at the flux emerges is the North Pole, the other side where the magnetic flux reenters is the South Pole. The strength of the magnetic field in the DC electromagnet can be increased by increasing the number of turns and/or current in the coil. The greater the number of turns the stronger the magnetic field will be. See fig (3.14) and (3.15)
Fig (3.14) Magnetic field in coils of different number of turns
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Fig (3.15)
Magnetic field in coils of different currents The magnetic flux density in the interior of a solenoid carrying an electric current depends on the current intensity passing through the coil (I) and number of turns per unit length (n), i.e. B is proportionally change with I an n .
B *n*I Where μ is the permeability of the core material. The equation can be written as follows
Where N is the number of turns of a solenoid and l is its length.
3.3.4. Changing polarity The magnetic field of an electromagnet has the same characteristics as a natural magnet, including a north and South Pole. However, when the direction of current flow through the electromagnet changes, the polarity of the electromagnet changes. The polarity of an electromagnet connected to an AC source will change at the same frequency as the frequency of the AC source. This can be demonstrated in the following illustration (fig. 3.7). At Time 1 current flow is at zero. There is no magnetic field produced around the electromagnet. At Time 2 current is flowing in a positive direction. A magnetic field builds up around the electromagnet. The electromagnet assumes a polarity with the South Pole on the top and the North Pole on the bottom. At Time 3 current flow is at its peak positive value. The strength of the electromagnetic field is at its greatest value. At Time 4 current flow decreases and the magnetic field 12
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begins to collapse, until Time 5 when current flow and magnetic field are at zero. Current immediately begins to increase in the opposite direction. At Time 6 current is increasing in a negative direction. The polarity of the electromagnetic field has changed. The north pole is now on top and the south pole is on the bottom. The negative half of the cycle continues through Times 7 and 8, returning to zero at Time 9. This process will repeat 50 times a second with a 50 Hz AC power supply (fig 3.16).
Fig (3.16)
3.3.5. Induced voltage A conductor moving through a magnetic field will have a voltage induced into it. This electrical principle is used in the operation of AC induction motors. In the following illustration an electromagnet is connected to an AC power source. Another electromagnet is placed above it. The second electromagnet is in a separate circuit. There is no physical connection between the two circuits. Voltage and current are zero in both circuits at Time 1. At Time 2 voltage and current are increasing in the bottom circuit. A magnetic field builds up in the bottom electromagnet. Lines of flux from the magnetic field building up in the bottom electromagnet cut across the top electromagnet. A voltage is induced in the top electromagnet and current flows through it. At Time 3 current flow has reached its peak. Maximum current is flowing in both circuits. The magnetic field around the coil continues to
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build up and collapse as the alternating current continues to increase and decrease. As the magnetic field moves through space, moving out from the coil as it builds up and back towards the coil as it collapses, lines of flux cut across the top coil. As current flows in the top electromagnet it creates its own magnetic field (fig 3.17).
Fig (3.17) magnetic field increases as the current increases
3.3.6. Electromagnetic attraction The polarity of the magnetic field induced in the top electromagnet is opposite the polarity of the magnetic field in the bottom electromagnet. Since opposite poles attract, the top electromagnet will follow the bottom electromagnet when it is moved (fig. 3.18)
Fig (3.18)
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3.4. Start coil arrangement The following schematic (fig 3.19) illustrates the relationship of the coils. In this example six coils are used, two coils for each of the three phases. The coils operate in pairs. The coils are wrapped around the soft iron core material of the stator. These coils are referred to as motor windings. Each motor winding becomes a separate electromagnet. The coils are wound in such a way that when current flows in them; one coil is a north pole and its pair is a south pole. For example, if A1 were a north pole then A2 would be a south pole. When current reverses direction the polarity of the poles would also reverse.
Fig (3.19)
3.5. Power supply The stator is connected to a 3-phase AC power supply. In the following illustration phase A is connected to phase A of the power supply. Phase B and C would also be connected to phases B and C of the power supply respectively.
Fig (3.20)
Phase windings (A, B, and C) are placed 120° apart. In this example, a second set of three-phase windings is installed. 15
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The number of poles is determined by how many times a phase winding appears. In this example, each phase winding appears two times. This is a two-pole stator. If each phase winding appeared four times it would be a four-pole stator.
Fig (3.21) 2 poles stator winding
When AC voltage is applied to the stator, current flows through the windings. The magnetic field developed in a phase winding depends on the direction of current flow through that winding. The following chart is used here for explanation only. It will be used in the next few illustrations to demonstrate how a rotating magnetic field is developed. It assumes that a positive current flow in the A1, B1 and C1 windings result in a north pole.
Winding A1 A2 B1 B2 C1 C2
Current Flow Direction Positive
Negative
North South North South North South
South North South North South North
3.5.1. Start It is easier to visualize a magnetic field if a start time is picked when no current is flowing through one phase. In the following illustration, for example, a start time has been selected during which phase A has no current flow, phase B has current flow in a negative direction and phase C has current flow in a positive direction. Based on the above chart, B1 and C2 are south poles and B2 and C1 are north poles. Magnetic lines of flux leave the B2 North Pole and enter the nearest South Pole, C2. Magnetic lines of flux also 16
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Electric Submersible Pumps Chapter 3
leave the C1 North Pole and enter the nearest South Pole, B1. A magnetic field results, as indicated by the arrow, fig (3.22).
Fig (3.22) start
3.5.2. Time 1 If the field is evaluated at 60° intervals from the starting point, at Time 1, it can be seen that the field will rotate 60°. At Time 1 phase C has no current flow, phase A has current flow in a positive direction and phase B has current flow in a negative direction. Following the same logic as used for the starting point, windings A1 and B2 are north poles and windings A2 and B1 are south poles, fig (3.23).
Fig (3.23) time 1 17
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3.5.3. Time 2 At Time 2 the magnetic field has rotated 60°. Phase B has no current flow. Although current is decreasing in phase A it is still flowing in a positive direction. Phase C is now flowing in a negative direction. At start it was flowing in a positive direction. Current flow has changed directions in the phase C windings and the magnetic poles have reversed polarity.
Fig (3.24) time 2
3.5.4. 360 degree rotation At the end of six such time intervals the magnetic field will have rotated one full revolution or 360°. This process will repeat 60 times a second on a 60 Hz power supply.
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Fig (3.25) time 1
3.6. Mathematical analysis of rotating magnetic field due to 3 phase current When a 3-phase winding is energized from a 3-phase supply, a rotating magnetic field is produced. This field is such that its poles do no remain in a fixed position on the stator but go on shifting their positions around the stator. For this reason, it is called a rotating magnetic field. It will be shown that magnitude of this rotating field is constant and is equal to 1.5 fm where fm is the maximum flux due to any phase. To see how rotating field is produced, consider a 2-pole, 3 phase winding as shown in fig 3.26(i). The three phases A, B and C are energized from a 3-phase source and currents in these phases are indicated as IA, IB and IC Referring to Fig 3.26 (ii), the fluxes produced by these currents are given by:
Here φm is the maximum flux due to any phase. We shall now prove that this 3-phase supply produces a rotating field of constant magnitude equal to 1.5 φm.
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(i)
(ii)
Fig (3.26) At start fig 3.26 (ii), and fig 3.27 (i) the current in phase A is zero and currents in phases B and C are equal and opposite. The currents are flowing outward in the top conductors and inward in the bottom conductors. This establishes a resultant flux towards right. The magnitude of the resultant flux is constant and is equal to 1.5 φm as proved under: At start, ωt = 0°. Therefore, the three fluxes are given by;
(i)
(ii) Fig (3.27) 20
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The phasor sum of - φB and φC is the resultant flux φR (fig. 3.28). It is clear that:
Fig (3.28) At time 1, fig (15 (ii)), fluxes are given by;
ωt = 60°. Therefore, the three
The phasor sum of - φB and φA is the resultant flux φR [See Fig. (3.18). It is clear that:
Fig (3.29) 21
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At time 2, fig (26 (ii), ωt = 120°. Therefore, the three fluxes are given by;
The phasor sum of – φC and φA is the resultant flux φR, (Fig. 3.30). It is clear that:
It follows from the above discussion that a 3-phase supply produces a rotating field of constant value (= 1.5 φm, where φm is the maximum flux due to any phase).
Fig (3.30) We shall now use another useful method to find the magnitude and speed of the resultant flux due to three-phase currents. The three-phase sinusoidal currents produce fluxes φ1, φ2 and φ3 which vary sinusoidally. The resultant flux at any instant will be the vector sum of all the three at that instant. The fluxes are represented vectors, fig (3.31).
by
three
variable
magnitude
In fig (3.31), the individual flux directions are fixed but their magnitudes vary sinusoidally as does the current that produces them. To find the magnitude of the resultant flux, resolve each flux into horizontal and vertical components and then find their vector sum.
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Fig (3.31)
The resultant flux is given by;
Thus the resultant flux has constant magnitude (= 1.5 φm) and does not change with time. The angular displacement of φR relative to the OX axis is,
So,
Thus the resultant magnetic field rotates at constant angular velocity ω(= 2πf) rad/sec. For a P-pole machine, the rotation speed (ωs) is
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Thus the resultant flux due to three-phase currents is of constant value (= 1.5 φm where φm is the maximum flux in any phase) and this flux rotates around the stator winding at a synchronous speed of 120f/P rpm. For example, for a 2-pole, 50 Hz, 3-phase induction motor, N, = 120x50/2 = 3000 rpm. It means that flux rotates around the stator at a speed of 3000 rpm.
3.7. Slip We have seen above that rotor rapidly accelerates in the direction of rotating field. In practice, the rotor can never reach the speed of stator flux. If it did, there would be no relative speed between the stator field and rotor conductors, no induced rotor currents and, therefore, no torque to drive the rotor. The friction and windage would immediately cause the rotor to slow down. Hence, the rotor speed (N) is always less than the suitor field speed (Ns). This difference in speed depends upon load on the motor. The difference between the synchronous speed Ns of the rotating stator field and the actual rotor speed N is called slip. It is usually expressed as a percentage of synchronous speed i.e.,
The quantity Ns-N is sometimes called slip speed. When the rotor is stationary (i.e., N = 0), slip, s = 1 or 100 %. In an induction motor, the change in slip from no-load to full-load is hardly 0.1% to 3% so that it is essentially a constant-speed motor.
3.8. Rotor current frequency The frequency of a voltage or current induced due to the relative speed between a vending and a magnetic field is given by the general formula;
Where: n = Relative speed between magnetic field and the winding P = Number of poles
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For a rotor speed N, the relative speed between the rotating flux and the rotor is Ns-N. Consequently, the rotor current frequency fr is given by;
i.e., Rotor current frequency = slip x Supply frequency. When the rotor is at standstill or stationary (i.e., s = 1), the frequency of rotor current is the same as that of supply frequency (fr = sf = 1, fr = f). As the rotor picks up speed, the relative speed between the rotating flux and the rotor decreases. Consequently, the slip S and hence rotor current frequency decreases.
3.9. Magneto-motive force and magnetic field strength An emf causes a current to flow in an electric circuit. Similarly a magneto motive force (mmf), symbol F, produces a magnetic flux in a magnetic circuit. The mmf of a coil is the product of current in the coil and the number of turns of the coil and has the unit of ampere turns (AT). The magnetic flux which can be set up in a magnetic circuit depends on the mmf and the length of the magnetic circuit. If the length is large, the mmf has to act over a long distance and the resulting magnetic flux is small. The magnetic field strength H is defined as the mmf per unit length of magnetic circuit, i.e.
Where I is the current in amperes, and N is the number of turns and l is the length of the magnetic circuit in meter. Example: The total flux of an electro magnet is 4x10-4 Wb. a. If the cross sectional area of the core is 1 cm2, find the flux density in the core. b. The electromagnet has 50 turns and a current of 1 A flow through the coil. If the length of the magnetic circuit is 20 cm, find the mmf and the magnetic field strength. Flux density:
B
A
4 104 Wb 4 T 1 104 m2 25
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F NI 50 1 50 AT NI 50 AT H 250 AT/m l 20 102 m
mmf: Field strength:
3.10. Force in current carrying conductor in magnetic field Figure 3.32 (a) shows a current carrying conductor (the current entering the page) laying in magnetic field flux density B. The current in the conductor sets up a flux in a clockwise direction around the conductor. When the external field is in the vertically downward direction, the field of the conductor assists the external field on the right hand side of the conductor. The effect of this is to produce a force that pushes the conductor to the lift. If the direction of the current is reversed as shown in figure 3.6 (b) the flux around the conductor is in counterclockwise direction and the resulting force pushes the conductor to the right. In both cases maximum force is generated if the conductor is at right angle to the direction of the magnetic flux. The force is always in a direction perpendicular to both the conductor and the field. The magnitude of the force F is given by: Where B is the flux density in telsas, I is the current in amperes, and l is the length of the conductor in meters. A force of one Newton is exerted on a 1 meter long conductor carrying a current of 1 ampere and situated at right angle to a magnetic field having a flux density of 1 tesla.
Force Flux set up by current in on conductor
Force
+
(a) Conductor current entering the page
Flux set up by current in on conductor
(b) Conductor current coming out of the page
Fig (3.32) Force on a current carrying conductor in a magnetic field 26
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If the conductor is not perpendicular to magnetic field but inclined at angle Ө to the magnetic field, the force is given by:
3.11. Torque on a current carrying coil in magnetic field Figure 3.33 shows a current carrying coil placed in magnetic field. From the discussion in section 3.5 it follows that a downward force is exerted on the left hand conductor and an upward force is exerted on the right hand conductor. The force on each conductor is given by the equation 3.3. The total force is given by:
I Flux
2r
S
N
Fig (3.33) Force on a coil carrying current in a magnetic field If the coil has N turns, the force is: Since the force is acting at a radius r meters, the torque on the coil is: The above simple arrangement is the basic part of electric measuring instrument. The operation of an electric motor is also based on this principle. Example: A 30 cm long conductor is carrying a current of 10 A and situated at a right angle to a magnetic field having flux density of 0.8 T. Calculate the force on the conductor. F = 0.8 x 10 x 30 x 10-2 = 2.4 N Example: A 200 turn coil having an axial length of 3 cm and radius of 1 cm is pivoted in magnetic field having a flux density of 0.8
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T. The coil carries a current of 0.5 A. Calculate the torque acting on the coil. Torque = 2 x 0.8 x 0.5 x 3 x10-2 x 200 x 1 x10-2 = 4.8 x 10-2 N.m
Figure 3.34 is the same as direction of the magnetic flux.
figure
3.33
but
with
other
Fig (3.34) Force on a coil in a magnetic field If the loop is in line with the magnetic field, the secondary magnetic field will be perpendicular to the main field. This will cause two equal and opposite forces (a torque) on the loop causing it to rotate until the forces balance (fig 3.35). The forces will reach a steady state and hold the magnet in place as long as current is applied. To cause rotation, the field must rotate. This is accomplished with the alternating current where the field is rotated. This is accomplished with the alternating current going through the windings in the stator of the induction motor.
Fig (3.35) Force on a coil in a magnetic field
3.12. Theory of operation An induction motor consists of a stator and rotor. The stator carries a 3 phase winding in its slots and is connected to a 3 phase supply. The rotor carries a cage winding and it is free to rotate within the stator. 28
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The 3 phase currents flowing in the stator winding produce a rotating field. The rotor winding cuts the rotating field and an emf is induced in the rotor winding. When the rotor is at rest, the frequency of this emf is the same as the supply frequency. If the rotor circuit is closed, a current flows in the rotor winding. The rotor current produces an mmf which rotates and is directed in opposition to stator mmf. The interaction of the stator and the rotor fields produces a torque which causes the rotor to rotate in the direction of the rotor field. If the rotor shaft is not loaded, the machine has only to rotate itself against the mechanical losses and the rotor speed is very close to the synchronous speed. However, the rotor speed cannot become equal to the synchronous speed because if it does so, the emf induced in the rotor winding would become zero and there will be no torque. Hence the rotor speed remains slightly less than the synchronous speed. If the motor shaft is loaded, the rotor will slow down and the relative speed of rotor with respect to the stator rotating field will increase. The emf induced in the rotor winding will increase and this will produce more rotor current which will increase the electromagnetic torque produced by the motor. Conditions of equilibrium are attained when the rotor speed has adjusted to a new value so that the electromagnetic torque is sufficient to balance the mechanical or load torque applied to the shaft. The speed of the motor when running under full load conditions is somewhat less than the no load speed. The speed of rotation of the field mmf is called synchronous speed is related to the frequency and number of poles by the expression:
p f ns 2 Where: f is the frequency in Hz, p is the number of poles and, ns is the synchronous speed in revolution per second. An alternative form of above expression is,
Ns
120 f p
Where: Ns is the synchronous speed in revolution per minute (rpm).
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3.13. Motor configurations Motors come in single sections (head and base) as well as tandem configurations. The tandems can include the UT (upper tandem - head but no base), the CT (center tandem - no head or base) or the LT (lower tandem - base but no head). An upper tandem motor can be used as a single section if it is completed on the bottom with either a Universal Motor Base (UMB) or Downhole Monitoring System (Sensor). If additional horsepower is required over what can be achieved in one piece, a CT or LT motor can be added. Submersible electric motors can be designed in tandem configuration to create the desired Horsepower required for each application. So, based on the sections, they are:
above,
there
are
four
different
motor
1. Single section Where the motor has head and base, a certain horsepower we cannot increase, and could not attach any equipment below.
2. Upper Tandem motor (UT) UT motor has head, no base (open circuit), and the horsepower can be increased by adding another central tandem (CT) or lower tandem (LT). The circuit must be closed either by Universal Motor Base (UMB) or sensor.
3. Central Tandem (CT) CT motor has no head, no base, and cannot use alone. UT must be attached on top of it. Another central tandem/s or LT can be attached on the bottom of it.
4. Lower Tandem (LT) Where the motor has a base, no head, and cannot use alone, it must be attached with upper tandem (as there is no head attached).
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Single
UT
CT
Fig (3.36) motor configurations Notes: 1. For any particular horsepower, the product of the volts and amps will be essentially constant. For example, in a particular Hp if we have a 1000 V, 50 A motor, a 2000 V motor would be 25 amps, and a 500V motor would be 100 amps. In other words KVA is constant. 2. When putting more than one motor together in tandem combinations, always keep the sections the same Hp and voltage. For example a 300 Hp 540 motor should be made of two 150 Hp motors. 3. With two motors we double the Hp (add the two Hp's together). We also double the voltage but the amperage remains the same. With three motors we triple the Hp and voltage but the amperage still does not change. For example, a 140 Hp, 1299 V, 69.5 A UT motor coupled to a 140 Hp, 1299V, 69.5 CT motor would give us a 280 Hp, 2598 V, 69.5 A motor. 4. Always take care when adding motors together so that the total voltage does not exceed the system limits i.e. do not try to put 3500 volts on a 3 kV cable. Surface controllers, transformers, wellhead feedthru mandrels, etc. will all have voltage limits we need to be concerned with. 5. For any given Hp there will be several voltages amperages available, why have more than one voltage? 31
and
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The answer is not in the motor but in the power cable. Lower voltage means higher current and this results in higher voltage lost in the power able. So even though the motor efficiency does not change, the overall system efficiency will decrease with higher amperage. If the amperage is too high, the motor may not even be able to start as we will see when we discuss power cable chapter. 6. Remember the higher voltage the better, but it cannot be so high that we exceed the control panel rating.
3.14. Motor current Induction motor current consists of reactive (magnetizing) and real (torque) components. The current component that produces torque (does useful work) is almost in phase with voltage, and has a high power factor close to 100% The magnetizing current would be purely inductive, except that the winding has some small resistance, and it lags the voltage by nearly 90°. The magnetizing current has a very low power factor, close to zero. The magnetic field is nearly constant from no load to full load and beyond, so the magnetizing portion of the total current is approximately the same for all loads. The torque current increases as the load increases. At full load, the magnetizing current.
torque
current
For a typical motor, the power current is between 85% and 90%.
is
factor
higher
than
of
resulting
the
the
As the load is reduced, the torque current decreases, but the magnetizing current remains about the same so the resulting current has a lower power factor. The smaller the load, the lower the load current and the lower the power factor. Low power factor at low loading occurs because the magnetizing remains approximately the same at no load as at full load.
3.15. Motor rating If we look in the manufacturer catalog we will find several pages of motors which give the horsepower rating and break down the various sizes into several voltages and amperages as 32
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Electric Submersible Pumps Chapter 3 the table below.
KMH 562 SERIES MOTORS HP
Volt / Amp
60 HZ
50 HZ
38
32
57
48
76
63
95
79
114
95
133
111
152
127
171
143
190
158
60 HZ 435/53 875/26 1315/18 430/81 870/40 1315/26 865/53 1360/34 840/69 1330/44 860/81 1300/53 2330/30 830/98 1345/60 2205/37 1340/69 2325/40 1290/81 2390/44 1185/98 1430/81 2415/48
50 HZ 363/53 729/26 1096/18 358/81 725/40 1096/26 721/53 1133/34 700/69 1108/44 717/81 1083/53 1942/30 692/98 1121/60 1838/37 1117/69 1938/40 1075/81 1992/44 988/98 1192/81 2013/48
Most conventional rating systems give a nameplate horsepower to the motor based on some assumptions which typically are bottom-hole temperature and fluid flow rate past the motor in this industry. The motor will put out exactly as much horsepower as the pump wants no more and no less! Most motors are designed to be "most efficient" and have an acceptable speed and power factor at the "design point“. In a standard application, the surface voltage is fixed and the amperage changes as the load on the motor changes. We, in fact, use this information in the form of an amp chart to see how the motor performs at downhole. We can very easily anticipate this relationship by simply looking at the equation for motor horsepower:
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Note that, if everything else is fixed, the amps will have to increase if the horsepower does and this should be a "linear" relationship. In reality this relationship is not linear since power factor and efficiency are not truly constant and the more they vary the greater will be the change in amperage. One problem with increasing the amperage too much is that higher amperage will give us greater copper losses in the motor winding which is not good for efficiency. There will be a practical limit to how far this can continue. We can look at the laminations to understand the basics of this concept. As we increase current on a motor, we increase the flux density induced in the laminations. Fig 3.37 shows an example of what the flux lines might look like for one phase winding based on a moderate loading of the motor.
Fig (3.37) If we place more load on the same motor, we get many more flux lines required to generate the necessary horsepower as shows in fig 3.38.
Fig (3.38) 34
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If we keep raising the amperage, we will eventually reach a point where we have all the flux lines we can handle this is called SATURATION. Any more horsepower beyond this point will severely overheat our motor. Another practical consideration on rating a motor is the speed. We know that the motor will slow down with load. If the motor speed is too low, we will lose pump performance, so, we must set the Hp at a point where the speed is acceptable. One of the most important considerations in rating the motor is temperature. Heat is generated in the windings which must be dissipated by the fluid flowing past the outside of the motor. Another limiting factor will be temperature differential. As the motor heats up, the components expand and they expand at different rates since not all the materials are the same. Even if the motor were all one material, expansion would vary since the internal temperature changes within the motor itself. The motor is designed with certain tolerances to allow this thermal expansion. If too much expansion occurs (such as with (overheating), tolerances might be exceeded and we could have bearing failures or other damage.
3.16. Motor protection In this discussion we will address proper protection for ESP motors operating down-hole. Motor controllers can provide very simple protection to very sophisticated protection. Simple controllers will look at overload and under-load conditions only. More advanced controllers look at all operating parameters including: 3 phase current and voltage, leg ground, power factor, kw, back-spin, and many others. In either type of controller, overload and under-load protection is of primary importance and it is critical that these both be set correctly in order to properly protect the motor from damage. Overloads and under-loads are usually set around +15% and 20% of running current as a rule-of-thumb. While it is not the intent here to give a complete guide to setting overloads and under-loads, we will look at some particular examples where the standard rule-of-thumb settings 35
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Electric Submersible Pumps Chapter 3 may not give adequate protection.
Let's take a particular case: Say we have a 145 stage DN1300 producing a fluid of 1.0 gravity on a 456 series 50 Hp, 885V, 35.5 Amp motor. Should we set the O/L at 115% of 35.5A and the U/L at 80% of 35.5A? Say, this pump will only require about 45 HP at BEP flow. This will place about a 90% load on the motor which means the motor should draw only 32 amps rather than the 35.5 of the nameplate. This should be the basis for our calculations and settings. If we set the under-load at 80% and the overload at 115%, these settings would be 25.6 amps and 36.8 amps respectively. Would this protect this unit? At shut-in, the pump will draw about 28 Hp which is a 56% load on the motor. The motor should draw about 62% of NP amps or 22 amps. Since our U/L is set at 25.6, the motor should trip on U/L if the well is shut-in. Note that a shut-in pump is NOT a "no-load condition" for the motor. What amperage would you expect for this motor if the pump shaft were broken at the intake (a true no-load condition)? For the motor to operate unloaded, it should draw about 32% of NP amps or, in this case, 11.4 amps. This is a good indication of a broken shaft. However, even if the current reads higher than this, it still could be a broken shaft so do not rule that out on this basis alone. Let's take another case: Say we have a 150 stage DN1750 producing a fluid of 0.86 gravity on a 456 series 50 Hp, 885V, 35.5A motor. Should we set the O/L at 115% of 35.5A and the U/L at 80% of 35.5A? This pump will only require about 47 HP at BEP flow. This will place about a 94% load on the motor which means the motor should draw only 34 amps rather than the 35.5 amp nameplate rating. This should be the basis for our calculations. If we set the under-load at 80% and the overload at 115%, these settings would be 27.2 amps and 39.1 amps respectively. Would this protect this unit? Note that at shut-in, the pump will draw about 37 Hp which is a 74% load on the motor. The motor should draw about 79% of NP amps or 28 amps. Since our U/L is set at 27.2, the motor may not trip on U/L if the well is shut-in. In this case we should set the U/L higher. The point is that these standard rules-of-thumb are not always perfect. Every application should be considered 36
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independently to ensure that the settings selected adequate to properly protect the down-hole equipment.
are
3.17. Application of ESP Motors Motors are in five different series 375, 450, 540, 562 and 738 for different casing sizes; see the following examples of Centrilift and Reda motors.
With all these choices, which motor should we use for a given application? The process to select the best motor for the application will depend on the economic compromises of the user, but in general, after defining the customers objectives and the pump horsepower load for the application, we can resume the process of selection of the motor as an iterative process which includes:
Motor Series Motor Type Motor configuration, Voltage and Amperage Actual motor performance & Operating compare against max. temperature
37
Temperature
and
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All motor Hp ratings are based on 1 ft per second minimum flow past the motor of water. Lower flow rates or higher oil cut can lower the effective Hp rating. After choosing the Motor Series and Type, we will know if it will be necessary for a single or tandem motor to match the HP requirement of the pump. Now we should look at Volts and Amps. For any given Hp there will be several voltages and amperages available.
For any particular horsepower, the product of the volts and amps will be essentially constant. For example, in a particular Hp if we have a 1000V, 50A motor a 2000V motor would be 25 amps and a 500V motor would be 100 amps. In other words KVA is constant. High voltage motors (single motors) are no more or less efficient than low voltage motors, so why have more than one voltage? The answer is not in the motor but in the power cable. Lower voltage means higher current and this results in higher voltage lost in the power able. So even though the motor efficiency does not change, the overall system efficiency will decrease with higher amperage. If the amperage is too high, the motor may not even be able to start as we will see when we discuss power cable. This explains why the various voltages, but why such odd voltages? Surface motors, for example, are rated at 460V, 4160V, 2300V, etc. These motors are made to "standard voltages". So why do the motor voltages turn out to be such strange numbers? 38
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In a way we have already answered this. The problem is the voltage drop in the cable. ESP motors have to contend with a very long length of power cable which surface motors do not. If we have a 460 V surface supply, we would probably only want about 430V down-hole (for a low Hp motor) to give us the necessary 460 V at the surface including the cable loss. So in determining motor voltages we are really limited by surface equipment. Motor control panels come in certain voltage ranges such as 600V, 1000V, 1500V, 2400V, etc. Motor voltages are selected assuming a length of cable such that the total voltage (motor plus cable loss) will fall just below one of the panel ratings. Remember the higher voltage the better, but it cannot be so high that we exceed the control panel rating. Higher voltage motors require smaller gauge wire and very low Hp motors simply cannot be wound at very high voltages because the wire would be too small to work with. As mentioned eerily motors come in single sections (head and base) as well as tandem configurations. The tandems can include the UT (upper tandem, head but no base), the CT (center tandem, no head or base) or the LT (lower tandem, base but no head). An upper tandem motor can be used as a single section if it is completed on the bottom with either a Universal Motor Base (UMB) or Downhole Monitoring System (Sensor). If additional horsepower is required over what can achieved in one piece, a CT or LT motor can be added.
be
3.18. Fundamentals of electricity This section is not an attempt to present a course in electricity, but is intended as a review of the terms and basic formulas associated with ESP applications.
Electricity Since the electrons are normally distributed evenly throughout a substance, a force called electromotive force (emf) is required to detach them from the atoms and make them flow in a definite direction. This force is also often called “potential” or “voltage”. The unit of measuring this electromotive force is the “volt”. The higher the voltage, the greater the number of electrons which will be caused to flow. It has been found experimentally that the charge on a single electron is 1.602e-19 coulomb. Hence, when a current of 1 39
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ampere (or 1 coulomb/second) flows in a conductor, the number of electrons passing any given point must be such that: 1.602x10-19 x no. of electrons/second = 1 coulomb/second So, no. of electrons/second = 6.24e18 I.e. when the current in a circuit is 1 ampere, electrons are passing any given point of the circuit at the rate of 6.24e18 per second. When a potential or voltage of sufficient strength is applied to a substance, it causes the flow of electrons. This flow of electrons is called an electric current. The rate of this flow of current is measured in amperes. An ampere is the rate of flow of electric current represented by movement of a unit quantity of electricity (coulomb) per second. Every substance is a conductor of electricity; but it flow very easily through some materials, such as copper, aluminum, iron, and called electric conductors. Wire and cables are the common forms of conductors. Materials such as rubber, glass, certain plastics, fibers, dry paper, and air allow almost no electricity to pass through them. Such materials are called non-conductors, insulators, or dielectrics. When an insulator is continuous, as for instance around a wire, it is commonly called insulation. The property of any material to oppose the flow of electricity through it is called impedance. The unit of the measurement of this impedance or opposition to the flow of current is the “ohm(Ω)”. Even the best conductors have some impedance; poor conductors have much impedance; insulators (dielectrics) have very high impedance. The unit for the measurement of very low impedance is the “microhm (µΩ)” and is equal to one-millionth of an ohm. The unit of very high impedance is “megaohm MΩ” and is equal to one million ohm. An element of impedance called resistance in a conductor varies directly as its length and inversely as its are (see item 5.6.9.). Resistance may be compared to the friction encountered by a flow of water through a pipe. A straight pipe, smooth inside, conducts water with little loss of pressure. If the pipe is rough inside and has many bends, the loss of pressure and the rate of flow will be greatly reduced. Similarly, a good conductor allows electricity to flow with small loss of voltage; a poor conductor offers a large resistance and so causes a corresponding large drop in voltage. The energy used in overcoming resistance is converted into heat.
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The voltage required to make a current flow depends upon the impedance of the circuit. A voltage of one volt will make one ampere flow through an impedance of one ohm I
E Z
or Z
E or E I * Z I
Where: I = Current in amperes E = Voltage in volts Z = Impedance in ohms
This formula is known as “ohm’s low”. Many alternating current circuits contain coils that produced magnetic effects. These magnetic effects in turn react upon the current. They retard the current and cause it to lag behind the voltage. It means that the voltage has reaches its maximum and started to fall some time before the current reaches a maximum. Some current will be flowing in the circuit at the instant when the voltage is zero This magnetic reaction is called “inductance” or “self inductance”. Another kind of influence on an alternating current is caused by the presence in the circuit of alternate plates of conducting material separated by insulation. This commonly referred to as “capacitor”, and its effect on the current is to cause it to lead ahead of the voltage. This reaction is called “capacitance”. It tends to counteract the inductance in a circuit and is useful in overcoming the inductive lag in the current inherent in most alternating current motors. Therefore, in an alternating current circuit there is resistance, inductance,, and reactance affecting the current. The combination of any two or all three of these effects is referred to as “impedance” of the circuit.
Power Power is defined as the rate of doing work. Electric power is measured in “horsepower”. One horsepower equals 746 watts. One watt is rather small unit of power, consequently, when speaking of power required by motors, the term “kilowatt” is used, one kilowatt being thousand watts. To obtain the power delivered to an alternating current apparatus, you can not merely multiply effective amperes by effective volts. If the circuit contains inductance, the apparatus circuits always contain it, the product of the effective current and effective voltage will be greater than the real power. This “apparent power” is measured in “volt-amperes” or more often in a unit 1000 times as large, “the kilovolt-ampere” usually abbreviated to “KVA”. In alternating current power system, the voltage and current follow an approximate sine wave. They build up from zero to a 41
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maximum in one direction then diminish to zero, build up again to a maximum but in the opposite direction and again diminish to zero, thus completing one cycle or two alternations and 360 electrical degree. The power factor is said to be 1.0 or unity if the voltage and current reach their respective maximum values simultaneously. However, as discussed previously, in most alternating current systems the voltage reaches its maximum value in a given direction before the current attains its maximum value, then the current is said to lag behind the voltage. This lag may measure in degree. The actual current drawn by apparatus of this class may be considered as having two components, one known as the magnetizing current, or that current which must overcome the choking effect produced by the characteristics of the apparatus, and which lags 90 electric degree behind the voltage. The value of this lagging current is zero when the voltage has reached its maximum value. This lagging or magnetizing current is called “reactive current”. The other component is known as “real current”, and it is in phase with the voltage. This real current and voltage reach maximum values simultaneously. The actual line current is therefore the vector sum of the reactive and real currents; furthermore, it is the current that would be registered if an ammeter is connected in the circuit. Since there one component lagging 90 electric degree or at right angles to the voltage, the resultant or actual line current of which this component is a part must, consequently, be out of phase with the voltage and lag behind it. The degree or amount that it lags depends upon the magnitude of this reactive current component and is a measure of power factor.
Resistance Consider a circuit having resistance R ohms connected across the terminals of an alternator A, as in the following figure and suppose the alternating voltage to be represented by the sine wave as in the next figure.
42
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Voltage + v
Vm i
Im Current
0
Time
-
V and I curve for a resistive Circuit If the value of the voltage at any instant is v volts, the value of the current at that instant is given by i v / R amperes. When the voltage is zero, the current is also zero; and since the current is proportional to the voltage, the wave from the current is exactly the same as that of voltage. Also the two quantities are in phase with each other; that they pass through their zero value at the same instant and attain their maximum values in a given direction at the same instant. If Vm and Im are the maximum values of the voltage and the current respectively, it follows that: I m V m / R If the instantaneous represented by:
value
of
the
applied
voltage
is
v V m sin Then the instantaneous value of current in resistive circuit is, V i m sin R Vector representing the voltage and the current in resistive circuit is as follows: V I
Inductive Reactance Let us consider the effect of alternating current flowing through a coil having an inductance of L henrys and negligible resistance as the following figure.
43
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Electric Submersible Pumps Chapter 3 i
v
A
Coil
L
Circuit with inductance only Suppose the instantaneous value of the inductance L henrys to be represented by
current
through
Where: t is the time in second. f is number of cycles per second. Suppose the current to increase by di ampere in dt second, then, Instantaneous value of induced emf is
Since f represents the number of cycles/second, the duration of 1 cycle = 1/f second, consequently,
Hence the wave of the induced emf is, represented by the curve in the figure below, lagging the current by a quarter of cycle (90O). Since the resistance of the circuit is assumed negligible, the whole of the applied voltage is absorbed in neutralizing the induce emf. So, instantaneous value of the applied voltage is,
44
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Comparison of expression (1) and (3), shows that the applied voltage leads the current by a quarter of cycle (90O). Also, from (3), it follows that that maximum value Vm of the applied voltage is, So that,
The inductive reactance represented by XL.
is
expressed
in
ohms
and
is
Hence,
Capacitance reactive The property of a capacitor to store an electric charge when its plates are at different potentials is referred to capacitance as the following figure. The unit of capacitance is termed the farad (F). Farad is defined as the capacitance of capacitor which required a potential difference (p.d) of 1 volt to maintain a charge of one coulomb on that capacitor. Let us consider the effect of alternating current flowing through a capacitor having a capacitance of C farad and negligible resistance.
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Circuit with capacitance only Suppose the instantaneous value of the voltage applied to the capacitance to be represented by:
If the applied voltage increases by dv volt in dt second as in the following figure:
Instantaneous value of current flow through capacitor is,
Comparison of (4)and (5) shows that current leads applied voltage by a quarter of cycle (90O)and the current and voltage can be represented vectorially as the following figure.
46
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Electric Submersible Pumps Chapter 3 I
90 0 v From expression (5) it follows that the maximum value Im of the current is, So,
The capacitive reactance is represented by the symbol Xc.
expressed
in
ohms
and
is
Resistance, Inductance and Capacitance in series An actual circuit may have resistance and inductance, or resistance and capacitance or resistance, conductance and capacitance in series. Hence, if we consider the general case of R, L, and C in series, we can adapt the results to the other two cases by merely omitting the capacitive or the inductance from the expression for general case. In following fig, OB vector represents L (2πfLI) leading the current by 900; and OC vector represents C (I/2πfC) lagging the current by 900. Since OD = OB-OC, OB being assumed greater than OC and supply voltage is the vectorial sum of OA and OD, namely OE.
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B
[L]
E
D φ
O
A
I
[C] C Vector diagram for above figure
From this expression it is seen that: Resultant reactance = inductive reactance,
Φ = phase voltage.
difference
between
48
the
reactance
current
and
-capacitive
the
supply
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Electric Submersible Pumps Chapter 3
If the inductive reactance is greater than the capacitive reactance, tan Φ is positive and the current lags the supply voltage by an angle Φ; if less, tan Φ is negative, signifying that the current leads the supply voltage by angle Φ. If a circuit consists of a coil having a resistance R ohms and inductance L henrys, such a circuit can be considered as a resistance and inductive in series and,
And the phase angle voltage is given by:
in
which
the
current
lags
the
supply
Example 1 A coil having a resistance of 12 Ω and inductance of 0.1 H is connected across a 100 V, 50 c/s supply. Calculate: a) The reactance and impedance of the coil. b) The current c) The phase difference between the current and the apply voltage. Solution (a) Reactance = XL = 2ПfL = 2x3.14159x50x0.1 = 31.4 Ω Impedance = Z = √R2+XL2 = √122+31.42 = 33.6 Ω (b) Current = V/Z = 100 / 33.6 = 2.975 A (c) Tan Φ = XL/R = 31.4 / 12 = 2.617 Φ = 690 Example 2 A metal filament lamp, rated at 750 watt, 100 v, is to be connected in series with a capacitance across a 230 v, 60 c/s supply. Calculate: a) The capacitance required b) The phase angle between the current and the supply voltage. Solution
i
From vector diagram below, (a) V2 = VR2+VC2 (230)2 = (100)2 + VC2 VC = 207 Volts Rated current of lamp = 750 w / 100 v = 7.5 A 7.5 = 2 x 3.14 x 60 x C x 207 C = 96 x 10-6 farad = 96 microfarad (μF) 49
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(b) If φ = phase angle between current I and supply V. Cos φ = VR / V = 100 / 230 = 0.435 Φ = 640 12’
Example 3 A resistance of 12 Ω, an inductance of 0.15 H and capacitance of 100 μF are connected in series across 100 V, 50 c/s supply. Calculate: a) The impedance b) The current c) The voltage across R. L, and C d) The phase difference between current and supply voltage Solution (a) Z
2 1 2 R 2fL 2fC 2
10 6 144 47.1 31.862 19.4 z 122 23.14159500.15 23.1415950100 (b) Current = V/Z = 100 / 19.4 = 5.15 A
(c) Voltage across R = VR = 12x5.15 = 61.8 V Voltage across L = VL = 47.1x5.15 = 242.5 V Voltage across C = VC = 31.85x5.15 = 164 V (d)
Phase difference between current and supply φ = cos-1(VR/V)= cos-1(61.8/100) = 510 48’
voltage
Or φ = tan-1 (VL-VC/VR ) = tan-1(242.5-164)/61.8) = 510 48’ Or alternatively, φ = tan-1 [2πfL-(1//2πfC)]/R = [47.1-31.85]/12 = 510 48’ See vector diagram below
50
=
Mohamed Dewidar 2013
Electric Submersible Pumps Chapter 3 242.5
VL
L
47.1
15.25 78.5
VR
12
61.8
164
R
φ
VC
31.85
C
Power in series RL circuit From the above analysis it seen that the voltage applied to a series RL circuit, the voltage leads the current by angle φ. It is equally valid to say that the current lags the voltage by an angle φ. If a voltage Emsinωt is applied to a series RL circuit, the current is Imsin(ωt-φ). The instantaneous power is, p = e x i = Emsin ωt x Imsin(ωt-φ) = E m I m cos E m I m cos2t 2 2 The second term of the right side has an average value of zero. p = E m I m cos EI cos -----------(11) 2 Where E and I are the r.m.s. (Root Mean Square) values of voltage and current and φ is the phase angle between the voltage and current. The following figure is the plot of instantaneous voltage, current and power. The instantaneous power is positive during the time interval when both e and i are simultaneously positive or negative. During the interval when one of the two quantities out of e and i is positive and the other is negative, instantaneous power is negative. The positive area is more than the negative area and therefore the average power is positive.
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i=ImSin( ωt-φ)
+ ve area
e=EmSin ωt
+ ve area p=EmSinωt*ImSin( ωt - φ) * c o s φ
0 0
-ve 20
φ 40
60
80
100
120
140
160
180
-ve 200
220
240
260
280
300
320
340
360
p=EmSinωt*ImSin( ωt - φ) * c o s φ
0
+ ve area
p=E*I *cosφ
+ ve area
φ 0
20
40
60
80
100
120
140
160
-ve
180
200
220
240
260
280
300
320
340
360
-ve
Power in RL circuit
Active power, reactive power and power factor (single phase) The average power in the circuit i.e. EIcosφ is the actual power supplied by the source to the circuit. This is known as active power of the circuit. The active power is measured in watts. The bigger units of active power are KW (kilowatt=103 watts) and MW (megawatt=106 watts). The product of voltage and current i.e. E*I called apparent power and is measured in volt-ampere (VA). The ratio of active power to apparent power equals cosφ. This term cosφ is called power factor of the circuit. It is the factor by which the apparent power (EI) must be multiplied to give the active power. The power factor for purely resistive circuit is 1. Therefore the apparent power and active power are equal for purely resistive circuit. A circuit may be 52
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characterized as having leading or lagging power factor. A leading power factor means that the current leads the voltage. This occurs in the capacitive circuits. An inductive circuit is described as having lagging power factor, since the current lags the voltage. The active power in ac circuit can also be written as I2R. Example A 10 ohm resistor and 20 mH (mille Henry) inductor are connected in series across a 230 volts, 50 c/s supply. Find the circuit impedance, current voltage across resistor, apparent power, active power, reactive power, and power factor. XL = ωL = 2πfL = 2x3.24x50x20x10-3 = 6.28 Ω Z = √R2+XL2 = √102+(6.28)2 = 11.8 Ω Φ = tan-1(XL/R) = 32.10 I = E/Z = 230/11.8 = 19.49 amperes Voltage across resistor = RI = 10x19.49 = 149.9 volts Voltage across inductor = IXL = 122.4 volts Apparent power = EI = 230x 19.49 = 4482.7 VA Power factor = cos(32.1) = 0.847 lagging Active power = EIcosφ = 230x19.49x0.847 = 3796.8 watts Reactive power = EIsinφ = 230x19.49xsin(32.1) = 2382.1 w
Practical importance of power factor (P.F.) If an alternator is rated at a given, say 2000 A at a voltage of 400 v, it means that these are the highest current and voltage values of the machine can give without the temperature exceeding a safe value. Consequently the rating of the alternator is given as 400x2000/1000=800 KVA. The phase difference between the voltage and the current depends upon the nature of the load and not upon the generator. Thus if the power factor of the load is unity, the 800 KVA are also 800 kW; and the engine driving the generator has to be capable of developing this power together with the losses in the generator. But if the p.f. of the load is, say 0.5, the power is only 400 kW; so that the engine is only developing about one half of the power which it is capable, through the alternator is supplying its rated output 800 KVA. Similarly the conductors connecting the alternator to the load have to be capable of carrying 2000 A without excessive temperature rise. Consequently they can transmit 800 kW if the power factor is unity, but only 400 kW at 0.5 p.f. for the same rise of temperature.
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It is therefore evident that the higher the p.f. of the load, the greater is the power that can be generated by a given alternator and transmitted by a given conductor. The matter may be put another way by saying that, for a given power, the lower the p.f. the larger must be the size of the alternator to generate that power and the greater must be the cross sectional area of the conductor to transmit it; in other words, the greater is the cost of generation and transmission of the electric energy. This is the reason why supply authorities do all they can to improve the p.f. for their loads (by ex. installing capacitors).
Three phase alternating current Three phase alternating current is the best suites for long distance transmission because it may be easily generated at low to moderately high voltages and can then have the voltage raised to very high values suitable for efficient transmission, and then the voltage can be reduced to a value suitable for general use by means of stationary device known as a transformer. The higher the voltage, the smaller the wire required to carry a given amount of power. The following figures are the curve and diagram representing three phase alternating current.
phase 1
phase 2
0
30
60
90
120
phase 3
150
180
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210
240
270
300
330
360
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Electric Submersible Pumps Chapter 3 EYNR
Line voltage (VL) Between any line conductors
phase voltage (VP) (voltage to neutral) EBN
ENR EYN
o
30
o
120
EBNY o
ENB
120
ENY
Vector diagram ERNB
Since the angle between ENR and EYN is 600, EYNR = 2ENRcos300 = √3ENR Line voltage = 1.73 x phase voltage In general: If VL = p.d. between any two line conductors = line voltage And VP = p.d. between a line conductor and neutral point = phase voltage And if IL and IP = line and phase then for a star connection system,
current
respectively,
VL = 1.73 VP IL = IP For delta connection: IL = 1.73 IP
Power in three phase system with balanced load If Ip = value of the current in each phase Vp = value of p.d. across each phase Power per phase = IpVp x power factor And total power = 3IpVp x power factor If IL and VL be the value of the line current and voltage respectively, then for star connection system, VP = VL/1.73 and IP = IL so, Total power in watts = 1.73 x ILVL x power factor For delta connection system, 55
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Electric Submersible Pumps Chapter 3 IP = IL/1.73 and VP = VL so,
Total power in watts = 1.73 x ILVL x power factor Example A three phase motor operating off 400 v system is developing 25 bhp at an efficiency of 0.87 and a power factor of 0.82. Calculate (a) the line current and (b) the phase current if the windings are delta-connected.
output power in watts input power in watts bhpx746 1.73 I L V L xP.F 25 x746 0.87 1.73 xI L x400 x0.82
(a) Efficiency
IL = 37.8 A (b) For delta-connected windings phase current
line current 37.8 21.8 A 1.73 1.73
Complex Numbers The mathematics used in Electrical Engineering to add together resistances, currents or DC voltages uses what are called "real numbers". But real numbers are not the only kind of numbers we need to use especially when dealing with frequency dependent sinusoidal sources and vectors. As well as using normal or real numbers, Complex Numbers were introduced to allow complex equations to be solved with numbers that are the square roots of negative numbers, √-1. In electrical engineering this type of number is called an "imaginary number" and to distinguish an imaginary number from a real number the letter "j" known commonly in electrical engineering as the j-operator. The letter j is used in front of a number to signify its imaginary number operation. Examples of imaginary numbers are: j3, j12, j100 etc. Then a complex number consists of two distinct but very much related parts, a "Real Number" plus an " Imaginary Number ". Complex Numbers represent points in a two dimensional complex or s-plane that are referenced to two distinct axes. The horizontal axis is called the "real axis" while the vertical axis is called the "imaginary axis". The rules and laws used in mathematics for the addition or subtraction of imaginary numbers are the same as for real numbers, j2 + j4 = j6 etc. The only difference is in multiplication because two imaginary numbers multiplied together becomes a positive real number, as two negatives make 56
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a positive. Real numbers can also be thought of as a complex number but with a zero imaginary part labeled j0. The j-operator has a value exactly equal to √-1, so successive multiplication of "j", (j x j ) will result in j having the following values of, -1, -j and +1. As the j-operator is commonly used to indicate the anticlockwise rotation of a vector, each successive multiplication or power of "j", j2, j3 etc, will force the vector to rotate through an angle of 90o anticlockwise as shown below. Likewise, if the multiplication of the vector results in a -j operator then the phase shift will be -90o, i.e. a clockwise rotation.
Vector Rotation of the j-operator So by multiplying an imaginary number by j2 will rotate the vector by 180o anticlockwise, multiplying by j3 rotates it 270o and by j4 rotates it 360o or back to its original position. Multiplication by j10 or by j30 will cause the vector to rotate anticlockwise by the appropriate amount. In each successive rotation, the magnitude of the vector always remains the same. There are different ways in Electrical Engineering to represent complex numbers either graphically or mathematically. One such way that uses the cosine and sine rule is called the Cartesian or Rectangular Form.
► Complex Numbers using the Rectangular Form Complex number is represented by a real part and an imaginary part that takes the general form of: Where: Z is the Complex Number representing the Vector x is the Real part or the Active component y is the Imaginary part or the Reactive component j is defined by √-1 In the rectangular form, a complex number can be represented as a point on a two-dimensional plane called the complex or splane. So for example, Z = 6 + j4 represents a single point whose coordinates represent 6 on the horizontal real axis and 4 on the vertical imaginary axis as shown.
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Complex Numbers using the Complex or s-plane But as both the real and imaginary parts of a complex number in the rectangular form can be either a positive number or a negative number, then both the real and imaginary axis must also extend in both the positive and negative directions. This then produces a complex plane with four quadrants called an Argand Diagram as shown below.
Four Quadrant Argand Diagram On the Argand diagram, the horizontal axis represents all positive real numbers to the right of the vertical imaginary 58
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axis and all negative real numbers to the left of the vertical imaginary axis. All positive imaginary numbers are represented above the horizontal axis while all the negative imaginary numbers are below the horizontal real axis. This then produces a two dimensional complex plane with four distinct quadrants labeled, QI, QII, QIII, and QIV. The Argand diagram can also be used to represent a rotating phasor as a point in the complex plane whose radius is given by the magnitude of the phasor will draw a full circle around it for every 2π/ω seconds. Complex Numbers can also have "zero" real or imaginary parts such as: Z = 6 + j0 or Z = 0 + j4. In this case the points are plotted directly onto the real or imaginary axis. Also, the angle of a complex number can be calculated using simple trigonometry to calculate the angles of right-angled triangles, or measured anti-clockwise around the Argand diagram starting from the positive real axis. Then angles between 0 and 90o will be in the first quadrant ( I ), angles ( θ ) between 90 and 180o in the second quadrant ( II ). The third quadrant ( III ) includes angles between 180 and 270o while the fourth and final quadrant ( IV ) which completes the full circle includes the angles between 270 and 360o and so on. In all the four quadrants the relevant angles can be found from tan-1(imaginary component/real component).
► Addition and Subtraction of Complex Numbers The addition or subtraction of complex numbers can be done either mathematically or graphically in rectangular form. For addition, the real parts are firstly added together to form the real part of the sum, and then the imaginary parts to form the imaginary part of the sum and this process is as follows using two complex numbers A and B as examples.
Example 1 Two vectors, A = 4 + j1 and B = 2 + j3 respectively. Determine the sum and difference of the two vectors in both rectangular (a + jb) form and graphically as an Argand Diagram. Addition
Subtraction
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Graphical Addition and Subtraction
► Multiplication and Division of Complex Numbers The multiplication of complex numbers in the rectangular form follows more or less the same rules as for normal algebra along with some additional rules for the successive 2 multiplication of the j-operator where: j = -1. So for example, multiplying together our two vectors from above of A = 4 + j1 and B = 2 + j3 will give us the following result.
Mathematically, the division of complex numbers in rectangular form is a little more difficult to perform as it requires the use of the denominators conjugate function to convert the denominator of the equation into a real number. This is called "rationalizing". Then the division of complex numbers is best carried out using "Polar Form", which we will look at later. However, as an example in rectangular form lets find the value of vector A divided by vector B.
Multiply top and bottom by conjugate (2-j3)
► The Complex Conjugate The Complex Conjugate, or simply Conjugate of a complex number is found by reversing the algebraic sign of the complex numbers imaginary number only while keeping the algebraic sign 60
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of the real number the same and to identify the complex conjugate of z the symbol z is used. For example, the conjugate of z = 6 + j4 is z = 6 - j4, likewise the conjugate of z = 6 - j4 is z = 6 + j4. The points on the Argand diagram for a complex conjugate have the same horizontal position on the real axis as the original complex number, but opposite vertical positions. Thus, complex conjugates can be thought of as a reflection of a complex number. The following example shows a complex number, 6 + j4 and its conjugate in the complex plane.
Conjugate Complex Numbers The sum of a complex number and its complex conjugate will always be a real number as we have seen above. Then the addition of a complex number and its conjugate gives the result as a real number or active component only, while their subtraction gives an imaginary number or reactive component only. The conjugate of a complex number is an important element used in Electrical Engineering to determine the apparent power of an AC circuit using rectangular form.
► Complex Numbers using Polar Form Unlike rectangular form which plots points in the complex plane, the Polar Form of a complex number is written in terms of its magnitude and angle. Thus, a polar form vector is presented as: Z = A ∠±θ, where: Z is the complex number in polar form, A is the magnitude or modulo of the vector and θ is its angle or argument of A which can be either positive or negative. The magnitude and angle of the point still remains the same as for the rectangular form above, this time in polar form the location of the point is represented in a "triangular form" as shown below.
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Polar Form Representation of a Complex Number As the polar representation of a point is based around the triangular form, we can use simple geometry of the triangle and especially trigonometry and Pythagoras's Theorem on triangles to find both the magnitude and the angle of the complex number. As we remember from school, trigonometry deals with the relationship between the sides and the angles of triangles so we can describe the relationships between the sides as:
Using trigonometry follows.
again,
the
angle θ of A is
given
as
Then in Polar form the length of A and its angle represents the complex number instead of a point. Also in polar form, the conjugate of the complex number has the same magnitude or modulus it is the sign of the angle that changes, so for example the conjugate of 6 ∠30o would be 6 ∠–30o.
► Converting between Rectangular Form and Polar Form In the rectangular form we can express a vector in terms of its rectangular coordinates, with the horizontal axis being its real axis and the vertical axis being its imaginary axis or j-component. In polar form these real and imaginary axes are simply represented by "A ∠θ". Then using our example above, the relationship between rectangular form and polar form can be defined as.
► Converting Polar Form into Rectangular Form, ( P→R ) ►
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We can also convert back from rectangular form to polar form as follows.
► Converting Rectangular Form into Polar Form, ( R→P )
► Polar Form Multiplication and Division Rectangular form is best for adding and subtracting complex numbers as we saw above, but polar form is often better for multiplying and dividing. To multiply together two vectors in polar form, we must first multiply together the two modulus or magnitudes and then add together their angles. Multiplication in Polar Form Multiplying us.
together 6 ∠30o and 8 ∠–
45o in
polar
form
gives
, Division in Polar Form Likewise, to divide together two vectors in polar form, we must divide the two modulus and then subtract their angles as shown.
3.19. Equivalent circuit of induction motor To analyze the operating and performance characteristics of an induction motor, an Equivalent Circuit can be drawn. We will consider a 3–phase, Y connected machine, the Equivalent Circuit for the stator is as shown below:
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Electric Submersible Pumps Chapter 3 V1 = Stator Terminal Voltage I1 = Stator Current R1 = Stator Effective Resistance X1 = Stator Leakage Reactance Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss component = Ic, and a magnetizing current = Im) The rotor winding can be represented as:
Rotor Circuit I2 = Rotor Current R2 = Rotor winding Resistance X2 = Rotor Leakage Reactance Z2 = Rotor Impedance (R1 + jX1) E2 = Induced EMF in the rotor (generated by the air gap flux) The EMF (E2) is equal to the stator terminal voltage less the voltage drop caused by the stator leakage impedance. Note: ● Never use three-phase equivalent circuit. Always use perphase equivalent circuit. ● The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor. ● Induction machine equivalent circuit is composed of stator circuit and rotor circuit.
3.19.1. Effective circuit of induction motor at standstill Standstill means rotor circuit is open. At open circuits, S = 1, accordingly, E1 = E2, fs = fr, X2 = 2πfrL2. Refer to section 3.7 and 3.8.
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Effective circuit at standstill E2 = V1 - I1 (Z1) Where E2 is rotor induced emf at standstill E2 = V1 - I1 (R1 + jX1) = I1 Z1
3.19.2. Effective circuit of induction motor under operating conditions (rotor winding is shorted) Now suppose induction-motor is loaded down. As motor’s load
increases its slip rises because of which the rotor speed falls. Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2. Then the rotor magnetic field φ also increases. Since rotor
slip is larger, rotor frequency rises (fr = s fs ) and rotor reactance increases (ωLr). Therefore, rotor current lags further behind the rotor voltage. Accordingly the parameters will be changed to be as follows:
The stator and rotor sides are, in the figure below, separated by an air gap. I2
= Rotor current in running condition
It is important to note that as load on the motor changes, the motor speed changes. Thus slip changes. As slip changes the 65
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reactance of standstill X2 changes to be sX2 which is shown variable. The rotor impedance can be represented by:
Effective circuit of induction motor under operating conditions In the running condition the variable resistance R2/s can be rearranged as follows:
So the variable rotor resistance R2/s has two parts. 1. Rotor resistance R2 itself which represents copper loss. 2. R2(1 - s)/s which represents load resistance RL. So it is electrical equivalent of mechanical load on the motor. Key Point : Thus the mechanical load on the motor is represented by the pure resistance of value R2(1 -s)/s. The effective circuit of the induction motor under operating condition shall be as following:
It means that R2/s = rotor copper resistance + load resistance.
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3.19.3. Power Relations 1. Input Power
2. Stator copper losses 3. Rotor copper losses 4. Air gap power
5. Mechanical power
6. Output power 7. Output torque
Power flow diagram
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Electric Submersible Pumps Chapter 3 Example 1
A 480-V, 60 Hz, 50-hp, three phase induction motor is drawing 60A at 0.85 PF lagging. The stator copper losses are 2 kW, and the rotor copper losses are 700 W. The friction and windage losses are 600 W, the core losses are 1800 W, and the stray losses are negligible. Find the following quantities: 1. The air-gap power PAG. 2. The power converted Pm. 3. The output power Pout. 4. The efficiency of the motor. Solution
Example 2 A 480V, 60 Hz, 6-pole, three-phase, delta-connected induction motor has the following parameters: R1=0.461 Ω, R2=0.258 Ω, X1=0.507 Ω, X2=0.309 Ω, Xm=30.74 Ω Rotational losses are 2450W. The motor drives a mechanical load at a speed of 1170 rpm. Calculate the following information: Synchronous speed in rpm Slip Line Current Input Power Air gap Power Torque Developed Output Power in Hp Efficiency Solution
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Note: the core losses is so small as the Io
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