Electro mechanical lecture 1
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1. The document discusses electro-mechanical energy conversion in plunger and rotating systems using a coil, plunger/rotor, and air gap. 2. It examines the change in stored magnetic energy and mechanical work done when the plunger/rotor moves due to a change in the air gap inductance. 3. The difference between the supplied electrical energy and change in stored energy equals the mechanical work done on the plunger/rotor.
Electro mechanical lecture 1
- 2. Consider the following plunger system x V R e i
- 3. • The voltage e exists only under dynamic conditions, ie when there is a change in current (transformer voltage) or a change in air-gap resulting in a change in inductance (mechanical voltage). Otherwise e = 0. • dt di dt dL dt λd e dt dLi dt di L dt dL i
- 4. dtie diLidLi2 Energy supplied through the coil Assume that initially that the air gap length is kept at x. A voltage is then impressed on the coil and the current allowed to grow until it stabilizes at value I. The stored energy is given by integrating above equation with no change in inductance diLiWfld 2 2 1 IL
- 5. Now release the plunger so that it moves by a distance . x A Nμ I NHA μ I φN I λ L 2 00 x Change in inductance x x A NμL 2 2 0 xSince is negative the change in inductance is positive (L increases).
- 6. Change in stored energy . The total energy supplied by the coil when the plunger has moved by distance iiLiLWfld 2 2 1 x diLidLi2 dtie iiLLi2 or
- 7. Comparing change in stored energy and energy supplied it is seen that more than half of the supplied energy is added to the stored energy. The rest is used as the work done to move the plunger through distance .x Difference between energy supplied and change in stored energy = 2 2 1 iL
- 8. 2 2 1 iL x dx dL i2 2 1 xF dx dL iF 2 2 1 2 22 02 1 x A INμ AHBF 2 1 dx d F 2 2 1 It is easily verified that other expressions for F are: Where is the reluctance of the air-gap.
- 9. Now consider a rotating device: r x l Effective area of air-gap = lr
- 10. x lr N x A NL 22 2 0 2 0Then, inductance If the rotor is rotated through an angle then x lr NL 2 2 0Change in inductance (Note that the angle decreases. Thus is negative and the inductance is less).
- 11. iiLiLWfld 2 2 1 As before change in stored energy iiLLi2 And energy supplied by the coil when the rotor was rotated by an angle is
- 12. Difference between energy supplied and change in stored energy is equal to work done 2 2 1 iL θ θd dL i2 2 1 θTBut work done θd dL iT 2 2 1 x lr INμ 2 22 02 1
- 13. The energy is actually extracted since is negative. The system acts as a generator supplying energy to the coil. The work done forces the rotor to move through an angle (against the reluctance torque). d d2 2 1 It can be verified that T can also be expressed as
- 14. It is important to note the following Motor action: Electrical energy supplied mechanical energy + change in field storage Generator action: Electrical energy delivered mechanical energy + change in field storage
- 15. The previous examples considered single excited system, ie only one coil. The resulting forces or torques are called reluctance forces and reluctance torques because they arise from the change of reluctance with gap distance or with angle dx d d d Double Excited Systems A double excited system has two energized coils. Most motors and generators are double excited. Consider the general case of two energized coils carrying currents I1, I2. e1 e2 i1 i2
- 16. Emf equations for both coils are )()( 2111 iM dt d iL dt d e )()( 1222 iM dt d iL dt d e To find the energy stored in the field let us first energize coil 1 up to current I1, with current in coil 2 = 0 (coils are stationary; hence inductances are constant). Stored energy dtedtieWfld 02111 2 112 1 IL
- 17. Next, we keep current in coil 1 fixed at I1 and energize coil 2 up to I2. Total energy stored in field dtiedtIeWfld 22112 2 222 1 21 ILIIM 21 2 222 12 112 1 21 IIMILILWWW fldfldfld
- 18. Now let coil 2 rotate by an angle . Any inductance which is position dependent will change with the angle. Hence the change in stored energy is: Or 21 2 222 12 112 1 iiMiLiLWfld 21221211 iiMiLiiMiL θii θd dM i θd dL i θd dL Wfld 21 2 2 2 2 12 1 1 2 1 21221211 iiMiLiiMiL
- 19. The energy supplied by the two coils during this rotation is tietieW 2211 MiiLiMiiLi 212 2 2211 2 1 21221211 iiMiLiiMiL θii θd dM i θd dL i θd dL 21 2 2 22 1 1 2 21221211 iiMiLiiMiL
- 20. The difference between energy supplied and change in stored energy constitutes the work done. Hence θii θd dM i θd dL i θd dL θT 21 2 2 2 2 12 1 1 2 1 21 2 2 2 2 12 1 1 2 1 ii θd dM i θd dL i θd dL T Or
- 21. The first two terms are known as the reluctance torques since they depend upon single excitation only. The third term is the excitation torque and requires that both coils are excited