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–Electrical losses: ohmic losses…

–Mechanical losses: friction, windage…

Fig. 3.2(b): a simple force-producing device with a single coil forming the electric terminal, and a movable plunger serving as the mechanical terminal.

  • The interaction between the electric and mechanical terminals, i.e. the electromechanical energy conversion, occurs through the medium of the magnetic stored energy.

Figure 3.2(a) Schematic magnetic-field electromechanical-energy-conversion device;

(b) simple force-producing device.

  • W fld size 12{W rSub { size 8{ ital "fld"} } } {} : the stored energy in the magnetic field

dW fld dt = ei f fld dx dt size 12{ { { ital "dW" rSub { size 8{"fld"} } } over { ital "dt"} } = ital "ei" - f rSub { size 8{"fld"} } { { ital "dx"} over { ital "dt"} } } {} (3.7)

e = dt size 12{e= { {dλ} over { ital "dt"} } } {} (3.8)

dW fld = id λ f fld dx size 12{ ital "dW" rSub { size 8{"fld"} } = ital "id"λ - f rSub { size 8{"fld"} } ital "dx"} {} (3.9)

  • Equation (3.9) permits us to solve for the force simply as a function of the flux and the mechanical terminal position x .
  • Equations (3.7) and (3.9) form the basis for the energy method.

§3.2 Energy Balance

  • Consider the electromechanical systems whose predominant energy-storage mechanism is in magnetic fields. For motor action, we can account for the energy transfer as

Energy input form electric source = Mechanical energy ouput + Increase in energy stored in magnetic field + Energy converted into heat size 12{ left [ matrix { "Energy input" {} ##"form electric" {} ## "source"} right ]= left [ matrix {"Mechanical" {} ## "energy" {} ##"ouput" } right ]+ left [ matrix { "Increase in energy" {} ##"stored in magnetic" {} ## "field"} right ]+ left [ matrix {"Energy" {} ## "converted" {} ##"into heat" } right ]} {} (3.10) {}

  • Note the generator action.
  • The ability to identify a lossless-energy-storage system is the essence of the energy method.
  • This is done mathematically as part of the modeling process.
  • For the lossless magnetic-energy-storage system of Fig. 3.3(a), rearranging (3.9) in form of (3.10) gives

dW elec = dW mech + dW fld size 12{ ital "dW" rSub { size 8{ ital "elec"} } = ital "dW" rSub { size 8{ ital "mech"} } + ital "dW" rSub { size 8{ ital "fld"} } } {} (3.11)

where

dW elec = id λ size 12{ ital "dW" rSub { size 8{ ital "elec"} } = ital "id"λ} {} differential electric energy input

dW mech = f fld dx size 12{ ital "dW" rSub { size 8{ ital "mech"} } =f rSub { size 8{ ital "fld"} } ital "dx"} {} differential mechanical energy output

dW fld size 12{ ital "dW" rSub { size 8{ ital "fld"} } } {} differential change in magnetic stored energy

  • Here e is the voltage induced in the electric terminals by the changing magnetic stored energy. It is through this reaction voltage that the external electric circuit supplies power to the coupling magnetic field and hence to the mechanical output terminals.

dW elec = eidt size 12{ ital "dW" rSub { size 8{ ital "elec"} } = ital "eidt"} {} (3.12)

  • The basic energy-conversion process is one involving the coupling field and its action and reaction on the electric and mechanical systems.
  • Combining (3.11) and (3.12) results in

dW elec = eidt = dW mech + dW fld size 12{ ital "dW" rSub { size 8{ ital "elec"} } = ital "eidt"= ital "dW" rSub { size 8{ ital "mech"} } + ital "dW" rSub { size 8{ ital "fld"} } } {} (3.13)

§3.3 Energy in Singly-Excited Magnetic Field Systems

  • We are to deal energy-conversion systems: the magnetic circuits have air gaps between the stationary and moving members in which considerable energy is stored in the magnetic field.
  • This field acts as the energy-conversion medium, and its energy is the reservoir between the electric and mechanical system.
  • Fig. 3.3 shows an electromagnetic relay schematically. The predominant energy storage occurs in the air gap, and the properties of the magnetic circuit are determined by the dimensions of the air gap.

Figure 3.3Schematic of an electromagnetic relay.

λ = L ( x ) I size 12{λ=L \( x \) I} {} (3.14)

dW mech = f fld dx size 12{ ital "dW" rSub { size 8{ ital "mech"} } =f rSub { size 8{ ital "fld"} } ital "dx"} {} (3.15)

dW fld = id λ f fld dx size 12{ ital "dW" rSub { size 8{ ital "fld"} } = ital "id"λ - f rSub { size 8{ ital "fld"} } ital "dx"} {} (3.16)

  • W fld size 12{W rSub { size 8{ ital "fld"} } } {} is uniquely specified by the values of λ size 12{λ} {} and x . Therefore, λ size 12{λ} {} and x are referred to as state variables.
  • Since the magnetic energy storage system is lossless, it is a conservative system. W fld size 12{W rSub { size 8{ ital "fld"} } } {} is the same regardless of how λ size 12{λ} {} and x are brought to their final values. See Fig. 3.4 where tow separate paths are shown.

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
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cm
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A mouse of mass 200 g falls 100 m down a vertical mine shaft and lands at the bottom with a speed of 8.0 m/s. During its fall, how much work is done on the mouse by air resistance
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Can you compute that for me. Ty
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Chemistry is a branch of science that deals with the study of matter,it composition,it structure and the changes it undergoes
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A ball is thrown straight up.it passes a 2.0m high window 7.50 m off the ground on it path up and takes 1.30 s to go past the window.what was the ball initial velocity
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2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
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you have been hired as an espert witness in a court case involving an automobile accident. the accident involved car A of mass 1500kg which crashed into stationary car B of mass 1100kg. the driver of car A applied his brakes 15 m before he skidded and crashed into car B. after the collision, car A s
Samuel Reply
can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
Joseph Reply
Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
Joseph
"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
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Magreth
progressive wave
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A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
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Source:  OpenStax, Intergrated library system management. OpenStax CNX. Jul 29, 2009 Download for free at http://cnx.org/content/col10801/1.1
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