1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Setting up the Transformer Efficiency as Objective Function

  1. Feb 4, 2017 #1
    1. The problem statement, all variables and given/known data
    To set-up an equation for the efficiency equation of single-phase toroidal transformer given the circuit with the following mesh equations

    (s*L1+R1)*I1 + (-s*M)*I2 = V1
    (-s*M)*I1 + (s*L2+R2+RL)*I2 = 0

    2. Relevant equations
    Efficiency equation(?), equation derived by The Electrician at thread https://www.physicsforums.com/threa...-parameter-i2-to-v1-non-ideal-updated.901475/.

    3. The attempt at a solution
    We're not sure about this but is it not V2*I2*cos(theta2)/(V1*I1*cos(theta1)) or M*di2/dt*cos(theta2)/(V1*I1*cos(theta1))? Or should I include the hysteresis and eddy current losses?

    P.S. We're Applied Mathematics students, not EE, however we actually needed these for our research. Thanks in advance.
  2. jcsd
  3. Feb 4, 2017 #2

    The Electrician

    User Avatar
    Gold Member

    I assume you are a university student; is this the case?

    You should visit your library and consult the Transactions of the IEEE on Magnetics. You can probably find a lot of information there about modeling losses in magnetics. The IEEE publications are available online at http://ieeexplore.ieee.org/Xplore/home.jsp. Usually a university library will give you free access.
  4. Feb 4, 2017 #3
    Yes, I am.

    Well, I think things will get a little bit more complex if I dig deeper into the losses. After all, the point of our thesis, is to simply apply mathematics in real-situations. Relating the electrical parameters in the previous thread has already applied enough of the serious and beautiful maths concepts. How about if I'll just assume the losses to be a portion of power? On your experience Sir, how much of the power (in %) is the combined value of the hysteresis and eddy current losses?
  5. Feb 4, 2017 #4

    The Electrician

    User Avatar
    Gold Member

    It is known that the optimum design (for continuous full power operation) for a transformer is to make the copper losses the same as the iron losses. The copper losses are the power loss due to the resistance of the copper wires used to make the primary and secondary windings. Those losses vary somewhat with temperature.

    The iron losses are mostly the hysteresis and eddy current losses.

    The optimum design where copper and iron losses are equal only applies when the transformer is operated at full power continuously, which is often not the case. If the transformer is not operated at full power continuously, the design for optimum becomes more difficult.

    A typical medium size to large size transformer will be on the order of 98% efficient, with copper and iron losses similar in value.

    You would be solving a problem close to real life if you chose your values for R1, R2 and RL so that the losses in R1 and R2 together are, say, 5% of the power in RL. Also, make the loss in R1 about the same as the loss in R2.

    The iron losses are almost invariant with the load on the transformer, so if you want to get more realistic you could add a third resistor, R3, to represent the combined iron losses. Connect R3 in parallel with the primary winding. This way, the loss in R3 will not increase when the power supplied to RL increases. Consult the "Real transformer equivalent circuit" here: https://en.wikipedia.org/wiki/Transformer. You would choose the value of R3 so that the loss in R3 is about the same as the loss in R1 and R2 at full load.
  6. Feb 4, 2017 #5

    The Electrician

    User Avatar
    Gold Member

    By the way, if you do choose to add another resistor R3 in parallel with the primary, don't be surprised when your expressions from both the phasor solution and the Laplace solution exhibit a fairly large increase in complexity. This phenomenon is the bane of Electrical Engineers. The very large expressions that occur when just a few more components are added to a circuit are called "high entropy" expressions. Before software like Mathematica became available, symbolic expressions were not derived. EEs learned how to make good approximations for real life circuits.
  7. Feb 4, 2017 #6

    The Electrician

    User Avatar
    Gold Member

    As I read your title for this thread, "Setting up the transformer efficiency as objective function", I'm wondering if you are hoping to come with a mathematical expression for the efficiency of the transformer, and further hoping that that function will have a maximum value as you vary certain transformer parameters? As your circuit stands now, its efficiency will vary with the values of R1, R2 and RL, but the maximum efficiency will occur when R1 and R2 are zero, not a very interesting outcome.
  8. Feb 4, 2017 #7
    Actually Sir, I intend to formulate a mathematical expression for the efficiency of the transformer that is explicitly expressed in terms of N1, N2, r2 (outer radius of the toroidal core), and h (height of the cross-sectional area, probably I may even assume a square cross-sectional area to make things simpler, so it may be just r2 - r1). Assuming the case of a step down transformer and the secondary winding will simply be coiled on top of the primary coil instead of placing it on the side of the secondary coil, we will know that Circumference of the Inner Circle of the Core = 2*pi*r1 = N1*d1, where r1 = radius of the inner circle and d1 = diameter of primary winding wire. Solving for r1, we get r1 = N1*d1/(2*pi), thus making r1 dependent to one of the parameters. The diameters of the wires will just be constant as the one who intends to design a certain transformer will just specify it. All the lengths of the wires and the cross sectional area of the core will be explicitly expressed in terms of the four parameters. As for the case of using the derived expression in the previous thread, I'll relate all the v's and i's to i2 and v1. After which I'll take the partial derivatives of the functions in terms of the four parameters, equate all of them to 0 and solve the system of equation for the values of the four parameters (Fingers crossed that Mathematica will be able to do these, and I do hope that v1 and i2 will not be cancelled out). These values will be the optimal N1, N2, r2, and h, given v1, i2, and the properties of the wire and core used, as per the application of multivariable optimization in calculus. Can you imagine what my plan looks like, Sir?
  9. Feb 5, 2017 #8

    The Electrician

    User Avatar
    Gold Member

    You should have a look at these:



    When using a given core, whether standard EI laminations, or a toroid, there is a winding window through which the turns of the winding must pass. The idea is to get as much wire through that window as possible. The area of the window is filled with wire, but since the cross section of the wire is round, there will be some wasted space. The ratio of the cross sectional area occupied by all the wire comprising the windings, to the window area is called the fill factor. If you change the number of turns in a winding, you must also change the wire diameter so that the fill factor is as large as possible.

    The voltage applied to the primary winding will be fixed. This means that as you change the number of turns, the flux density in the core will change, and that will change the core loss.

    So when you make the wire diameter larger, and thus that winding will have fewer turns, the resistance of that winding will decrease, BUT the flux density will go up, so the core loss will increase.

    This is the major trade off in transformer design. Fewer turns of bigger wire means less copper loss, but more core loss. You have to find the balance between those two giving minimum total loss, and therefore maximum efficiency.
    Last edited: Feb 5, 2017
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted