- Member since
December, 2001
- From: Austin, TX
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9,824 posts
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Posted by lionelsoni
on Thursday, February 07, 2019 1:56 PM
Here is my analysis at last:
A universal motor comprises an armature and a field winding wired in series. Each of these elements has an ohmic resistance due to the wire of which it is constructed. I lumped these series resistances together and modeled them separately as a fixed resistance Rw in series with an ideal zero-resistance motor.
The ideal motor consumes electrical power Pi that is the product of the motor voltage Ei and current I. It puts out the same amount of mechanical power that is the product of the torque T and the velocity V:
Pi = Ei * I = T * V
The torque is proportional to the field current and to the armature current. In a series-connected motor, these are the same, so the torque is proportional to the square of the motor current:
T = k * sq(I)
where sq() denotes the square. I substituted this expression for T in the previous equation and then divided both sides of the equation by sq(I):
Ei / I = k * V
This is to say that the motor looks electrically like a resistance proportional to the velocity, where k is the constant of proportionality:
Ri = k * V
I added to this ideal motor's resistance the previously set-aside ohmic resistance:
Rm = Rw + Ri = Rw + kV
The worst case for the rectifier resistor is for the whistle control to be operated with the transformer voltage Et, including the compensating winding, at its highest value, or 25 volts RMS for the ZW, and with the lowest motor resistance that does not draw the RMS current Ib that will trip the circuit breaker, or 15 amperes for the ZW:
sq(Ib) = sq(Et) * (.5 / sq(Rm + Rr) + .5 / sq(Rm))
where Rr is the resistance of the "rectifier resistor". Note that the RMS currents for the two waveform polarities are orthogonal and therefore can be combined as the root-sum-square (RSS), with the coefficient .5 applied to the squares to reflect the relative durations of the half-cycles. I solved this equation iteratively for Rm:
Rm = 1.299 ohm
I plugged that value into this expression for the square of Ir, the RMS current in the rectifier resistor:
sq(Ir) = sq(Et) * .5 / sq(Rm + Rr)
Then the power Pr dissipated by the rectifier resistor is simply:
sq(Ir) * Rr = 59.82 W
Sixty watts seems like a lot of power, but it comes from working with the worst-case transformer voltage (25 volts) and load current (15 amperes), which correspond to 375 watts before the circuit breaker trips. I'm sure that Lionel hoped that it was unlikely the whistle would ever be blown under those conditions or held on long enough to overheat the resistor. But, like some other weaknesses of old Lionel transformers, there is no hard and fast limit to prevent such a thing.
Nevertheless, a 60-watt resistor is not all that large nor expensive. Mouser has in stock Ohmite 284-HS75-1.5F 1.5-ohm 75-watt aluminum-case wirewound resistors for $10.26 (284-HS75-1.5F). Here is the data sheet:
https://www.mouser.com/datasheet/2/303/HS-Datasheet3-779273.pdf
As for the resistance value, both 1.5 and 1.8 are preferred numbers in the E12 series. So Lionel may have changed the resistor value at some time, to increase the DC component, but never updated the manuals. In any case, I repeated the computation and found that 1.8 ohms makes little difference, changing the power dissipation very slightly, to 59.45 watts.
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