Replacing the resistor wires in a Lionel zw postwar transformer

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  • Member since
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  • From: PERRY, GEORGIA
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Replacing the resistor wires in a Lionel zw postwar transformer
Posted by Sierra1 on Thursday, January 31, 2019 3:22 PM

I was able to determine from a circuit diagram that the lamp resistor is rated at 62 ohms with a tolerance of +/- 10 %. Form a previous post , power rating calculations of greater that 5 watts were made by lionelsoni. One down and one to go. The circuit diagram for the zw list the whistle rectifier resistor wire as 1.5 ohms,although all I've tested were 1.8 ohms. I can't locate a power rating for this resistor nor do I have the knowledge to calculate it. I've replaced the rectifier disc with a 40 amp stud diode. I read on the forum that had I used a zener diode the resistor wire wouldn't be necessary. The resistor wires are getting harder to find and the possibility of an asbestos covering is not too appealing . It is my intention to replace the wires with modern al housed wirewind resistors . Any help in solving the power rating calculations will be greatly appreciated .

 

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Posted by ADCX Rob on Friday, February 01, 2019 10:34 PM

20-30 watts should be plenty. Lionel's rheostats were only 20 watts.

Rob

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Posted by Sierra1 on Saturday, February 02, 2019 10:32 AM

Thanks for your replies. After restoring  a Lionel 1033, kw and zw transformer for my grandchildren I'm  hooked. I've purchased a 1.8 ohm 25 watt al housed resistor to check the fit. I also have 1.5 ohm  10 watt  resistors which are a nice size but  not adequate in their power rating. I'm still looking for a 1.5 ohm 25 watt resistor. I would still be very interested in seeing how the power calculations are done for the whistle circuit on the zw transformer. Just don't have the skill set to do it. Paul 

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Posted by Sierra1 on Saturday, February 02, 2019 2:09 PM

Thanks for your replies. After restoring  a Lionel 1033, kw and zw transformer for my grandchildren I'm  hooked. I've purchased a 1.8 ohm 25 watt al housed resistor to check the fit. I also have 1.5 ohm  10 watt  resistors which are a nice size but  not adequate in their power rating. I'm still looking for a 1.5 ohm 25 watt resistor. I would still be very interested in seeing how the power calculations are done for the whistle circuit on the zw transformer. Just don't have the skill set to do it. Paul 

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Posted by Penny Trains on Saturday, February 02, 2019 6:59 PM

Welcome aboard!

Big Smile  I'm Cuckoo For Choo Choo Stuffs!  Big Smile

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Posted by lionelsoni on Monday, February 04, 2019 7:07 PM

Paul, would you mind posting a link to that earlier post of mine that you mentioned?

Bob Nelson

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Posted by Sierra1 on Tuesday, February 05, 2019 6:52 PM

This link is to the post you made calculating the power dissipation  value of the 62 ohm  short circuit lamp resistor in the kw transformer. 18 volts was used in this calculation. The whistle circuit with it's 5 volt compensating winding and superimposed dc current stopped me dead in my tracts as far as determining the power dissipation of the 1.5 ohm resistor wire. I've added 40 amp diodes to replace the rectifier disc which still requires the 1.5 ohm resistor. I understand if I had used a 12 volt 50 watt zener diode this would eliminated the need for the resistor . Not sure. The link is http://cs.trains.com/ctt/f/95/t/86436.aspx. Thanks again for your help. Paul 

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Posted by lionelsoni on Wednesday, February 06, 2019 10:27 PM

I'm working on it, Paul.  It's not an easy computation, but it is interesting.

Bob Nelson

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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.

Bob Nelson

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Posted by Sierra1 on Thursday, February 07, 2019 8:23 PM

Bob I appreciate the hard work. I had no idea of the complexity of the computations. After seeing them, I don't feel quite as dumb either . Admittedly, I was hoping for 25 watts or less but this is doable and should handle the worse case scenario . Thanks again. Paul 

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Posted by lionelsoni on Friday, February 08, 2019 9:40 AM

You're welcome.  I enjoyed the exercise!

Take a look at the data sheet.  It warns that you won't get the full power rating without a heat sink.  There's probably some substantial part inside that you can mount the resistor on for that purpose.  Use heat-sink grease:

https://en.wikipedia.org/wiki/Thermal_grease

Bob Nelson

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