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Sam Brown

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Viewing 15 posts - 166 through 180 (of 461 total)
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  • in reply to: Troubleshooting Computer-Controlled Appliances #20182
    Sam Brown
    Keymaster

      Good question! There’s an important (and slightly unintuitive) rule for situations with parallel loads: equivalent resistance. If you don’t remember how it works, you should review Module 3, unit 5 of Core. We have a video there that explains how it works.

      Sam Brown
      Keymaster

        Yes, one timer motor operates multiple cams. That’s how they manage to use one motor to operate all the timer switches in the machine.

        Sam Brown
        Keymaster

          Scott just threw that out as an example of a clamp-on thermocouple — he didn’t mean for it to be a hard and fast recommendation. In fact, he doesn’t use that kind of thermocouple himself. Usually what we use are K-type thermocouple probes — like this one: https://www.grainger.com/product/1T322?gclid=EAIaIQobChMI8MPL5cbs6wIVivHjBx2XzAt9EAQYASABEgLugvD_BwE&cm_mmc=PPC:+Google+PLA&ef_id=EAIaIQobChMI8MPL5cbs6wIVivHjBx2XzAt9EAQYASABEgLugvD_BwE:G:s&s_kwcid=AL!2966!3!264955916540!!!g!436873882546!&gucid=N:N:PS:Paid:GGL:CSM-2295:HRLBWZ:20500731

          The point with thermocouples is not the specific type you use, but rather that whichever you use, you make sure it makes solid contact with whatever you’re trying to measure. Whether that be with a clamp or with a thermocouple probe that’s securely taped on to the condenser coil is up to you.

          Sam Brown
          Keymaster

            A cam is a very simple mechanical device that you find in many different applications, from sewing machines to car engines to timer motors. All it is is an oval-shaped disc attached to a shaft. If the shaft rotates so that the long part of the oval is facing an actuator, the actuator will get pushed down. In the case of a timer motor in an appliance, this actuator would be a switch contact.

            This demonstrates their action pretty well:

            Cams

            in reply to: Mod 3 Unit 5 Open leg in Parallel Circuit. #20166
            Sam Brown
            Keymaster

              You’re on the right track — let me just see if I can help you zero in on the target.

              So, my question is. Since the open circuit “A” NO LONGER can flow. Do the electrons, because of less demand with “A” being open, just not flow because of the less need for current.

              You’re correct that the voltage to each of the parallel branches would be the same — 120 VAC. And you’re correct that with branch A open, current would not flow through that branch. With branch A open, the total circuit current (meaning the sum of he current flowing through branches A, B, and C) would decrease, because no current is flowing through branch A.

              There’s still 120 volts sitting on that open in branch A, but because there’s no path for the electrons to flow through, they simply can’t get to where they want to go. Therefore, current doesn’t flow through that branch.

              Let me know if you need any further clarification.

              in reply to: Module 3 Unit 6 confusion #20151
              Sam Brown
              Keymaster

                i dont understand what he means by “L2 is missing” missing as in theres a break in a wire somewhere?

                “Missing” simply meaning that L2 is not reaching the load in question. It could be a broken wire or any other component along that line that has gone open.

                i also dont fully understand how he came to that conclusion or why he decided to disconnect L1.

                He came to that conclusion because, in this scenario, he’s reading 120 VAC from each line with respect to neutral, and yet and yet there’s 0 voltage drop across the element and it isn’t heating. If both lines were reaching the element, there would be 240 volts of voltage drop across it.

                In order to determine which line is missing, he disconnected one of them. This is a powerful troubleshooting technique called “half-splitting”. You’re literally splitting the circuit in half by doing this, isolating them from each other and allowing you to determine which is the bad actor. Since L2 is now disconnected from L1, L1’s 120 volts can no longer feed through the element into L2, and so our meter on L2 wrt neutral reads 0 volts, confirming that there is an open somewhere in L2.

                in reply to: Capasitor #20146
                Sam Brown
                Keymaster

                  Hi Stuart,

                  Could you clarify what you mean? What kind of printout are you talking about?

                  in reply to: Have trouble understanding on oven door lock assembly’s #20134
                  Sam Brown
                  Keymaster

                    There aren’t any “hi-limit thermostat switches” on a door lock assembly. They’re simply called limit switches. They don’t sense the temperature themselves. All they’re there for is to tell the control board what state the door lock is in.

                    If you go to around 1:55 in the first video in module 3, unit 3, you’ll see the cam on the bottom of the door lock motor. That cam actuates one of the limit switches, indicating which position the door lock is in.

                    Does that makes sense?

                    Sam Brown
                    Keymaster

                      .I heard Scott mention a Fluke TT80..or TP80

                      I don’t know exactly what he says there, but I think he was referring to a clamp-on thermocouple like this: https://www.fluke.com/en-us/product/accessories/probes/fluke-80pk-8

                      Also, I went to the Danfoss web page..which app am I looking for..? Is it the Ref Slider…

                      Yep, they have renamed the app to Ref Slider. Still the same app!

                      in reply to: pract-se and aaled-system-thermodynamics-part 1 #19524
                      Sam Brown
                      Keymaster

                        Sorry for the delay in getting back to you!

                        At 20:00 in the video, he shows at point #8 a measured temperature of 49 degrees. So, on my Danfoss app, I get a temperature of 38.30 degrees. The difference in subtraction is 49-38=11. He has 10. I understand as he stated that my current app will be off. I checked my measurements with his and have the same results except the one he has labeled #1. He subtracted the shown temperature (60) from the absolute pressure (99), not the temperature difference according to the Danfoss app. Please explain.

                        The Samurai didn’t actually subtract the measured temp from the absolute pressure — that’s a nonsensical calculation to do, since it’s completely different units. There was actually an error in the app’s data back when this video was recorded. The superheat should just be 21 degrees, but the app was showing 41 degrees back then. Fortunately, that’s been corrected now. So that’s why you’re seeing that big discrepancy.

                        Good spot! Glad you’re working through the scenarios yourself.

                        in reply to: practical-sealed-system-thermodynamics-part 2 #19522
                        Sam Brown
                        Keymaster

                          I don’t think we ever covered those terms specifically because they’re not too important to know, but curiosity is never a bad thing!

                          The process tube refers to a short, dead-ended tube sticking off of the compressor. It’s used during manufacturing when the sealed system is first charged. Heat exchanger is just a more general term for an evaporator — essentially, any component that’s designed to transfer heat between the exterior and the sealed system.

                          in reply to: Entropy #19490
                          Sam Brown
                          Keymaster

                            Hi Ronny,

                            Could you post the timestamp of exactly the part of the video that you’re referring to? That will help me clarify it for you best I can.

                            in reply to: entropy #19461
                            Sam Brown
                            Keymaster

                              Isentropic as it’s used in thermodynamics does not actually mean equal or unchanged entropy. It’s one of those confusing misnomers that you sometimes run into in engineering.

                              An isentropic process refers to a thermodynamic process that is completely unaffected by conditions outside of the system (we call that adiabatic) and reversible, meaning that the process makes no changes to the system or the surrounding environment that could not be undone by reversing the process.

                              Isentropic processes don’t actually exist. They’re just an idealized concept that’s close enough to reality to be useful for engineers when they’re creating these systems. And they’re useful to us techs who want to understand thermodynamics, since an idealized system, while not completely true to life, is a lot easier to wrap your head around than the hairy reality.

                              in reply to: refrigerant’s #19438
                              Sam Brown
                              Keymaster

                                Rewatch what the Samurai says around 36:45 in that video. When a refrigerant is at saturation, pressure and temperature are in lockstep. When one increases or decreases, so does the other.

                                So, if you have a 100 F refrigerant entering the evaporator (as shown on the P-H diagram in the video), what happens to the temperature when you have a dramatic pressure drop, like what happens when the refrigerant leaves the narrow capillary tube and enters the relatively wider evaporator coils?

                                in reply to: Compressor windings. #19409
                                Sam Brown
                                Keymaster

                                  1) is there a tool or method for testing the windings on these compressors? They are 2 inches from the left wall of the machine compartment, very difficult to see the windings let alone test them with leads.

                                  You run into that every now and then, and it is tricky. Clip-on leads are probably the best way to deal with it, but those configurations are always a bit of a pain.

                                  2) all was good with the refrigeration course, and what I’ve learned, right up until last week when I got a call for an LG refer, and the compressor only had 2 windings.

                                  I assume you mean that it had one winding with two terminals. Your typical split phase compressor has two windings (but three terminals), and that doesn’t sound like what you’re talking about.

                                  If you mean one winding with two terminals, that would be a linear compressor. I’m fairly certain that we cover them in the course. But in any case, we have an in-depth webinar recording that covers the technology:

                                  https://player.vimeo.com/video/237970331

                                Viewing 15 posts - 166 through 180 (of 461 total)