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You’re correct that the main control sends a DC PWM signal to the inverter. But the inverter sends electronically commutated AC voltage to the BLDC compressor. The PWM signal tells the inverter what frequency and amplitude the voltage it sends out should be.
Yes, that’s correct — you cannot look at the quiz answers again unless you start another attempt. But I’m happy to tell you want the possible answers for that question were! They are:
A. variable frequency
B. pulse width modulated
C. split phaseHa! Thanks for posting that — I was curious what you were talking about. That’s certainly the kind of thing you would have seen over at Appliantology. A little too extracurricular for the courses at this site. 😀
As mentioned in the video, there’s a lot of terminology that manufacturers throw around when it comes to direct spark ignition systems, so don’t feel bad about getting a little turned around. Fortunately, it’s not as confusing as the terminology makes it out to be.
A direct spark ignition system is just any burner system that ignites gas with a spark. As simple as that. There are a lot of variations depending on whether there’s flame sensing or whether all the burners spark at the same time, but the fundamental principle of ignition is the same.
Now go back to that question that was confusing you with this in mind. How many of the listed systems use a spark to light the gas coming out of a burner? Some of them? None of them? All of them?
Let me know if it’s still not clear.
Consider yourself familiarized with the forums!
As for that video, I’m not sure which one you’re referring to. Did you see it here at Master Samurai Tech, or was it over at Appliantology? If you can find it on YouTube, post the link to it here — I’ll see if it rings a bell.
That’s an excellent observation and a great question!
The beef is this: there’s more that can impede the flow of current than just resistance. In fact, there’s a whole other category of impedance that’s called reactance.
While reactance isn’t present to a significant degree on all loads, a lightbulb is an example of a reactive load. So that explains why you’re getting lower amperage than you would expect.
We’ve got a great video on the topic of reactance that you should check out. It uses a motor as an example of a reactive load, but the principles in play are the same as what you’re seeing with your lightbulb. Here’s a link to the video: https://www.youtube.com/watch?v=n9gaaKtOfDQ&feature=youtu.be
“Voltage drop” and “voltage difference” are actually two distinct terms for two distinct things. Here are a couple of examples.
You have a circuit supplied with 120 VAC. In this circuit, there is a switch that is currently open. If you put one meter probe on either side of the switch, you would read 120 VAC, because there is 120 volts on one side of the switch and 0 volts on the other side. There is clearly a voltage difference here, but no voltage drop, because there’s no current flowing through that open switch.
Now, imagine that you have another circuit, this one also supplied with 120 VAC. This circuit is closed, with current flowing through it. There’s a heating element in the circuit, and you place your meter probes on either side of the element. Like in the previous example, you read 120 VAC, but it’s not because the element is open — it’s because there’s a voltage drop of 120 volts across the element, and that’s what you’re reading with your meter probes.
It may seem like a subtle distinction, but it’s important to understand, because those are two very different situations electrically. For further clarity, I recommend that you click here to review the videos in the Voltage Drop and Load unit of the Fundamentals course.
In the fundamentals course, we learned that if you check the voltage at the load on a 120 VAC circuit, you should get a 120 VAC voltage drop or a reading of 0 VAC on the meter.
You’re a little bit off on your understanding of voltage drop there. Let me see if I can clear things up.
Remember that when you measure voltage, what you’re actually doing is measuring the voltage difference between two points. Those two points are where you’ve placed your meter probes.
When a load has current flowing through it, there’s going to be voltage drop across it, as you said. But that means that you’re going to read 120 VAC when you put your leads on either side of the load, not 0 VAC. The difference between those two points (a difference created by the voltage drop across the load) is 120 volts, and so that’s what your meter shows.
Does that help explain it?
Good question, John!
The diagram that you’re looking at in the Kleinert text is for one specific compressor design. The phrasing of the quiz question is meant to be more general, which is why we don’t mention a connecting rod.
Thanks for asking!
when you have a working microswitch and its normal State and a live 12 volt circuit. Answer is youre reading 12 volts DC on the normally open and common terminals.
What you need to know is right there in the wording of the question — the premise is that you’re measuring from the normally open terminal to the common terminal. Since the switch is in its normal state, you know that the normally open terminal isn’t connected to the common terminal. Therefore, you would read a voltage difference of 12 VDC between the two points.
Hi John,
Excellent question! You actually caught a typo in the text — it should definitely say “warm liquid” there. Great catch!
By the way, we go into much more specific detail about the thermodynamics of the refrigeration cycle in Module 3, so look forward to that! I think you’ll enjoy it.
Hi Abe,
The point of this question isn’t to jump straight to theorizing about the root cause. What we want the student to do is simply recognize that, when you have freezing in the fresh food compartment of a single-evaporator unit, it’s always because too much air is flowing from the freezer into the fresh food compartment. It is, therefore, the air distribution system that’s not behaving how it’s supposed to.
In a real troubleshooting scenario, you would then move on to finding the component that’s causing the problem. It might be one of the parts of the air distribution system, or it might be part of the control system, as you pointed out. But that goes beyond the scope of this question.
Hope that makes sense!
July 17, 2018 at 8:03 pm in reply to: APPLINCE MOTORS MODULE 8 VARIOUS FREQUENCY DRIVE SYSTEMS #14646Hi Papa John,
Sorry for the delay in getting back to you! Team Samurai was out of the country for a little bit, but we’re back now and getting things rolling again.
While 2 is the correct answer for that question, your reasoning isn’t quite right. And this is partly to do with the wording of the question being a little confusing. It said “energized”, when what we really meant was “polarized to do work”. I’ve changed the question to clear this up.
If you recall from the video on electronic commutation, it’s not enough for a motor winding to just have current flowing through it. The current has to be flowing in the direction that will create a properly aligned magnetic field to push on the permanent magnets in the rotor.
As shown about 14 minutes into that video, even if all three windings have current flowing through them, only two of them will be polarized correctly.
Thanks for asking!
Sure thing! Here are those two images. You can just right click on them and download them to your computer.
Yes, that’s correct that the ice would be absorbing the heat in that case. However, something else to notice is that, in this theoretical situation, you would not be able to freeze anything. The coldest that you’re able to get something with ice is 32 F. Think about putting ice cubes in a glass of water. If you really stuff the glass with ice, the water would get close to 32 degrees, but it would not actually freeze.
If you want a really thorough explanation of thermodynamics in refrigeration, I recommend you watch our series of webinar recordings on the subject. You can find them either in the bonus module in the Refrigeration course or on the Webinar Recordings Index page at Appliantology.
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