basic motor circuit / control cabinet questions

Hello,

As a mechanical engineer moving into the automation world i have a few simple hardware questions for the more electrically inclined than I. I have a conveyor project aimed at becoming one subject of an automation project for students. It was donated to us by a local factory and will be controlled by a Siemens s7-1200 series cpu. The conveyor is driven by a 3phase Motor with no plans for vfd, just contactors controlling the forward and backward directions. I'm overseeing a couple interns who are working On developing a small control cabinet. My questions:

1) the s7-1200 has dc 24v supply voltage, which means we need an extra power supply from 220v to 24v. We have some ps307 5A power supplies laying around. Is there a General rule for calculating the power supply amperage required for a specific plc setup and i/o setup? The motors will be switched On and off by contactors, and there are a few electropneumatic solenoids and lights.

2) How do you size circuit breakers properly for a 3 phase motor? We have a bunch laying around for this purpose, but want to be sure they are right for the application.

3) i was told by a colleague that the voltage coming into the control cabinet is usually 480v or something like this and is stepped down to control voltage. Does this take place within the control cabinet generally by a step-down transformer, or does it take place outside of it?

4) besides circuits breakers and contactors, are there any other components i should have in the motor circuit? We have a switch somewhere that I was told sums current flowing in and out of the motor, then trips if there is significant leakage. Should something like this also be in there?

5) how do i size the contactors for the motor to make sure nothing gets damaged by inrush current when switched on?

I'm thinking these are pretty rudamentary questions, but I'd really appreaciate any help from the experts On the board.

Thanks!

I strongly suggest that you get some help from one of the EE faculty, since electricity kills. Here are some intital guidelines, but you should really get some help from someone with eyeballs on site.

1) The catalog or operations manual for the PLC will tell you the power requirements.

2) The breaker should be sized for 140% of the motor full load amps, but local codes may vary. They should be investigate.

3) In your country I would expect 380 VAC 50 Hz, not 480. Particularly if you have students running around, I would keep the 380 VAC in a completely separate lockable panel. Get a 24 VDC power supply in that cabinet, and have everything in the PLC enclosure at 24 VDC.

4) Instead of a contactor you want a reversible starter. A starter is a contactor plus overload in one device.

5) If you size the contactor for the motor kW and voltage it will handle the inrush. Similarly, if you use a breaker rated for motor protection circuits it can handle the inrush.

Again, you don't want to learn on your own with students hanging around. Get some one local to give you a hand.

I have to concur with Tom. Since this is apparently an actual project and you are asking very basic questions I very strongly encourage you to get help. This isn't disparaging your ability to learn the material - I have no doubt that you can. Mechanical Engineers can make really good automation engineers once they get the electrical part down. However safety is paramount and your students should see in you an example of safe behavior so please get help. Unfortunately in my own education I saw too many examples of unsafe behavior from the instructors but didn't know it until I made it out into industry. Don't discount the help and advice of tradesmen in the school's maintenance department - they may already be familiar with the IEC regulations and know the equipment and control methods.

Most manufacturers of breakers and motor starters publish the maximum motor sizes for which their product is intended in their catalogs. They usually account for inrush loads when doing so, however in some extreme duty applications it is advisable to use a larger starter. A conveyor is sometimes an extreme duty application because accelerating a heavily loaded conveyor (such as one loaded with rocks in a mine) can put severe demands on a motor. In your student lab that is unlikely but this is one example of where your mechanical background can be applied to determine what the starting load is.

The switch to which you refer that trips if there is any leakage is what we call a Ground Fault Circuit Interrupter and is sometimes also called an Earth Leakage Relay. The conductors pass through the middle of a toroidal transformer coil. Since the sum of currents in and currents out should be zero no voltage is induced in the torroidal winding. If there is even a small bit of leakage to Earth then the sum is non-zero and voltage is induced in the toroid, tripping the contacts to open the circuit.

I agree with every thing Jenkins and Connelly said.
What size motor ie kW or HP are you using?

Re one statement Tom made
Instead of a contactor you want a reversible starter. A starter is a contactor plus overload in one device.
Let me add you need two relays to reverse a 3 phase motor. What Tom is referring to is a factory package that has the two relays AND they are mechanically locked - nearly impossible to energize in both foward and reverse. This is the ideal and should be seriously considered if you are going to let students work with teh control wiring ie 24V.
I agree with Tom the "high voltage" should be in completely locked cabinet EXCEPT
1. Label each relay and component for its function ie OVERLOADS FOWARD RELAY etc etc.
2. Take off the steel door and replace with a good durable plastic ie Lexan - in this way students can visually see what the relays are doing and how it is all put together.

Since students and lab I would tend to put a line disconnect at each end of conveyer - students have a tendancy to get fingers near things that can bite them ie conveyers - yes I know they should be guarded and all that. I say this because I assume students can change conrol wiring thus they may change an E stop circuit. They would not be allowed to change the hard wired hi power and disconnects.

If you need help sizing stuff get your local vendor out to have look see - some of them really know their stuff and are willing to help.

Dan Bentler

Guys,

Thank you so much for the responses. A little background on the project and why I am asking these things: I am a mechE with some mechatronics background, but more microprocessor control, nothing industrial. I work at a technical school where we teach future engineering bachelors students. We teach almost exclusively using real systems, control using mostly Siemens equipment. Our automation guy left last week to go back to industry and I have been assigned to take over his automation course starting in about 5 months. I am quite familiar with PLC programming as I spent a lot of time working with him in the past on specific projects. However, it's the industrial hardware and electronics that I am unfamiliar with. We have tons of equipment laying around here for teaching purposes, so I would like to use some of what we already have, as opposed to buying new stuff.

We have 3 phase outlets in a number of our classrooms and the conveyors run just fine plugged in directly (they are smaller conveyors that were used for a PCB assembly line). The question about the Industry 380V being present in the cabinet was not because I am intending on running 380V into the cabinet for this project, it was simply to enhance my own understanding of common industry control cabinet setups. The S7-1200 to be used has +24V DC supply inputs, I was curious if the common method of sizing a power supply for an application like this would be to just look at the max current drawn by everything if on all at once (PLC, Contactors, LED Buttons, Lights) and then use the next size up (in terms of amperage). After I posted I started thinking about it, and it seems that in a real control cabinet the power source for all peripherals is not going to come from a small supply like the PS307s we have, but from stepped down power from the mains and converter to DC. Right?

The purpose of this project is really just for students to learn PLC programming and a starting point for the basic structure of a control cabinet, nothing fancy or wildly dangerous. I am trying to keep the equipment as safe as I can, so thanks for the input.

The reason I am asking these questions here is because I was hoping for some formulas or selection guidelines to see if we can use the equipment we already have. Most of my colleagues are electrical engineers or electronics guys, but everyone is so overloaded that pulling them aside for help is tough. I've already learned a ton reading from this board the last few weeks and have been impressed at how helpful everyone has been, also regarding my plc posts, so I figured I'd ask the questions here.

We do have a set of contactors that are mechanically interlocked as you mentioned that came from another industrial robot project that was donated to us. I was hoping for any formulas or guidelines to make sure these will handle the required loads.
Thanks!

Btw. the rating on the motor tags:

Bosch
50Hz, 0.18kW.
50Hz Delta/Wye 230/400V 1.2/0.7A
1325/1590 min-1
cos phi 0.7/0.7

That's a pretty small motor, much smaller than what I anticipated. The ratings on the motor tag tell you what the motor will draw. At 230V it will draw 1.2 A and at 400V it will draw .7A. For the benefit of those of us who haven't joined the 21st century and are still using imperial units its roughly a 1/4 HP motor. At 60 HZ and 480V it would draw .59 amps at full load. Inrush to the motor will be somewhere in the 4 to 5 amp range for most normal mechanical loads.

Look on the side of your contactor and you should see a table that indicates the current rating of the contacts. That will tell you if its OK to use (I suspect that for such a small motor its very likely that your contactor is much larger than what is required so using it won't present any problems).

Can you give us the part numbers of the interlocked contactors you have and can you tell us if the contactors are set up for Forward/Reverse starting or are they set up for Wye-Delta starting. A wye-delta starter is a motor starting scheme that lowers the inrush current and is rarely used on fractional horsepower motors. Wye-Delta starters will sometimes have a pneumatic timer attached to one of the contactors, which may help in identifying how the dual contactor starter is configured. If its set up for wye-delta starting as opposed to forward/reverse starting the jumpers between contacts will need to be rewired.

The reason I am asking these questions here is because I was hoping for some formulas or selection guidelines to see if we can use the equipment we already have.

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For basic information about sizing wiring, fuses, and circuit breakers, get a copy of the National Electrical Code Handbook.

For motor starters, there are NEMA tables available, which lists the NEMA standard sizes 00 through 6, and the motor size range that each starter size can operate. Here is a link to one such table. It gives the motor sizes in Horsepower for which each motor starter (contactor, contacts, motor overloads all in a package) can be used.

http://www.southlandelectric.com/motor_starter_sizing_chart_for_s.htm

I was curious if the common method of sizing a power supply for an application like this would be to just look at the max current drawn by everything if on all at once (PLC, Contactors, LED Buttons, Lights) and then use the next size up (in terms of amperage).

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You also have to factor in motor Starting Current, which is much higher than running or operating current (FLA on motor nameplates). Starting current varies with motor usage, and can range from 1.25 to 2 times the normal operating current.