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Old March 20th, 2017, 03:57 PM   #16
dmargineau
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Quote:
Originally Posted by ceilingwalker View Post
Hello All.
I have a project that is pretty easy however, it involves over 1,000,000 analog inputs. It is a solar panel farm that has 200 inverters, that contain 25 panels each. The customer is requesting Allen-Bradley products, however it is designed. I have never worked with this much I/O in the past and I am hoping someone here has. I am not even sure I want to use a PLC here, since it isn't going to be controlling anything, merely reading data. I am not sure which SCADA system would be best, with this many tags. A one-minute sample rate is all they are requesting. Any and all advice would be greatly appreciated.

Thank you for your time.
The math does not quite add up:

200 X 25 = 5000 panels

1000000 / 5000 = 200 sensors/panel

Do you really need 200 temperature sensors for each panel?!

There are only 200 inverters and each VFD will simultaneously move 25 panels at a time; 5000 analog inputs for defining a single inverter application it is at least a 2 order of power overkill.

IMHO 10000 analog inputs will more than suffice.
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Old March 20th, 2017, 04:21 PM   #17
janner_10
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Pick your local reps brains, let him do the costing for you. Although it does sound quite pie in the sky.
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Old March 20th, 2017, 04:30 PM   #18
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For that application Mebbe you need one analog input and a lot of multiplexing.
Look at pc data acquisition systems. Rockwell hardware will break the bank for no real benefit.
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Old March 20th, 2017, 04:47 PM   #19
AustralIan
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awesome project, if some of those numbers work out. Use proposal works to put a spreadsheet together of cost per point for various AB products, maxing out points per adapter and per PLC. Then have an informed discussing with rockwell about it.

Did they want a complete Rockwell solution, or can you use someone else's SCADA/historian?

SCADA straight to io with AB is either impossible or rare enough that it is not worth persuing. Definitely use a PLC for that bit.
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Old March 20th, 2017, 11:39 PM   #20
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Old March 21st, 2017, 07:22 AM   #21
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While the title indicates "analog inputs" I wonder if the customer literally means 1M physical sensors, or if it is really 1M process variables, some of which may be acquired by communication to inverters or other "smart" devices. In that case, communication protocols may have a significant impact on technology choice.

Even so, I agree the math pointed out in reply #16 still does not add up.
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Old March 21st, 2017, 08:16 AM   #22
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Originally Posted by ceilingwalker View Post
I have a project that is pretty easy however, it involves over 1,000,000 analog inputs. It is a solar panel farm that has 200 inverters, that contain 25 panels each.
Wow. I'd say you need to figure out what kind of analog signal you are going to be using. If you used a 4-20 ma signal powered by 24V, 0.02 x 24 x 1000000 = 480 KW. The average home uses about 11,000 KWH. 480 * 24 * 365 / 11000 = 380. If the average signal was 50%, you'd still be using enough power to power 190 homes for just for the inputs. If your signal was a voltage readings the power usage would be much less.
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Old March 21st, 2017, 08:37 AM   #23
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Quote:
Originally Posted by proof View Post
Wow. I'd say you need to figure out what kind of analog signal you are going to be using. If you used a 4-20 ma signal powered by 24V, 0.02 x 24 x 1000000 = 480 KW. The average home uses about 11,000 KWH. 480 * 24 * 365 / 11000 = 380. If the average signal was 50%, you'd still be using enough power to power 190 homes for just for the inputs. If your signal was a voltage readings the power usage would be much less.
I don't think that quite works out correctly. 4-20mA signals are powered by 24V, but they don't apply 24V directly across the load (usually approximately a short circuit).

The 4-20mA driver automatically adjusts its voltage to maintain 4-20mA, so the actual voltage across the load is very low.
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Old March 21st, 2017, 09:00 AM   #24
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The 4-20mA driver automatically adjusts its voltage to maintain 4-20mA, so the actual voltage across the load is very low.
A 4-20 transmitter adjusts its resistance to control the signal current. The voltage drop must match the voltage source, some of the drop is in the input card, and some the transmitter, and some in the wiring. Using a lower voltage source will allow a lower power usage from the system. AB 4-20 ma inputs are 250 ohm. This means the input card with a 20 ma input dissipates 0.02 * 0.02 * 250 = 0.1 Watt. The voltages across the card is 0.02 * 250 = 5V. This is still 100 KW for the system. If the average signal was 50%, then the power usage would be .012 * .012 * 250 = 0.036 Watt (not half as implied in my last post). This is 36 KW, which doesn't include the power used by the transmitter.
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Old March 21st, 2017, 09:39 AM   #25
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^Good point- I stand corrected.
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Old March 21st, 2017, 11:52 PM   #26
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Quote:
Originally Posted by proof View Post
Wow. I'd say you need to figure out what kind of analog signal you are going to be using. If you used a 4-20 ma signal powered by 24V, 0.02 x 24 x 1000000 = 480 KW. The average home uses about 11,000 KWH. 480 * 24 * 365 / 11000 = 380. If the average signal was 50%, you'd still be using enough power to power 190 homes for just for the inputs. If your signal was a voltage readings the power usage would be much less.
Why has the plant power output gone down 10%?!

Er... we installed a a SCADA system.

This is the first time I've ever thought about the power consumption of analog inputs!!
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Old March 22nd, 2017, 04:07 AM   #27
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Quote:
Originally Posted by ceilingwalker View Post
Hello All.
I have a project that is pretty easy however,...
Every time I hear a project description start this way, pain shoots up my back, I then feel like I am getting electrocuted, then the seizures.
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Old March 22nd, 2017, 04:17 AM   #28
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Quote:
Originally Posted by proof View Post
A 4-20 transmitter adjusts its resistance to control the signal current. The voltage drop must match the voltage source, some of the drop is in the input card, and some the transmitter, and some in the wiring. Using a lower voltage source will allow a lower power usage from the system. AB 4-20 ma inputs are 250 ohm. This means the input card with a 20 ma input dissipates 0.02 * 0.02 * 250 = 0.1 Watt. The voltages across the card is 0.02 * 250 = 5V. This is still 100 KW for the system. If the average signal was 50%, then the power usage would be .012 * .012 * 250 = 0.036 Watt (not half as implied in my last post). This is 36 KW, which doesn't include the power used by the transmitter.
This excludes the actual power supply for the transmitter (if it is a 2 wire device). The actual voltage across the transmitter and the load will be higher. Most 2 wire transmitters require something like 10Vdc to operate correctly which is why cheap loop calibrators with 9V batteries often fail to work correctly.
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Old March 22nd, 2017, 11:34 AM   #29
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Are we all sure that the analogue inputs are indeed analogue in the sense of 0-10V or 4-20mA as we know them ?

The OP hasn't specified that and as he says he doesn't get told much, was it a comment in passing "we need to pick up 1000000 points" or "we have to pick up 1000000 4-20mA signals" ? There is a massive difference !

We see specs that detail analogue but in fact just mean anything that isn't digital, if it's just 1000000 data points with a refresh of 1 minute then that's well withing the realms of a well designed PLC/SCADA system.
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Old March 22nd, 2017, 12:23 PM   #30
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Depending on the inverters being used, they may have an Ethernet communication available to view each inverter's data... then no analog inputs required. That would be the best/easiest solution. However, the lack of information from the customer is problematic.
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