Mitsubishi Q-series: Converting integer to engineering units...

Hi,
Our PLC measures WATER FLOW RATE as an analog input.
The PLC reads it as an integer value (D34) in the range 0-4000 as follows:
Range: 0-100 m3/hr
Current: 4-20mA
Linear relationship:
4mA= 0 m3/hr (PLC value 0000)
12mA= 50 m3/hr (PLC value 2000)
20mA= 100 m3/hr (PLC value 4000)

The conversion is as follows:
Flow Rate = [ (PLC value/200,000) * 6250] -25

What's the best way to do this conversion? In the ladder logic or "linked" to the "script" in the GUI?

The data analysis person has asked for the exported data (csv file) to be shown in engineering units, not the raw value.

At the moment he gets the raw value and has to do the conversion himself, but prefers engineering units

Can the PLC (Q-series) handle the division without losing too much accuracy?

Should the conversion be done in the ladder logic? Is that the best place to do it? Thanks

Edit: Just realised I started a similar thread elsewhere, but feel free to reply to this one.


Edit 2: Please answer on this thread - the other thread got completely de-railed with off-topic comments.

Do it in the PLC, this way you can do any further maths on it and if you are controlling a pump with a PID it makes sense.
So first convert the analogue value to a real, then do the maths. to display it on an HMI although the value will have 6 decimal places you can limit them on the HMI to x number of decimal places.
Also your calculation is incorrect 2000 will not equal 50 Cm3 that is 2400
All you need to do is convert the analogue to a float divide that by 40 that will give you your answer.
your calc works for 4000 but not for 2000
((2000/200,000) *6250) -25 = 37.500
((4000/200,000) *6250) -25 = 100.000
Think about it: if 4000 = 100 cm3 then 40 = 1 cm3
2000 / 40 = 50.000
4000 / 40 = 100.000
Edit: The Q series will have a resolution to 3 decimal places at least

parky said:

Do it in the PLC, this way you can do any further maths on it and if you are controlling a pump with a PID it makes sense.
So first convert the analogue value to a real, then do the maths. to display it on an HMI although the value will have 6 decimal places you can limit them on the HMI to x number of decimal places.
Also your calculation is incorrect 2000 will not equal 50 Cm3 that is 2400
All you need to do is convert the analogue to a float divide that by 40 that will give you your answer.
your calc works for 4000 but not for 2000
((2000/200,000) *6250) -25 = 37.500
((4000/200,000) *6250) -25 = 100.000
Think about it: if 4000 = 100 cm3 then 40 = 1 cm3
2000 / 40 = 50.000
4000 / 40 = 100.000
Edit: The Q series will have a resolution to 3 decimal places at least

Click to expand...

Brilliant - thanks for this.
Yes - my mistake : 50m3/hr is "2400". Turns out the original formula given to me was from another system - and they were reading a different mA range.
I shall implement the above logic. Clear explanation too, thanks.

parky said:

Do it in the PLC, this way you can do any further maths on it and if you are controlling a pump with a PID it makes sense.
So first convert the analogue value to a real, then do the maths. to display it on an HMI although the value will have 6 decimal places you can limit them on the HMI to x number of decimal places.
Also your calculation is incorrect 2000 will not equal 50 Cm3 that is 2400
All you need to do is convert the analogue to a float divide that by 40 that will give you your answer.
your calc works for 4000 but not for 2000
((2000/200,000) *6250) -25 = 37.500
((4000/200,000) *6250) -25 = 100.000
Think about it: if 4000 = 100 cm3 then 40 = 1 cm3
2000 / 40 = 50.000
4000 / 40 = 100.000
Edit: The Q series will have a resolution to 3 decimal places at least

Click to expand...

One question:
In your ladder logic, if I wanted to display D104 on the GUI, what 'Data Type' do I need to specify in the Numerical Display?
The options are:
Signed BIN16
Unsigned BIN16
Signed BIN32
Unisigned BIN32
BCD16
BCD32
Real


...I think it should be REAL. Do you agree?

You have to use a real, remember, real (float) numbers take up two registers so in actual fact D104 is D104 and D105 but you only reference them as D104, it knows to use D105. If you tried to display it as an integer the value would not make sense.
That goes for converting an integer so in the logic I posted where we convert the integer to a float D102 uses D102 and D103 that is why I did not use D103 for the result.

parky said:

You have to use a real, remember, real (float) numbers take up two registers so in actual fact D104 is D104 and D105 but you only reference them as D104, it knows to use D105. If you tried to display it as an integer the value would not make sense.
That goes for converting an integer so in the logic I posted where we convert the integer to a float D102 uses D102 and D103 that is why I did not use D103 for the result.

Click to expand...

Brilliant, that makes sense. Thanks again for a clear explanation.

UPDATE: The data analysis person now wants the units in litres/second....ok...
So this means I need to multiply my m3/hr value (D104) by 0.278 to convert it to litres/second (if my maths is correct).


How can I represent 0.278 in ladder logic so I can do this? (D106 to hold the value in litres/second):


--------------[E* D104 "E0.278" D106]---


---------------------------------------------[END]




(yes, I could multiply by 278 , then divide by 1000, but was wondering if there was a neater way).

Yes its just [E* D104 E0.278 D106]

parky said:

Yes its just [E* D104 E0.278 D106]

Click to expand...

Cheers again for the help.

parky said:

Yes its just [E* D104 E0.278 D106]

Click to expand...

I did the update on the PLC just now and it's correctly showing flow in litres/sec. Cheers.

For floating point data registers, are they stored in 2 registers, so is that 32 bits?

Yes that correct so for example if you use D106 as a floating point number then it will automatically use D106 & D107 so do not use D107 for anything else.
It also makes sense to start floating point numbers with even registers this makes it more likely that you will not overlap by mistake for example if you have D106 to D110 as 32 bit floating point numbers then you know D110 is a float then the next 16 bit register is D112 not 111. Just makes it a little easier.

parky said:

Yes that correct so for example if you use D106 as a floating point number then it will automatically use D106 & D107 so do not use D107 for anything else.
It also makes sense to start floating point numbers with even registers this makes it more likely that you will not overlap by mistake for example if you have D106 to D110 as 32 bit floating point numbers then you know D110 is a float then the next 16 bit register is D112 not 111. Just makes it a little easier.

Click to expand...

Good advice that, thanks.