Derivation of standard 4-20 mA equation and 4-20 mA formula

 This article is about standard 4-20 mA equation and 4-20 mA formula focusing to the engineers, technicians and supervisors. You will find lot of documents related to this article. Just navigate our website www.paktechpoint.com  and find more articles. Please! Do not forget to subscribe our Youtube channel also. Thanks

PAKTECHPOINT YOUTUBE CHANNEL SUBSCRIBE BY CLICKING

Derivation of standard 4-20 mA equation and 4-20 mA formula

4-20 mA scale is linear with 0-100 scale. So can use linear function on this 4-20 mA. As shown in figure

As you can use slope intercept form on this linear function
Y=mx+b    ………………(1)
You can consider
Y= transmitter current signal output
B= y-intercept point (zero point live in instrument range )
X= percentage of process variable
We can predict any value of current signal just by finding the slope (m)  and y-intercept point  as shown in figure

M= rise/run
=20-4/100-0= 16/100
Putting this value in slope intercept equation
Y=(16/100)x+b
Now we have to find slope-intercept point (b) . For this we will select any coordinates x,y
In this calculation we are using x=0 and y=4
Now the equation is
4= (16/100)*0+b
b=4
Now the final formula for converting percentage in to mA is
Y=(16/100)X+ 4
For any given percentage we can find the current equivalent mA.
For example one control valve is 65 % open what will be equivalent current signal?
Y=(16/100)X+ 4
Y = (16/100) 65+4
=14.4 mA
Output current will be 14.4 if the control valve is 65 % open.
We use this linear equation for value of Y when x is given and vice versa. We can calculate the value of current signal mA of any process transmitter i.e pressure, temperature, flow and control valve.

Please read also:  Basic Control Valve Principles

paktechpoint.com

You may also like...

Leave a Reply

%d bloggers like this: