Question

You produce cryogenice fridges in your factory. You must design a control system for the fridge, but, using the same sensor, must also supply the end-user with a temperature monitoring. Unlike a commercial freezer with a somewhat daunty electromechanical thermostat, a cryo freezer is controlled in a tight band around minus 85^oC.
Therefore, in controlling a cryogenic fridge, you need to have the most accurate information available that you can get around the operating range of the fridge (nominally at -85^oC, but in monitoring the fridge, it can run away (colder than -85^oC) or defrost if faulty and reach ambient temperature (say max 50^oC to be safe).
Due to the economy of RTD sensors, it is decided by management to use a PT100 RTD sensor (range -400 ^oF to 1200 ^oF) with a smart transmitter and it need to be set up correctly to a 4-20mA input on the PLC, controlling the fridge.
As motivated above, the measuring range should be -100^oC to 50^oC (which should cater for the worst ambient temperature).
(i) Make sure the RTD is the correct choice,
(ii) set the smart transmitter up correctly (span, offset, type),
(iii) find the straight line equation (in the form y = ax+b, y in mA, x in ^oC) for the temperature in ^oC to mA output of the smart transmitter,
(iv) test your equation at the extremes (-85^oC to 50^oC)

Answer 1

PART 1.

Conversion from ⁰F to ⁰C :

T(^oC) = ( T(^oF) - 32 )* 5/9

Range of RTD Sensor = -400F to 1200F
In degree Celsius, -400 F   = (-400-32)*5/9
                           = -432*5/9
                           = -240^oC

                   1200 F = (1200-32)*5/9
                           = (1168)*5/9
                           = 648.889^oC  

So, Range of RTD Sensor = -240^oC to 648.889^oC
Temperature range of fridge = -100^oC to 50^oC

Since temperature range of fridge lies within the temperature range of RTD Sensor, Hence RTD Sensor is the correct choice.

Part 2.

The set up can be designed in the following manner.
We will use PT100 RTD of Class A (Type A) because its temperature deviation is less. Temperature Deviation from -100C to 50C is ±0.35°C.
Thermal coefficient for PT100 class A = α = 0.00385±0.000063/C
So the span of the PT100 RTD (from -100C to 50C) will be about 0.25%, since the deviation is ±0.35C which is 0.234% (about 0.25% ) of teh span of 150C.
Offset required: No offset is required as the Complete temperature range of frridge is already covered by the span of the PT100 RTD.

Part 3:

We will consider there is a direct relation in Current and Temperature to find a straight line equation.
range of fridge = -100C to 50C
The current required is 4-20mA.

slope(m) = (y2-y1) / (x2-x1) = 16/150
equation : y-y1 =m(x-x1)

        

        

Part 4

Testing the linear equation at the extremes.

At extreme temp = -85^oC

y = 5.607 mA

At extreme temp 50^oC,

y=20mA.