A thermocouple provides an output of between 4 and 20 mA, corresponding linearly

Question

A thermocouple provides an output of between 4 and 20 mA, corresponding linearly with temperature. The thermocouple is first placed in boiling water at 1.0018 atm, and provides an output of 12.32 mA. The thermocouple is then placed in boiling toluene at the same pressure and provides an output of 12.59 mA.

A.   Derive a formula that converts the output from the instrument into a reading in degrees C.

B.   Convert the formula to provide an answer in degrees F.

C.   Why do you think the thermocouple has an output range of 4 to 20 mA instead of 0 to 16 mA?

D.   What do you think are the operating limits of the instrument? Report your answer in degrees F.

Explanation / Answer

a)at 100 degrees -->12.32mA

at 110.6 degrees-->12.59mA.

Hence converting it into temperature gives

T-100 =((110.6-100)/(0.27x10^-3))(R-12.32x10^-3)

which gives T=39256.26R -383.67

here R is to be given in absolute terms(like for 12.59mA R=12.59x10^-3),T is in degrees celcius

b)Puting F=9C/5 +32 ,gives F=70661.27R -658.606,here again R is to be given in absolute terms(like for 12.59mA R=12.59x10^-3),T is in degrees farenheit.

c)The output range is taken in this way because the absolute error can be found in this case.If there was an error in 0 to 16 mA case,and if reading was below 0mA,it still shows 0mA,but whereas in the present case,if the reading is below 4mA,we can know that there is an error which can be rectified.

puting R=4mA in (b) gives T=-375.96F,

puting R=20mA gives T=754.62 F.

hence operating limits are [-375.96F ,754.62F]

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