Question

Assume a thermocouple, which supplies the input to an analog module, generates a linear voltage from 20 to 50mV when the temperature changes from 750 to 1250 degrees F. How much voltage will be generated when the temperature of the thermocouple is at 1100 degrees F?

Answer 1

Answer:

41 mV of voltage would be generated.

Explanation:

The voltage generated at 1100 °F is obtained by interpolation.

Let the voltage generated at 1100 °F be y

Voltage (mV). Temp. (°F)

20. 750

y. 1100

50. 1250

(y - 20)/(50 - 20) = (1100 - 750)/(1250 - 750)

(y - 20)/30 = 350/500

y - 20 = 0.7 × 30

y - 20 = 21

y = 21 + 20 = 41 mV

Answer 2

Answer:

41mV

Explanation:

Using the principle of interpolation, the voltage of the thermocouple at the specified temperature of 1100°F  can be found as follows;

At;

Voltage = 50mV, temperature = 1250°F

Voltage = X , temperature = 1100°F

Voltage = 20mV, temperature = 750°F

Calculating the value of X will give the voltage of thermocouple at 1100°F.

Apply the interpolation principle as follows;

=> $\frac{50 - X}{X - 20}$ = $\frac{1250 - 1100}{1100 - 750}$

=> $\frac{50 - X}{X - 20}$ = $\frac{150}{350}$

=> $\frac{50 - X}{X - 20}$ = $\frac{3}{7}$

Cross multiply;

7(50 - X) = 3(X - 20)

Expand the equation;

350 - 7X = 3X - 60

Solve for X;

350 + 60 = 3X + 7X

410 = 10X

X = $\frac{410}{10}$

X = 41

Therefore, the voltage at 1100 degrees F is 41mV