A load cell with the following manufacturer specifications is used in an environ

Question

A load cell with the following manufacturer specifications is used in an environment where ambient temperature is not controlled, and can vary between 55°F and 95°F. Assume the calibration temperature is 75°F.

Input Range 0-100 lbf

F.S.O. 30 mV

Linearity ±0.1% FS

Hysteresis ±0.08% FS

Repeatability ±0.03% FS

Zero balance ±2% FS

Temperature Effect on Zero ±0.002% FS/ °F

Temperature Effect on Span (change in sensitivity) ±0.002% of reading / °F

a) For an input of 10 lbf, what will be the nominal output voltage?

In an experiment, the load cell was used to measure the output voltage 10 times to determine an average value. The following data were collected:

Voltage [mV] 19.072 19.112 19.124 19.064 19.202 19.105 19.177 19.086 19.138 19.090

Answer parts (b)-(h) based on the above data, and load cell’s specification table:

b) Calculate the average, standard deviation, and standard deviation of the mean for the output voltage. c) Determine the number of degrees of freedom and the t statistic (95%). d) Determine random uncertainty associated with the measurement of the mean. e) Estimate uncertainty due to each error source that can be determined from the manufacturer specifications. Express uncertainties in both mV and % of reading. Do not combine uncertainties. f) What is the total random uncertainty associated with the output voltage? g) What is the total systematic uncertainty associated with the output voltage? h) What is the total/combined uncertainty associated with the output voltage?

Explanation / Answer

a. F.S.O=30mV

Full load = 100lb

so diflection in mV/lb=30/100=0.3mV/lb

For an input of 10 lbf, the nominal output voltage be=0.3*10+Linearity error+Hysteresis error+Repeatability error+Zero balance=3+0.001*3+0.0008*3+0.0003*3+0.002*3=3+0.003+0.0024+0.0009+0.006

=3.0123mV to 2.9877mV

b. mean or average of o/p voltage =sum of all readings /no of samples =(19.072 +19.112 +19.124 +19.064+ 19.202 19.105+ 19.177+ 19.086+ 19.138 +19.090)/10=19.117mV

Standard deviation =(((19.072-19.117)^2 +1(9.112-19.117)^2 +(19.124-19.117)^2 +(19.064-19.117)^2+ (19.202-19.117)^2 +(19.105-19.117)^2+( 19.177-19.117)^2+ (19.086-19.117)^2+ (19.138-19.117)^2 +(19.090-19.117)^2)/10)^(1/2)=0.042

Standard deviation of mean=SD/(10^(1/2))=0.042/(10^(1/2))=0.013

c.

no of degree of freedom is =sample-1=10-1=9

calculate the t statistics depends population means of samples above 95% is = mean of values between 19.077 and19.1569=19.109

t statistics = [ mean- population mean ] / [ SD / sqrt( sampleas ) ] =(19.77-19.109)/(0.043/( 0.042)^(1/2))=3.15

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