A strain gauge, diaphragm pressure transducer (accuracy: <0.1% reading) is subjected to a pres- sure differential of 10 kPa. If the output is meas- ured using a voltmeter having a resolution of 10 mV and accuracy of better than 0.1% of the reading, estimate the uncertainty in pressure at the design stage. How does this change at 100 and 1,000 kPa

the design stage uncertainty for pressure is   (√[ 25 + 0.10⁻⁴ P² ]) / 10 kPa

the change in sensitivity is  0.01 mV/kPa

Explanation:

Given that;

Differential pressure in the transducer ΔP = 10 kPa

Accuracy in the reading of diaphragm pressure transducer, α < 0.1%.

Resolution of the voltmeter, Rv = 10 mV  

Now we write the expression for the design stage uncertainty

Us = √[(Us)₁² + (Us)₂²]  equ 1

nex we write the expression for design stage uncertainty for the voltmeter having resolution 10 mV .  

(Us)₁² = ± 1/2Rv

we substitute the values

(Us)₁² = ± 1/2(10 mV) = ±5 mV

we write the expression for design stage uncertainty with respect to strain gauge.

(Us)₂ = aKtΔP

Kt is the strain gauge constant, Rs is the resolution for the voltmeter when accuracy in strain gauge is less than 0.1% .  

Substitute the values.  

(Us)₂ = (0.1%) (P mV / kPa) 10 kpa

= 0.01P mV

Here, P is a constant.  

Substitute the values in equation 1

Us = √[(±5 mV)² + (0.01P mV)²]  

= √[ 25 + 0.10⁻⁴ P² ]

Write the expression for the design stage uncertainty for pressure

Up = Us / ΔP

Substitute the values.  

Up =   (√[ 25 + 0.10⁻⁴ P² ]) / 10 kPa

therefore the design stage uncertainty for pressure is   (√[ 25 + 0.10⁻⁴ P² ]) / 10 kPa

Write the value of the design stage uncertainty for pressure, ΔP is 100 kPa .

(Up)₁ =   (√[ 25 + 0.10⁻⁴ P² ]) / 100 kPa

Write the value of the design stage uncertainty for pressure ΔP is 1000 kPa

(Up)₂ =   (√[ 25 + 0.10⁻⁴ P² ]) / 1000 kPa

Write the expression for the sensitivity of the pressure transducer.

s = Rv / 1000 kPa  

Substitute the values

s = 10 mV / 1000 kPa

= 0.01 mV/kPa  

therefore the change in sensitivity is  0.01 mV/kPa