Answer:
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
