[ \frac\partial V\partial \rho = C \cdot \frac12 \left( \frac2 \Delta P\rho \right)^-1/2 \cdot \left( -\frac2 \Delta P\rho^2 \right) ] Simplify: ( \frac\partial V\partial \rho = -C \sqrt\frac\Delta P2 \rho^3 ) Plug numbers: ( \sqrt\frac2502 \times (1.20)^3 = \sqrt\frac2502 \times 1.728 = \sqrt\frac2503.456 = \sqrt72.338 \approx 8.505 ) Then ( \frac\partial V\partial \rho \approx -0.99 \times 8.505 \approx -8.420 , \textm/s per kg/m^3 ) Given ( u_C = 0.01 ), ( u_\Delta P = 5 ), ( u_\rho = 0.01 )
A pitot-static tube is used to measure air velocity in a duct. The measured differential pressure is ( \Delta P = 250 \pm 5 , \textPa ), air density ( \rho = 1.20 \pm 0.01 , \textkg/m^3 ), and the calibration coefficient ( C = 0.99 \pm 0.01 ). Find the velocity and its overall uncertainty at 95% confidence. Step 1 – Known equation For a pitot-static tube, [ V = C \sqrt\frac2 \Delta P\rho ] Step 2 – Nominal value [ V_\textnom = 0.99 \sqrt\frac2 \times 2501.20 = 0.99 \times \sqrt416.6667 ] [ \sqrt416.6667 \approx 20.412 ] [ V_\textnom \approx 0.99 \times 20.412 \approx 20.208 , \textm/s ] Step 3 – Uncertainty propagation (fixed systematic uncertainties) Use the root-sum-square method with sensitivity coefficients.
[ \frac\partial V\partial C = \sqrt\frac2 \Delta P\rho = 20.412 ] [ \frac\partial V\partial \Delta P = C \cdot \frac1\sqrt2 \Delta P / \rho \cdot \frac1\rho = \fracC\sqrt2 \rho \Delta P = \frac0.99\sqrt2 \times 1.20 \times 250 ] Better: ( \frac\partial V\partial \Delta P = \fracC\sqrt2 \rho \Delta P ) Check: ( \sqrt2 \rho \Delta P = \sqrt2 \times 1.20 \times 250 = \sqrt600 = 24.4949 ) So ( \frac\partial V\partial \Delta P = 0.99 / 24.4949 \approx 0.04041 , \textm/s per Pa )
It sounds like you’re looking for the for Theory and Design for Mechanical Measurements , 7th Edition, by Richard S. Figliola and Donald E. Beasley.
[ \frac\partial V\partial \rho = C \cdot \frac12 \left( \frac2 \Delta P\rho \right)^-1/2 \cdot \left( -\frac2 \Delta P\rho^2 \right) ] Simplify: ( \frac\partial V\partial \rho = -C \sqrt\frac\Delta P2 \rho^3 ) Plug numbers: ( \sqrt\frac2502 \times (1.20)^3 = \sqrt\frac2502 \times 1.728 = \sqrt\frac2503.456 = \sqrt72.338 \approx 8.505 ) Then ( \frac\partial V\partial \rho \approx -0.99 \times 8.505 \approx -8.420 , \textm/s per kg/m^3 ) Given ( u_C = 0.01 ), ( u_\Delta P = 5 ), ( u_\rho = 0.01 )
A pitot-static tube is used to measure air velocity in a duct. The measured differential pressure is ( \Delta P = 250 \pm 5 , \textPa ), air density ( \rho = 1.20 \pm 0.01 , \textkg/m^3 ), and the calibration coefficient ( C = 0.99 \pm 0.01 ). Find the velocity and its overall uncertainty at 95% confidence. Step 1 – Known equation For a pitot-static tube, [ V = C \sqrt\frac2 \Delta P\rho ] Step 2 – Nominal value [ V_\textnom = 0.99 \sqrt\frac2 \times 2501.20 = 0.99 \times \sqrt416.6667 ] [ \sqrt416.6667 \approx 20.412 ] [ V_\textnom \approx 0.99 \times 20.412 \approx 20.208 , \textm/s ] Step 3 – Uncertainty propagation (fixed systematic uncertainties) Use the root-sum-square method with sensitivity coefficients.
[ \frac\partial V\partial C = \sqrt\frac2 \Delta P\rho = 20.412 ] [ \frac\partial V\partial \Delta P = C \cdot \frac1\sqrt2 \Delta P / \rho \cdot \frac1\rho = \fracC\sqrt2 \rho \Delta P = \frac0.99\sqrt2 \times 1.20 \times 250 ] Better: ( \frac\partial V\partial \Delta P = \fracC\sqrt2 \rho \Delta P ) Check: ( \sqrt2 \rho \Delta P = \sqrt2 \times 1.20 \times 250 = \sqrt600 = 24.4949 ) So ( \frac\partial V\partial \Delta P = 0.99 / 24.4949 \approx 0.04041 , \textm/s per Pa )
It sounds like you’re looking for the for Theory and Design for Mechanical Measurements , 7th Edition, by Richard S. Figliola and Donald E. Beasley.
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