Question

Determine the resolution of a manometer required to measure the velocity of air at 50 m/s using a pitot-static tube and a manometer fluid of mercury (density: 13,600 kg/m3) to achieve uncertainty of 5% (i.e., 2.5 m/s) and 1 % (0.5 m/s).

Answer 1

Answer:

a)  Δh = 2 cm,  b) Δh = 0.4 cm

Explanation:

Let's start by using Bernoulli's equation for the Pitot tube, we define two points 1 for the small entry point and point 2 for the larger diameter entry point.

            P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

Point 1 is called the stagnation point where the fluid velocity is reduced to zero (v₁ = 0), in general pitot tubes are used  in such a way that the height of point 2 of is the same of point 1

           y₁ = y₂

subtitute

           P₁ = P₂ + ½ ρ v₂²

           P₁ -P₂ = ½ ρ v²

where ρ is the density of fluid  

now we measure the pressure on the included beforehand as a pair of communicating tubes filled with mercury, we set our reference system at the point of the mercury bottom surface

           ΔP =ρ_{Hg} g h - ρ g h

           ΔP =  (ρ_{Hg} - ρ) g h

as the static pressure we can equalize the equations

          ΔP = P₁ - P₂

         (ρ_{Hg} - ρ) g h = ½ ρ v²

         v = $\sqrt{\frac{2 (\rho_{Hg} - \rho) g}{\rho } } \ \sqrt{h}$

in this expression the densities are constant

        v = A  √h

       A =$\sqrt{\frac{2(\rho_{Hg} - \rho ) g}{\rho } }$

 

They indicate the density of mercury rhohg = 13600 kg / m³, the density of dry air at 20ºC is rho air = 1.29 kg/m³

we look for the constant

        A = $\sqrt{\frac{2( 13600 - 1.29) \ 9.8}{1.29} }$

        A = 454.55

we substitute

       v = 454.55 √h

to calculate the uncertainty or error of the velocity

         h = $\frac{1}{454.55^2} \ v^2$

       Δh = $\frac{dh}{dv}$   Δv

       $\frac{\Delta h}{h } = 2 \ \frac{\Delta v}{v}$

Suppose we have a height reading of h = 20 cm = 0.20 m

             

a) uncertainty 2.5 m / s ( 0.05)

        $\frac{\delta v}{v} = 0.05$

       $\frac{\Delta h}{h}$ = 2 0.05  

       Δh = 0.1 h

       Δh = 0.1  20 cm

       Δh = 2 cm

b) uncertainty 0.5 m / s ( Δv/v= 0.01)

        $\frac{\Delta h}{h}$ =  2 0.01

        Δh = 0.02 h

        Δh = 0.02 20

        Δh = 0.1 20 cm

        Δh = 0.4 cm = 4 mm