Bernoulli's Equation — Question 3

Question 3

A Pitot tube measures stagnation pressure rather than static pressure

A Pitot-static probe in a steady air stream connects to a water manometer. The water levels differ by $h=25\,\mathrm{mm}$. The air-filled pressure lines connect at equal elevation. Take $\rho_w=1000\,\mathrm{kg/m^3}$, $\rho_a=1.25\,\mathrm{kg/m^3}$ and $g=10\,\mathrm{m/s^2}$. Treat air as incompressible at the inferred speed and neglect probe losses.

1. Find the stagnation-minus-static pressure and the stream speed, retaining the air-density correction in the manometer.

2. If the static absolute pressure is $98.0\,\mathrm{kPa}$, find stagnation pressure. Why would using the water density in the Bernoulli speed formula be wrong?

Question 3 - Solution

A Pitot tube measures stagnation pressure rather than static pressure

(a) Derivation and calculation

$$\boxed{p_t-p_s=(\rho_w-\rho_a)gh=(1000-1.25)(10)(.0250)=249.6875\,\mathrm{Pa}}.$$$$p_s+\tfrac12\rho_av^2=p_t\Rightarrow\boxed{v=\sqrt{\frac{2(249.6875)}{1.25}}=19.9875\,\mathrm{m/s}}.$$

(b) Result and interpretation

$$\boxed{p_{t,\mathrm{abs}}=98.0+.2496875=98.2496875\,\mathrm{kPa}}.$$The stream kinetic energy uses air density, while the manometer head chiefly uses water density. Substituting water in the speed formula would give$$v_{\mathrm{wrong}}=\sqrt{2(249.6875)/1000}=.706665\,\mathrm{m/s}.$$

Physical check

The inferred air speed is consistent with the stipulated low-speed incompressible approximation.