Question 5.8: Velocity Measurement by a Pitot Tube A piezometer and a Pito...
Velocity Measurement by a Pitot Tube
A piezometer and a Pitot tube are tapped into a horizontal water pipe, as shown in Fig. 5–42, to measure static and stagnation (static + dynamic) pressures. For the indicated water column heights, determine the velocity at the center of the pipe.
The static and stagnation pressures in a horizontal pipe are measured. The velocity at the center of the pipe is to be determined.
Assumptions 1 The flow is steady and incompressible. 2 Points 1 and 2 are close enough together that the irreversible energy loss between these two points is negligible, and thus we can use the Bernoulli equation.
Analysis We take points 1 and 2 along the streamline at the centerline of the pipe, with point 1 directly under the piezometer and point 2 at the tip of the Pitot tube. This is a steady flow with straight and parallel streamlines, and the gage pressures at points 1 and 2 can be expressed as
P_1 = ρg(h_1 + h_2)
P_2 = ρg(h_1 + h_2 + h_3)
Noting that z_1 = z_2, and point 2 is a stagnation point and thus V_2 = 0, the application of the Bernoulli equation between points 1 and 2 gives
\frac{P_1}{\rho g} + \frac{V^2_1}{2g} + \cancel{z_1} = \frac{P_2}{\rho g} +\overset{0}{\frac{\cancel{V^2_2}}{2g} } + \cancel{z_2} → \frac{V^2_1}{2g} = \frac{P_2 – P_1}{\rho g}
Substituting the P_1 and P_2 expressions gives
\frac{V^2_1}{2g} = \frac{P_2 – P_1}{\rho g} = \frac{\rho g(h_1 + h_2 + h_3 ) – \rho g(h_1 + h_2)}{\rho g} = h_3
Solving for V_1 and substituting,
V_1 = \sqrt{2gh_3} = \sqrt{2(9.81 m/s^2)(0.12 m)} = 1.53 m/s
Discussion Note that to determine the flow velocity, all we need is to measure the height of the excess fluid column in the Pitot tube compared to that in the piezometer tube.