Question

A piezometer and a Pitot tube are tapped into a 4-cm-diameter horizontal water pipe, and the height of the water columns are measured to be 26 cm in the piezometer and 32 cm in the Pitot tube (both measured from the top surface of the pipe). Determine the velocity at the center of the pipe.

Answer 1

Answer:

1.08m/s

Explanation:

To develop this problem it is necessary to resort to the concept developed by Bernoulli in his equations in which

describes the behavior of a fluid along a channel.

Bernoulli's equation is given by

$\frac{P_1}{\rho g}+\frac{V_1^2}{2g}+z_1 = \frac{P_2}{\rho g}+\frac{V_2^2}{2g}+z_2$

Where,

$P_i =$ Pressure at determinated point

$\rho =$density (water in this case)

$V_i =$ Velocity at determinated point

g = Gravity acceleration

z = Pressure heads at determinated point.

We know that the problem is given in an horizontal line, then the pressure heads is zero.

And for definition we know that,

$P = \rho g (h+R)$

Where h means the heights of water column measured by the pitot tube at the top and the piezometer. Then replacing both pressure with the previous values we have:

$P_1 = \rho g (h_{Piezometer}+R)$

$P_2 = \rho g (h_{Pitot}+R)$

Then replacing

$\frac{\rho g(h_{Piezometer}+R)}{\rho g}+\frac{V_1^2}{2g}-\frac{\rho g(h_{Pitot}+R)}{\rho g}-\frac{V_2^2}{2g}=0$

$\frac{\rho g(h_{Piezometer}+R)}{\rho g}+\frac{V_1^2}{2g}=\frac{\rho g(h_{Pitot}+R)}{\rho g}+\frac{V_2^2}{2g}$

$(h_{piezometer}+R)+\frac{V_1^2}{2g}=(h_{pitot}+R)+\frac{V_2^2}{2g}$

$h_{piezometer}+\frac{V_1^2}{2g}=h_{pitot}\frac{V_2^2}{2g}$

At the end of the pipe the speed is zero, since there is stagnation then:

$h_{piezometer}+\frac{V_1^2}{2g}=h_{pitot}+\frac{0}{2g}$

$h_{pitot}-h_{piezometer}=\frac{V_1^2}{2g}$

Re-arrange for $V_1$ =

$V_1 =\sqrt{2g(h_{pitot}-h_{piezometer})}$

Replacing the values where $h_{pitot}= 32*10^{-2}m$ and $h_{piezometer}=26*10^{-2}m$ we have,

$V_1 = \sqrt{2*9.8(0.32-0.26)}$

$V_1 = 1.08m/s$

Therefore the velocity at the centerline is 1.08m/s