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
[tex]\frac{P_1}{\rho g}+\frac{V_1^2}{2g}+z_1 = \frac{P_2}{\rho g}+\frac{V_2^2}{2g}+z_2[/tex]
Where,
[tex]P_i =[/tex] Pressure at determinated point
[tex]\rho =[/tex]density (water in this case)
[tex]V_i =[/tex] 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,
[tex]P = \rho g (h+R)[/tex]
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:
[tex]P_1 = \rho g (h_{Piezometer}+R)[/tex]
[tex]P_2 = \rho g (h_{Pitot}+R)[/tex]
Then replacing
[tex]\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[/tex]
[tex]\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}[/tex]
[tex](h_{piezometer}+R)+\frac{V_1^2}{2g}=(h_{pitot}+R)+\frac{V_2^2}{2g}[/tex]
[tex]h_{piezometer}+\frac{V_1^2}{2g}=h_{pitot}\frac{V_2^2}{2g}[/tex]
At the end of the pipe the speed is zero, since there is stagnation then:
[tex]h_{piezometer}+\frac{V_1^2}{2g}=h_{pitot}+\frac{0}{2g}[/tex]
[tex]h_{pitot}-h_{piezometer}=\frac{V_1^2}{2g}[/tex]
Re-arrange for [tex]V_1[/tex] =
[tex]V_1 =\sqrt{2g(h_{pitot}-h_{piezometer})}[/tex]
Replacing the values where [tex]h_{pitot}= 32*10^{-2}m[/tex] and [tex]h_{piezometer}=26*10^{-2}m[/tex] we have,
[tex]V_1 = \sqrt{2*9.8(0.32-0.26)}[/tex]
[tex]V_1 = 1.08m/s[/tex]
Therefore the velocity at the centerline is 1.08m/s
