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](https://tex.z-dn.net/?f=%5Cfrac%7BP_1%7D%7B%5Crho%20g%7D%2B%5Cfrac%7BV_1%5E2%7D%7B2g%7D%2Bz_1%20%3D%20%5Cfrac%7BP_2%7D%7B%5Crho%20g%7D%2B%5Cfrac%7BV_2%5E2%7D%7B2g%7D%2Bz_2)
Where,
Pressure at determinated point
density (water in this case)
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)](https://tex.z-dn.net/?f=P%20%3D%20%5Crho%20g%20%28h%2BR%29)
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)](https://tex.z-dn.net/?f=P_1%20%3D%20%5Crho%20g%20%28h_%7BPiezometer%7D%2BR%29)
![P_2 = \rho g (h_{Pitot}+R)](https://tex.z-dn.net/?f=P_2%20%3D%20%5Crho%20g%20%28h_%7BPitot%7D%2BR%29)
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](https://tex.z-dn.net/?f=%5Cfrac%7B%5Crho%20g%28h_%7BPiezometer%7D%2BR%29%7D%7B%5Crho%20g%7D%2B%5Cfrac%7BV_1%5E2%7D%7B2g%7D-%5Cfrac%7B%5Crho%20g%28h_%7BPitot%7D%2BR%29%7D%7B%5Crho%20g%7D-%5Cfrac%7BV_2%5E2%7D%7B2g%7D%3D0)
![\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}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Crho%20g%28h_%7BPiezometer%7D%2BR%29%7D%7B%5Crho%20g%7D%2B%5Cfrac%7BV_1%5E2%7D%7B2g%7D%3D%5Cfrac%7B%5Crho%20g%28h_%7BPitot%7D%2BR%29%7D%7B%5Crho%20g%7D%2B%5Cfrac%7BV_2%5E2%7D%7B2g%7D)
![(h_{piezometer}+R)+\frac{V_1^2}{2g}=(h_{pitot}+R)+\frac{V_2^2}{2g}](https://tex.z-dn.net/?f=%28h_%7Bpiezometer%7D%2BR%29%2B%5Cfrac%7BV_1%5E2%7D%7B2g%7D%3D%28h_%7Bpitot%7D%2BR%29%2B%5Cfrac%7BV_2%5E2%7D%7B2g%7D)
![h_{piezometer}+\frac{V_1^2}{2g}=h_{pitot}\frac{V_2^2}{2g}](https://tex.z-dn.net/?f=h_%7Bpiezometer%7D%2B%5Cfrac%7BV_1%5E2%7D%7B2g%7D%3Dh_%7Bpitot%7D%5Cfrac%7BV_2%5E2%7D%7B2g%7D)
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}](https://tex.z-dn.net/?f=h_%7Bpiezometer%7D%2B%5Cfrac%7BV_1%5E2%7D%7B2g%7D%3Dh_%7Bpitot%7D%2B%5Cfrac%7B0%7D%7B2g%7D)
![h_{pitot}-h_{piezometer}=\frac{V_1^2}{2g}](https://tex.z-dn.net/?f=h_%7Bpitot%7D-h_%7Bpiezometer%7D%3D%5Cfrac%7BV_1%5E2%7D%7B2g%7D)
Re-arrange for
=
![V_1 =\sqrt{2g(h_{pitot}-h_{piezometer})}](https://tex.z-dn.net/?f=V_1%20%3D%5Csqrt%7B2g%28h_%7Bpitot%7D-h_%7Bpiezometer%7D%29%7D)
Replacing the values where
and
we have,
![V_1 = \sqrt{2*9.8(0.32-0.26)}](https://tex.z-dn.net/?f=V_1%20%3D%20%5Csqrt%7B2%2A9.8%280.32-0.26%29%7D)
![V_1 = 1.08m/s](https://tex.z-dn.net/?f=V_1%20%3D%201.08m%2Fs)
Therefore the velocity at the centerline is 1.08m/s