Respuesta :
Answer:
a) [tex]-7.4\frac{lb}{ft^3}[/tex]
b) [tex]-69.8\frac{lb}{ft^3}[/tex]
c) [tex]55 \frac{lb}{ft^3}[/tex]
Explanation:
1) Notation
[tex]\tau=1.85\frac{lb}{ft^2}[/tex] represent the shear stress defined as "the external force acting on an object or surface parallel to the slope or plane in which it lies"
R represent the radial distance
L the longitude
[tex]\theta=0\degree[/tex] since at the begin we have a horizontal pipe, but for parts b and c the angle would change.
D represent the diameter for the pipe
[tex]\gamma=62.4\frac{lb}{ft^3}[/tex] is the specific weight for the water
2) Part a
For this case we can use the shear stress and the radial distance to find the pressure difference per unit of lenght, with the following formula
[tex]\frac{2\tau}{r}=\frac{\Delta p -\gamma Lsin\theta}{L}[/tex]
[tex]\frac{2\tau}{r}=\frac{\Delta p}{L}-\gamma sin\theta[/tex]
If we convert the difference's into differentials we have this:
[tex]-\frac{dp}{dx}=\frac{2\tau}{r}+\gamma sin\theta[/tex]
We can replace [tex]r=\frac{D}{2}[/tex] and we have this:
[tex]\frac{dp}{dx}=-[\frac{4\tau}{D}+\gamma sin\theta][/tex]
Replacing the values given we have:
[tex]\frac{dp}{dx}=-[\frac{4x1.85\frac{lb}{ft^2}}{1ft}+62.4\frac{lb}{ft^3} sin0]=-7.4\frac{lb}{ft^3}[/tex]
3) Part b
When the pipe is on vertical upward position the new angle would be [tex]\theta=\pi/2[/tex], and replacing into the formula we got this:
[tex]\frac{dp}{dx}=-[\frac{4x1.85\frac{lb}{ft^2}}{1ft}+62.4\frac{lb}{ft^3} sin90]=-69.8\frac{lb}{ft^3}[/tex]
4) Part c
When the pipe is on vertical downward position the new angle would be [tex]\theta=-\pi/2[/tex], and replacing into the formula we got this:
[tex]\frac{dp}{dx}=-[\frac{4x1.85\frac{lb}{ft^2}}{1ft}+62.4\frac{lb}{ft^3} sin(-90)]=55 \frac{lb}{ft^3}[/tex]
