Answer:
7.12 m
Explanation:
Bernoulli's principle can be used to relate the pressure (P), speed (v), and elevation (h) of an incompressible fluid with constant density (ρ) at different points.
[tex]\Large \text {$ P_1+\frac{1}{2}\rho {v_1}^2+\rho gh_1 = P_2+\frac{1}{2}\rho {v_2}^2+\rho gh_2 $}[/tex]
The speed of the fluid is equal to the volumetric flow rate (V) divided by the cross sectional area (A). For a round pipe, the cross sectional area is one fourth pi times the square of the diameter (d).
[tex]\Large \text {$ v= $}\huge \text {$ \frac{V}{A} $}\\\\\Large \text {$ v= $}\huge \text {$ \frac{V}{\frac{1}{4} \pi d^2} $}\\\\\Large \text {$ v= $}\huge \text {$ \frac{4V}{\pi d^2} $}[/tex]
Substituting into the Bernoulli equation and solving for the elevation change, we get:
[tex]\Large \text {$ h_2-h_1= $}\huge \text {$ \frac{P_1-P_2}{\rho g} $}\Large \text {$ + $}\huge \text {$ \frac{8V^2}{\pi^2 g}(\frac{1}{{d_1}^4}-\frac{1}{{d_2}^4} ) $}[/tex]
Converting all units to SI and plugging in, we find that the elevation change is approximately 7.12 meters.
