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A pipe carrying oil of specific gravity 0.877 changes in size from 150 mm at section 1 and 450 mm section 2. Section 1 is below section 2 and the pressures are 90 kPa and 60 kPa respectively. If the discharge is 150 L/s, determine the elevation from section 1 to section 2.

Sagot :

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.

View image MathPhys
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