Question

Water "bubbles up" h2 = 9 in. above the exit of the vertical pipe attached to three horizontal pipe segments. The total length of the 1.75-in.- diameter galvanized iron pipe between point (1) and the exit is 21 inches. Determine the pressure needed at point (1) to produce this flow.
Assume h3 = 18 in.

Answer 1

The pressure needed at point (1) to produce the flow where water "bubbles up above the exit of the vertical pipe attached to three horizontal pipe segments is 0.750 psi.

What is Bernoulli's equation?

The Bernoulli's equation for incompressible fluid can be given as,

$P_1+\dfrac{1}{2}\rho v_1^2+\rho gh_1=P_2+\dfrac{1}{2}\rho v_2^2+\rho gh_2$

Here, ρ is density of fluid, g is acceleration due to gravity, (P) is the pressure v is velocity, h is height of elevation and subscript (1 and 2) is used for point 1 and 2.

Water "bubbles up" h2 = 9 in.  Above the exit of the vertical pipe attached to three horizontal pipe segments.

The total length of the 1.75-in.- diameter galvanized iron pipe between point (1) and the exit is 21 inches. h3 = 18 in.

Reynolds number is,

$R_e=\dfrac{V_3D}{v}\\R_e=\dfrac{4.01\dfrac{0.75}{12}}{1.21\times10^{-15}}\\R_e=2.07\times10^4$

The formula for the ratio of pressure to radius, when pressure and velocity at point 2 is zero can be given as,

$\dfrac{p_1}{r}=z_2\left(f\dfrac{l}{d}+\sum k_L\right)\dfrac{v^2}{2g}-\dfrac{v_1^2}{2g}$

Here, f=0.039 and ∑K(L)=4.5. Put the values,

$\dfrac{p_1}{r}=\dfrac{7}{12}\left(0.03\dfrac{21}{0.75}+4.5-1\right)\dfrac{4.01^2}{2\times32.2}\\\dfrac{p_1}{r}=1.73\rm\; ft$

Put the value of r we get,

$p_1=62.4\times1.73\\p_1=108\rm\; ib/ft^2\\p_1=0.750\rm\; psi$

The pressure needed at point (1) to produce the flow where water "bubbles up above the exit of the vertical pipe attached to three horizontal pipe segments is 0.750 psi.

Learn more about the Bernoulli's equation here;

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