Calculating Flow Rate in Horizontal & Vertical Pipes

AI Thread Summary
To determine the volume flow rate that equalizes pressure in two horizontal pipes connected by a vertical section, the Bernoulli equation is applied. The equation considers the pressures, velocities, and heights at both ends, simplifying to p1 + 0.5ρv1² + ρgy1 = p2 + 0.5ρv2² + ρgy2. By setting y1 to 0, y2 to 10m, and assuming incompressible flow with v1 = v2/4, the pressures and densities cancel out, leaving a solvable equation with one unknown. This approach effectively calculates the required flow rate for pressure equilibrium. Understanding these principles is crucial for fluid dynamics in engineering applications.
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Homework Statement


A liquid is flowing through a horizontal pipe whose radius is 0.02m. The pipe bends staight upward through a height of 10.0m and joins another horizontal pipe whose radius is 0.04m.


Homework Equations


what volume flow rate will keep the pressures in the two horinontal pipes the same?

The Attempt at a Solution


i tried to use p1+1/2xpxv1^2+pgh1=p2+1/2xpxv2^2+pgh2..
any idea?
 
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raywang5 said:

Homework Statement


A liquid is flowing through a horizontal pipe whose radius is 0.02m. The pipe bends staight upward through a height of 10.0m and joins another horizontal pipe whose radius is 0.04m.


Homework Equations


what volume flow rate will keep the pressures in the two horinontal pipes the same?

The Attempt at a Solution


i tried to use p1+1/2xpxv1^2+pgh1=p2+1/2xpxv2^2+pgh2..
any idea?


I think you are dealing with the same formula I found in a Physics text:

p_1 + rho*(v_1)^2/2 +rho*g*y_1 =
p_2 + rho*(v_2)^2/2 +rho*g*y_2

Set y_1 = 0, y_2 = 10m, set p_1 = p_2 = p, and we know v_1 = v_2/4 (assuming incompressible flow)

and we are left with one equation and one unknown. Note the p's cancel along with the rho's.
 
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