What Volume Flow Rate Maintains Equal Pressure in Two Connected Pipes?

[biophys]

Please help me with the following:
"A liquid is flowing through a horizontal pipe whose radius is 0.0200 m. The pipe bends straight upward through a height of 8.2 m and joins another horizontal pipe whose radius is 0.0410 m. What volume flow rate will keep the pressures in the two horizontal pipes the same?"



Here is my reasoning, that doesn't seem to be working out.
- first of all, the problem just says "liquid" so I know that I can't possibly need to know specific density.
-The equation for volume flow rate [ pi*R^4*(P2-P1) ] / (8*viscosity*L)
doesn't really make sense for this problem, how would I find the vidcosity and wouldn't (P2-P1) be zero anyway?

?? Thanks

 

[jamesrc]

Use the Bernoulli equation (you'll have to assume steady, frictionless, incompressible flow along streamlines):

$$\frac p \rho + \frac {v^2} 2 + gz = {\rm constant}$$

You want to solve for the volumetric flow rate (which I'll call Q) which, by mass continuity and our assumptions above can be shown to be:

$$Q = v_1A_1 = v_2A_2$$

(subscripts denote first position and second position; $$A_i = \pi r_i^2$$)

So:

$$\frac{p_1}{\rho_1} + \frac{Q^2}{2A_1^2} + gz_1 = \frac{p_2}{\rho_2} + \frac{Q^2}{2A_2^2} + gz_2$$

since we're solving for the case where p₁ = p₂ (and ρ₁ = ρ₂ because it's the same liquid), the first term on the left hand side of the equation cancels with the first term on the RHS. Solve the equation for Q and plug in the known quantities (z is the height of the fluid in the equations).

 

[Clausius2]

Use the Bernoulli equation (you'll have to assume steady, frictionless, incompressible flow along streamlines):

$$\frac p \rho + \frac {v^2} 2 + gz = {\rm constant}$$

You want to solve for the volumetric flow rate (which I'll call Q) which, by mass continuity and our assumptions above can be shown to be:

$$Q = v_1A_1 = v_2A_2$$

(subscripts denote first position and second position; $$A_i = \pi r_i^2$$)

So:

$$\frac{p_1}{\rho_1} + \frac{Q^2}{2A_1^2} + gz_1 = \frac{p_2}<br /> {\rho_2} + \frac{Q^2}{2A_2^2} + gz_2$$

since we're solving for the case where p₁ = p₂ (and ρ₁ = ρ₂ because it's the same liquid), the first term on the left hand side of the equation cancels with the first term on the RHS. Solve the equation for Q and plug in the known quantities (z is the height of the fluid in the equations).

Yes, I'm with jamesrc. It has no sense thinking of how to solve this problem using viscous flow, because in it a loss of pressure is assured notwithsanding what volume flux is used.