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Engineering
Mechanical Engineering
Multi-Suction Point Pipe Flow
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[QUOTE="ATKrank, post: 5489062, member: 595099"] By "assume a pump suction pressure", are you referencing what would be known as the NPSHR for a specific pump? I think this concept of suction pressure is where I am stuck. Below I am showing two examples: [ATTACH=full]101530[/ATTACH] To evaluate example 1: Let's assume its a 6" pipe and the medium is water. By continuity, my flow rate in is equal to my flow rate out and the velocity can be determined by Q=V*A. So my velocity is 20 gal/min * .13368ft[SUP]3[/SUP]/gal / .19635 ft[SUP]2[/SUP] = 13.6165 ft/min. Got it, that's easy. Now to find my pressure at point 2 I need to use Bernoulli's and Darcy-Weisbach equations. So the equation should look like this: p[SUB]1[/SUB]+1/2*rho*v[SUB]1[/SUB][SUP]2[/SUP]+rho*g*h[SUB]1[/SUB]=p[SUB]2[/SUB]+1/2*rho*v[SUB]2[/SUB][SUP]2[/SUP]+rho*g*h[SUB]2[/SUB]+h[SUB]f[/SUB] but v[SUB]1[/SUB]=0 and h[SUB]1[/SUB]=0 So it simplifies to: p[SUB]1[/SUB]=p[SUB]2[/SUB]+1/2*rho*v[SUB]2[/SUB][SUP]2[/SUP]+rho*g*h[SUB]2[/SUB]+h[SUB]f[/SUB] But h[SUB]f[/SUB] is a function of velocity, pipe length, pipe diameter, and a friction factor (estimated by the moody chart or by the Colebrook-White equation which is a function of the Reynolds number, relative roughness, pipe diameter and the Reynolds number is a function of velocity, Pipe length and velocity) which looks like: h[SUB]f[/SUB]=f(L/D)*(v[SUP]2[/SUP]/2g) substituting in: p[SUB]1[/SUB]=p[SUB]2[/SUB]+1/2*rho*v[SUB]2[/SUB][SUP]2[/SUP]+rho*g*h[SUB]2[/SUB]+f(L/D)*(v[SUP]2[/SUP]/2g) looks like we have everything except for friction factor f, to use the moody diagram we need to Reynolds number Re=(rho*V*d)/(mu) So we would now be able to calculate p[SUB]2[/SUB] as long as it is in turbulent flow. Which after calculating the Reynolds number I realized I chose poor numbers but just substitute the laminar head loss equation instead of the turbulent one and we are good to go. So now if I want to evaluate example 2: let's say our header from points 4-2 is 6" and branches 1-2 and 3-4 are both 3". Now the velocity at point 2 is the same as example 1 since the flow rate and pipe size are still the same. But this is where I get stuck and seems like I just keep going back and forth that I need to calculate the flow rate to get the velocity through each branch. But to calculate the velocity, I need the pressure drop through the leg. But to get the pressure drop I need the velocity. Can you please explain how to evaluate this in a similar manner as I did for example 1? That is the easiest way I can understand these concepts and I think most people feel the same. Thanks! [/QUOTE]
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Multi-Suction Point Pipe Flow
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