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#55
Jun611, 10:55 PM

P: 688

If there was a pressure multiplying effect with parallel reservoirs, we could create some pretty cool pressure and flow control loops. Imagine a ring array of reservoirs and actuated valves: if you need more or less flow, just open or close valves. If this were true, there would probably be products like this on the market. 


#56
Jun611, 11:39 PM

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#57
Jun711, 12:14 AM

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#58
Jun711, 01:29 AM

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Regards, Harald 


#59
Jun711, 05:43 AM

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harrylin,
Except in the case of very fast flow or very small tubes, the friction losses would be very small. Regardless, if we approach this from the eyes of a mechanical engineer, we could easily introduce a head loss term to fairly accurately account for the effect of viscous losses. You could theoretically do the same for the loss associated with the Tjoint. The important thing is that neither of those additional loss terms will depend on the number of reservoirs or amount of fluid in each reservoir. 


#60
Jun711, 07:29 AM

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#61
Jun711, 08:17 AM

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#62
Jun711, 08:32 AM

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#63
Jun711, 10:12 AM

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The important takeaway here is that you use a Poiseuille flow solution to get the velocity profile, which you don't need. The volumetric flow rate that is quoted on Wikipedia is just one of the takeaways from integrating the velocity profile across a plane of interest. [tex]\boldsymbol{\tau} = \nabla p \;\mathbb{I} + \mu \nabla^2 \vec{v}[/tex] The pressure term shows up as a gradient, and has nothing to do with the velocity gradient and everything to do with externally enforced pressure gradients or hydrostatic pressure. It has no explicit dependence on velocity. What I mean is this: imagine you have a garden hose and you are dripping water through it with a turkey baster. That is essentially what you are doing with the stopcock. Your mass flow rate is dictated by the stopcock, and for the same conditions, the stopcock admits less flow than the tubes would. Your stopcock is throttling your flow, in essence. Based on your results, I infer that even the combined mass flow of both stopcocks is likely not as much as the tube could admit under the given flow conditions. In all likelihood, that means there is a bit of unused space in the tubes when the water is draining from your funnels, or in other words, the entire cross section is not filled with water. You are only getting as much water through that system as the stopcocks will allow, which is apparently less than what the tubes would allow. If you did take [itex]\Delta h[/itex] to be time varying, you would notice that with two bags, the fluid level falls more slowly, so the pressure would stay marginally higher, but since we are only talking a matter of inches here, it still won't speed the bags up enough to pass the single bag, especially because even if the two bags did drain faster, they would catch up to the height of the single bag and end up draining at half the rate again. In other words, you would expect that in practice, neglecting the effect of the junction downstream, the two bags would probably drain in something like 1.75 times as long as the single bag rather than a perfect 2. Finding the actual value is nontrivial and would need to be numerically integrated, so I don't really plan to do that since I ought to be finishing this paper that needs to be submitted to AIAA by next week. 


#64
Jun711, 01:57 PM

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Bernoulli's law is simply conservation of mechanical energy. Let's compare two control volumes, one at the top of the reservoir (say, a large diameter thin cylinder) and one within the needle, terminating at the outlet (a long slender cylinder). In some time dt the volume dV exits the needle. Each control volume dV is the same (since conservation of mass holds and we assume an incompressible fluid) and has a mass dm. Conservation of mechanical energy gives: P + 1/2 dm v^2 + dm gy = constant. replacing dm = [itex]\rho \Delta V[/itex] and v = dl/dt = 1/A dV/dt = Q/A (since dV = A dl, and Q = dV/dt), we get for the upper control volume: P[itex]_{atm}[/itex] + [itex]\frac{1}{2} \rho\Delta V\frac{Q^{2}}{A^{2}_{1}}+\rho\Delta Vgh[/itex], where h is the height above the needle outlet, and we assumed that l_1 was so thin that the entire control volume is held at atmospheric pressure. Writing the expression for the lower volume is not so simple, as we will see since we want A_1 >> A_2 (the diameter of the reserviour is much larger than the needle), the corresponding height l_2 >> l_1, and so we must be careful about assigning a pressure and potential energy. For now, let's do the obvious linearization: [itex]P_{atm} +\delta P + \frac{1}{2} \rho\Delta V\frac{Q^{2}}{A^{2}_{2}}+\rho\Delta Vg \frac{l_{2}}{2}[/itex] and thus Bernoulli's law gives: [itex]\delta P = \rho\Delta V [ \frac{Q^{2}}{2}(\frac{1}{A^{2}_{1}}\frac{1}{A^{2}_{2}}) + g(h\frac{l_{2}}{2})][/itex] Now we take some limits: A_1 >> A_2, h >> l_2: [itex]\delta P = \rho\Delta V [ \frac{Q^{2}}{2}\frac{1}{A^{2}_{2}} + gh)][/itex]. From this, we can generate Torricelli's result v = [itex]\sqrt{2gh}[/itex] by setting dP = 0. This corresponds to putting a hole in the side of the container, so there is no pressure difference across the ends of the lower control volume. But, our control volume in the needle is vertical, and so there is a pressure difference between the top and bottom face. Since we can't use the result from Poiseuille flow, it's not entirely obvious what to put for this pressure difference. The bottom line, Bernoulli's law is not appropriate to model viscous flow. All discussions about "maximum flow capacity", the roles of "flow restrictors" and the like only arise when trying to force inviscid flow equations to handle viscous effects. 


#65
Jun711, 02:28 PM

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Anyway, all that friction loss will do is change the total pressure drop between the bag and the needle, meaning a lower exit velocity. It doesn't change the fundamental dependencies of the solution. Also, the flow restriction talk has nothing to do with viscous or inviscid flow. It has to do with the stopcock not physically being able to physically pass the same amount of mass through it for a given flow condition. That has nothing to do with viscosity and everything to do with simply the size of the hole. Stopcocks are designed to restrict flow rate. YOur stopcocks are doing their jobs. 


#66
Jun711, 03:14 PM

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http://arxiv.org/pdf/0809.1309 http://www.google.com/url?sa=t&sourc...cG55Sw&cad=rja http://www.engineersedge.com/fluid_f..._bernoulli.htm The second reference has the fluid exiting out of a hole in the side, which makes direct comparison to our situation questionable. The third reference requires empirical input to solve which begs the question why it is preferred in our simple system. 


#67
Jun711, 06:21 PM

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A book I happen to have on hand here that I can cite is "Fundamentals of Fluid Mechanics" by Munson, Young and Okiishi (http://www.amazon.com/FundamentalsF.../dp/0471675822). Particularly check the section about pipe flow. You essentially get a corrected Bernoulli equation. Other assumptions necessary, such as fullydeveloped flow, can also be accounted for in terms of an energy loss term that shows up as a head loss. The same can be said for the T where the IV bags come together, and that is where the slowup comes from in the OP's example, since friction is present in both cases. Still, none of them would support your original claim that two bags would create twice the mass flow rate. The only way that you can get more flow rate is by raising the bags higher, putting pressure on the bags or increasing the exit diameter of the needle. That goes for the viscous case, the inviscid case, and any other incompressible case you can think of that involves obeying the laws of physics. 


#68
Jun711, 07:45 PM

P: 688

In practice, the Bernoulli Equation is corrected for viscous flow with a "discharge coefficent." Similar to the equation boneh3ad derived earlier, we can express the volumetric flow rate as:
Vdot = Cd * Aorf * sqrt (2gh) where, Vdot = volumetric flow rate Cd = discharge coefficient of opening = actual discharge flow / theoretical discharge flow Aorf = area of discharge opening h = fluid level. As was derived earlier by boneh3ad and Andy Resnick, the surface area of the reservoir does not appear in this expression. 


#69
Jun711, 10:00 PM

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Look up the threereservoir problem. The flow rates add the flow out of the nozzle Q_3 is equal to the flow in from both funnels: Q_1 + Q_2 = Q_3. http://cee.engr.ucdavis.edu/faculty/...ir_problem.pdf http://personalpages.manchester.ac.u...s/threeres.htm http://excelcalculations.blogspot.co...rproblem.html 


#70
Jun711, 10:09 PM

P: 1,023

Of course the flow rates add. That's simple continuity. That's not the same thing as saying that the flow rates from each reservoir while both are flowing are the same as they would be with only one reservoir supplying the pressure (or, to put it another way, the flow rates are found in a way which is dependent on the overall arrangement  they can't be found independently and then summed). It is also, as has been explained to you extensively, highly dependent on the details of the problem, specifically where the majority of the pressure loss occurs.
(I could also note that you haven't addressed my experimental results yet) 


#71
Jun711, 11:12 PM

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#72
Jun811, 05:17 AM

P: 3,060

One can be dead wrong, it's OK, even being a professor, but being wrong and brag about it ... BTW, very informative thread, thanks cjl and the rest. 


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