Specialty Chemistry Forums > Other Sciences Question Forum
liquid flow calculation - question for Eugene :)
Borek:
Here it goes:
There is a pump and a system of tubes - something similar in the idea to the heart and veins. How do you calculate flow through the individual tubes?
I have a gut feeling that one may use some analogy to Kirchoff laws using pressure instead of voltage, mass (or volume) flow instead of current and assigning a resistance to every tube (depending - if the tubes are otherwise identical - on the tube diameter).
Am I right? Or is there some other approach? Any web based sources?
eugenedakin:
Hello Borek...
Wow... do you ever have great intuition... here is a website that has some of the mathematical formula's
http://www.pmengineer.com/CDA/ArticleInformation/features/BNP__Features__Item/0,2732,106650,00.html
And yes, it is the same mathematical formula used in calculating electrical parallel circuits. Since the surface area of the contact of air/liquid between increases due to the increased pipe cross sectional area, the resistance/drag will also increase.
The electrical parallel circuit can be modified to adjust for added drag.
Well done!!!
Eugene Dakin Ph.D., P.Chem.
Borek:
--- Quote from: geodome on May 12, 2005, 02:55:52 AM ---sounds like a chemical engineering problem. assuming steady state and constant density, the sum of volumetric flowrate out = sum of volumetric flowrate out. I would use the electrical circuit analogy to solve the problem too.
--- End quote ---
From my technology classes I remember there is a loss of pressure on the tube, which sounds like an analogy to loss of potential on the resistance in the electrical circuit. First Kirchoffs law is just about charge balance, so the mass balance applies as well. Thus electric circuit analogy is not very difficult.
Howeve, Eugene posted a link which I have checked - and there are some discrepancies. Resistance of parallely connencted tubes (heaters to be precise, but I don't think it matters) doesn't follow simple rule that reciprocal of resistance is sum of reciprocals - there are some additional nasty powers in the formula (formula 3 to be exact). I suppose that's because there is no wide-range linearity in the dependence between flow, tube resistance and pressure applied (as opposed to Ohm law).
eugenedakin:
Hello Borek,
You are right about the wide-range linearity.
In the diagram, most pipelines have an intake in which the liquid lowers its initial horizontal directional velocity and is split (relatively) evenly in the header (a) where the vertical directional velocity increases. The vertical directional velocity changes once again to horizontal. Each velocity change increases the resistance.
The human body is significantly more efficient, and will split two liquid flow rates (b) with lower resistant values. I tend to believe that resistance values for an artery that contracts and expands with each heart beat will increase this resistance value. I am not sure how much greater the resistance will increase.
Formula 3 in the previously mentioned website would be a starting formula for resistance balance. Other factors such as viscosity, changing pressures, changing flowrates (acceleration and deceleration of blood flow with each heartbeat) would be significant factors that change these resistance values are factors which immediately come to mind.
Sincerely,
Eugene
eugenedakin:
Chuckle... I don't know how to get the image to show... I have attached the image, and clicked 'insert image' .. Simple question.. where are the instructions? ..
Eugene
Navigation
[0] Message Index
[#] Next page
Go to full version