Hey koverot.
Just to clarify which variables are you keeping constant, which ones are you keeping varied and what constraints are you keeping in terms of the formulaic constraints over time?
Hi guys,
first time on this site. In my current work, I deal with mainly hydraulic related calculations, and the remainder of my math skills have become pretty lack luster. On looking through a spreadsheet in work, designed to calculate the size of a drainage basin (like an upside down pyramid, with the top chopped off) I spotted a volume error, but my math eludes me how to fix it. I have attached a crude sketch to help illustrate the problem.
We have a rectangular basin of a pyramidal frustum shape. Q represents the water flowing in, D (Depth) is the depth of water in the basin, and A (area) is the surface area of the surface of the water, both which increase over time. I ( Infiltration) is a constant of the water infiltrating (being removed from basin volume), and acts over the area A. so A*I*t= Volume removed, and Q*t = volume added. (Q*t)-(A*I*t) = Volume in basin; t=time.
I need to find a relationship that can compute the surface Area, A, at a step in time, with a varying flowrate. As (A*I*t) increase with depth, I realize some voodoo magic integration is required, but I haven't touched it since uni, and I'm unsure where to start.
I apologize if this is the wrong forum, mods please move if it is.
Anyway any help is very much appreciated.
koverot
Hey koverot.
Just to clarify which variables are you keeping constant, which ones are you keeping varied and what constraints are you keeping in terms of the formulaic constraints over time?
Yes Chiro, but bearing in mind that I*A = Qout, so the area which is a factor of volume and depth would look like
V= (Qin - Qout)*t , and A=function of frustum volume equation, which can be easily computed.
Sorry I didn't expand on this in my OP. I had originally included it, but my initial attempt at a post failed, and I hurriedly put together the re-draft
Is there a constraint on how "thick" the volume is for the frustum to determine its weight? (i.e. through a relationship to the flow rate)?
If so then you can solve for the boundary of the frustum and get the area of the lowest cross section towards the apex of the frustum.
Hi chiro,
Im afraid you've lost me. if you could dumb down your question for me that would be great. just to reiterate so theres no confusion;
I need to find the area (A) of the surface of the water at a given time (t) in a basin that is essentially an upside down frustum.
The area (A) can be derived from the volume (V), and the volume is a function of Qin (flow in,m3/s) minus the Qout(flow out m3/s) for a given time (t).
Qout = A*I, I being a constant.
To simplfy, Q in can be kept constant.
Is integration required, or is this a simple equation. seems like integration is involved as the equation requires solving volumes over time.
Thanks again bud
Basically your frustum is filled with water that has two cross sections: a small one and a big one.
The perpendicular distance between those two cross sections is the distance I am referring to (you call it Depth on your diagram).
If you know the equations for the angle of the frustum sides then you can set up an integral equation to solve for the two cross sections which will indirectly give you a way to get the area.
What you have to do is given the depth, solve a volume integral that is enclosed by the frustum that has bounds a and a+D where D is the depth and then solve for a.
You then take a and get the area by using a formula for the cross sectional area at a particular point.
Do you know how to calculate the volume of a frustum? If so you simply have the volume and then solve for a where you have Integral [a,a+D] {Region}dV = Volume and calculate the cross sectional area from getting a.
Hi Chiro,
As I said, my integration is pretty weak at present. how does the arranged equation look? are the limits [a, a+D] correct? what is the region?
I am struggling somewhat in getting my head around this.