Volume flow rate DE  Help needed
Hi all,
So I've got a fairly straight forward problem to solve here regarding flow rate out of a tank with uniform crosssectional area. I am treating this is a volumetric flow problem since there is assumed to be volume flow out of the tank, and volume flow into the tank. I have two Qout terms (one out of a hole in the bottom of the tank, and one from fluid consumed by an engine), as well as one Qin term (via a pump feeding fluid into the tank at a constant rate). Qouthole (Qh) = a*C*sqrt(2*g*z) Qoutengine (Qe) = constant Qin (Qi) = constant NOTE: a = exit hole area C = energy loss coefficient g = gravity z = head height in tank My DE looks like: A*(dh/dt) = Qh  Qi + Qe Separating the terms and solving for time by integrating h from Hi to Hf and t from 0 to T, I get this... ((a*C)/A)*sqrt(2*g)*(2/3)*((Hf^(3/2))(Hi^(3/2))) = (1/A)q*t I combined the Qi and Qe terms early on since they are constants (easier to integrate), therefore, q in the above equation equals QiQf. The problem is, solving for t eliminates the tank crosssectional area (A) since it is divided out. I feel confident with my general equation, but may have made a mistake somewhere in solving for t. Any help would be GREATLY appreciated! Thank you 
Re: Volume flow rate DE  Help needed
Could you outline how did you arrive at the expression you found after integrating?

Re: Volume flow rate DE  Help needed
Quote:
Qleak = a*C*sqrt(2*g*h) Qi = constant Qe = constant integrate dH from Hi to Hf integrate dt from 0 to T  1. A * (dH/dt) = Qleak  Qi + Qe 2. ... simplify by combining constants... Qi + Qe = Qtot 3. dH = (Qleak/A + Qtot/A)dt 4. dH = 1/A*(Qleak + Qtot)dt 5. (A*dH) / (Qleak + Qtot) = dt 6. int((A*dH) / (Qleak + Qtot)) = int(dt) 7. ... integrate dt from 0 to T 8. int((A*dH) / (Qleak + Qtot)) = T  This is where I stopped since I could not find an easy way to integrate the left side. The basic form of the integral would be: dH/(h^1/2 + constant) Any tips on where I may have gone wrong, or how to solve that integral would be appreciated! Thanks 
Re: Volume flow rate DE  Help needed
Quote:

All times are GMT 5. The time now is 01:08 PM. 
Powered by vBulletin Copyright ©2000  2014, Jelsoft Enterprises Ltd.
© 2014 Physics Forums