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Volume flow rate DE - Help needed |
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| Feb9-12, 11:45 PM | #1 |
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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 cross-sectional 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 Qi-Qf. The problem is, solving for t eliminates the tank cross-sectional 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 |
| Feb11-12, 04:49 AM | #2 |
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Could you outline how did you arrive at the expression you found after integrating?
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| Feb12-12, 10:46 AM | #3 |
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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 |
| Feb13-12, 02:16 PM | #4 |
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Volume flow rate DE - Help needed |
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