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Derivative in mass flow rate equation - Hydrology

  1. Sep 14, 2016 #1
    • Member advised to use the homework template for posts in the homework sections of PF.
    Hello,

    I am working with the mass flow rate equation which is:


    $$\frac{d \dot{m}}{dt}=\dot{m}_{in}-\dot{m}_{out}$$

    To determine the change of the height of water in a reservoir. Assuming m_in = 10 and m_out = sqrt(20h), then :

    $$\frac{d (\rho \cdot Q) }{dt}=\rho \cdot Q_{in} - \rho\cdot Q_{out}$$

    $$\frac{d ( Q) }{dt}=Q_{in} - Q_{out}$$
    $$\frac{d ( Q) }{dt}=10 - \sqrt{20h}$$

    The final form of the formula is:

    $$Area \cdot \frac{dh}{dt}=10 - \sqrt{20h}$$

    How do we get to the right hand side?

    I know that Q=A*v , and since the cross sectional area is independent of height, then it is a constant and is independent of the derivative. That leaves dv/dt - but the equation has dh/dt....

    Any input is appreciated.

    Thank you!
     
  2. jcsd
  3. Sep 14, 2016 #2
    The symbol Q is usually used for volumetric flow rate. But in this example, on the left hand side of the equation, you seem to also be using it for volume of liquid in the tank. So, in this case, $$\frac{dQ}{dt}=A\frac{dh}{dt}$$
     
  4. Sep 14, 2016 #3
    Hi Chestermiller, I agree with your equation.

    Here is what is bothering me though, what do you make of the following:

    $$Q=Av$$
    $$\frac{dQ}{dt}=\frac{d(Av)}{dt}$$
    $$\frac{dQ}{dt} = A \frac{d}{dt}(v) = A \frac{dh}{dt}??$$

    I don't see the flaw in my current logic. Why would dv/dt become dh/dt.

    I do agree, by unit analysis that A*dv/dt = m^2 * m/s^2 = m^3/s^2 - and what I had derived is incorrect.
    I just don't see where the mistake is in my math.

    Thank you!
     
  5. Sep 14, 2016 #4
    Your starting equation is incorrect. It should read $$\frac{dV}{dt}=Q_{in}-Q_{out}$$where the volume of liquid in the tank is given by V=Ah.
     
  6. Sep 14, 2016 #5
    Omg!
    Thank you very, very much!
     
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