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Derivative in mass flow rate equation - Hydrology
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[QUOTE="CivilSigma, post: 5566820, member: 463654"] 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! [/QUOTE]
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Derivative in mass flow rate equation - Hydrology
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