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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!

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!