# What is the continuity equation

## Fluid mechanics

We idealize the garden hose as a pipe with an even flow, a so-called **Flow conduit**. Such a current is called **steady flow**, because no matter when we take a picture of this current, it will always be approximately the same picture. If, on the other hand, the garden hose were supplied with water by a hand pump, the flow would change with every up and down movement of the pump, the flow would not be stationary.

If we at the left end of the filled flow tube in **Fig. 3** add a quantity of liquid of mass \ (m_1 \) into the flow tube, then a quantity of liquid of mass \ (m_2 \) must come out at the right end of the flow tube. From our everyday experience we now know that matter is not lost and that the mass of a certain amount of matter also remains constant. We adopt this experience in physics and refer to it as **Mass conservation**. This conservation of mass of matter leads us to the equation \ [{m_1} = {m_2} \ quad (1) \] for our garden hose. For the two masses \ (m_1 \) and \ (m_2 \), with the well-known formula \ (m = \ rho \ cdot V \) the relationships \ ({m_1} = {\ rho _1} \ cdot {V_1} \) and \ ({m_2} = {\ rho _2} \ cdot {V_2} \). Why not the same \ (\ rho \) and the same \ (V \)? The fluid in the flow tube may have been compressed and therefore has a smaller volume and greater density. If we put these relationships into the first equation, we get \ [{\ rho _1} \ cdot {V_1} = {\ rho _2} \ cdot {V_2} \ quad (2) \] If the flow tube on the left has the cross-sectional area \ (A_1 \), the amount of liquid flowing into the flow tube on the left has the length \ (l_1 \) with \ (V_1 = A_1 \ cdot l_1 \). Correspondingly, the relation \ (V_2 = A_2 \ cdot l_2 \) applies to the right with the cross-sectional area \ (A_2 \) and the length \ (l_2 \). This gives us \ [{\ rho _1} \ cdot {A_1} \ cdot {l_1} = {\ rho _2} \ cdot {A_2} \ cdot {l_2} \ quad (3) \] We now assume that we Bring the amount of liquid \ (m_1 \) on the left into the flow tube in a time \ (t \) and in the same time \ (t \) the amount of liquid \ (m_2 \) leaves the flow tube on the right. Since the flow tube flows evenly (i.e. without acceleration at the inlet and outlet), the following applies to the velocities of the fluid at the left and right end \ [v_1 = \ frac {l_1} {t} \; \; \ rm {and} \; \; v_2 = \ frac {l_2} {t} \ quad (4) \] Let us now divide equation \ ((3) \) by \ (t \) \ [\ frac {m_1} {t} = \ frac {{\ rho_1 \ cdot {l_1} \ cdot {A_1}}} {t} = \ frac {{\ rho_2 \ cdot {l_2} \ cdot {A_2}}} {t} = \ frac {m_2} {t } \] and use the equations in \ ((4) \), we get \ [\ frac {m_1} {t} = \ rho_1 \ cdot {v_1} \ cdot {A_1} = \ rho_2 \ cdot {v_2} \ cdot {A_2} = \ frac {m_2} {t} \ quad (5) \] Since the size \ (\ frac {m} {t} = \ rho \ cdot v \ cdot A \) at all points of the Is constant, we make the following definition:

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