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```
derive continuity equation

```
6 years ago

```							p1a1v1=p2a2v2  or A1V1=A2V2
```
6 years ago
```							Let us consider a tube AB of varing cross-section with water is flowing through that tube.
At side A, area of cross-section be `A1`
velocity of fluid be `v1`
At side B, area of cross-section be `A2`
velocity of fluid be `v2`
and density of water be `d` and let `A1`>`A2`

Now v=distance travelled/time=l/t --->l=vt
---> Mass of fluid entered the tube at A in `t`seconds, m1=dV1=dAl=dA1v1t ----->(1)
---> Mass of fluid leaved the tube at B in `t`seconds, m2=dV2=dAl=dA2v2t ----->(2)

But mass of fluid which entered into the  tube is equal to the mass of fluid leaved the fluid.
Therefore from equations (1) and (2)
m1=m2
dA1v1t=dA2v2t
Therefore  A1v1t=A2v2t
Therefore Av=constant
```
6 years ago
```							Consider the mass of a fluid entering a at A in `t` seconds
M1=density*A1*V1*t
Mass of the fluid leaving the tube at B in `t` seconds
M2=A2*density*V2*t
We know that M1=M2,
density*A1*V1*t=density*A2*V2*t
A1*V1=A2*V2
Therefore A1*V1=constant
```
6 years ago
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