Question

Water flows from a large tank through a large pipe that splits into two smaller pipes as shown in figure below. Height, H=15m.
The density of water is 1000kg/m3, atmospheric pressure patm = 101.325kPa and gravitational acceleration g = 9.81m/s2.

Answer 1

Regarding all mentioned information and the figure file attached, whilst viscous effects are negligible, determine:
- the velocity (in m/s) through pipe exit 2.
- The velocity (in m/s) through pipe exit 3.
- The pressure (in kPa) at point 1 in the large pipe.

Answer 2

I'm wondering if we can use bournelli's law for this.

considering that the total mechanical energy is conserved. delta E = 0.

\[KE_{1}+PE_{1} = KE_{f}+PE_{f}\]

This translates to the following:

\[\frac{ 1 }{ 2 }mv^{2}_{1}+mgh_{1} = \frac{ 1 }{ 2 }mv^{2}_{2}+mgh_{2}\]

we need to express this in terms of density:

\[\rho = \frac{ m }{ v }\]

let's express it this way.

\[m = \rho*v \]

\[\frac{ 1 }{ 2 }rho*v*v^{2}_{1}+rho*v*gh_{1} = \frac{ 1 }{ 2 } rho*v*v^{2}_{2}+rho*v*gh_{2}\]

which gives us

\[\rho*\frac{ 1 }{2 }m_{1}v^{2}_{1}+\rho*g*h + p_{i} = \rho*\frac{ 1 }{ 2 }m_{2}v^{2}_{2}+\rho*gh_{2}+p_{2}\]