10 peles A 3-m-diameter tank is initially filled with water 2 m above the center of a sharp-edged 10 cm diameter orifice. The
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We will take the point 1 at the free surface of the tank and point 2 at the exit of the pipe. after this centerline is taken at the centerline of the orific where Z2=0. Positive direction of z is taken upwards

Fluid velocity at free surface is very low and almost 0

Also the fluid at both the points is open to atmosphere so in that case


\frac{P_{1}}{\rho g}+\alpha _{1}\frac{V_{1}^{2}}{2g}+z_{1}+h_{pump,u}=\frac{P_{2}}{\rho g}+\alpha _{2}\frac{V_{2}^{2}}{2g}+z_{2}+h_{turbine}+h_{L}

After putting the above mentioned conditions

z_{1}+h_{pump,u}=\alpha _{2}\frac{V_{2}^{2}}{2g}+h_{L}..................1

where \alpha _{2}=1 Given

So from here the total head can be defined as

h_{L}=h_{L,total}=h_{L,major}+h_{L,minor}=\left ( f\frac{L}{D}+\sum K_{L} \right )\frac{V^{2}}{2g}

h_{L}=\left ( f\frac{L}{D}+ K_{L} \right )\frac{V^{2}}{2g}..................2

Since the dia of the piping system is constant and the initial velocity is given as 4 m/s so from here required mass flow rate and head can be calculated as

\dot{m}=\rho A_{c}V_{2}

\dot{m}=\rho (\frac{\Pi D^{2}}{4})V_{2}

\dot{m}=996 kg/m^{3} (\frac{\Pi .1m^{2}}{4})4m/s

\dot{m}=31.3 Kg/m^{3} Answer

Now from equation 1 and 2

h_{pump,u}=\left ( 1+f\frac{L}{D}+ K_{L} \right )\frac{V^{2}}{2g}-z_{1}

h_{pump,u}=\left (1+ (.015)\frac{100m}{.1m}+ .5 \right )\frac{4m/s^{2}}{2\times 9.81m/s^{2}}-2m

h_{pump,u}=11.46 m Answer

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