Question

Question 2 A very large rainwater harvesting reservoir, Tank A, is situated at a height above the ground. Water is siphoned out of the tank at a rate of th = 11kg/s into a second tank, Tank B, which is sitting on the ground. Water flows out of Tank B from an outlet situated at the bottom of the tank. The system is shown in Figure 1 Figure 1: 1. Deduce an expression for the maximum height that the water level will reach in Tank B in terms of the flow rate through the siphon and the outlet dinmeter. (10) 2. Determine the height of the water surface, h, in the reservoir (Tank A) if the diameter of the outlet on Tank B and the water siphon are both equal to 4 cm (10) (20)

Answer 1

Tank A h Tank B V

1)

If we want the maximum height to be stable in the Tank B, then the entire water mass which comes into the Tank B has to be expelled out through the outlet of tank B at a Velocity "V".

Thus, Mass in = Mass out.

Now, we know,

Mass Volumetric flow rate x Density

where

• Volumetric flow rate - represented by Q
• Density of water = 1000 kg/m3

Mass Volumetric flow rate x Density

Mass = Q x 1000 1000

Mass in = Mass flow rate through the syphon = 1000 Q

Now,

Mass out Volumetric flow rate x Density

where

• Volumetric flow rate = $Area\times Velocity=\frac{\prod Dia^{2}}{4}\times V$
  • Area = $\frac{\prod Dia^{2}}{4}$ ​​​​​​​
    • Dia = Diameter of outlet pipe
  • Velocity = V
• Density = 1000 kg/m3

Thus, $Mass\: \: out =\frac{\prod Dia^{2}}{4}\times V\times 1000$

Now, Mass in = Mass out

$1000Q=\frac{\prod Dia^{2}}{4}\times V\times 1000$

$=>Q=\frac{\prod Dia^{2}}{4}\times V$

$=>V=\frac{4\times Q}{\prod Dia^{2}}$

Now, we will apply Bernoulli's theorem between points B and C.

A Tank A h B HTank B V с

$\frac{P_{B}}{\omega }+\frac{V_{B}^{2}}{2g}+Z_{B}=\frac{P_{C}}{\omega }+\frac{V_{C}^{2}}{2g}+Z_{C}$

where

• P(B) = 0 (open surface)
• P(C) = 0 (water flows into atmosphere)
• V(B) = 0 (open surface)
• V(C) =
  4xQ V= II Dia?
  (assumed)
• Z(B) = H (Maximum height of water in tank B) (our answer)
• Z(C) = 0 (considered datum)
• g = 9.81 m/s2

$\frac{P_{B}}{\omega }+\frac{V_{B}^{2}}{2g}+Z_{B}=\frac{P_{C}}{\omega }+\frac{V_{C}^{2}}{2g}+Z_{C}$

$0+0+H=0+\frac{\left (\frac{4\times Q}{\prod Dia^{2}} \right )^{2}}{2g}+0$

$\mathbf{H=\frac{0.0826\times Q^{2}}{Dia^{4}}}$

The above equation has maximum height in Tank B, H in terms of Q and Diameter of outlet pipe.

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2)

Apply Bernoulli's Theorem across points A and C.

$\frac{P_{A}}{\omega }+\frac{V_{A}^{2}}{2g}+Z_{A}=\frac{P_{C}}{\omega }+\frac{V_{C}^{2}}{2g}+Z_{C}$

where

• P(A) = 0 (open surface)
• P(C) = 0 (datum)
• V(A) = 0 (open surface)
• V(C) = ?
• Z(A) = h ( height of water in Tank A when steady state is achieved in Tank B) (our answer)
• Z(C) = 0 (datum)
• g = 9.81 m/s2

Now, we will apply continuity equation across syphon and the outlet pipe. The syphon has 4 cm dia and so does th eoutlet pipe. And since water is incompressible, no matter what, the velocity in both the syphon and the outlet pipe have to be same. Because if the velocity changes, then water is being compressed, which is not the case. This is becuase, due to law of conservation of mass.

$\rho _{syphon}\times Area_{syphon}\times V_{syphon}=\rho _{outlet}\times Area_{outlet}\times V_{outlet}$

where

• $\rho _{syphon} = \rho _{outlet}\: \: \left [ \because Water \: is \: incompressible \right ]$
• Area(syphon) = Area (outlet) = $\frac{\prod Dia_{syphon}^{2}}{4}=\frac{\prod 0.04^{2}}{4}=1.2566e-3 \: m^{2}$ $\left [ \because Diameter\: of \: syphon\: \: and \: \: outlet\: \: pipe \: \: same \right ]$
  • D(syphon) = 4 cm = 0.04 m (given in question)
• V(syphon) = V(outlet) = Velocity of water in syphon = Velocity of water in outlet pipe = ?

$\dot{m}=\rho _{syphon}\times Area_{syphon}\times V_{syphon}$

where

• m = mass flow rate of water = 11 kg/s
• $\rho _{syphon}$ = Density of water in syphon = 1000 kg/m3
• Area(syphon) = 1.2566e-3 m2
• V(syphon) = ?

$\dot{m}=\rho _{syphon}\times Area_{syphon}\times V_{syphon}$

$11=1000\times 1.2566e-3\times V_{syphon}=>V_{syphon}=V_{outlet}=8.7535\: \: m/s$

Now, $\frac{P_{A}}{\omega }+\frac{V_{A}^{2}}{2g}+Z_{A}=\frac{P_{C}}{\omega }+\frac{V_{C}^{2}}{2g}+Z_{C}$

$0+0+h=0+\frac{8.7535^{2}}{2g}+0$

$=>\mathbf{h=3.905\: m}$

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NOTE-

This height and the height of water in the Tank B may seem to be the same, and yes they are the same. The reason being, if the syphon and the outlet pipe have the same diameter, they both MUST have the same velocities. In such a case, no water will be stored in Tank B and all the water with Velocity 8.7535 m/s falling into Tank B must be removed by the outlet pipe with the same velocity. Thus, water essentially does not store inside Tank B, so essentially, the water level in Tank B will be ZERO (0) when water is at level 3.905 m in Tank A.

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