Page: Flow through syphon:
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It is defined as a long bent pipe which is used to transfer liquid from a reservoir which is at a higher elevation to another reservoir which is at the low level when both the reservoirs are separated by a hill.

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The highest point of the syphon i.e.,'G 'is known as Summit.

Uses of Syphon:-

1) To transfer water or liquid from one reservoir to another reservoir separated by a hill.

2) To take out the liquid out from a tank without an outlet.

3) The channels which are not provided with any sluice can be emptied.


Numericals

Q1) A Syphon of diameter 200 mm to Reservoir having a difference in elevation of 15 m. The total length of the syphon is 600 m and the Summit is 400 m above the water level in the upper reservoir. If the separation takes place at 2.8 M of water absolute, find the maximum length of Siphon from the upper reservoir to the summit. Take f=0.004 and atmospheric pressure=10.3 of water.

Solution: Given:-

Diameter of syphon, d=200mm=0.2m

Difference of level in two reservoirs = 15m

Total length of pipe = 600m

Height of summit from upper reservoir = 4m

Sep 1:

Pressure head at summit, $\dfrac {P_C}{\rho g}=2.8m$ of water absolute

Atmospheric pressure head, $\dfrac {P_G}{\rho g}=10.3m$ of water absolute

coefficient of friction, f=0.004

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Step 2: Applying Bernoulli's equation to points A and G and taking the datum line passing through A,

$\dfrac {P_A}{\rho g}+\dfrac {V_A^2}{2g}+z_A=\dfrac {P_C}{\rho g}+\dfrac {V_C^2}{2g}+z_C+\text{losses of head due to friction (from A to G)}$

Substituting the values of pressure in terms of absolute, we have

$10.3+0+0=2.8+\dfrac {V_2^2}{2g}+4.0+hf_1$

[$\because V_c=$ velocities in pipe]

$\therefore hf_1=10.3-2.8-4.0-\dfrac {V^2}{2g}=3.5-\dfrac {V^2}{2g}..............(1)$

Step 3: Applying Bernoulli's equation to points A and B and taking datum line passing through B,

$\dfrac {P_A}{\rho g}+\dfrac {V_A^2}{2g}+z_A=\dfrac {P_B}{\rho g}+\dfrac {V_B^2}{2g}+z_B+\text{losses of head due to friction (from A to G)}$

But $\dfrac {P_A}{\rho g}=\dfrac {P_B}{\rho g}=$ atmospheric pressure.

And $V_A=0, V_B=0, z_A=15, z_B=0$

$0+0+15=0+0+0+hf$

$\therefore hf=15$

or $15=\dfrac {4fLV^2}{d\times 2g}$

$15=\dfrac {4\times0.004\times600\times V^2}{0.2\times 2\times 9.81}$

$\therefore V^2=\dfrac {15\times 0.2\times 2\times 9.81}{4\times 0.004\times 600}$

$V=2.47m/s$

Substituting this value in equation (1), we get

$hf_1=3.5-\dfrac {2.47^2}{2\times 9.81}=3.5-0.311$

$hf_1=3.189m..............(2)$

But $hf_2=\dfrac {4fLV^2}{d\times 2g}$

when $L_1=$ inlet leg of syphon

$hf_1=\dfrac {4\times 0.04\times L_1\times (2.47)^2}{0.2\times 2\times 9.81}=0.0248\times L_1$

Substituting this value in equation (2), we get

$0.248 L_1=3.189$

$L_1=\dfrac {3.189}{0.0248}=128.58m$


Q2) A syphon of diameter 200 mm connect to reservoir whose water surface level differ by 40 m. Total length of pipe is 8000 m. The pipe crosses a bridge. The summit of the ridge is 8 m above the level of water in the upper reservoir. Determine the minimum depth of the pipe below the summit of the ridge, if the absolute pressure head at the summit of the syphon is not to fall below 0.3 M of the water. Take f=0.006 and atmospheric pressure = 10. 3 mm of water. The length of syphon from the upper reservoir to the summit is 500 m. Find the discharge also.

Solution: Given:-

Diameter of syphon, d=200mm=0.20m

Difference in levels of two reservoirs, H=40m

Total length of the pipe, L=8000m

Height of ridge summit from water level in upper reservoir = 8m

Let the depth of the pipe below the summit of ridge = x m

Therefore the height of syphon from water surface in the upper reservoir = (8-x) m

Step 1: Pressure head at C, $\dfrac {P_C}{\rho g}=3.0m$ of water absolute

Atmospheric pressure, $\dfrac {P_a}{\rho g}=10.3m$ of water

coefficient of friction, f=0.006

Length of syphon, $L_1=500m$

Step 2:

Applying Bernoulli's equation to points A and B, and taking datum line passing through B, we have

$\dfrac {P_A}{\rho g}+\dfrac {V_A^2}{2g}+z_A=\dfrac {P_B}{\rho g}+\dfrac {V_B^2}{2g}+z_B+\text{loss due to friction from A to B}$

$\therefore 0+0+40=0+0+0+hf$

$40=\dfrac {4fLV^2}{d\times 2g}$

$40=\dfrac {4\times 0.006\times8000\times V^2}{0.2\times 2\times 9.81}$

$V^2=\dfrac {0.2\times 2\times 9.81\times 40}{4\times 0.006\times 8000}$

V=0.904m/s

Step 3:

Now applying Bernoulli's equation to points 'A' and 'C' and assuming datum line passing through, we have

$\dfrac {P_A}{\rho g}+\dfrac {V_A^2}{2g}+z_A=\dfrac {P_C}{\rho g}+\dfrac {V_C^2}{2g}+z_C+\text{loss due to friction from A to C}$

Substituting $\dfrac {P_A}{\rho g}$ and $\dfrac {P_C}{\rho g}$ in terms of absolute pressure,

$10.3+0+0=3.0+\dfrac {V^2}{2g}+(8-x)+\dfrac {4fL_1V^2}{d\times 2g}$

$\therefore 10.3=3.0+\dfrac {(0.904)^2}{2\times 9.81}+(8-x)+\dfrac {4\times 0.006\times 500\times (0.904)^2}{0.2\times 2\times 9.81}$

$\therefore 10.3=3.0+0.041+(8-x)+2.499$

$\therefore 10.3=13.54-x$

$\therefore x=13.54-10.3=3.24m$

Step4: Discharge

Q=Area $\times$ Velocity

$Q=\dfrac \pi4\times (0.2)^2\times 0.904$

$Q=0.0283 \ m^3/s$

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modified 4 months ago by gravatar for Sanket Shingote Sanket Shingote250 written 4 months ago by gravatar for Syedahina Mohi Syedahina Mohi0
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