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.

 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, PCρg=2.8m${\displaystyle \frac{P_C}{\rho g}}=2.8m$ of water absolute

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

coefficient of friction, f=0.004

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

PAρg+V2A2g+zA=PCρg+V2C2g+zC+losses of head due to friction (from A to G)${\displaystyle \frac{P_A}{\rho g}}+{\displaystyle \frac{V_A^2}{2g}}+z_A={\displaystyle \frac{P_C}{\rho g}}+{\displaystyle \frac{V_C^2}{2g}}+z_C+\text{losses of head due to friction (from A to G)}$

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

10.3+0+0=2.8+V222g+4.0+hf1$10.3+0+0=2.8+{\displaystyle \frac{V_2^2}{2g}}+4.0+h{f}_1$

[∵Vc=$\because V_c=$ velocities in pipe]

∴hf1=10.3−2.8−4.0−V22g=3.5−V22g..............(1)$\therefore h{f}_1=10.3-2.8-4.0-{\displaystyle \frac{V^2}{2g}}=3.5-{\displaystyle \frac{V^2}{2g}}..............(1)$

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

PAρg+V2A2g+zA=PBρg+V2B2g+zB+losses of head due to friction (from A to G)${\displaystyle \frac{P_A}{\rho g}}+{\displaystyle \frac{V_A^2}{2g}}+z_A={\displaystyle \frac{P_B}{\rho g}}+{\displaystyle \frac{V_B^2}{2g}}+z_B+\text{losses of head due to friction (from A to G)}$

But PAρg=PBρg=${\displaystyle \frac{P_A}{\rho g}}={\displaystyle \frac{P_B}{\rho g}}=$ atmospheric pressure.

And VA=0,VB=0,zA=15,zB=0$V_A=0,V_B=0,z_A=15,z_B=0$

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

∴hf=15$\therefore hf=15$

or 15=4fLV2d×2g$15={\displaystyle \frac{4fL{V}^2}{d\times 2g}}$

15=4×0.004×600×V20.2×2×9.81$15={\displaystyle \frac{4\times 0.004\times 600\times V^2}{0.2\times 2\times 9.81}}$

∴V2=15×0.2×2×9.814×0.004×600$\therefore V^2={\displaystyle \frac{15\times 0.2\times 2\times 9.81}{4\times 0.004\times 600}}$

V=2.47m/s$V=2.47m/s$

Substituting this value in equation (1), we get

hf1=3.5−2.4722×9.81=3.5−0.311$h{f}_1=3.5-{\displaystyle \frac{{2.47}^2}{2\times 9.81}}=3.5-0.311$

hf1=3.189m..............(2)$h{f}_1=3.189m..............(2)$

But hf2=4fLV2d×2g$h{f}_2={\displaystyle \frac{4fL{V}^2}{d\times 2g}}$

when L1=$L_1=$ inlet leg of syphon

hf1=4×0.04×L1×(2.47)20.2×2×9.81=0.0248×L1$h{f}_1={\displaystyle \frac{4\times 0.04\times L_1\times (2.47{)}^2}{0.2\times 2\times 9.81}}=0.0248\times L_1$

Substituting this value in equation (2), we get

0.248L1=3.189$0.248L_1=3.189$

L1=3.1890.0248=128.58m