Three reservoir A, B, C are connected by a pipe system. Find the discharge into or from the reservoirs Bull and Cif the rate of flow from reservoir A is 60 litres per second. Find the height of water level in the reservoir C. f = 0.06 for all pipes.

 Three reservoir A, B, C are connected by a pipe system. Find the discharge into or from the reservoirs Bull and Cif the rate of flow from reservoir A is 60 litres per second. Find the height of water level in the reservoir C. f = 0.06 for all pipes.

Solution: 

Given:-

Length of pipe AD, L1=1200m$L_1=1200m$

Diameter of pipe AD, d1=30cm=0.30cm$d_1=30cm=0.30cm$

Discharge through 'AD', Q1=60lit/s=0.06m3/s$Q_1=60lit/s=0.06m^3/s$

height of water level in A from reference line, zA=40m$z_A=40m$

For pipe DB length,

L2=600m$L_2=600m$, diameter, d2=20cm=0.20cm$d_2=20cm=0.20cm$, zB=38$z_B=38$

For pipe DC

Length, L3=800m$L_3=800m$, diameter d3=30cm=0.30m$d_3=30cm=0.30m$,

Step 1:

Applying Bernoulli's eqaution to point E and D

zA=zD+PDρg+hf1$z_A=z_D+{\displaystyle \frac{P_D}{\rho g}}+h{f}_1$

where hf1=4fL1V21d1×2g$h{f}_1={\displaystyle \frac{4f{L}_1{V}_1^2}{d_1\times 2g}}$

V1=Q1A=0.06π4(0.3)2=0.848m/sec$V_1={\displaystyle \frac{Q_1}{A}}={\displaystyle \frac{0.06}{{\displaystyle \frac{\pi }{4}}(0.3{)}^2}}=0.848m/sec$

∴hf1=4×0.006×1200×(0.848)20.3×2×9.81=3.518m$\therefore h{f}_1={\displaystyle \frac{4\times 0.006\times 1200\times (0.848{)}^2}{0.3\times 2\times 9.81}}=3.518m$

∴zA=zD+PDρg+3.518$\therefore z_A=z_D+{\displaystyle \frac{P_D}{\rho g}}+3.518$

40=zD+PDρg+3.518$40=z_D+{\displaystyle \frac{P_D}{\rho g}}+3.518$ zD+PDρg=40−3.518=36.482m$z_D+{\displaystyle \frac{P_D}{\rho g}}=40-3.518=36.482m$

Hence piezometric head at D=36.482

But zB=38m$z_B=38m$

Hence water flows from 'B' to 'D'.

Step 2:

Applying Bernoulli's equation to point 'B' and 'D'

zB=zD+PDρg+hf2$z_B=z_D+{\displaystyle \frac{P_D}{\rho g}}+h{f}_2$

38=36.482+hf2$38=36.482+h{f}_2$

∴hf2=38−36.482=1.518m$\therefore h{f}_2=38-36.482=1.518m$

But hf2=4fL2V22d2×2g$h{f}_2={\displaystyle \frac{4f{L}_2{V}_2^2}{d_2\times 2g}}$

∴,1.518=4×0.006×600×V220.2×2×9.81$\therefore ,1.518={\displaystyle \frac{4\times 0.006\times 600\times V_2^2}{0.2\times 2\times 9.81}}$

V2=0.2×2×9.81×1.5184×0.006×600−−−−−−−−−−−−−−−−−−√$V_2=\sqrt{{\displaystyle \frac{0.2\times 2\times 9.81\times 1.518}{4\times 0.006\times 600}}}$

V2=0.643m/s$V_2=0.643m/s$

Q2=V2×π4(d2)2$Q_2=V_2\times {\displaystyle \frac{\pi }{4}}(d_2{)}^2$

Q2=0.643×π4×(0.2)2$Q_2=0.643\times {\displaystyle \frac{\pi }{4}}\times (0.2{)}^2$

Q2=0.0202m3/s=20.2lit/s$Q_2=0.0202m^3/s=20.2lit/s$

Step 3:

Applying Bernoulli's equation to 'D' and 'C'

zD+PDρg=zC+hf3$z_D+{\displaystyle \frac{P_D}{\rho g}}=z_C+h{f}_3$

∴36.482=zC+4fL3V23d3×2g$\therefore 36.482=z_C+{\displaystyle \frac{4f{L}_3{V}_3^2}{d_3\times 2g}}$

where, V3=Q3π4d23$V_3={\displaystyle \frac{Q_3}{{\displaystyle \frac{\pi }{4}}{d}_3^2}}$

but from continuity Q1+Q2=Q3$Q_1+Q_2=Q_3$

∴Q3=0.06+0.0202=0.0802m3/s$\therefore Q_3=0.06+0.0202=0.0802m^3/s$

∴V3=Q3π4(0.3)2$\therefore V_3={\displaystyle \frac{Q_3}{{\displaystyle \frac{\pi }{4}}(0.3{)}^2}}$

∴V3=0.0802π4(0.3)2=1.134m/s$\therefore V_3={\displaystyle \frac{0.0802}{{\displaystyle \frac{\pi }{4}}(0.3{)}^2}}=1.134m/s$

∴36.482=zC+4×0.006×800×1.13420.3×2×9.81$\therefore 36.482=z_C+{\displaystyle \frac{4\times 0.006\times 800\times {1.134}^2}{0.3\times 2\times 9.81}}$

36.482=zC+4.194$36.482=z_C+4.194$

zC=36.482−4.194=32.288m