Question 7.P.24: Two adjacent city centres B and D receive water from separat......
Two adjacent city centres B and D receive water from separate sources A and C. The water level in A is 4 m above that in C. Reservoir A supplies city centre B by 0.4 m diameter pipe of 3000 m length with a level difference of 10 m. City centre D’s is supplied by reservoir C through a 4000 m long pipe of 0.45 m diameter, with a level difference of 15 m. After sometime it is found that centre B has excess water while centre D is staraved. So it is proposed to interconnect these lines and draw 100 l/s from the line A to B. The junction on AB is at a distance of 2000 m from A. The junction CD is at 3000 m from C. Determine the original supply rates and supply rates with interconnection to centres B and D. Also determine the diameter of the interconnecting pipe, if the length is 1500 m Friction factor, f = 0.01 in all cases.
The arrangement is shown in Fig. P. 7.24
(i) Without interconnection : using equation 7.11.15
h_{f}={\frac{8\;f L Q^{2}}{\pi^{2}\;g D^{5}}}
Drop in Line AB is 10 m
10={\frac{8\times0.01\times3000\times Q^{2}}{\pi^{2}\times9.81\times0.4^{5}}}
∴ {\bf Q}_{\mathrm{AB}}=0.20325~{\bf m^{3}}/{\bf s}\quad\mathrm{or}\quad203.25~{\mathrm{l/s}}Drop in line CD is 15 m
15={\frac{8\times0.01\times4000\times Q^{2}}{\pi^{2}\times9.81\times0.45^{5}}}
∴ \mathbf{Q}_{\mathrm{CD}}=0.2894\mathrm{~m^{3}/s}\quad\mathrm{or}\quad289.4\ { \mathrm{l/s}}
After interconnection: line AB:
Let the flow up to R be Q and then in RB (Q – 0.1)
Total frictional loss =10={\frac{8\times0.01\times2000\times Q^{2}}{\pi^{2}\times9.81\times0.4^{0.5}}}+{\frac{8\times0.01\times1000(Q-0.1)^{2}}{\pi^{2}\times9.81\times0.4^{5}}}
This reduces to 3Q² – 0.2Q – 0.11393 = 0. Solving Q = 0.231 m³/s or 231 l/s
Now the centre B will receive 131 l/s (previous 203 l/s)
Line CD: Let the flow upto S be Q and then (Q + 0.1) upto D
Total head loss =15=\frac{8\times0.01\times3000\times Q^{2}}{\pi^{2}\times9.81\times0.45^{0.5}}+\frac{8\times0.01\times1000(Q+0.1)^{2}}{\pi^{2}\times9.81\times0.45^{5}}
This reduces to 4Q² + 0.2Q – 0.32499 = 0
Solving Q = 0.261 m³/s or 261 l/s
Now the city center C will receive 361 l/s (previous 289.4 l/s)
To determine the diameter of the connecting pipe RS:
Head drop from A to R
h_{f1}={\frac{8\times0.01\times2000\times0.231^{2}}{\pi^{2}\times9.81\times0.4^{5}}}=8.61~{\mathrm{m}}
Head drop from S to E
h_{f2}={\frac{8\times0.01\times1000\times0.36{1}^{2}}{\pi^{2}\times9.81\times0.45^{5}}}=5.84\ \mathrm{m}
Head drop from A to E = 4 + 15 = 19 m
∴ Head available between
RS = 19 – 8.61 – 5.84 = 4.55 m
Considering Pipe RS
4.55={\frac{8\times0.01\times1500\times0.1^{2}}{\pi^{2}\times9.81\times D^{5}}} Solving D = 0.307 m.
