A pump draws water from reservoir A and lifts it to reservoir B as shown. The loss of head from A to 1 is 3 times the velocity head in the 150 mm pipe and the loss of head from 2 to B is 20 times the velocity head in the 100 mm pipe. The discharge is 20 L/s.

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To solve the problem of determining the total head loss and the power required by the pump to move water from reservoir A to reservoir B, we need to apply the principles of fluid mechanics, particularly the Bernoulli equation and head loss equations. Here’s the step-by-step approach:

Given Data:Calculate Velocity in Each Pipe:The velocity $VV$ in a pipe can be calculated using the continuity equation:$V=\frac{Q}{A}V\; =\; \backslash frac\{Q\}\{A\}$V=AQwhere $AA$A is the cross-sectional area of the pipe ($A=\pi {D}^{2}\mathrm{/}4A\; =\; \backslash pi\; D^2\; /\; 4$A=πD2/4).

For the 150 mm pipe:

${A}_{1}=\frac{\pi (0.15{)}^{2}}{4}=0.0177\hspace{0.17em}{2}^{}A\_1\; =\; \backslash frac\{\backslash pi\; (0.15)^2\}\{4\}\; =\; 0.0177\; \backslash ,\; \backslash text\{m\}^2$A1=4π(0.15)2=0.0177m2 ${V}_{1}=\frac{0.02}{0.0177}\approx 1.13\hspace{0.17em}m/sV\_1\; =\; \backslash frac\{0.02\}\{0.0177\}\; \backslash approx\; 1.13\; \backslash ,\; \backslash text\{m/s\}$V1=0.01770.02≈1.13m/sFor the 100 mm pipe:

${A}_{2}=\frac{\pi (0.10{)}^{2}}{4}=0.00785\hspace{0.17em}{2}^{}$${V}_{2}=\frac{0.02}{0.00785}\approx 2.55\hspace{0.17em}m/s$

Calculate Velocity Heads:The velocity head ${h}_{v}h\_v$hv is given by:${h}_{v}=\frac{{V}^{2}}{2g}h\_v\; =\; \backslash frac\{V^2\}\{2g\}$hv=2gV2where $g=9.81\hspace{0.17em}{2}^{}g\; =\; 9.81\; \backslash ,\; \backslash text\{m/s\}^2$g=9.81m/s2.

For the 150 mm pipe:

${h}_{v1}=\frac{(1.13{)}^{2}}{2\times 9.81}\approx 0.065\hspace{0.17em}mh\_\{v1\}\; =\; \backslash frac\{(1.13)^2\}\{2\; \backslash times\; 9.81\}\; \backslash approx\; 0.065\; \backslash ,\; \backslash text\{m\}$

For the 100 mm pipe:

${h}_{v2}=\frac{(2.55{)}^{2}}{2\times 9.81}\approx 0.331\hspace{0.17em}mh\_\{v2\}\; =\; \backslash frac\{(2.55)^2\}\{2\; \backslash times\; 9.81\}\; \backslash approx\; 0.331\; \backslash ,\; \backslash text\{m\}$

Calculate Head Losses:${h}_{L1}=3\times {h}_{v1}=3\times 0.065\approx 0.195\hspace{0.17em}mh\_\{L1\}\; =\; 3\; \backslash times\; h\_\{v1\}\; =\; 3\; \backslash times\; 0.065\; \backslash approx\; 0.195\; \backslash ,\; \backslash text\{m\}$

${h}_{L2}=20\times {h}_{v2}=20\times 0.331\approx 6.62\hspace{0.17em}mh\_\{L2\}\; =\; 20\; \backslash times\; h\_\{v2\}\; =\; 20\; \backslash times\; 0.331\; \backslash approx\; 6.62\; \backslash ,\; \backslash text\{m\}$

Total Head Loss:The total head loss ${h}_{L}h\_L$hL is the sum of ${h}_{L1}h\_\{L1\}$hL1 and ${h}_{L2}h\_\{L2\}$hL2:${h}_{L}={h}_{L1}+{h}_{L2}=0.195+6.62\approx 6.815\hspace{0.17em}mh\_L\; =\; h\_\{L1\}\; +\; h\_\{L2\}\; =\; 0.195\; +\; 6.62\; \backslash approx\; 6.815\; \backslash ,\; \backslash text\{m\}$

Total Head to be Supplied by the Pump:To lift the water from reservoir A to B, the pump must overcome the elevation difference (static head ${H}_{s}H\_s$Hs) and the total head loss. Assume ${H}_{s}H\_s$Hs is the vertical distance between A and B.${H}_{p}={H}_{s}+{h}_{L}H\_p\; =\; H\_s\; +\; h\_L$Since ${H}_{s}H\_s$Hs is not provided, let’s denote it as ${H}_{s}H\_s$Hs.

Power Required by the Pump:The power $PP$P required by the pump is given by:$P=\rho gQ{H}_{\mathrm{p}}$where $\rho \backslash rho$ρ is the density of water (1000 kg/m³).

$P=1000\times 9.81\times 0.02\times ({H}_{s}+6.815)$$P=196.2({H}_{s}+6.815)\hspace{0.17em}W$

Final Expression:The power required by the pump depends on the static head ${H}_{s}H\_s$Hs and the calculated head loss. The formula for the power required by the pump is:$P=196.2({H}_{s}+6.815)\hspace{0.17em}W$

If the static head ${H}_{s}H\_s$Hs is provided, you can substitute it into this formula to find the exact power required by the pump.