# A Pump Draws Water From a Reservoir

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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1. 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:

1. Given Data:
• Discharge ($QQ$Q): 20 L/s = 0.02 m³/s
• Diameter of pipe from A to 1 (${D}_{1}D_1$D1): 150 mm = 0.15 m
• Diameter of pipe from 2 to B (${D}_{2}D_2$D2): 100 mm = 0.10 m
• Head loss from A to 1:
• Head loss from 2 to B:
2. Calculate Velocity in Each Pipe: The velocity $VV$ in a pipe can be calculated using the continuity equation:

$V=\frac{Q}{A}V = \frac\left\{Q\right\}\left\{A\right\}$V=AQwhere $AA$A is the cross-sectional area of the pipe ($A=\pi {D}^{2}\mathrm{/}4A = \pi D^2 / 4$A=πD2/4).

For the 150 mm pipe:

${A}_{1}=\frac{\pi \left(0.15{\right)}^{2}}{4}=0.0177 {2}^{}A_1 = \frac\left\{\pi \left(0.15\right)^2\right\}\left\{4\right\} = 0.0177 \, \text\left\{m\right\}^2$A1=4π(0.15)2=0.0177m2 ${V}_{1}=\frac{0.02}{0.0177}\approx 1.13 m/sV_1 = \frac\left\{0.02\right\}\left\{0.0177\right\} \approx 1.13 \, \text\left\{m/s\right\}$V1=0.01770.021.13m/sFor the 100 mm pipe:

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

3. Calculate Velocity Heads: The velocity head ${h}_{v}h_v$hv is given by:

${h}_{v}=\frac{{V}^{2}}{2g}h_v = \frac\left\{V^2\right\}\left\{2g\right\}$hv=2gV2where $g=9.81 {2}^{}g = 9.81 \, \text\left\{m/s\right\}^2$g=9.81m/s2.

For the 150 mm pipe:

${h}_{v1}=\frac{\left(1.13{\right)}^{2}}{2×9.81}\approx 0.065 mh_\left\{v1\right\} = \frac\left\{\left(1.13\right)^2\right\}\left\{2 \times 9.81\right\} \approx 0.065 \, \text\left\{m\right\}$
For the 100 mm pipe:

${h}_{v2}=\frac{\left(2.55{\right)}^{2}}{2×9.81}\approx 0.331 mh_\left\{v2\right\} = \frac\left\{\left(2.55\right)^2\right\}\left\{2 \times 9.81\right\} \approx 0.331 \, \text\left\{m\right\}$

${h}_{L1}=3×{h}_{v1}=3×0.065\approx 0.195 mh_\left\{L1\right\} = 3 \times h_\left\{v1\right\} = 3 \times 0.065 \approx 0.195 \, \text\left\{m\right\}$
${h}_{L2}=20×{h}_{v2}=20×0.331\approx 6.62 mh_\left\{L2\right\} = 20 \times h_\left\{v2\right\} = 20 \times 0.331 \approx 6.62 \, \text\left\{m\right\}$

5. Total Head Loss: The total head loss ${h}_{L}h_L$hL is the sum of ${h}_{L1}h_\left\{L1\right\}$hL1 and ${h}_{L2}h_\left\{L2\right\}$hL2:

${h}_{L}={h}_{L1}+{h}_{L2}=0.195+6.62\approx 6.815 mh_L = h_\left\{L1\right\} + h_\left\{L2\right\} = 0.195 + 6.62 \approx 6.815 \, \text\left\{m\right\}$

6. 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.

7. Power Required by the Pump: The power $PP$P required by the pump is given by:

$P=\rho gQ{H}_{\mathrm{p}}$where $\rho \rho$ρ is the density of water (1000 kg/m³).

$P=1000×9.81×0.02×\left({H}_{s}+6.815\right)$$P=196.2\left({H}_{s}+6.815\right) W$

8. 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:

$P=196.2\left({H}_{s}+6.815\right) W$

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.