A Pump Draws Water From a Reservoir

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: $h_{L1}=3\times velocity head in 150 mm pipeh\_\{L1\}\; =\; 3\; \backslash times\; \backslash text\{velocity\; head\; in\; 150\; mm\; pipe\}$
     
   • Head loss from 2 to B: $h_{L2}=20\times velocity head in 100 mm pipeh\_\{L2\}\; =\; 20\; \backslash times\; \backslash text\{velocity\; head\; in\; 100\; mm\; pipe\}$
     
2. 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=AQ​where $AA$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$

3. 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​=2gV2​where $g=9.81\hspace{0.17em}{2}^{}g\; =\; 9.81\; \backslash ,\; \backslash text\{m/s\}^2$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:

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

4. Calculate Head Losses:

   $h_{L1}=3\times h_{v1}=3\times 0.065\approx 0.195\hspace{0.17em}mh\_\{L1\}\; =\; 3\; \backslash times\; h\_\{v1\}\; =\; 3\; \backslash times\; 0.065\; \backslash approx\; 0.195\; \backslash ,\; \backslash text\{m\}$
   $h_{L2}=20\times h_{v2}=20\times 0.331\approx 6.62\hspace{0.17em}mh\_\{L2\}\; =\; 20\; \backslash times\; h\_\{v2\}\; =\; 20\; \backslash times\; 0.331\; \backslash approx\; 6.62\; \backslash ,\; \backslash text\{m\}$
   

5. 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​:

   $h_L=h_{L1}+h_{L2}=0.195+6.62\approx 6.815\hspace{0.17em}mh\_L\; =\; h\_\{L1\}\; +\; h\_\{L2\}\; =\; 0.195\; +\; 6.62\; \backslash approx\; 6.815\; \backslash ,\; \backslash text\{m\}$
   

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:

   $P=\rho gQ{H}_{\mathrm{p<span></span>}}$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$

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:

   $P=196.2(H_s+6.815)\hspace{0.17em}W$

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.