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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 (QQQ): 20 L/s = 0.02 m³/s
      • Diameter of pipe from A to 1 (D1D_1D1): 150 mm = 0.15 m
      • Diameter of pipe from 2 to B (D2D_2D2): 100 mm = 0.10 m
      • Head loss from A to 1: hL1=3×velocity head in 150 mm pipeh_{L1} = 3 \times \text{velocity head in 150 mm pipe}
      • Head loss from 2 to B: hL2=20×velocity head in 100 mm pipeh_{L2} = 20 \times \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=QAV = \frac{Q}{A}V=AQwhere AAA is the cross-sectional area of the pipe (A=πD2/4A = \pi D^2 / 4A=πD2/4).

      For the 150 mm pipe:

      A1=π(0.15)24=0.0177m2A_1 = \frac{\pi (0.15)^2}{4} = 0.0177 \, \text{m}^2A1=4π(0.15)2=0.0177m2 V1=0.020.01771.13 m/sV_1 = \frac{0.02}{0.0177} \approx 1.13 \, \text{m/s}V1=0.01770.021.13m/sFor the 100 mm pipe:

      A2=π(0.10)24=0.00785m2V2=0.020.007852.55 m/s

    3. Calculate Velocity Heads: The velocity head hvh_vhv is given by:

      hv=V22gh_v = \frac{V^2}{2g}hv=2gV2where g=9.81m/s2g = 9.81 \, \text{m/s}^2g=9.81m/s2.

      For the 150 mm pipe:

      hv1=(1.13)22×9.810.065 mh_{v1} = \frac{(1.13)^2}{2 \times 9.81} \approx 0.065 \, \text{m}
      For the 100 mm pipe:

      hv2=(2.55)22×9.810.331 mh_{v2} = \frac{(2.55)^2}{2 \times 9.81} \approx 0.331 \, \text{m}

    4. Calculate Head Losses:

      hL1=3×hv1=3×0.0650.195 mh_{L1} = 3 \times h_{v1} = 3 \times 0.065 \approx 0.195 \, \text{m}
      hL2=20×hv2=20×0.3316.62 mh_{L2} = 20 \times h_{v2} = 20 \times 0.331 \approx 6.62 \, \text{m}

    5. Total Head Loss: The total head loss hLh_LhL is the sum of hL1h_{L1}hL1 and hL2h_{L2}hL2:

      hL=hL1+hL2=0.195+6.626.815 mh_L = h_{L1} + h_{L2} = 0.195 + 6.62 \approx 6.815 \, \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 HsH_sHs) and the total head loss. Assume HsH_sHs is the vertical distance between A and B.

      Hp=Hs+hLH_p = H_s + h_LSince HsH_sHs is not provided, let’s denote it as HsH_sHs.

    7. Power Required by the Pump: The power PPP required by the pump is given by:

      P=ρgQHpwhere ρ\rhoρ is the density of water (1000 kg/m³).

      P=1000×9.81×0.02×(Hs+6.815)P=196.2(Hs+6.815) W

    8. Final Expression: The power required by the pump depends on the static head HsH_sHs and the calculated head loss. The formula for the power required by the pump is:

      P=196.2(Hs+6.815) W

    If the static head HsH_sHs is provided, you can substitute it into this formula to find the exact power required by the pump.