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Rope Pump – An alternative for developing country like India

Rope Pump – An alternative for developing country like India. By Kishan Majethia Aasif Momin Raviraj Parmar Ritesh Patel. Design Contents. Power and Torque, Pipe size and Head. Load on the pistons, Piston size. Volume of water column. Velocity of rope, length.

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Rope Pump – An alternative for developing country like India

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  1. Rope Pump – An alternative for developing country like India By Kishan Majethia Aasif Momin Raviraj Parmar Ritesh Patel

  2. Design Contents • Power and Torque, Pipe size and Head. • Load on the pistons, Piston size. • Volume of water column. • Velocity of rope, length. • Pulley diameter, Shaft diameter, bearing. • Flow rate. • Truss analysis, for structure, supporting pulley and shaft

  3. Torque Calculation • A single human can provide an effort of about 50 W and can give rotation of about 30-100 rpm to the handle. • So, we should design a pulley according to this capacity of human. • Let P = 50 W N = 30 rpm P = (2∏NT) / 60 So, T = 15.91 N m

  4. Weight Relationship • Let H = 2 m head • Pipe size is affected by weight of water because, W = V * ₰ * g Where V = Volume ₰ = density So, W ∞ d2 H • So, as the diameter and head increases the weight of water also increases. Data says the size of the pipe available in the market is 0.75”, 1”, 1.25”, 2.5”. Let us choose size of diameter of pipes as 1” for head of 2 m

  5. Spacing between two pistons • As the space between two pistons increases the load on pistons also increases the load on pistons also increases. P = ₰gh So, P ∞ h So, 20-30 mm should be selected. Weight of water column overall, W = (3.14/4) D2 H ₰ g W = (3.14/4) (0.0381)2 * 2 * 9810 = 22.368 N

  6. Standard data • D = diameter of pulley = 250 mm Therefore, (T1 – T2) = 22.368 N T = (T1 – T2) D2 T = 22.368 * (0.250/2) T = 2.796 N m • This is less than torque limit of human so it is on the safer side. • P = (2 * ∏ * N * T) / 60 For W = 22.369 N H = 2 m d = 1” D = 250 mm

  7. Design of shaft T1 / T2 = e(µαcosecβ) T1 / T2 = e(0.3 * ∏ cosec 25) = 11.8 Where µ = 0.3 for nylon and steel (T1 – T2) = 22.368 = net weight 11.8 T2 - T2 = 22.368 T2 = 2.071 N T1 = 24.438 N F = T1 + T2 = 26.509 N R1 = R2 = 13.254 N

  8. Design of shaft Moment max M = 26.509 * 150 Mo = 3976.35 N mm Now T = 2.796 N m M = 3.976 N m Te = √(km*M)2 + (kt*T)2 Where km = 1.5 kt = 1 So, Te = 6.587 Nm Te = (∏ / 16) * T * d3 D3 = 798.74 D = 10 mm

  9. Load on pistons Piston diameter = 0.037 m Weight of water = (∏/4) * d2* h * ₰ * g = (∏/4) * (0.037)2 * 0.2 * 9810 = 2.236 N Now h = 20 cm So, number of pistons in a pipe at a moment is, H = N h N = 2000 / 200 = 10

  10. Load on pistons Load on piston P = ₰ * g * h = 9810 * 0.2 = 1962 N / m2 This much pressure should be sustainable by each piston. So, piston material should be selected based on this.

  11. Stress distribution in each pistons

  12. Flow rate obtained Q = Velocity * Area Velocity of rope V = (∏*d*N) / 60 = (∏ * 250 * 30) / 60,000 = 0.392 m/s Area of the pipe = 1.14 * 10-3 m2 = (∏/4) d2 So, d = 0.0381 m = 1.5 inch

  13. Flow rate obtained • Q = 4.469 * 10-4 m3/s • = 0.4469 lit/ min • So, Q = 26.814 lit/min • We have Q ∞ DN ∞ Velocity

  14. Result table

  15. Final design for 2 m head

  16. Thanks

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