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Local Head Losses

Local Head Losses. Local head losses are the “loss” of energy at point where the pipe changes dimension (and/or direction). Pipe Expansion Pipe Contraction Entry to a pipe from a reservoir Exit from a pipe to a reservoir Valve (may change with time) Orifice plate Tight bends.

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Local Head Losses

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  1. Local Head Losses • Local head losses are the “loss” of energy at point where the pipe changes dimension (and/or direction). • Pipe Expansion • Pipe Contraction • Entry to a pipe from a reservoir • Exit from a pipe to a reservoir • Valve (may change with time) • Orifice plate • Tight bends • They are “velocity head losses” and are represented by

  2. Value of kL • For junctions and bends we need experimental measurements • kL may be calculated analytically for • Expansion • Contraction • By considering continuity and momentumexchangeandBernoulli

  3. Losses at an Expansion • As the velocity reduces (continuity) • Then the pressure must increase (Bernoulli) • So turbulence is induced and head losses occur Turbulence and losses

  4. Value of kL for Expansion • Apply the momentum equation from 1 to 2 1 2 • Using the continuity equation we can eliminate Q • From Bernoulli

  5. Value of kL for Expansion • Combine and - Borda-Carnot Equation • Using the continuity equation again

  6. Losses at Contraction • Flow converges as the pipe contracts • Convergence is narrower than the pipe • Due to vena contractor • Experiments show for common pipes • Can ignore losses between 1 and 1’ • As Convergent flow is very stable

  7. Losses at Contraction • Apply the general local head loss equation between 1’ and 2 • And Continuity The value of k depends on In general

  8. Losses: Junctions

  9. Losses: Sharp bends

  10. kL values Bell mouth Entry T-branch kL= 1.5 kL= 0.1 Sharp Entry/Exit kL= 0.5 T-inline kL=0.4

  11. Pipeline Analysis • Bernoulli Equation • equal to a constant: Total Head, H • Applied from one point to another (A to B) • With head losses

  12. Bernoulli Graphically • Reservoir • Pipe of Constant diameter • No Flow A

  13. Bernoulli Graphically • Constant Flow • Constant Velocity • No Friction A

  14. Bernoulli Graphically • Constant Flow • Constant Velocity • No Friction Change of Pipe Diameter A

  15. Bernoulli Graphically • Constant Flow • Constant Velocity • With Friction A

  16. Reservoir Feeding Pipe Example 5.0m

  17. Reservoir Feeding Pipe Example • Apply Bernoulli with head losses pA=pC = Atmospheric uA = negligible

  18. Find pressure at B: Apply Bernoulli A-B pA= Atmospheric u = uB = 2.41 m/s Negative i.e. less than Atmospheric pressure

  19. Pipes in series • Consider the situation when the pipes joining two reservoirs are connected in series 1 2 • Total loss of head for the system is given as A1U1=A2U2=AnUn=Q Q1=Q2=Qn=Q

  20. Pipes in Series Example • Two reservoirs, height difference 9 m, joined by a pipe that changes diameter. For, 15m d=0.2 m then for 45 m, d=0.25 m. f = 0.01 for both lengths. • Use kL,entry= 0.5, kL,exit = 1.0. Treat the joining of the pipes as a sudden expansion. • Find the flow between the reservoirs.

  21. Apply Bernoulli A-B

  22. Pipes in parallel • The head loss across the pipes is equal • Diameter, f, length L,Q, andu may differ • Total flow is sum in each pipe

  23. Knowing the loss of head, the discharges in each pipe can be obtained • The values of U1, U2 and U3 may be obtained from the above equations and hence the discharges Q1, Q2 and Q3 obtained.

  24. The discharge Q is given, the distribution of discharge in different branches is required • The discharges Q1, Q2, Q3 can be expressed in terms of H Where

  25. Similarly Therefore: As the discharge is known and K1,K2 and K3 are constants, the value of H is obtained and the discharge in individual pipes obtained

  26. Pipes in Parallel Example • Two pipes connect two reservoirs which have a height difference of 10 m. Pipe 1 has diameter 50 mm and length 100 m. Pipe 2 has diameter 100 mm and length 100 m. Both have entry loss kL,entry = 0.5 and exit loss kL,exit =1.0 and Darcy f of 0.008. • Find Q in each pipe • Diameter D of a pipe 100 m long and same f that could replace the two pipes and provide the same flow.

  27. Apply Bernoulli for each pipe separately Pipe 1:

  28. Pipe 2:

  29. Flow required in new pipe = • Replace u using continuity Must solve iteratively

  30. Must solve iteratively for D • Get approximate answer by leaving the 2nd term • Increase deqa little, say to 0.106 • … a little more … to 0.107 So

  31. Flow through a by-pass • A by-pass is a small diameter pipe connected in parallel to the main pipe Meter By-pass Main pipe The ratio of q in the by-pass, to the total discharge is known as theby-pass coefficient. by-pass coefficient

  32. Head loss in main pipe = head loss in by-pass Minor losses in the by-pass due to bend and the meter etc Divide by Implies

  33. Using continuity equation Adding one on both sides

  34. By-pass coefficient • Knowing the by-pass coefficient, the total discharge in the main pipe can be determined

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