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Piping Systems

Piping Systems. or (a). or (b). Piping Systems- Example 1-1. Type I (explicit) problem : Given: L i , D i , Q i ; Find H i. Type II (implicit) problem : Given: L i , D i , H i ; Find Q i. Type III (implicit) problem : Given: L i , H i Q i ; Find D i. Example 1-1 (Continue).

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Piping Systems

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  1. Piping Systems

  2. or (a) or (b) Piping Systems- Example 1-1 Type I (explicit) problem : Given: Li, Di, Qi; Find Hi Type II (implicit) problem : Given: Li, Di, Hi; Find Qi Type III (implicit) problem : Given: Li, Hi Qi; Find Di

  3. Example 1-1 (Continue) May be used to find any variable if the others are given, i.e. for any type of problem (I, II, III)

  4. Example 1-2 with MathCAD For valve (55) & elbows (2*30), NOTE: K=CfT For inlet (0.78) & exit (1), p.18

  5. No pump Given Re & fT

  6. Solution Satisfying conservation of mass and energy equations we may solve or “guess and check” any piping problem …!

  7. Iteration # • Since a node pressure must be unique, then net pressure loss head around any loop must be zero: • If we assume Qb0 to satisfy Eq. 1, the Eq. 2 will not be satisfied. • So we have to correct Qb0 for DQloop, so that Eq. 2 is satisfied, i.e.: Hardy-Cross Method & Program • For every node in a pipe network:

  8. Hardy-Cross Method & Program (2)

  9. Example 1-13 (cont.) ...loops ...pipes ...guess (LX1) …while tolerance is not satisfied Hardy-Cross subroutine …correction (LX1) …new Qs (PX1)=(PXL)(LX1) …result (PX1) The results after Hardy-Cross iterations

  10. Loop 1 Loop 2 This 3rd loop is not independent (no new pipe in it) Example 1-13

  11. 2 loops 3 pipes “Connection” matrix N Example 1-13 (cont.)

  12. Example 1-13 (cont.) d’s L’s units & g Pump and reservoirs dhd/dQ derivative roughnesses Kin. viscosity

  13. Example 1-13 (cont.) Re & fT Laminar & turbulent f No minor losses

  14. About constant “Connection” matrix N 2 loops 3 pipes Example 1-13 (cont.) Loss & device heads Derivative of h(Q) Qi guesses from conservation of mass

  15. Example 1-13 (cont.) ...loops ...pipes ...guess …while tolerance is not satisfied Hardy-Cross subroutine Since assumed Q>0 …correction The results after Hardy-Cross iterations …new Qs …result

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