1. Piping Systems

2. Piping Systems- Example 1-1 or (a)or (b) 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

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 inlet (0.78) & exit (1), p.18 For valve (55) & elbows (2*30), NOTE: K=Cf T

5. Re & f T No pump Given

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

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

8. Hardy-Cross Method & Program (2)

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

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

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

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

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

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

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