# V 0 04 2000 0 3 2 9 81 2 2 v therefore v 1 0767 ms v

• Notes
• 44

This preview shows pages 25–31. Sign up to view the full content.

v = 0 04 2000 0 3 2 9 81 2 2 . * . * . v Therefore, v 1 = 0.767 m/s v 2 = 0.664 m/s Hence Q 1 = A 1 v 1 = π * . * . 0 6 4 0 767 2 = 0.217 m 3 /s Q 2 = A 2 v 2 = π * . * . 0 3 4 0 664 2 = 0.047 m 3 /s Total discharge, Q = Q 1 + Q 2 = 0.217 + 0.047 m 3 /s = 0.264 m 3 /s

This preview has intentionally blurred sections. Sign up to view the full version.

Fluid Mechanics Chapter 7 – Steady Flow in Pipes P.7-26 7.5.2 Branched-pipe Problem h 1 h 3 h 2 J reservoir 1 reservoir 3 reservoir 2 pipe1, k 1 pipe2, k 2 pipe 3, k 3 Q 1 Q 2 Q 3 Assume h 1 > h 2 > h 3 and the 3 pipes intersect at junction J. As h 1 is the highest head, the flow in pipe 1 must be toward J. As h 3 is the lowest head, Q 3 is flowing from J to the reservoir 3. The flow Q 2 ’s direction is unknown because it depends on the head at junction J. If h J be the head at junction J. There are two possible cases (i) h 1 > h J > h 2 , or (ii) h 2 > h J > h 3 For case (i), h 1 > h J > h 2 , Q 2 is from J to reservoir 2. Q 1 - Q 2 - Q 3 = 0 h 1 – h J = k 1 *Q 1 2 h J - h 2 = k 2 *Q 2 2 (7.24) h J - h 3 = k 3 *Q 3 2 For case (ii), h 2 > h J > h 3 , Q 2 is from reservoir 2 to J. Q 1 + Q 2 - Q 3 = 0 h 1 – h J = k 1 *Q 1 2 h 2 – h J = k 2 *Q 2 2 (7.25) h J - h 3 = k 3 *Q 3 2
Fluid Mechanics Chapter 7 – Steady Flow in Pipes P.7-27 Both sets of equations have 4 unknowns Q 1 , Q 2 , Q 3 and h J . We have to determine which case controls the problem. It is determined by assuming h J = h 2 , i.e. no flow from J to reservoir 2. Therefore Q 1 = h h k 1 2 1 (7.26) Q 3 = h h k 2 3 3 (7.27) If Q 1 ’ > Q 3 ’, Q 2 is from J to reservoir 2 - case (i). If Q 1 ’ < Q 3 ’, Q 2 is from reservoir 2 to J - case (ii).

This preview has intentionally blurred sections. Sign up to view the full version.