Page 76 - ebook fluid mechanics_finalize

FLUID MECHANICS

b) Discharge at section A = Discharge at section B

Q = Q
B
A

A v = A v
A A B B

A v
v = A A
B
A B

. 0 318 ( ) 4
=
 ( ) 3.0 2

4= 5 . m s /

For continuity of flow

Q = Q + Q
C
B
D
Q = Q − Q
C
D
B
(A = B v B ) (A− C v C )

 ( ) 3.0 2   ( ) 2.0 2 
4
.

=   5  −   4
 4   4 
0= . 192 m 3 s /

For pipe BD

Q = A  v
D
D
D
0= . 192 m 3 s /

Q
v = D
D
A D

. 0 192 4
= 10= . 86 m s /
 ( 15.0 ) 2

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