Page 110 - ebook fluid mechanics_finalize

FLUID MECHANICS

This equation can therefore be written,

A 2  p − p 
v 1 = 2g  1 2  ——————(4)
(A 1 2 − a 2 2 )   

So,

Actual discharg e = coefficien t of discharg e theoretica l discharg e

Q actual = Cd  A 1 1 v ——————(5)

Putting v1 into (5)

A 2  p − p 
A
Q actual = Cd  1  2g  1 2  ——————(6)
(A 1 2 − a 2 2 )   

A 2
but, m = 1
A 2 2

so putting m into (6),

C A  p − p 
Q actual = Cd  A 1  d 1 2g  1 2 
(m 2 −1 )   

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