Page 91 - ebook fluid mechanics_finalize

FLUID MECHANICS

From Bernoulli’s Equation,

v 2 p v 2 p
z + 1 + 1 = z + 2 + 2
1
2g  1 2g 

 p − p  

v 2 2 − v 1 2 = 2g  1 2  + ( − zz 1 2 ) ——————(1)
   

For continuity of flow,

A
A v = A v or v = 1 v = mv
2
2
2
1
1
A 2 1 1
where
A
m = area ratio = 1
A 2

Substituting in equation (1) and solving for v1

 p − p  
v 2 2 − v 1 2 = 2g  1 2  + ( − zz 1 2 )

   

1   p − p   
v 1 =  2g  1 2  + ( − zz 1 2 )  
(m 2 −1 )      

Actual discharge, Qactual = C  A  v
1
1
d

C d  A 1   p 1 − p 2  )  
Q actual =  2g   + (z 1 − z 2   ——— (2)
(m 2 − ) 1      

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