In: Mechanical Engineering
Solution:
Given,
Flowrate of Q=0.70ft3/s = 1209.6 in3/s.
Diameters D1=3 in and
D2=1 in.

At section 1,
A1 = 
 = 
 = 7.069 in2
P1 = ?
At section 2,
A2 = 
 = 
 = 0.7854 in2
P2 = The nozzle discharges to atmosphere at point 2 = 0
And
z1 = z2
Now,
Q = A1V1 = A2V2
V1 = 
 = 
 = 171.11 in/s
And
V2 = 
 = 
 = 1540.11 in/s
Applying bernoulli’s equation at sections 1 and 2, we get
= 
Assume density of water 
 = 0.03612729 lb/in3
And gravity g = 386.220472 in/s2

= 
   
  
= 1171330.09
  P1
= 1171330.09*0.03612729
  P1
= 42316.98 lb/in2