In: Mechanical Engineering
Given,
Dia. of pipe, (D) = 6 in.
A1 = D2 = *62 = 28.27 in2
Dia. of Nozzle, (d) = 3 in.
A2 = d2 = *32 = 7.068 in2
Pressure drop (p1 - p2) = 0.525 psi
From continuity equation,
A1V1 = A2V2
28.27V1 = 7.068V2
V1 = 0.25V2
Now,
Applying Bernoulli's equation at sections 1 & 2, we get
+ + z1 = + + z2
But,
z1 = z2
+ = +
- = -
=
=
0.03768 = 1.209*10-3*
= 31.16
V2 = 5.583 in/s
Theoretical discharge (Q) = V2*A2 = 5.583*7.068
Q = 39.46 in3/s
Q = 10.249 gallons per minute
Assume ,
Actual discharge (Qact) = V1*A1
But,
We know,
V1 = 0.25*V2 = 0.25*5.583 = 1.395 in/s
(Qact) = 1.395*28.27 = 39.44 in3/s
(Qact) = 10.243 gallons per minute.
Now,
Cd = Coefficient of discharge
We have,
Cd =
Cd = = 0.99