Determine the loss in available energy from section 1 to sec

Determine the loss in available energy from section (1) to section (2) of the nozzle in the following figure. Density of water is 62.4 lb/ft^3 and gravity of acceleration is 32.174 ft^2/s.

Solution

Due to the conservation of mass the relation for mass flow at the inlet and outlet is given by,

A1 x V1 = A2 x V2

V2 = A1 x V1 / A2

V2 = (3.14 x (12x 0.83)^2 / 4) x 5 = 3.89 ft/s or 46.72 in/s

Applying Bernoulli\'s principle for the pipe flow,

p1 + 1/2 rho V1^2 + rho x g x h1 = p2 + 1/2 rho x V2^2 + rho x g x h2

Assuming p2 = 0 and h1 = , the loss in the pipe given by,

p1 + 1/2 rho (V1^2 - V2^2) = rho x g x (h2 - h1)

p1/ (rho x g) + 1/2g (V1^2 - V2^2) = del h

[15 x 144 / 62.4 x 32.17] + 1/ 2 x 32.17 x (5^2 - 3.89^2) = del h

del h = 1.22 ft

The head loss per unit volume is rho x g x del h = 62.4 x 32.17 x 1.22 = 2455.8 lb / ft-s