Consider a curved pipe and nozzle design as sketched in gure

Consider a curved pipe and nozzle design as sketched in gure 1. Suppose water is supplied at a velocity V1, with gauge pressure p1, from the left-hand-side opening, which has area A1. After the 180 bend, the pipe cross sectional area reduces to A2. Assume that head losses are negligible, such that Bernoulli\'s equation may be used.

(i) Obtain an expression for the horizontal force required to hold the pipe in place;

(ii) Evaluate the force if V1 = 0:1 m/s, A1 = 6 cm2, A2 = 2 cm2;

(iii) What is the vertical force component?

Please box all answers and write legibly.

(e) Consider a curved pipe and nozzle design as sketched in figure 1. Suppose water is supplied at a velocity Vi, with gauge pressure p1, from the left-hand-side opening, which has area A1. After the 180o bend, the pipe cross sectional area reduces to A2. Assume that head losses are negligible, such that Bernoulli\'s equation may be used. (i) obtain an expression for the horizontal force required to hold the pipe in place; (ii) Evaluate the force if Vi 0.1 m/s, A1 6cm2, A2 2cm2; (iii) hat is the vertical force component? Figure 1: Nozzle geometry for problem le.

Solution

Assuming Steady, incompressible flow and negligible height difference.

Taking control volume as a rectangle enclosing the whole pipe.

Mass in = mass out -------- conservation of mass - continuity

Density*V₁A₁ = Density*V₂A₂

V₁A₁ = V₂A₂

0.1*6 = V₂*2

V₂ = 0.3 m/s in -x direction

Now applying momentum equation

Fx + density*V₁A₁*V₁ - density*V₂A₂*V₂ = 0

1) F_x + density*(A₁V₁² - A₂V₂² ) = 0

2) F_x = 1000*(0.0002*0.3² - 0.0006*0.1²)

F_x = 0.012 N      ---------------- Horintal Force

3) Since no change in velocity in y direction, No momentum change in y and hence Vertical component of force is 0.

F_y = 0.