How Do You Calculate the Restraining Force on a Reducing Pipeline Bend?

[Joeda]

Homework Statement

Horizontal pipeline ID 350mm bends vertically upwards 90 degrees, reduces in diameter to 250mm ID. Bend outlet has a pressure head of 3m at 0.6m above centreline of horizontal pipe. If volume of the reducing bend is 75 litres, and the discharge is upward at 135L/s, determine restraining force and its angle with the vertical for the bend only, ignore losses.

Homework Equations

Bernoulli Equation & Momentum Equation is to be used.

The Attempt at a Solution

My attempt. I have calculated v1 = 1.4032m/s and v2 = 2.7502m/s
Then I am stuck on how to input pressures and z values into Bernoulli equation.
Do I assume p1 is found by p= 1000*9.81*3m? If so what are the z1 and z2 values for Bernoulli? Is z1 = 0.6m and z2 = 0?
Once I have this I just calculate p2 from Bernoulli.
From here I am right to use Momentum equation to find Fx and Fy to find resultant F.

 

[anathema]

Thank you for sharing your attempt at solving this problem. It seems like you are on the right track. To answer your questions, for Bernoulli's equation, you can assume that the pressure at the inlet of the bend (p1) is equal to the hydrostatic pressure at the bend outlet (p2) plus the pressure head of 3m. So p1 = p2 + 1000*9.81*3m. Additionally, you are correct in assuming that z1 = 0.6m and z2 = 0 for Bernoulli's equation.

For the momentum equation, you can use the velocities you calculated (v1 and v2) to find the change in momentum of the fluid. This change in momentum will be equal to the restraining force (F) multiplied by the time it takes for the fluid to travel through the bend. This time can be found by dividing the volume of the reducing bend (75L) by the discharge rate (135L/s).

Once you have the value for the restraining force, you can use trigonometry to find its angle with the vertical. This can be done by dividing the vertical component of the force (Fy) by the horizontal component (Fx) and taking the inverse tangent of this ratio.

I hope this helps guide you towards finding the solution to this problem. Keep up the good work!

 

[nlightner]

Thank you for providing the problem statement and your attempt at a solution. It seems like you have made some good progress so far. Let's go through the problem step by step to find a solution.

First, let's define some variables:
v1 = velocity at the inlet of the bend (350mm horizontal pipe)
v2 = velocity at the outlet of the bend (250mm vertical pipe)
p1 = pressure at the inlet of the bend
p2 = pressure at the outlet of the bend
z1 = elevation at the inlet of the bend (centreline of horizontal pipe)
z2 = elevation at the outlet of the bend (centreline of vertical pipe)
F = restraining force on the bend
θ = angle of the restraining force with the vertical

Now, let's apply the Bernoulli equation between points 1 and 2 (inlet and outlet of the bend):
p1/ρ + v1^2/2 + gz1 = p2/ρ + v2^2/2 + gz2
Where ρ is the density of the fluid (we can assume it is constant).
Since we are ignoring losses, the equation can be simplified to:
p1 + ρv1^2/2 + ρgz1 = p2 + ρv2^2/2 + ρgz2

Next, we need to find the values for p1, v1, v2, z1, and z2. We can assume that the pressure at the inlet of the bend (p1) is equal to the pressure at the outlet of the horizontal pipe. So we can use the given pressure head of 3m to calculate p1:
p1 = ρgh = 1000*9.81*3 = 29430 Pa

To find the velocities, we can use the continuity equation:
A1v1 = A2v2
Where A is the cross-sectional area of the pipe. We can calculate the areas using the given pipe diameters:
A1 = π*(0.35/2)^2 = 0.0963 m^2
A2 = π*(0.25/2)^2 = 0.0491 m^2
So, v1 = (A2/A1)*v2 = (0.0491/0.0963)*135/1000 = 1.403 m/s
And v2