Conical diffuser calculation using Bernoulli's Equation

[Michael V]

Homework Statement

The following information applies to a conical diffuser:
Length = 750 mm
Inlet diameter = 100 mm
Outlet diameter = 175 mm
Water flow = 50 l/s
Pressure at inlet = 180 kPa

Friction loss = $\frac{k(V_{1} - V_{2})^{2}}{2g}$ where k = 0.15

Calculate the pressure at exit. (10)

The same diffuser is then installed in a vertical pipeline where the flow is now downward and the quantity of water flow is to be double the above flow. The inlet pressure in the diffuser inlet (smaller diameter) is 150 kPa. Consider friction loss and calculate the pressure at the exit of the
diffuser. (10)

Homework Equations

Bernoulli's Equation : $Z_{1}+\frac{P_{1}}{ρg}+\frac{V_{1}^{2}}{2g}$ = $Z_{2}+\frac{P_{2}}{ρg}+\frac{V_{2}^{2}}{2g}+H_{loss}$

The Attempt at a Solution

The 1st part of the question is not a problem where $Z_{1}$= $Z_{2}$. The problem I'm having is when it becomes vertical does $Z_{1}$or $Z_{2}$ = 0.75m.

 

[Chestermiller]

What do the symbols r₁ and r₂ represent in your equation, and what are they doing there?

 

[Michael V]

Sorry, Pr is pressure.

 

[Chestermiller]

To see what to do about Z1 and Z2, consider the simpler case where the flow rate is zero, so all you have is hydrostatic equilibrium. Z1 and Z2 will be the same values when you have flow.

Chet

 

[Michael V]

So what happens when there is flow? Please could just have a look at my answer that's attached. Thanks.

 

[Chestermiller]

So what happens when there is flow? Please could just have a look at my answer that's attached. Thanks.

When there is flow, you just add the velocity and friction terms back into the equation. For hydrostatic equilibrium, if p1 is the pressure at the top and p2 is the pressure at the bottom, and z1 is the height of the top above the bottom datum, while z2 is the height of the bottom datum (z2 =0), Then the pressure at p2 would have to be higher than the pressure p1. So

p2 = p1 + ρg (z1 - z2)

So p1 + ρgz1 = p2 +ρgz2

This should automatically tell you what to use for z1 and z2 in the flow problem.

In your calculations for part 1, you made an error in calculating the velocities. They are a factor of 100 smaller.

 

[Chestermiller]

It seems to me also that you are using the wrong units for pressure in the equation. You are substituting kPa, but it should be expressed in Pa.

 

[Michael V]

When there is flow, you just add the velocity and friction terms back into the equation. For hydrostatic equilibrium, if p1 is the pressure at the top and p2 is the pressure at the bottom, and z1 is the height of the top above the bottom datum, while z2 is the height of the bottom datum (z2 =0), Then the pressure at p2 would have to be higher than the pressure p1. So

p2 = p1 + ρg (z1 - z2)

So p1 + ρgz1 = p2 +ρgz2

Its starting make sense now, I was just struggling with where the bottom datum should be but I can see it has nothing to do with flow rate or its direction.
Thank you so much for your time.

 

[Chestermiller]

This is great news Michael. Sometimes stuff like this can drive you crazy. Keep up the good work.

Chet