Conical diffuser calculation using Bernoulli's Equation

AI Thread Summary
The discussion focuses on calculating the pressure at the exit of a conical diffuser using Bernoulli's Equation, considering both horizontal and vertical flow scenarios. Key points include the importance of correctly identifying the height terms (Z1 and Z2) in relation to hydrostatic equilibrium and flow conditions. Participants emphasize the need to adjust velocity calculations and ensure pressure units are consistent, converting kPa to Pa as necessary. The conversation highlights common misconceptions about flow direction and its impact on pressure calculations. Overall, the thread provides guidance on applying Bernoulli's principles to real-world fluid dynamics problems.
Michael V
Messages
25
Reaction score
0

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.
 
Last edited:
Physics news on Phys.org
What do the symbols r1 and r2 represent in your equation, and what are they doing there?
 
Sorry, Pr is pressure.
 
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
 
So what happens when there is flow? Please could just have a look at my answer that's attached. Thanks.
 

Attachments

Michael V said:
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.
 
  • Like
Likes 1 person
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.
 
Chestermiller said:
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.
 
This is great news Michael. Sometimes stuff like this can drive you crazy. Keep up the good work.

Chet
 
Back
Top