Calculating Pressure Drop in a Fluidized Bed of Spheres Using the Ergun Equation

In summary, the problem involves air flowing at 100 F through a packed bed of spheres with a void fraction of 0.38, a bed diameter of 2 feet, and a height of 8 feet. The air enters at 16.2 psia and a rate of 47.5 lb/min. Using the Ergun equation, the pressure drop of the bed was found to be 104.3883 lbf/ft^2 after iterations that took into account the changes in density and velocity. It was also clarified that this is a packed bed and not a fluidized bed.
  • #1
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Homework Statement


Air at 100 F is flowing through a packed bed of spheres having a diameter of 0.5 inches. The void fraction (ε) of the bed is 0.38 and the bed diameter is 2 feet. The bed height is 8 feet. If the air enters at 16.2 psia at a rate of 47.5 lb / min calculate the pressure drop of the bed in psi.

Homework Equations


Ergun Equation
http://en.wikipedia.org/wiki/Ergun_equation

The Attempt at a Solution


For this problem, I want to use the Ergun equation. In our lecture slide, it said that for gases, the ergun equation can be used if the average density of the gas is used in the density term (slide attached). I have a couple questions about this problem. First, do I assume that the gas is isothermal and remains at 100 F upon entry and exit? Second, I don't think I can use the density of the air at the inlet as the density term, but I did it anyway because I didn't know what to do. If I don't know the pressure at the outlet, then how can I find the average density? This is twofold because the air density can be affected by the temperature, which is uncertain right now if its 100 F at the outlet, as well as the pressure at the outlet.

The viscosity of air was found online
http://www.lmnoeng.com/Flow/GasViscosity.php
 

Attachments

  • 4.1 attempt 1.pdf
    249 KB · Views: 238
  • Ergun Equation.pdf
    65 KB · Views: 286
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  • #2
You can assume that the temperature is constant if the bed is adiabatic. There are two reasons for this. (1) The amount of viscous heating is very little, so that wouldn't cause much of a temperature rise, even for a liquid. (2) From the steady flow version of the first law, the change in enthalpy of the gas is zero for a throttling operation; this is because, for an ideal gas, the viscous heating is "exactly" canceled by the expansion cooling.

You could take into account the density variation very accurately by applying the ergun equation differentially (i.e., dp/dL) to the problem, but that probably isn't necessary in this case because the density change is so small.

Now that you have an estimate of the pressure change, calculate the density at the exit of the bed with this exit pressure. Then average the exit density with the inlet density, and redo the calculation (taking into account that uo must also be averaged, say, using the average density). This is going to be an iterative solution. Keep iterating until the exit pressure and exit density have converged enough to satisfy you.

Incidentally, this is not a fuildized bed, it's a packed bed. In a fluidized bed the particles are levitated by the gas flow.

Chet
 
  • #3
I did the iterations, and got 104.3883 lbf/ft^2 as the pressure drop
 

Attachments

  • 4.1.xlsx
    9.4 KB · Views: 254
  • #4
Maylis said:
I did the iterations, and got 104.3883 lbf/ft^2 as the pressure drop
Wow. I'm a little confused. What happened to that 3.84 psi you got originally?

Chet
 
  • #5
Yes, my R gas constant was wrong. I did it as Fahrenheit when it should have been Rankine.
 

Related to Calculating Pressure Drop in a Fluidized Bed of Spheres Using the Ergun Equation

1. What is a fluidized bed?

A fluidized bed is a type of reactor or container where a solid material, such as a powder or granular substance, is suspended and mixed with a gas or liquid. This creates a fluid-like behavior and allows for efficient mixing and reactions.

2. What causes pressure drop in a fluidized bed?

The pressure drop in a fluidized bed is caused by the resistance of the gas or liquid passing through the bed and the friction between the particles. As the fluidization velocity increases, the pressure drop also increases due to increased particle movement and collisions.

3. How is the pressure drop in a fluidized bed calculated?

The pressure drop in a fluidized bed can be calculated using the Ergun equation, which takes into account the fluid velocity, particle size, and bed density. Other factors such as bed height and particle shape can also affect the pressure drop and should be considered in the calculation.

4. How can pressure drop be reduced in a fluidized bed?

Pressure drop in a fluidized bed can be reduced by optimizing the process parameters, such as fluid velocity and particle size, to minimize particle collisions and friction. The use of additives or anti-agglomeration agents can also help reduce pressure drop by preventing particles from sticking together.

5. What are the consequences of high pressure drop in a fluidized bed?

High pressure drop in a fluidized bed can lead to reduced efficiency and performance of the reactor or container. It can also cause issues such as back-mixing, channeling, and non-uniform gas or liquid distribution, which can affect the quality of the final product. Therefore, it is important to monitor and control pressure drop in a fluidized bed to ensure optimal operation.

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