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

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SUMMARY

The discussion centers on calculating the pressure drop in a packed bed of spheres using the Ergun equation. The problem involves air flowing at 100°F through a bed with a diameter of 2 feet and a height of 8 feet, with a void fraction of 0.38 and sphere diameter of 0.5 inches. The final pressure drop calculated after iterations was 104.3883 lbf/ft², correcting an earlier miscalculation due to the incorrect use of the gas constant in Fahrenheit instead of Rankine.

PREREQUISITES
  • Understanding of the Ergun equation for pressure drop calculations
  • Knowledge of fluid dynamics, particularly in packed beds
  • Familiarity with gas properties, including density and viscosity
  • Basic thermodynamics principles, especially regarding isothermal processes
NEXT STEPS
  • Study the application of the Ergun equation in packed versus fluidized beds
  • Learn about iterative methods for solving fluid dynamics problems
  • Research the effects of temperature and pressure on gas density
  • Explore the significance of using Rankine versus Fahrenheit in gas calculations
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Students and professionals in chemical engineering, mechanical engineering, and anyone involved in fluid dynamics and pressure drop calculations in packed beds.

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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

Physics news on Phys.org
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
 
I did the iterations, and got 104.3883 lbf/ft^2 as the pressure drop
 

Attachments

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

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