# Ergun Equation reulting to zero pressure ratio

• maistral
In summary, the pressure near the inlet and outlet of the bed decreases as the volumetric flow rate is increased. This is equivalent to a physical phenomenon where the exit pressure becomes zero.

#### maistral

I was trying to emulate a packed bed reactor, and apparently the pressure midway at the packed column became zero, until my numerical method failed altogether (as the pressure variable in the Ergun is at the denominator).

I'd like to be enlightened about one thing - does this phenomena (zero pressure midway the packed column) has something to tell me at what will happen at the end? No gas flow or something, perhaps? Thanks.

EDIT: Sorry for the title spelling error, it's 'resulting', not reulting.

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maistral said:
I was trying to emulate a packed bed reactor, and apparently the pressure midway at the packed column became zero, until my numerical method failed altogether (as the pressure variable in the Ergun is at the denominator).

I'd like to be enlightened about one thing - does this phenomena (zero pressure midway the packed column) has something to tell me at what will happen at the end? No gas flow or something, perhaps? Thanks.

EDIT: Sorry for the title spelling error, it's 'resulting', not reulting.
The pressure should not go to zero if you are using the ergun equation. Please write your form of the ergun equation so we can see what you are dealing with.

Chet

I was toying around the flowrate, Vo. When I assign increasing values for Vo, I found out that the pressure is going down.

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maistral said:
I was toying around the flowrate, Vo. When I assign increasing values for Vo, I found out that the pressure is going down.

What determines the pressure at the inlet and outlet of the bed?

I'm really sorry, I didn't understand the question :|

Actually what's happening is that, the inlet pressure is at 7.5MPa, and I am using the Ergun equation to find out the outlet pressure. What's happening is that if I toy around the Vo by increasing it (volumetric rate), the pressure decreases rapidly until it approaches a certain high Vo value and the pressure approaches 0.

I was wondering if this is equivalent to a physical phenomenon when the exit pressure becomes zero (ie. the gas won't flow anymore, the packings get destroyed, or something).

Last edited:
maistral said:
I'm really sorry, I didn't understand the question :|

Actually what's happening is that, the inlet pressure is at 7.5MPa, and I am using the Ergun equation to find out the outlet pressure. What's happening is that if I toy around the Vo by increasing it (volumetric rate), the pressure decreases rapidly until it approaches a certain high Vo value and the pressure approaches 0.

I was wondering if this is equivalent to a physical phenomenon when the exit pressure becomes zero (ie. the gas won't flow anymore, the packings get destroyed, or something).

What you are saying is that, as you increase the throughput rate through a packed bed, the pressure drop through the bed increases. Why does this surprise you?

Uh, I do know that the pressure should go down, but what I mean is that the pressure ratio at the end given a certain high flowrate is zero; does this signify something as a physical phenomenon? No effluent or something?

This is my first time to see a zero pressure ratio in my calculations, I'm sorry. XD

maistral said:
Uh, I do know that the pressure should go down, but what I mean is that the pressure ratio at the end given a certain high flowrate is zero; does this signify something as a physical phenomenon? No effluent or something?

This is my first time to see a zero pressure ratio in my calculations, I'm sorry. XD

I don't know what you mean by the pressure ratio.

Chet

I mean P/Po is going towards zero. The P there at the denominator is continually decreasing, causing the entire thing to approach zero apparently.

I was wondering what does this have to do as a physical phenomenon, when the exit pressure is zero.

Arr, nevermind. Fluidization velocity is what I was looking for, apparently. Thanks.

## What is the Ergun Equation and how does it relate to zero pressure ratio?

The Ergun Equation is a mathematical formula that describes the pressure drop in a packed bed or fluidized bed. It is commonly used in chemical engineering and fluid mechanics. When the pressure ratio is zero, it means that there is no change in pressure between the inlet and outlet of the bed, resulting in a pressure drop of zero.

## What factors affect the Ergun Equation when the pressure ratio is zero?

The Ergun Equation is affected by several factors, including the particle size and density, fluid velocity, bed porosity, and fluid viscosity. When the pressure ratio is zero, the bed porosity and fluid velocity become the dominant factors in determining the pressure drop.

## How is the Ergun Equation used in practical applications with a zero pressure ratio?

The Ergun Equation can be used to determine the minimum fluid velocity required to maintain a packed bed or fluidized bed at a constant pressure when the pressure ratio is zero. It can also be used to determine the maximum pressure that can be applied to the bed before it becomes fluidized.

## What are some limitations of the Ergun Equation when the pressure ratio is zero?

One limitation of the Ergun Equation at a zero pressure ratio is that it assumes a constant bed porosity. In reality, the porosity may change depending on the fluid velocity and pressure. Additionally, the equation may not accurately predict pressure drops at extremely low fluid velocities or high viscosities.

## Are there any alternative equations or models that can be used instead of the Ergun Equation at a zero pressure ratio?

Yes, there are alternative equations and models that can be used in place of the Ergun Equation when the pressure ratio is zero. These include the Blake-Kozeny Equation and the Forchheimer Equation, which take into account factors such as particle shape and interparticle forces that may affect the pressure drop in a bed. Experimental data and empirical correlations can also be used to estimate pressure drops in these situations.