## EXAIR Adjustable Air Amplifiers Helps Clear the Smoke

An overseas customer had a problem with their coal blasting furnace. As the workers would open a 1.2 meter by 1.2 meter door to shovel in coal, the foundry would fill with smoke.  This was a hazard and a nuisance for the crew.  They saw articles about how EXAIR Air Amplifiers were used in smaller ovens for exhausting hot flue gases, and they wondered if the EXAIR Air Amplifiers could be used for something much larger.

He sent me an email with some additional details about their furnace system. They had a fan that was mounted in the stack that had a capacity of 50 m^3/min.  This was fed into a filtration collection system to remove the residue byproducts.  The temperature inside the furnace was approximately 450 deg. C.  From this information, I could calculate the required velocity to keep the smoke inside the furnace.

In sizing this application, I determined that I could use an equation from Heskestad and Spaulding. This equation was developed to find the minimum velocity required to keep smoke from egressing into corridors during fires.  In this case, we were keeping the smoke from egressing into the foundry.  The formula looks like this:

V = 0.64 * Sqrt(g * H * (T – To)/T)      Equation 1

V – Velocity (m/s)

g – Gravitational acceleration (9.8 m/s^2)

H – Height of Opening (meters)

T – Avg. Fire Temperature (Kelvin)

To – Avg. Space Temperature (Kelvin)

In this equation, we are mainly fighting the forces of the temperature difference from inside the hot furnace area to the outside cooler area.  The outside area was near 40 Deg. C, and this gave me the temperature difference.  In converting these temperatures to the absolute temperature, Kelvin.  I calculated the fire temperature, T, to be 450 Deg. C + 273 = 723 Kelvins; and the space temperature, To, to be 40 Deg. C + 273 = 313 Kelvin.

In placing the given information into Equation 1, the minimum velocity could be found.

V = 0.64 * Sqrt(9.8 m/s^2 * 1.2m * (723K – 313K)/ 723K)

V = 1.65 m/s

If the velocity could be maintained at this mark of 1.65 m/s, then the smoke could not egress into the plant.  They had a stack fan that was flowing 50 m^3/min, or 0.83 m^3/sec.  We can determine the velocity that the stack fan was producing by calculating the flow over an area:

V = Q/A      Equation 2

V – Velocity (m/s)

Q – Flow (m^3/sec)

A – Area (m^2)

With a door opening of 1.2m by 1.2m, or 1.44m^2, the velocity can be calculated by placing the known values into Equation 2:

V = (0.83 m^3/s) / (1.44m^2)

V = 0.58 m/s

Now we can see why they were getting smoke pluming from the coal furnace into their facility. They required a minimum of 1.65 m/s, and the stack fan was only drawing 0.58 m/s.  If we take the difference, we can determine how much additional velocity will be required to keep the smoke within the furnace: 1.65 m/s – 0.58 m/s = 1.07 m/s.

To determine how much air flow would be needed to create a velocity of 1.07 m/s through the door opening, I just had to rearrange Equation 2 to determine the flow, Q.

Q = V * A = 1.07 m/s * 1.44 m^2 = 1.54 m^3/s

To better correlate the flow data, I converted 1.54 m^3/s to 92.4 m^3/min of air flow.