fluidized bed processing of steel shot

8/3/2019 Fluidized Bed Processing of Steel Shot

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Introduction

In this fluidized bed processing scenario, molten steel is poured from atundish into a cylindrical chamber (diameter = 0.5 m). The molten steel flows

through high velocity inert gas jets, creating small droplets of molten metal. Thedroplets have a diameter of 1 mm and solidify during free fall in the chamber. Forthis example, the chamber has a possibility of four different inert gas atmospheres(Helium, Neon, Argon, and Krypton) at 27°C. However, in an attempt to save space,the droplets are only allowed to cool to 1000°C, well into the Austenite phase fieldfor any range of steel compositions, by the time they reach the end of the chamber.It is assumed that the steel shot has cooled enough that it does not deform when ithits the bottom of the chamber. At the end of the chamber the steel shot rolls ontoa conveyor belt which passes over a set of fans blowing room temperature air upthrough the bed of steel shot, fluidizing the bed. This is where the rest of the coolingoccurs for the steel shot. The conveyor is 0.5 m wide, moves at 5 cm/sec, and is tohave the shortest length possible to conserve space. Since the conveyor moves at

such a low speed, it is assumed that all the cooling is due to the fluidization of thebed, not from the movement of the conveyor. Figure 1 gives a visual representationfor the entire process, which is expected to produce 1 million pounds of steel shotper day.



Fig. 1: A visual depiction of the entire fluidized bedprocessing scenario.



Approach and Calculations

The first step to solving this processing example was to determine the heightof the chamber required to cool the droplets from the tundish temperature (1727°C)to 1000°C. The height of the chamber depended upon how much time is required tocool the shot and the velocity at which the shot is falling. The shot is small enoughthat it will reach its terminal velocity very quickly, so the assumption is made thatthroughout freefall the shot is traveling at terminal velocity. The terminal velocityof spheres in a fluidized bed is given in Equation 1.

where d is the sphere diameter, ρs and ρf are the densities of the sphere and fluid,respectively, g is the acceleration due to gravity, and f is the friction factor. Whilethe diameter, densities, and acceleration due to gravity are all known, the frictionfactor (found in Table 1) depends on the Reynolds number (Re, Equation 2), whichgives an indication of how the fluid is flowing (i.e., laminar, semi-turbulent, or fullyturbulent). The fluid viscosity is denoted ηf .

Table 1: Values for the friction factor, f, based on theReynolds number

Friction Factor, f Reynolds Number, Re

24/Re Re < 118.5/Re3/5 1 < Re < 500

0.44 500 < Re < 20000

Since Re cannot be found without the viscosity, the assumption was madethat Re would fall into one of the three ranges listed in Table 1 and the appropriate

Eq. 1

Eq. 2



friction factor was substituted into Equation 1 as a function of the terminal velocity. The resulting equation looked as follows

Rearranging and combining terms allowed a solution for vt to be found. Thisvalue was substituted into Equation 2 and if the resulting Re fell into the range thatwas assumed, the terminal velocity found was correct. However, if Re fell outside of the assumed range, calculations had to be done again assuming a different frictionfactor. This was done for the 4 different atmospheric gases in the chamber.

The next step to this process was to determine the heat transfer coefficientfor each inert gas. The heat transfer coefficient, h, gives an indication, in this case,of how well heat transfers between the solidifying drop and inert atmosphere andwas found using the Nusselt number. The Nusselt number (Nu) is a dimensionlessnumber which relates the heat transfer coefficient, the sphere diameter, and thethermal conductivity of the fluid (kf ). This relationship can be found in Equation 3.

The Nusselt number also has a relationship between the Reynolds number and thePrandlt number (Pr), another dimensionless number, which can be found in Equation4. The Prandlt number is related to the fluid’s viscosity, specific heat capacity at aconstant pressure (Cpf ), and thermal conductivity by Equation 5.

Substitution of Equation 3 into Equation 4 and rearranging the variablesyielded Equation 6, one which gives h in terms of known variables.

Eq. 3

Eq. 4

Eq. 5

Eq. 6



Heat transfer coefficients were found for each inert gas and the Biot number(Bi) for each gas was found. The Biot number, another dimensionless number,indicates if there are any significant temperature gradients across the solidifyingsolid. If the Biot number (found using Equation 7) is below 0.1, no significant

temperature gradients are found across the solid, making the calculations forcooling time relatively simple. If Bi > 0.1, the calculations for cooling can becomepretty complicated.

In Equation 6, L is the characteristic length associated with the solidifyingsolid and is found as the ratio of volume/surface area and ks is the thermalconductivity of the solid. In the case of a sphere, the characteristic length is L = d/6.Since each gas has a different heat transfer coefficient, each gas will also have adifferent Biot number; therefore, Bi values were calculated for each atmosphericgas to ensure that the following equations for heat removal were valid.

Determining the time to solidify and cool the steel droplets was the next step. The following calculations were made using equations which are based on therebeing no significant temperature gradients across the droplets (Bi >0.1). Thedroplets are at 1727°C (superheated) when they first fall from the tundish. Beforethey can solidify the droplets must cool to the melting temperature for the steel,which was given as 1515°C. Equation 8 gives the equation that was used indetermining the time required to remove the superheat from the molten droplets,

where t is the time required to go from temperature T0 to temperature T1 in a fluidat temperature Tf and Cps is the specific heat capacity at a constant pressure for thesolidifying material. All temperatures entered into the equation were in units of degrees Kelvin.

Even after the superheat is removed the molten drops will not solidify untilthe heat of fusion (characteristic of the material) is removed. The time required to

remove the heat of fusion was found using equation 9. In the equation Vs and As arethe volume and surface area of the solid, respectively, ΔHs is the latent heat of fusion for the solid, and Tmelt is the melting temperature of the solidifying material.

Eq. 7

Eq. 8

Eq. 9



Once the heat of fusion is removed from the droplets, they are finally solidand can continue cooling. The cooling is now exactly the same as outlined in coolingfrom the tundish temperature to the melting temperature and therefore Equation 8was used again. The calculations performed for this report were done assuming that

Cps and ρs stayed constant between the liquid and solid phases. The total coolingtime required to reach 1000°C in the chamber is simply the sum of the three timesmentioned above: time to remove superheat, time to remove the heat of fusion,and time to cool from the melting temperature to 1000°C.

Knowing the velocity of the falling droplets and the amount of time requiredfor them to cool allowed for the determination of the required chamber height.Velocity is a measure of distance/time and multiplying it by a time gives a distance.

The terminal velocity for each gas was multiplied by the total cooling time for thatgas and that returned the distance the droplets would fall in the given atmosphere.

This distance is the required height of the chamber.

It was known that this process was expected to produce one million poundsof steel shot per day. In order to satisfy the principles of the conservation of energy,the volume flow rate entering the chamber had to be the same as that exiting thechamber. Volume flow rates are calculated by multiplying the velocity of an objectby the cross section area that it passes through. The volume flow rate at the topwas the production rate (1,000,000 lbs/day = 5.261 kg/s) divided by the density of the material

However, at the exit of the chamber, the steel shot will not form one solidmass of metal. Instead there will be voids as can be seen in Figure 2. To determinethe area of the voids the size of the square, characterized by the side of length a,had to be determined. This was done using the Pythagorean Theorem with thehypotenuse of the triangle being four times the radius of the shot. Knowing a it waseasy to get the area of the square. The area of the circles inside the square wasalso easily determined and the area of the voids was found by subtracting the areaof the circles from the area of the square. The void area fraction, ω, was determinedusing Equation 10.

Eq. 10



Knowing that the cooling bed was 0.5 m wide, moved at 5 cm/sec, and thatthe volume flow rate at the exit was the same as at the entrance, a bed thicknesscould be determined. Since the spheres only occupied a fraction of 1 – ω of the bed,the cross sectional area which it traveled through had to be multiplied by 1 – ω asseen in Equation 11.

In Equation 10 vbed is the velocity of the conveyor belt, w is the bed width,and z is the bed thickness. Inputting the known variables allowed for a solution for zto be determined. The bed thickness does not factor into any of the remainingequations, but was determined as a sort of reality check to make sure that thecooling bed would not be unreasonably thick.

Figure 2: Schematic diagram that was used indetermining the void area fraction of the shot on theconveyor belt.

The last hurdle to this processing example was to determine how long thecooling conveyor belt needed to be. Cooling of the shot on the conveyor belt is doneby fluidizing the bed of shot with room temperature air. In order to fluidize the bed,the air velocity must be greater than or equal to the minimum fluidization velocityfor the shot. Minimum fluidization is outlined through Equations 12 and 13.

Eq. 11

Eq. 12



The Galileo number, Ga, was first determined because it relies only onphysical constants. Inserting the calculated Galileo number into Equation 12 gavethe Reynolds number required for minimum fluidization. The minimum fluidizationvelocity was then determined using Equation 2. Now that the minimum velocity tofluidize the bed was determined, the terminal velocity for the shot had to bedetermined because when the air velocity reaches the terminal velocity of the shot,the shot would be elutriated, or blown out of the bed. The terminal velocity for theshot being fluidized by air was found in the same fashion as the terminal velocity of the shot in the chamber, using Equation 1 and the correct friction factor.

The velocity of the air has an effect on how fast the shot will cool in thefluidized bed, so determining the heat transfer coefficient as a function of the airvelocity was done next. Equation 6 was used to determine the heat transfercoefficient, with the Reynolds number defined as a function of air velocity, in turndefining the heat transfer coefficient as a function of air velocity. With the heattransfer coefficient defined relative to the air velocity, the Biot number was definedas a function of the air velocity as well. Plots of the heat transfer coefficient and theBiot number versus air velocity can be seen in Figure 3.

Figure 3: Plots of the heat transfer coefficient (left) andBiot number (right) for cooling in the fluidized bed. Note:Bi > 0.1 for all velocities, meaning there are never any significant temperature gradients across the shot.

0 5 10 155 .10

4

0.001

0.0015

0.002

Biair

vair( )

vair

0 5 10 15100

200

300

400

500

hair

vair( )

vair

Eq. 13



Figure 4: A plot of how the cooling time in the fluidizedbed is affected by the air velocity. Note: Larger air velocities result in lower cooling times.

0 5 10 1510

15

20

25

30

35

tcoolair

vair( )

vair



Determining the time for the shot to cool in the fluidized bed was the laststep before a conveyor belt length could be determined. Using Equation 8 the timeto cool the shot from 1000°C to room temperature as a function of air velocity wasfound and a plot of the cooling time can be found in Figure 4. The temperature of the air was assumed to be 25°C and the shot was assumed to cool to 27°C becauseif the shot were to cool to 25°C the equation would be invalid as ln(0) is undefined.

Similar to finding the height required for the chamber, the length of the conveyorbelt was found by multiplying the conveyor velocity by the time required to cool thesteel shot. The conveyor belt length plotted against the velocity of the air can befound in Figure 5. For the complete set of calculations and notes that accompanythem, refer to Appendix A.

Figure 5: Conveyor belt length plotted against airvelocity. Note: Since higher velocities lower the cooling

times, it makes sense that the conveyor belt length would decrease with increasing velocity.

0 5 10 150.5

1

1.5

2

ConveyorLengthvair( )

vair

fluidized bed processing of steel shot

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