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Suspension and nucleation power

The hydrostatic pressure exerted by the crystals at the base of the bed on the liquid is:

Assuming the following mean values: void fraction of bed: s = 0.85

crystal density: ps = 1,500 kg.m-3

gravity field: g = 9.81 m.s 2

bed height: h = 2 m

Assuming, with respect to the liquid and the volume of the body:

  • - the velocity of the liquid in the empty vat, UL = 10-2 m.s-1;
  • - the diameter of the body, DC = 3 m.

The suspension power is:

The volume of the body is:

Hence, 312/14 = 23 W.m-3 for the suspension volume power.

The suspension volume power for a CFC is considerably greater: 200-400 W.m-3.

Thus, the secondary nucleation is negligible and the primary nucleation is low.

Calculation procedure

Here, we base our method on that proposed by Miller and Seaman [MIL 47].

We estimate the porosity ?, giving it a constant value that is not unreasonable. At the top of the bed, the crystals are small and their slip velocity is low. At the bottom of the bed, the number of crystals is the same but these crystals are larger and their slip velocity is higher. It follows that porosity e is the result of two values of opposite effect, LP and Vgl.

The empty vat velocity U of the liquid varies between certain limits according to the product treated:

On the other hand, if we accept:

Void fraction of bed: e = 0.8

Real density of solid: pS = 1,500kg.m-3

Maximum mass flowrate density: 150kg.m-2.h-1

Fall speed of crystals relative to the workshop is:

Therefore, we can disregard the velocity of crystals relative to that of the liquid. Thus, assuming immobile crystals, the velocity of the liquid in the empty vat is:

Vgl: slip velocity of the liquid relative to immobile crystals (m.s-1)

n: number of crystals per cubic meter of the slurry (m-3)

L: size of crystals at the level in question (m).

From the above, we can deduce that:

The crystal surface per cubic meter of the bed is:

The growth kinetic is written as:

CE : mass of free water per cubic meter of the bed (kg.m-3)

AC : crystal area per cubic meter of the bed (m-1).

We can write out:

By knowing the desired production P (in kg/h), together with the desired product size Lp, we can deduce the product JV of the nucleation rate J by the volume V of the bed:

We establish the material balance for supersaturation to disappear completely through a fluidized bed:

WE : free water flow (kg.s-1)

AXP : supersaturation at the bed base.

From this equation, we deduce the value of WE corresponding to AXp. In addition, on a certain level:

Differentiating relative to height z in the bed:

If we replace d^X and AX in the kinetic equation with their values, we dx

obtain:

Write:

The kinetic equation becomes:

We integrate this equation in order to have the profile of L and that of AX .

In the calculation of a, we see that CEo intervenes, which is calculated as follows:

e : fraction of fluid volume in the bed pL : liquor density (kg.m-3)

Xo: solute ratio (kilogram of solute per kilogram of liquor).

EXAMPLE.-

Consider the crystallization of K2 SO4 with the following data (taken from an example treated differently by Mullin [MUL 72]).

Assume that, at the base of the bed, the volume fraction of liquid is:

So:

The kinetic equation is written:

So:

We will call Lo the size of crystals at height x that will be defined as the bed summit. After integration, we obtain, for a height of 6 m, the size limit at the bed summit:

Laboratory study of a fluidized bed

We use a mass M of crystals that all have a size equal to the size limit set for the crystallizer summit. The crystals are fluidized with the saturated liquor at empty vat velocity U.

The bed cross-section is A and its measured height is H. Its volume is, accordingly, AH.

If ps is the density of the crystals:

According to Richardson and Zaki:

Or rather:

The straight line obtained in logarithmic coordinates provides Vt and n. We know that:

V : limit fall velocity of an isolated particle of size L (m.s-1) n: exponent function of V (see section 4.2.4).

We can write out:

and:

 
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