4.4 Mixing, Agitation, and Agitator Scale-Up

4.4 Mixing, Agitation, and Agitator Scale-Up

The NCEES PE Chemical specification lists Mixing as one of three named subtopics under Fluids Applications, alongside rotating machinery and flow measurement. It is the fluids topic candidates most often skip, and it is examined with the same dimensionless-group machinery as pipe flow — just built from impeller diameter and rotational speed instead of pipe diameter and velocity.

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1. Impeller Classification

Impellers are classified by the direction of the discharge stream they produce.

The engineering trade is shear versus flow. A Rushton turbine spends its power creating intense local shear behind the blades, which is what you want to break bubbles or droplets. A hydrofoil spends its power moving bulk liquid, which is what you want to blend a large tank or keep solids off the floor. Selecting a Rushton to blend a low-viscosity storage tank wastes an order of magnitude of shaft power.

The standard geometrically similar configuration assumed by most correlations is:

           |<--------- T --------->|
        ___|_______________________|___
       |   :                       :   |   H = T          (liquid depth = tank diameter)
       |   :        baffle ->|     :   |   D/T = 1/3      (impeller diameter)
       |   :     _______     |     :   |   C/T = 1/3      (off-bottom clearance)
       |   :    |__|__|  <-- impeller  |   4 baffles, width T/10 to T/12
       |   :       |   D     |     :   |
       |___:_______|_________|_____:___|
             C  ___|___

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2. The Dimensionless Groups

Agitation Reynolds number. The characteristic length is the impeller diameter and the characteristic velocity is (ND):

$Re = \frac{\rho N D^2}{\mu}$

(N) must be in revolutions per second for the standard power-number charts. Regimes: laminar below (Re \approx 10), transitional between (10) and (10^4), fully turbulent above (Re \approx 10^4).

Power number. The dimensionless shaft power:

$N_P = \frac{P}{\rho N^3 D^5} \quad \Longrightarrow \quad P = N_P \rho N^3 D^5$

• Fully turbulent, baffled: (N_P) is a constant set only by impeller and tank geometry. Power scales as (N^3 D^5) and is proportional to density but independent of viscosity.
• Laminar: (N_P = K_L / Re), so (P = K_L \mu N^2 D^3). Power is proportional to viscosity and independent of density.

That inversion is a favorite exam target: doubling the viscosity of a turbulent low-viscosity blend changes shaft power essentially not at all, while doubling the viscosity of a laminar polymer blend doubles it.

Flow number. The pumping capacity of the impeller:

$N_Q = \frac{Q}{N D^3}$

where (Q) is the volumetric discharge from the impeller. The tank turnover time is (V/Q), and blending requires several turnovers.

Blend time. In the turbulent regime the dimensionless blend time (N\theta_{95}) is approximately constant for a given geometry. A widely used turbulent correlation is:

$N \theta_{95} \approx \frac{5.2}{N_P^{1/3}} \left( \frac{T}{D} \right)^{2}$

For a Rushton turbine at (D/T = 1/3): (N\theta_{95} \approx (5.2/1.77)(9) \approx 26). At (N = 2\text{ rev/s}), (\theta_{95} \approx 13\text{ s}).

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3. Worked Example: Turbulent Power Draw

Problem. A baffled vessel holds water ((\rho = 1{,}000\text{ kg/m}^3), (\mu = 1.0 \times 10^{-3}\text{ Pa}\cdot\text{s})). A Rushton six-blade disk turbine of diameter (D = 1.00\text{ m}) turns at (100\text{ rpm}). Take (N_P = 5.5). Find the regime and the shaft power.

Step 1 — convert speed. (N = 100/60 = 1.667\text{ rev/s}).

Step 2 — regime. $Re = \frac{(1{,}000)(1.667)(1.00)^2}{1.0 \times 10^{-3}} = 1.67 \times 10^{6}$ This is far above (10^4), so the flow is fully turbulent and (N_P = 5.5) is a constant.

Step 3 — power. $P = N_P \rho N^3 D^5 = (5.5)(1{,}000)(1.667)^3(1.00)^5 = (5.5)(1{,}000)(4.632) = 25{,}500\text{ W} \approx 25.5\text{ kW}$

Note the sensitivity: a 10% speed increase raises power by (1.1^3 = 1.33), a 33% increase. Motor sizing on agitators is dominated by the cube law.

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4. Scale-Up: The Criteria Conflict

Scale-up from a pilot vessel to production under geometric similarity (all length ratios held constant) forces a choice, because the common criteria are mutually incompatible.

Worked scale-up. A pilot agitator has (D_1 = 0.30\text{ m}) at (N_1 = 300\text{ rpm}) ((5.00\text{ rev/s})). Scale up geometrically by a factor of 4 to (D_2 = 1.20\text{ m}) at constant (P/V).

$N_2 = N_1 \left( \frac{D_1}{D_2} \right)^{2/3} = 5.00 (0.250)^{2/3} = 5.00(0.3969) = 1.98\text{ rev/s} = 119\text{ rpm}$

Check that power per volume really is constant. Volume scales as (D^3), so (V_2/V_1 = 4^3 = 64). Power scales as (N^3D^5):

$\frac{P_2}{P_1} = (0.3969)^3 (4)^5 = (0.06252)(1{,}024) = 64.0 \quad \checkmark$

Now check tip speed: (v_{tip,1} = \pi(5.00)(0.30) = 4.71\text{ m/s}) and (v_{tip,2} = \pi(1.98)(1.20) = 7.47\text{ m/s}). Tip speed rose by (4^{1/3} = 1.59). Holding (P/V) constant necessarily increases tip speed and therefore maximum shear. If the product is a shear-sensitive crystal slurry, this scale-up will damage it, and you must instead hold tip speed constant — which then drops (P/V) by a factor of (4^{2/3} \approx 2.5) and lengthens blend time.

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5. Solids Suspension and Gas Dispersion

Just-suspended speed. The Zwietering correlation gives the impeller speed (N_{js}) at which no particle rests on the tank floor for more than about one second:

$N_{js} = S\, \nu^{0.1} d_p^{0.2} \left[ \frac{g(\rho_s - \rho_L)}{\rho_L} \right]^{0.45} X^{0.13} D^{-0.85}$

where (X) is the mass ratio of solid to liquid expressed as a percentage and (S) is a geometry constant. The practical lesson is the weak exponents: doubling particle size raises (N_{js}) by only (2^{0.2} = 15%), and quadrupling solids loading raises it by only (4^{0.13} = 19%), while the impeller diameter term (D^{-0.85}) is strong. A larger, slower impeller suspends solids far more efficiently than a small, fast one.

Gas dispersion. For a sparged vessel the aeration number is (N_A = Q_g/(N D^3)). As gas rate rises at fixed speed, gas cavities form behind the blades, the effective density seen by the impeller falls, and the gassed power (P_g) drops below the ungassed power (P_0) — typically to (P_g/P_0 \approx 0.4) to (0.6) for a heavily loaded Rushton turbine. Push further and the impeller floods: gas rises straight up the shaft without being dispersed, and mass transfer collapses. Sizing a motor on ungassed power and then operating gassed leaves the motor oversized; sizing on gassed power and then losing sparger flow overloads the motor, which is why agitator drives on sparged vessels are specified for the ungassed condition.