Figure 3.5 Typical pumping number curve.
(3.14)
The range between laminar and turbulent is called the transition range. It is a very broad range. That is because turbulence commences first near the impeller blade tips somewhere in the region of NRe = 10, but does not reach all of the far corners of the tank until NRe > 20 000. The power number gradually levels out in this range. For some impellers, it goes through a minimum in the transition range.
Figure 3.5 represents a typical pumping number curve, as a function of Reynolds number and having D/T as a parameter.
In laminar flow, the pumping number becomes a constant. This means that under such conditions, the impeller pumping is independent of viscosity, though the power draw is directly proportional to viscosity.
In turbulent flow, the pumping number again becomes constant at a given D/T, though it is much higher than the laminar pumping number. As D/T increases, the pumping number generally decreases. This is because recirculating flow within the tank creates a velocity opposite the impeller discharge, impeding its flow. Nonetheless, larger impellers pump more for a given power input than smaller ones.
When vendors quote a pumping number, it is usually in turbulent flow and at a D/T of 1/3 and a C/T of 1/3. Many consider this geometry to be “standard”, but there is no fundamental reason to adhere to this geometry when designing agitation equipment.
Figure 3.6 represents a typical generic blend time curve.
We can observe that blend time becomes constant in both laminar and turbulent flow. However, the laminar dimensionless blend time is often several orders of magnitude greater than the turbulent blend time. Specialized impellers have been developed for laminar flow mixing. We need not delve into these in this book, as fermenters never operate in the laminar range, because it is basically impossible to both incorporate gas into highly viscous liquids and have the gas exit in a reasonable amount of time after the gas is depleted.
Some versions of the dimensionless blend time curve incorporate the D/T effect into the Y‐axis expression, as stated under the dimensionless blend time definition given earlier.
Figure 3.7 depicts gassing factor as a function of Aeration number for three different impeller types.
Gassing factor depends on Aeration number, Froude number, D/T, Reynolds number, and impeller type. Therefore, it is impossible to show the entire relationship on a simple two dimensional graph. Figure 3.7 is based on turbulent flow, a D/T of 1/3, and a Froude number of 0.5.
We can readily see that the traditional impeller for gas dispersion, the Rushton (or disk) turbine, has a rather severe drop in power at high gas flow rates. As we will see later, this is a problem for maintaining maximum mass transfer. That is why it has been pretty much replaced by modern, deeply concave impellers such as the BT‐6 by Chemineer, the Phasejet by Ekato, and several others that have less effect of gas flow on power draw.
Figure 3.6 Dimensionless blend time.
Figure 3.7 Gassing factors.
We will cover other dimensionless parameters, such as for heat transfer, as needed in subsequent chapters. For now, we will show some example calculations.
Example 1: Power Draw Calculation
A tank has an impeller of 1000 mm diameter, rotating at a shaft speed of 125 rpm, in a fluid with a specific gravity of 1.2. It has a known power number of 0.75. How much power will it draw?
Answer
It is best to first convert all parameters to SI units. So, we have a 1m impeller diameter, turning at a shaft speed of 125 rpm/60, or 2.0833/s, in a fluid with a density of 1200 kg/m3. Since the power number is dimensionless, it remains at 0.75. The definition of power number, from Eq. 3.3, is:
(3.15)
Rearranging to calculate power:
(3.16)
Since it is more common to think of power in kW, this would become 8.13 kW. An 11 kW motor (a standard motor rating) would be ample to power this impeller under these conditions.
Example 2: Pumping Calculation
A tank has a hydrofoil impeller of 6 ft. diameter rotating at 30 rpm. It has a known pumping number of 0.5. How much fluid will it pump?
Answer
We do not have to do a unit conversion for this one. The units used merely determine the resultant pumping units. Equation (3.5) defines pumping number and is reproduced below:
(3.17)
Rearranging to calculate Q (pumping rate), we get:
(3.18)
Had we used SI units, the calculated pumping capacity would have been expressed as m3/s, but the physical quantity represented would have been the same.
Example 3: Blend Time Calculation
This really only gets interesting in transition flow, where dimensionless blend time varies with Reynolds number. But let’s suppose we have an impeller with a dimensionless blend time of 10. How quickly would it blend the tank operating at 30 rpm?
Answer
If τ*N =10, τ = 10/N = 10/30(1/min) = 1/3 min = 20 s. Note that one can use any units for shaft speed; the resultant blend time units will be determined by this. Note also that as long as the dimensionless blend time is fixed, the result is independent of tank size. This means, for a given D/T and Reynolds number, the blend time at 30 rpm will be the same in any size tank. However, the P/V goes up exponentially when trying to keep the same D/T and the same shaft speed. So, designing for the same blend time in a large tank as a small one can be problematic.
A Primer on Rheology
It is not the purpose of