I'm using a 2-inch sawtooth high shear impeller at rpm of 10K-20K for particle size reduction, tanks diameters are between 130-140mm with aspect ratio of 1.3-1.4. I want to improve the particle size reduction while minimizing the heating since the product is heat sensitive. The parameters I can control are the RPM, mixing time and impeller diameter. The formulas I know: P_imp=Np*rho*N^3*D^5 Q=N*D^3 –represents circulation H=N^2*D^2 - represents shear P is the power, N is the RPM, D is the impeller diameter. I also read that at high Reynolds numbers, the concepts of shear, mean shear, and the impeller rotational speed, N, become unimportant relative to the impeller tip velocity, ND, since viscosity is no longer the mechanism by which momentum is transferred.
1.At high Reynolds numbers does the shear proportional to the tip velocity, ND, or to N^2*D^2?
2. Why does viscosity is no longer the mechanism by which momentum is transferred at high Reynolds numbers?
3. If I increase the rpm by 50% and reduce the mixing time by factor of (1.5)^3=3.3 (P_imp=Np*rho*N^3*D^5), would it improve the particle size reduction while keeping the same heating?
4. Or should increase the RPM while reducing the impeller diameter and keep the same mixing time? For example, if I decrease the impeller diameter by 30% and increase the rpm by 50%,would I have better particle size reduction and less heating of the product? P=Np*Rho*(1.5)^3*(0.7)^5=0.56 –less heating and higher shear. What would give the best results while minimize the heating?
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Re: How do I improve the particle size reduction while minimizing the heating?
6 September 2012 at 3:11pmLet's take your problem statement, assumptions and questions a step at a time:
Dimensions: Impeller diameter [D] 2" (50.8 mm) diameter
Vessel diameters [T] 130mm D/T = 0.39 140mm D/T = 0.36 Impeller to tank diameter ratios [D/T] look reasonable for high-shear dispersion with sawtooth impeller
Aspect ratios 1.3 to 1.4 (liquid height to vessel diameter ratios (?)
Aspect ratios between 0.8 and 1.0 would give better mixing in most applications.
However, high impeller speed in a small vessel is likely to form a deep vortex and draw air into the impeller.
Gas dispersion through the impeller destroys flow, reduces power, and hurts liquid-liquid dispersion. Operating assumptions:
Impeller power: P_imp = Np * rho * N^3 * D^5
Correct description for impeller power, Np is impeller power number which will depend on sawtooth design.
Impeller pumping: Q = Nq * N * D^3
Description for impeller pumping capacity, if Nq (pumping number) is known.
Pumping capacity has limited value, since it does not tell about flow direction, impact on fluid volume, or effect on mixing and shear.
Head: H related to N^2 * D^2
Head is an attempt to define pressure force on fluid in impeller region. Tells almost nothing about effective shear.
Other important assumptions:
Heat = P * t All power input by the impeller adds heat to the fluid.
So heat input (temperature increase) equals power times time.
Tip speed = pi * N * D
Ultimate drop diameter in typical liquid-liquid dispersions is roughly inversely proportional to impeller tip speed.
Ultimate drop diameter may take different amounts of time to achieve in different systems.
Times should be very short in very small vessels. Shorter dispersion time will reduce heat input and may not change drop size.
Questions and answers:
1. Average shear is inversely proportional to rotational speed [1/N]. Local shear is a function of local velocity gradients, and closely related to tip speed [ND]. Ultimate drop size is inversely proportional to tip speed [1/ND]. Batch uniformity is closely related to N^2 * D^2 (torque per volume), but time to uniformity will be very fast in small vessels. Blend time is inversely proportional to rotational speed [1/N] with geometric similarity.
2. High values of Reynolds number just describe turbulent conditions. Turbulence occurs when inertial forces (pressure forces) dominate over viscous forces (drag forces). Turbulence results in small eddies dissipating energy to the molecular level (heat). Viscous forces are so small that they have little influence on mixing results at high Reynolds numbers.
3. Increase impeller speed with same diameter impeller means that N=1.5x (x means times), so ND=1.5x, drop size = 1/(1.5ND) = 67% of original size. Power increase = (1.5)^5 = 3.4x. Reducing time by 1/(3.4) means equal energy input, so equal temperature increase. Reduced time may not reach ultimate drop size, but probably will in small vessel. Larger vessel may be different. Ambient heat loss in a small vessel may be proportionately much greater than in larger vessel.
4. Increase rpm and reduce impeller diameter: 1.5x N and 0.78x D result in equal power. Tip speed = 1.5 N * 0.78 D = 1.17 ND, a 17% increase in tip speed should result in 85% drop size. Higher speed, smaller impeller could increased tip speed and reduce drop size for the same power input. Practical limit set by impeller ability to rapidly mix all of the vessel contents.
The answers are simple in a small vessel, because mixing and approach to ultimate drop size can be rapid. The disadvantage is the possibility of drawing air into the impeller and having dispersed gas hurting mixing and dispersion. Scale-up effectively reverses the situation. A large vessel will take longer to mix and dispersion may take longer to achieve ultimate drop size. Rate may affect results. Surface vortexing will be less in large scale and same tip speed will be less likely to draw air into the dispersion. Larger vessels will have smaller surface to volume ratios. Heat removal at vessel walls will be less and impeller heat input will result in higher batch temperatures.
The answers by this expert are based on the best available interpretation of the information provided. The consequences of the application of this information are the responsibility of the user. If clarification is needed, please submit a further question.
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