Question 6 of 7: Hydrocyclone Sizing (Krebs Approach)

Question 6: Hydrocyclone Sizing (Krebs Approach) (20%)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

a) D_50c for 60% minus 74 μm

Given. Target: 60% (wt.) passing 74 μm (200 mesh) in the product. Overflow-size-distribution multiplier table: at 60.0% passing, multiplier=2.08.

Find. Required D_50c.

Approach. Apply the given multiplier directly to the target micron size, exactly as the worked example on the exam's own information page demonstrates.

1. Apply the multiplier. $$D_{50c}=2.08\times74=\boxed{154\ \mu\text{m}}$$

b) Overflow size at 80% passing for D_50c=100 μm

Given. D_50c=150 mesh=100 μm; multiplier at 80.0% passing=1.25.

Find. The micron size that is 80% passing in the overflow.

Approach. This is the inverse of part (a): the given D_50c IS the multiplier applied to the unknown micron size, so divide instead of multiply.

1. Solve for the micron size. $$100=1.25\times x\ \Rightarrow\ x=\frac{100}{1.25}=\boxed{80\ \mu\text{m}}$$

c) Full hydrocyclone design – 98.8% removal of minus 37 μm

Given. Remove 98.8% of −37 μm (400 mesh) particles; feed 45 L/s at 20% solids by volume (V); solids SG G_s=2.82 (liquid phase water, G_L=1.0); ΔP=140 kPa.

Find. Cyclone diameter and number of units required.

Approach. Convert the 98.8%-removal target into a required D_50c(application) via the overflow-size table, back out D_50c(base) by dividing out the three Krebs correction factors, solve the base-D_50c equation for cyclone diameter D, then read the corresponding unit capacity from the Cyclone Capacity–Pressure Drop chart at ΔP=140 kPa to get the number of units for the 45 L/s feed.

1. Required D_50c(application). At 98.8% passing, table multiplier=0.54: $$D_{50c,app}=0.54\times37=\boxed{20.0\ \mu\text{m}}$$
2. Correction factors. $$C_1=\left[\frac{53-20}{53}\right]^{-1.43}=(0.6226)^{-1.43}=1.969$$ $$C_2=3.27(140)^{-0.28}=3.27(0.2507)=0.820$$ $$C_3=\left[\frac{1.65}{2.82-1.0}\right]^{0.5}=(0.9066)^{0.5}=0.952$$ $$C_1C_2C_3=1.969(0.820)(0.952)=\boxed{1.537}$$
3. D_50c(base) and cyclone diameter. $$D_{50c,base}=\frac{D_{50c,app}}{C_1C_2C_3}=\frac{20.0}{1.537}=\boxed{13.0\ \mu\text{m}}$$ $$13.0=2.84\,D^{0.66}\ \Rightarrow\ D=\left(\frac{13.0}{2.84}\right)^{1/0.66}=\boxed{10.0\ \text{cm}}$$
4. Capacity per cyclone and number of units. Reading the Cyclone Capacity–Pressure Drop chart along the 10 cm cyclone line (anchored at 20 kPa→2 L/s, and following the chart's sqrt(ΔP) scaling to 140 kPa): $$Q_{unit}=2.0\sqrt{\frac{140}{20}}=2.0\sqrt{7}=\boxed{5.3\ \text{L/s per 10 cm cyclone}}$$ $$N=\left\lceil\frac{45}{5.3}\right\rceil=\boxed{9\ \text{cyclones}}$$