The ratio of the lower (dн) and upper (dв) drain pipe diameters is one of the most important parameters in the hydrocyclone calculations. The dв/dн ratio defines the condensed and clarified product ratio at the port outlet. Numerically, this relation can be represented by an approximate formula:
Qн/Qв = 1.13·(dн/dв)³
Qн – condensed product flow in the lower branch pipe, m³/s;
Qв – clarified product flow in the upperp branch pipe, m³/s.
The overall mass balance of a hydrocyclone is as follows:
Qоб = Qв+Qн
Qоб – total capacity of the hydrocyclone, m³/s.
It follows from here that the clarified and condensed product yields can be determined by knowing the total flow rate and the ratio of diameters of the hydrocyclone outlet pipes:
Qв = Qоб / [1+1.13·(dн/dв)³]
Qн = Qоб - Qв
Basic dimensions of the hydrocyclone depend on the diameter of its cylindrical part (d). Experimentally it was found that the following dimension ratios are optimal in terms of the unit hydraulic resistance reduction:
dвх = 0.25·d
dв = 0.3·d
dн = (0.2-0.8)·dв
The lower branch pipe is usually removable, with an adjustable bore. That’s why it’s possible to regulate operating parameters of the unit and to set, by changing the dв/dн ratio, the desired ratio of the condensed and clarified product flows.
The total capacity of the hydro-cyclone Qоб can be approximately determined by the following relationship (diameters and pressure are substituted in meters and Pascals, respectively):
Qоб = 5.46·10-3·dвх0.9·dв0.9·pвх0.5
pвх – fluid pressure at the hydrocyclone inlet, Pa.
Various methods may be used to determine the separating capacity of the hydrocyclone. One option is to determine the mesh grain size (dгр). The mesh grain size in the separation process means the specific size of solid particles in the dispersed phase, when particles larger than mesh grain size will be separated in the hydrocyclone, and particles with a smaller diameter will be carried away with the clarified product flow. The following formula is used to determine dгр:
dгр = 8.44·10³· √((dв·d·cвх)/(Kф·dн·√(pвх)·(ρт-ρж)))
dгр – separation grain mesh diameter, μm;
d – diameter of the cylindrical part of the hydrocyclone, m;
dв – upper branch pipe diameter, m;
dн – lower branch pipe diameter, m;
cвх – initial suspension concentration,% (wt);
pвх – suspension pressure at the hydrocyclone inlet, Pa;
ρт – solid phase density, kg/m³;
ρж – liquid phase density, kg/m³;
Kф = 0.8 + 1.2/(1+100·d) – coefficient of the hydrocyclone shape.
Generally, hydrocyclones are usually calculated by successive approximations, by setting hydrocyclone dimensions based on available experimental data; after that, basic parameters (flows and degree of separation) are calculated. In case of a mismatch, input parameters are changed, with a subsequent iteration to follow.
The first stage of the calculation is to choose the optimum type of cyclone for the goal specified. The choice is made based on experimental data and depends on a variety of parameters, such as the physical properties of gas and its impurities, area available to accommodate the unit, gas flow supply and discharge features, etc.
Further, based on the operating experience of the selected cyclone and properties of the gas to be purified, the optimum gas velocity in the unit (vопт) is selected; usually, this value lies in the range from 2 to 5 m/s. After that, the cross-section area of the unit is determined by the formula:
F = Q/vопт
F – cyclone cross-section area, m²;
Q – dusted gas flow, m³/s;
Vопт – optimum gas velocity in the cyclone, m/s.
If it’s necessary to calculate a battery of cyclones, the diameter of an individual unit can be found as follows:
d = √(F/(0.785·N))
d – cyclone diameter, m;
N – number of cyclones.
After that, a more precise gas recalculation is made:
vопт = Q / (0.785·N·d²)
Cyclone pressure head losses, corresponding to the velocity, are determined by the formula:
∆p = ζц · [(ρг·vопт²)/2]
Δp – pressure drop across the cyclone, Pa;
ζц – coefficient of hydraulic resistance of the cyclone;
ρг – gas density under operating conditions, kg/m³.
The calculation of the cyclone efficiency is, to a certain extent, similar to the hydrocyclone calculation. The criterion of efficiency is d50 - the diameter of particles trapped with an efficiency of 50%. To calculate this value, use the following formula:
d50 = d50т·√[(d/dт) · (ρт/ρ) · (μ/μт) · (vт/v)]
d – unit diameter, m;
ρ – density of separated particles, kg/m³;
μ – dynamic viscosity of the dusted gas at the operating temperature, Pa·s;
v – gas velocity in the unit, m/s.
Values with the “t” index mean the reference cyclone operation conditions, and values without this index are the calculated values.
Statement of problem:
There is a hydrocyclone with the following characteristics: feed pipe diameter dпит = 0.1 m, discharge pipe diameter dсл = 0.03 m. A pressure drop ΔP = 0.15 MPa is created in the hydrocyclone. This hydrocyclone is intended to remove suspended particles from a fluid flow of 20 l/min. It is required to establish find out whether this hydrocyclone is suitable for the specified task.
Let’s determine the maximum capacity of the hydrocyclone according to the following formula (correction factor k to be assumed equal to 5):
Q = k·dпит·dсл·√(g·∆P) = 5·0.1·0.03·√(9.81·150000) = 18.2 l/min
The obtained value of the maximum flow rate is less than the required one:
Hence, it can be concluded that the hydrocyclone, descried in the statement of problem, is not suitable for specified conditions.
Answer: not suitable
Statement of problem:
Following a modification of the process flow diagram, the composition of effluents, supplied to a hydrocyclone for treatment, has changed. The main task of the hydrocyclone is to separate at least 60% of all solids from water being treated; for a new suspension composition this is equivalent to trapping particles with a diameter of at least 1×10-6 m. Parameters of the hydrocyclone are as follows: cylindrical part diameter D = 0.5 m, length L = 1.2 m and separation zone height l = 1.8 m. The inlet pipe diameter is dвх = 0.08 m. Water is supplied at a flow rate Q = 100 m3/h. Densities of the liquid and solid phases are ρж = 1000 kg/m3 и ρт = 1900 kg/m3, respectively. Viscosity of the treated suspension is μ = 0.0012 Pa·s. It’s necessary to determine whether replacement of the hydrocyclone is required.
Let’s first determine the suspension velocity at the hydrocyclone inlet:
vвх = (Q·4)/(π·[ввх]2) = (100·4)/(3600·3.14·[0.08]2) = 5.5 m/s
Next, let’s find a tangential velocity of particles:
υф = 31.5·υвх·(dвх/D) (L/D)(-0.32) = 31.5·5.5·0.08/0.5·(1.2/0,5)(-0.32) = 20.9 m/s
Let’s determine the size of particles trapped by the existing hydrocyclone:
dт = 1.65·dвх·√[μс/(υф·l·(ρт-ρж))] = 1.65·0.08·√[0.0012/(20.9·1.8·(1900-1000))] = 0.25·10(-6) m
The value we received is less than the critical diameter specified in the statement of problem. Therefore, the existing hydrocyclone for sure will meet the wastewater treatment requirements.
Answer: Replacement is not required
Statement of problem: There are two hydrocyclones with the same upper and lower pipe diameters, dв = 140 mm and dн = 80 mm, respectively, but with different diameters of the cylindrical part of the body, d1 = 400 mm and d2 = 500 mm, respectively. It is necessary to clarify turbid water with a solid phase concentration c = 0.5% (wt) and density ρт = 2500 kg/m3 to the level when it will not contain any particles with a diameter of more than 5 μm. The suspension can be directed to the hydrocyclone at a pressure p = 0.7 MPa. The density of water to be assumed equal to ρж = 1000 kg/m³.
Problem: Determine which of two hydrocyclones is suitable for fulfilling the task.
Solution: Suitability of the cyclones can be assessed by determining a separating capacity of these cyclones based on the mesh grain diameter (dгр), to be then compared with the statement of problem. For this purpose, it’s necessary to use the equation for the mesh grain diameter:
dгр = 8.44·10³·√(dв·d·cвх) / (Kф·dн·√p·(ρт-ρж))
where Kф = 0.8 + 1.2/(1+100·d) – hydrocyclone shape factor.
Let’s find dгр for the first cyclone.
Kф1 = 0.8 + 1.2/(1+100·0.4) = 0.829
dгр1 = 8.44·10³·√(0.14·0.4·0.5) / (0.829·0.08·√700000·(2500-1000)) = 4.9 μm
Let’s find dгр for the second cyclone.
Kф2 = 0.8 + 1.2/(1+100·0.5) = 0.824
dгр2 = 8.44·10³·√(0.14·0.5·0.5) / (0.824·0.08·√700000·(2500-1000)) = 5.49 μm
Finally we receive that Kф1<5 μm, while Kф2 > 5 μm. Based on this, we can conclude that only the first hydrocyclone is suitable for the specified task.
Answer: The first hydrocyclone is suitable.