# Examples of centrifuge calculations

**Contents:**

## Sedimentation centrifuges

In order to determine the basic design characteristics of a sedimentation centrifuge, let’s consider a cylindrical rotor section having a certain length (L) and inner radius (R) transferring liquid with layer thickness (h) and throughput Q. Let the inner radius of the liquid ring be designated as (r). Let us also introduce average throughput value along rotor axis (v_{axis}), which may be represented as volumetric flow rate of slurry divided by the cross sectional area of the flow perpendicular to the rotor axis (F):

v_{axis} = Q/F = Q/(2·π·h·r_{av})

where:

*r _{av} = (R+r)/2* – average radius of liquid layer, m.

Then, just like for regular sedimentation, let’s determine sedimentation velocity of particles from the area that is most unfavorable for sedimentation - the inner surface of liquid ring. In other words, the particles that are the farthest from the inner rotor surface, which is the settlement area for them. This velocity (v_{rad}) may be expressed from the sedimentation velocity of a similar particle in the gravity field (v_{gf}) determined from Stoke’s law:

v_{gf} = [d²·(ρ_{p}-ρ_{l})·g] / (18·μ)

where:

d – particle diameter, m;

ρ_{p} – particle density, kg/m³;

ρ_{l} – liquid density, kg/m³;

g – gravitational acceleration, m/s²;

μ – dynamic viscosity of liquid, Pa·s.

The correlation between sedimentation velocities in the gravity field and centrifugal field is described via Froude number as follows:

v_{rad} = (v_{gf}/g) · Fr

where:

*Fr = (ω²·R)/g* – Froude number;

ω – angular velocity of rotor, s^{-1}.

Similar to regular sedimentation, a prerequisite for a full separation of dispersed phase is also that time values for sedimentation of particles farthest from side walls (t_{sed}) and for their stay in the centrifuge (t_{stay}) should be equal:

t_{sed} = t_{stay}

The equation may be rewritten as:

L/v_{sed}= h/v_{rad}

Let’s proceed with replacing the variables according to previous equations:

(L·2·π·h·r_{av})/Q = (h·g)/(v_{gf} ·Fr)

Usually liquid layer thickness in a centrifuge is low, so we can assume that R = r_{av}. Thus it is possible to make a replacement, where F is sedimentation area of a centrifuge. So, we can derive an equation to determine throughput of a sedimentation centrifuge:

Q = (F·v_{gf }·Fr)/g

In practice, however, in centrifuge calculations there are a lot of factors that are difficult to take into account, which, nevertheless, can have a significant impact, so β coefficient is usually added into equations for Q calculations, which takes into account different specific factors for a particular case:

Q = β·[(F·v_{gf} ·Fr)/g]

F·Fr is often replaced with Σ – throughput index. Experimentally it was determined that the throughput index also depends on the flow pattern:

*Σ = F·Fr* – laminar;

*Σ = F·Fr ^{0,73}* – transient;

*Σ = F·Fr*– turbulent.

^{0,5}This implies that the most preferable flow pattern providing for the highest throughput index, is laminar flow.

## Filtering centrifuge calculation

As with sedimentation centrifuges the calculations for filtering centrifuges have some common patterns with filter calculations considering the same principle of operation, however centrifugal field conditions bring about some differences.

The common equation for determining theoretical throughput of centrifuges is shown below:

Q = a·Σ

where:

Q – centrifuge throughput, m³/s;

a – adjustment factor depending on the type of centrifuge (for a filtering centrifuge а is replaced by the filtering constant k, determined experimentally);

Σ – throughput index.

In its turn, the throughput index for a centrifuge is calculated as follows:

Σ = F_{av}·K_{av}

where:

F_{av} = 2·π·L·(R+r) – is the average separation surface, m²;

L – drum length, m;

R – inner radius of a centrifuge rotor, m;

r – inner radius of slurry ring in a centrifuge, m;

K_{av} = [ω²·(R+r)] / [2·g] – average separation factor of a centrifuge;

ω – angular velocity of a centrifuge rotor, s^{-1};

g – gravitational acceleration, m/s.

However, the actual throughput is often less than the theoretical one due to some factors, such as friction of the liquid layer against the centrifuge drum, etc. To account for all those factors an adjustment factor (ζ) called an efficiency indicator is introduced into throughput equation for a filtering centrifuge. Thus, the final throughput equation is as follows:

Q = ζ·a·Σ

For a batch type filtering centrifuge the formula is different:

Q = a·√t_{f}·V_{w}·Σ

where:

a – adjustment factor characterizing the sediment resistance;

t_{f} – slurry feed time, s;

V_{w} = π·L·(R²-r²) – working volume of the drum, m³.

To obtain maximum average throughput of a filtering centrifuge t_{f} is usually assumed to be equal to the aggregate time of centrifugation (t_{c}) and sediment discharge (t_{d}) processes:

t_{f} = t_{c}+ t_{d}

In the calculations of centrifuge power one should distinguish between starting power (N_{start}) and operation period power (N_{op}). The starting power is a sum of the following values:

N_{start} = N_{i}+N_{b}+N_{a} [kW]

where:

N_{i} – initial power of a centrifuge, W;

N_{b} – power consumed for bearing losses, W;

N_{a} – power consumed for air friction, W.

In its turn, the operation period power is composed of the following elements:

N_{op} = N_{l}+N_{s}+N_{b}+N_{a}; [kW]

where:

N_{l} – centrifuge power for transferring kinetic energy to the liquid phase of the slurry, W;

N_{s} – centrifuge power for transferring kinetic energy to the solid phase of the slurry, W.

The initial power of a centrifuge takes into account all the moments of inertia occurring at the start:

N_{i} = (I·ω²) /(2·10³·t_{s}); [kW]

where:

I – total inertia moment of rotor and load with respect to rotation axis, kg·m²;

ω – angular velocity of a centrifuge rotor, s^{-1};

t_{s} – starting time of a centrifuge, s.

Power lost to friction in bearings:

N_{b} = [f·ω·Σ(P·d)] / [2·10³]; [kW]

where:

f – friction coefficient in bearings;

Σ(P·d) – a sum of products of dynamic loads on bearings (P, Н) by respective shaft diameters (d, m).

Power lost to air friction:

N_{a} = 12·10^{-6}·ρ_{a}·R_{aor}·ω²; [kW]

where:

ρ_{a} – air density, kg/m³;

R_{aor} – average outer radius of a rotor, m.

### Example 1

Selection of a centrifuge and calculation of its throughput

**Conditions:** A sedimentation centrifuge is available capable of reaching operating angular velocity of ω = 600 rpm. Specification of the drum: inner radius R = 300 mm, length L = 500 mm. The centrifuge is used for clarification of water from suspended solid particles having diameter d_{p} = 0.5 mm and density ρ_{p} = 2,100 kg/m³. For this problem dynamic viscosity is assumed to be equal to μ = 0.001 Pa·s, and density ρ_{l} = 1,000 kg/m³.

**Problem:** Calculate the centrifuge throughput Q.

**Solution:** The sought quantity may be calculated using the formula below:

Q = (F·v_{gf }·Fr)/g

v_{gf} is particle sedimentation velocity in the gravity filed, which may be determined as follows (g = 9.81 m/s – gravitational acceleration):

v_{gf} = [d_{p}²·(ρ_{p}-ρ_{l})·g] / [18·μ] = [0.0005²·9.81·(2,100-1,000)] / [18·0.001] = 0.15 m/s

The settlement area of the drum F may be found from its geometry according to:

F = 2·π·R·L = 2·3.14·0.3·0.5 = 0.942 m2

Fr is the Froude number characterizing correlation between particle sedimentation velocities in the centrifugal field and gravity field:

Fr = (ω²·R) / g = ((600/60)²·0.3) / 9.81 = 30.58

From here, particle sedimentation velocity in the centrifugal field equals to:

v_{c} = (v_{gf}/g)·Fr = (0.15/9.81)·30.58 = 0.47 m/s

F·Fr is usually replaced with Σ – throughput index, which may be adjusted depending on the particle precipitation pattern, which, in its turn, is determined by Reynolds number:

Re = (ρ_{l}·v_{c}·d_{p}) / μ = (1000·0.47·0.0005) / 0.001 = 235

The resulting Re value is in the range of 2<Re<500, consequently, the sedimentation pattern is transient, for which the adjusted formula for the throughput index is as follows:

Σ = F·Fr^{0.73} = 0.942·30.58^{0.73} = 11.44

Let’s insert the resulting values into the initial equation and calculate the target value:

Q = (F·v_{gf}·Fr)/g = (v_{gf}/g)·Σ = (0.15/9.81)·11.44 = 0.17 m³/s.

**Result:** centrifuge throughput is 0.17 m³/s.

### Example 2

Calculation of starting power of a filtering centrifuge

**Conditions:** A filtering centrifuge is available for separating slurry having the density of ρ_{s} = 1,100 kg/m³. A drum weighing m_{b} = 200 kg has inner radius R = 0.5 m, wall thickness b = 0.005 m and length L = 0.4 m. The initial load of the drum is 50% of its internal volume. The centrifuge ramp-up time to the operating velocity is t_{stay}= 7 s. The centrifuge angular velocity is ω = 1,000 rpm. In the calculations air density ρ_{a} is assumed to be equal to 1.3 kg/m³ and bearing friction coefficient f = 0.05. Shaft neck diameter d_{n} = 80 mm.

**Problem:** Calculate the starting power N_{start}.

**Solution:** The starting power (N_{start}) is a sum of powers for bearing friction losses (N_{b}), for air friction (N_{a}) and for initial overcoming of inertia at start (N_{i}):

N_{start} = N_{b}+N_{a}+N_{i}

In order to determine the power consumed for bearing friction losses let’s use the formula based on the mass of rotating parts of a centrifuge. Let’s assume that only the drum and slurry weights are involved into the rotation:

N_{b} = f·g·M·v_{n}

М is the total mass of the rotating parts of a centrifuge. The drum mass is known and we only need to determine the initial mass of the slurry. Since the initial filling of the drum is 50%, we can determine the slurry mass m_{s} by finding its volume and multiplying it by density:

m_{s} = 0.5·2·π·R·L·ρ_{s} = 0.5·2·3.14·0.5·0.4·1100 = 691 kg

Then the total mass is equal to:

M = m_{b}+m_{s} = 200+691 = 891 kg

The circumferential velocity of the shaft neck v_{n} is determined from the formula:

v_{n} = ω·d_{n}/2 = (1,000/60)·(0.08/2) = 0.66 m/s

Let’s calculate the power N_{b}:

N_{b} = f·g·M·v_{n} = 0.05·9.81·891·0.66 = 288.4 W

Let’s also calculate the power N_{a} assuming that the outer radius of the drum is R_{o} = R+b:

N_{a} = 0.012·ρ_{a}·R_{o}·ω² = 0.012·1.3·(0.5+0.005)·(1,000/60)² = 2.2 W

We will calculate N_{i} assuming that the entire rotating mass is concentrated on the inner radius of the drum R, then the total moment of inertia may be represented as I = M·R²:

N_{i} = (I·ω²)/(2·t_{stay}) = (M·R²·ω²)/(2·t_{stay}) = (891·0.5²·(1,000/60)²)/(2·7) = 4,419.6 W

Now we can find the target value:

N_{start} = N_{b}+N_{a}+N_{i} = 288.4+2.2+4,419.6 = 4,710.2 W

**Result:** The starting power is 4.71 kW

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