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__Condition:__

A suspension with flow rate Q_{c} = 10 m³/h is separated in a filter while filtrate flow rate is Q_{ф} = 9,5 m³/h. Accordingly, density of hard phase and liquid phase are ρ_{т} = 1700 kg/m³, ρ_{ж} = 1000 kg/m³. The calculations have shown that density of filtrate and settlement are ρ_{ф} = 1020 kg/m³ and ρ_{о} = 2100 kg/m³ accordingly. It is required to determine density and mass content of hard phase in the suspension.

__Solution:__

Let us compose the equation of process material balance:

Q_{с}·ρ_{с} = Q_{о}·ρ_{о}+Q_{ф}·ρ_{ф}

Settlement consumption Q_{о} can be expressed through flow volumes of suspension and filtrate:

Q_{о} = Q_{с}-Q_{ф} = 10-9,5 = 0,5 m³/h

Let us express and determine the density of suspension from the equation of process material balance:

ρ_{с} = (Q_{о}·ρ_{о}+Q_{ф}·ρ_{ф})/Q_{с} = (0,5·2100+9,5·1020)/10 = 1074 kg/m³

Let us denote the content of hard phase in the suspension as *m* and compose the following equation to determine the density of suspension:

1/ρ_{c} = (1-m)/ρ_{ж} +m/ρ_{т}

Let us substitute the numeric values and find the unknown m:

1/1074 = (1-m)/1000+m/1700

then we receive the value of hard phase content in the suspension:

m = 0,17

Answer: the suspension density is equal to 1074 kg/m³, the content of hard phase in the suspension is equal to 0,17.

__Condition:__

To calculate the required filtration area of a drum-type vacuum filter which can be operated under load of a suspension Q = 32 m³/h. Rotation speed of the drum is n = 0,2 rpm. It was found on a laboratory model that the ratio of settlement volume to filtrate volume was x = 0,07 while height of settlement layer was h = 0,02 m on conversion to an operating model.

__Solution:__

Let us determine the time of total filtration cycle for the drum-type vacuum filter:

τ = 1/n = 60/0,2 = 300 sec.

then we calculate the specific filtrate volume acc. to the formula;

v_{уд} = h/x = 0,02/0,07 = 0,29

Finally, we determine the value to be found using the correction coefficient К_{п} equal to 0,8:

F = (Q·τ)/(υ_{уд}·K_{п}) = (32·300)/(3600·0,29·0,8) = 11,5 m²

Answer: 11,5 m²

__Condition: __

There is the nutsch filter filtrating V_{С} 3,2 m³ (suspension volume) for one load. The suspension being filtrated includes x = 15% of hard phase as per mass and has density ρ_{C} = 1100 kg/m³. After the filtration process is finished, the settlement with humidity w = 74% and density ρ_{ОС} 1185 kg/m³ collects. It is required to find the volume of the forming filtrate V_{ф} when y = 2% of hard phase passes through the filter not stopping.

__Solution: __

Let us find the quantity of hard phase coming with the suspension being treated into the filter

G_{тф}_{1} = V_{c}·ρ_{c}·x/100 = 3,2·1100·15/100 = 528 kg

To determine the quantity of hard phase that is not indicated by the nutsch filter:

G_{тф}_{2} = G_{тф}_{1}·y/100 = 528·2/100 = 10,56 kg

The quantity of hard phase that remains on the filter is equal to:

G_{тф}_{3} = G_{тф}_{1}-G_{тф}_{2} = 528-10,56 = 517,44 kg

Having humidity of the forming settlement, we will find the total settlement weight:

G_{ос} = G_{тф}_{3}/w·100 = 517,44/74·100 = 699,24 kg

Accordingly, volume of the forming settlement is equal to:

V_{ос} = G_{ос}/ρ_{ос} = 699,24/1185 = 0,59 m³

then we receive volume of the forming filtrate:

V_{ф} = V_{с}-V_{ос} = 3,2-0,59 = 2,61 m³

Answer: 2,61 m³

__Condition: __

By testing start of the a filter there was found that V_{1} = 1 m³ of filtrate formed in t_{1} = 4,5 min and V_{2} = 2 m³ of filtrate in t_{2} = 12 min., the total filtration area was F = 1,6 m². The required day capacity of the filter for filtrate is Q = 16 m³. It is required to calculate the time period for filter’s day operation.

__Solution: __

To determine the relative values of the collected filtrate at the testing start:

V_{1F} = V_{1}/F = 1/1,6 = 0,625 m³/m²

V_{2F} = V_{2}/F = 2/1,6 = 1,25 m³/m²

Based on the data of the testing start, let us compose the filtration equation system and determine the filtration constants:

{ | [V_{1F}]²+2·V_{1F}·C = K·t_{1} |
= > | { | [0,625]²+2·0,625·С = K·4,5 | = > | { | C = 0,62 |

[V_{2F}]²+2·V_{2F}·C = K·t_{2} |
[1,25]²+2·1,25·С = K·12 | K = 0,26 |

Using the received equation for filtration, let us determine the target value substituting the relative volume of the required filtrate:

(16/1,6)²+2·16/1,6·0,62 = 0,26·t_{об}

then we receive the value t_{об} = 7,2 hours. Regard to the whole filtration surface.

Answer: 7,2 h.

__Condition:__

There is the drum-type vacuum filter with the following data: Angles of filtration/flushing/drying sectors are equal to φ_{ф} = 110^{0}, φ_{п} = 130^{0} and φ_{с} = 60^{0}. The time for these operations is t_{ф} = 4 min, t_{п} = 6 min and t_{с} = 2 min accordingly. To calculate the rotation frequency of the drum is required.

__Solution: __

Having these data, it is better to use two calculating equations for rotation frequency by following choosing the most minimal one from the received values.

The first frequency of the drum is calculated acc. to the formula:

n_{1} = φ_{ф}/(360·τ_{ф}) = 110/(360·4·60) = 0,00127 с^{(-1)}

The first frequency of the drum is calculated acc. to the formula:

n_{2} = (φ_{п}+φ_{с})/(360·(τ_{п}+τ_{с})) = (130+60)/(360·(6+2)·60) = 0,0012 с^{(-1)}

Comparing two received values of rotation frequency we receive as follows:

n_{1}>n_{2}

It means the target value is 0,0012 с^{-1}. - cycles per second ?

Answer: 0,0012 с^{-1}

__Condition: __

The locking mechanism of the press filter can reach effort P = 2·10^{4} H. Dimensions of the working plate surface are 300х300 mm, and width of the packing line is 20 mm. To calculate the maximal pressure of suspension supply is required.

__Solution:__

Preliminary, we have calculated filtration area and mesh packing. Mesh filtration area is as follows:

F_{ф} = 0,3·0,3 = 0,09 m²

Packing surface (in form of a frame):

F_{у} = (0,3+2·0,02)·(0,3+2·0,02)-0,3·0,3 = 0,0256 m²

then we consider the equation to determine the required sealing pressure:

P = Q_{д}+R_{пр}

where

Q_{д} = p·F_{ф}

R_{пр} = m·p·F_{у}

Generally we receive the equation of sealing pressure is as follows:

P = p·F_{ф}+m·p·F_{у}

Regard to the correction coefficient m = 3 we substitute the known values and find the main working load p:

40000 = p·0,09+3·p·0,0256

then we receive:

p = 0,24·[10]^{6} H

Further, we should determine the maximum possible pressure of suspension at inlet:

P_{max} = p/F_{ф} = (0,24·[10]^{6})/0,09 = 2,7 MPa

Answer: 2,7 MPa

__Condition: __

It is required to find capacity of the enclosed sandy filter with diameter of the cylindrical part D = 2 m (pay no regard to pore clogging), Sand as a filter medium has the following properties. Sand grain diameter is d = 0,5 mm. Porosity of sand layer is x = 0,42. Thickness of sand layer is l = 1,6 m. Filtration is taking place at temperature T = 20 °C. It has been found out that head loss in the filter is h = 4,5 mH2O.

__Solution:__

Let us calculate filtration rate (the correction coefficient as equal to 40):

w = 3600·c·d²·h/l·(0,7+0,03·t) = 3600·40·[0,0005]²·4,5/1,6·(0,7+0,03·20) = 0,13 m/sec

then we find passage area of filtrating layer (where F = filter area):

F_{пр} = F·x = (π·D²)/4·x = (3,14·2²)/4·0,42 = 1,32 m²

Based on the known values, we can determine the value to be find:

Q = w·F_{пр} = 0,13·1,32 = 0,17 m³/sec

Answer: 0,17 m³/sec

__Condition: __

It is scheduled to apply the filters with the following data for sewage treatment at the rate of Q = 1000 m³/day. Design filtration rate is v = 10 m/h. The filter requires flushing each seven (7) hours, with flushing time is t = 0,2 h. q = 10 m³ of water is used for one flushing. Operation is carried out 24 hours a day, i.e. the total working time is t_{об} = 24 h. It is required to calculate the required number of filters.

__Solution:__

As the filter shall be flushed each seven hours, it will be for one hour as follows:

n = 24/7≈3

Let us calculate the required filtration area:

F = Q/(t_{об}·v-n·q-n·t·v) = 1000/(24·10-3·10-3·0,2·10) = 4,9 m²

then let us determine the required number of filters acc. to the formula:

N = 0,5·√F = 0,5·√4,9 = 1,1

We round to the biggest integer and receive the search value =2.

Answer: 2 filters.

**Condition:** Quartz sand particles are settled in water at temperature t = 20 °C, sand density is ρ_{п} = 2600 kg/m³. Within the problem to be solved, shape of particle is to be considered as sphere with diameter d = 1,2 mm.

**Problem:** To determine particle settling velocity v_{ос}.

**Solution: **To solve this problem is to use the criterial equation for settling process:

Re²·ζ = 4/3·Ar

First of all, we should calculate Archimedes criterion (Ar). We will consider water density as ρ_{в} = 1000 kg/m and its dynamic viscosity as μ = 0,01 cPs when t = 20 °C, and then we will substitute the known values in the calculating formula (g = 9,81 m/sec – free-fall acceleration):

Ar = [g·ρ_{ж}·d³·(ρ_{т}-ρ_{ж})] / μ² = (9,81·1000·0,0012³·(2600-1000)) / 0,001² = 27123

The received value for Archimedes criterion falls within 36<Ar<83000 that is in accordance with the transient settling mode where the resistance coefficient (ζ) shall be calculated as per the formula:

ζ = 18,5/Re^{0,6}

We substitute the received dependence and value Ar in the preliminary criterial equation to determine the criterion value Re:

Re² · (18,5/Re^{0,6}) = (4/3)·27123

Re^{1,4} = 1955

Re = 224,3

The equation may be written for Reynolds criterion and the target value may be found:

Re = (ρ_{в}·v_{ос}·d) / μ

v_{ос}= (Re·μ) / (ρ_{в}·d) = (224,3·0,001) / (1000·0,0012) = 0,187 m/sec

**Answer:** 0,187 m/sec

**Condition: **A settler is required to treat a flow of muddy water. It is clear that the dispersed phase is mainly formed by solids with unknown shape and mass m_{ч} = 2 mg and density ρ_{т} = 1800 kg/m³. Consumption of water being supplied for treatment is Q = 0,6 m³/h. While calculating for water density is to be considered as ρ_{в} = 1000 kg/m³ and dynamic viscosity as μ = 0,001 cPs. It is understood also that settling is taking place under confined conditions at volume ratio of the dispersed phase ε = 0,5.

**Problem:** To determine the required settling area of the settler.

**Solution:** Design value for settling area may be determined acc. to the formula:

F = Q/v_{ст}

where v_{ст} – velocity of hindered particle settling.

To determine v_{ст} is to calculate Archimedes criterion (g = 9,81 m/sec2 – free-fall acceleration):

Ar = [ρ_{ж}·g·d_{ч}³·(ρ_{т}-ρ_{ж})] / μ²

In the calculation formula of Archimedes criterion d_{ч} =diameter of settling particle. Particle’s shape of hard phase is unknown, so the following formula is to be used;

d_{ч} = [(6·V_{ч})/π]^{1/3}

V_{ч} – particle volume that may be used through the ration of the known mass of particle to its density V_{ч} = m_{ч}/ρ_{ч}. Making this substitution we calculate the value d_{ч}:

d_{ч} = [(6·m_{ч}) / (π·ρ_{ч})]^{1/3} = [(6·0,000002) / (3,14·1800)]^{1/3} = 0,00128 m

Now Archimedes criterion can be calculated:

Ar = [ρ_{ж}·g·d_{ч}³·(ρ_{т}-ρ_{ж})] / μ² = (1000·9,81·0,00128³·(1800-1000)) / 0,001² = 16458

Using the criterial equation that joins Archimedes and Reynolds (Re_{ст}) criterions for hindered settling we calculate the value Re_{ст}:

Re_{ст} = (Ar·ε^{4,74}) / (18+0,6·√(Ar·e^{4,75})) = (16458·0,5^{4,74}) / (18+0,6·√16458·0,5^{4,75}) = 18,8

When Reynolds criterion for hindered settling is known, other formula may be used for its calculation where velocity of hindered particle settling is applied. Then v_{ст}should be calculated as follows:

Re_{ст} = (ρ_{ж}·v_{ст}·d_{ч}) / μ

v_{ст} = (Re_{ст}·μ) / (ρ_{ж}·d_{ч}) = (18,8·0,001) / (1000·0,00128) = 0,015 m/sec

Having all required values we determined the target value:

F = Q/v_{ст} = 0,6/0,015 = 40 m²

**Answer:** Settling area is 40 m².

**Condition: **A filter operating in the mode of constant pressure difference has been delivered to a factory without any accompanying documents. After testing the filter for filtrating a suspension it has been clear that the filter can obtain the filtrate V_{1} = 7,8 liter in τ_{1} = 5 min but in τ_{2} = 10 min already V_{2} = 12,1 l of filtrate.

**Problem:** To determine how much time is required to obtain V_{0} = 50 l of filtrate for the similar suspension.

**Solution:**

We will use the equation for filtration at constant pressure difference (Δp = const):

V² + 2·[(R_{фп}·S)/(r_{о}·x_{о})]·V = 2 [(∆p·S²)/(μ·r_{о}·x_{о})]·τ

Let us denote a = (R_{фп}·S)/(r_{о}·x_{о}) и b = (∆p·S²)/(μ·r_{о}·x_{о}). The values a and b are constant, and to determine them is to make up and solve the set of equations based on the test data

{ | V_{1}²+2·a·V_{1} = 2·b·τ_{1} |
= | { | 7,8²+2·a·7,8 = 2·b·5 | = | { | a = 3,53 |

V_{2}²+2·a·V_{2} = 2·b·τ_{2} |
12,1²+2·a·12,1 = 2·b·10 | b = 11,59 |

In total we receive, the filtration equation for this case and dimensions may be written as:

V²+7,06·V = 23,59·τ

Let us substitute the value V_{0} in the received equation and find the value τ corresponding to him:

τ = (50²+50·7,06) / 23,59 = 121 min

**Answer:** To receive 50 liter of filtrate is required to spend 121 min.

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