What is the first test you should do when checking the charging system?

A wastewater treatment plant has two primary clarifiers, each 20m in diameter with a 2-m side-water depth. the effluent weirs ar

jasenka [17]

Answer:

overflow rate 20.53 m^3/d/m^2

Detention time 2.34 hr

weir loading 114.06 m^3/d/m

Explanation:

calculation for single clarifier

$sewag\ flow Q = \frac{12900}{2} = 6450 m^2/d$

$surface\ area =\frac{pi}{4}\times diameter ^2 = \frac{pi}{4}\times 20^2$

$surface area = 314.16 m^2$

volume of tank $V = A\times side\ water\ depth$

$=314.16\times 2 = 628.32m^3$

$Length\ of\ weir = \pi \times diameter of weir$

$= \pi \times 18 = 56.549 m$

overflow rate = $v_o = \frac{flow}{surface\ area} = \frac{6450}{314.16} = 20.53 m^3/d/m^2$

Detention time $t_d = \frac{volume}{flow} = \frac{628.32}{6450} \times 24 = 2.34 hr$

weir loading $= \frac{flow}{weir\ length} = \frac{6450}{56.549} = 114.06 m^3/d/m$

6 0

3 years ago

I have a stream with three components, A, B, and C, coming from another process. The stream is 50 % A, and the balance is equal

tigry1 [53]

Answer:

$X_{A} = \frac{N_{Ao}-N_{A}}{N_{Ao}}$

Nₐ₀-Nₐ = 1.33

Nₐ₀ = 2.5

Conversion X = 1.33/2.5 = 0.533

Explanation:

A + 2B + 4C ⇒ 2X + 3Y

Given a stream containing 50% A, 25% B and 25% C, to get the limiting reactant, lets take a simple basis

Say stream is 10 moles, this give

A = 5moles

B = 2.5mole

C = 2.5moles

from the balanced equation above,

1mole of A ⇒ 4moles of C

∴ 5moles of A ⇒ (5x4)/1 ⇒ 20moles of C

also;

2mole of B ⇒ 4moles of C

∴ 2.5moles of B ⇒ (2.5x4)/2 ⇒ 5moles of C

so clearly from above reactant C is the limiting reactant.

Note: To get conversion of a process, we must use the limiting reactant. this is because ones it is used up, the reaction comes to an end

Formula to obtain conversion is:

Conversion = (Amount of A used up)/(Amount of A fed into the system)

$X_{A} = \frac{N_{Ao}-N_{A}}{N_{Ao}}$

where, Nₐ₀-Nₐ = is the amount in moles of A used up

Nₐ₀ = amount in moles of A fed into the system

The next question is what mole of reactant C will give 0.1mole fraction of Y

Recall our basis = 10moles

from conservation of mass law, 10mole of product must come out which 0.1 moles fraction is Y

therefore amount Y in the product is = 0.1x10 = 1mole

if 3moles of Y ⇒ 4mole of C

∴ 1mole of Y ⇒ (1x4)/3 ⇒ 1.33moles of C

calculating the conversion of limiting reactant C that will give 0.1mole fraction of Y

Nₐ₀-Nₐ = 1.33