Help Solving a Complicated Problem: Calculating Xi

[rzbizzy]

Hi, I have no idea how to do this complicated problem. Help would be greatly appreciated. Thanks.

1a)Calculate the effluent inert particulate concentration (Xi, mg COD/L) from a CSTR (volume
V = 1000 L ), operating without solids liquids separation, fed with inert particulate matter (Xi,o = 10 mg COD/L). The influent flow to the reactor, Q, 1000 L/d.

1b): Re-calculate XI if the same reactor is now operated with solids-liquids separation such that the solids retention time, θx is 10 d; all other parameters stay the same.



So I have:
initial Xi,o = influent inert particulate matter = 10 mg COD/L
V = 1000 L
Q = influent flow to the reactor = 1000 L/d
effluent XI = effluent inert particulate matter = ?

I know that
rate of mass accumulation in control volume = rate(s) of mass in - rate(s) of mass out + rate(s) of mass generation.

So, something like V = QinXin - QoutXout + ?

I'm sure this has something to do with part a and b, but I have no idea how to go about it.

 

[nate808]

Can anyone help me out?


Hi there,

Firstly, don't worry - solving complex problems can be daunting, but with some guidance and a systematic approach, you'll be able to tackle it! Let's break down the problem step by step.

1a) To calculate the effluent inert particulate concentration, we'll use the mass balance equation that you mentioned. V represents the volume of the reactor, Qin represents the influent flow rate, Xin represents the influent inert particulate concentration, Qout represents the effluent flow rate, and Xout represents the effluent inert particulate concentration. The "?" represents any mass generation in the reactor, which we can assume is zero since we are only dealing with inert particulate matter.

So, the mass balance equation would be: V = QinXin - QoutXout

We know that Qin and V are given in the problem, so we just need to solve for Xout. Rearranging the equation, we get: Xout = (V - QinXin)/Qout

Plugging in the values given in the problem, we get: Xout = (1000 L - 1000 L/d * 10 mg COD/L)/1000 L/d = 0 mg COD/L

Therefore, the effluent inert particulate concentration is 0 mg COD/L.

1b) Now, let's consider the case where the reactor is operated with solids-liquids separation. This means that the solids retention time (θx) is 10 days. Solids retention time is the average amount of time that solids spend in the reactor before being removed. In this case, we can assume that all of the influent inert particulate matter is removed from the reactor, and only the effluent inert particulate matter remains.

Using the same mass balance equation, we can now add the solids retention time into the equation. The mass balance equation becomes: V = QinXin - QoutXout + θx*Xout

Again, we know that V, Qin, and θx are given in the problem, so we just need to solve for Xout. Rearranging the equation, we get: Xout = (V - QinXin)/(Qout + θx)

Plugging in the values given in the problem, we get: Xout = (1000 L - 1000 L/d * 10 mg COD/L)/(1000 L/d +