Thread: Air + gas polution in furnace flush rate problem

I have a problem with flush rate differential equation for the following situation.

A furnace has volume V(oven) [m3]. During thermical treatment a polution comes into the furnace with a volume rate Vg [m3/min].
To keep the polution under certain concentration the furnace is flushed with air at volume rate Vf [m3/min].
The flush rate flows at the same rate out with a certain volume of polution. Assume ideal mixing in the oven between air and polution gas.
At t=0 volume of oven is 100% air.
What is the differential equation which describes equilibrium at t=? when volume of polution flushed out is the same as the volume of incoming polution?
In other words: at what time the concentration of polution is not changed any more by flush air? I need equation that describes concentration or volume as a function of time inside the oven.

Any ideas? Thanks!

Originally Posted by adosana

I have a problem with flush rate differential equation for the following situation.

A furnace has volume V(oven) [m3]. During thermical treatment a polution comes into the furnace with a volume rate Vg [m3/min].
To keep the polution under certain concentration the furnace is flushed with air at volume rate Vf [m3/min].
The flush rate flows at the same rate out with a certain volume of polution. Assume ideal mixing in the oven between air and polution gas.
At t=0 volume of oven is 100% air.
What is the differential equation which describes equilibrium at t=?

Differential equation for what? What variable do you want to to solve for? Probably the amount of pollution in the room. If you let, say, P(t), be the amount of pollution in the air at time t, in, say, kg, P/V is the amount of pollution in each cubic meter of air. The pollution is flushed out at Vf cubic meters per minute and each of those carries P/V kg of pollution so Vf(P/V) kg of pollution go out each minute. Vg cubic meters per minute of pollution goes out each minute so the dV/dt= Vg- Vf(P/V).

when volume of polution flushed out is the same as the volume of incoming polution?
In other words: at what time the concentration of polution is not changed any more by flush air? I need equation that describes concentration or volume as a function of time inside the oven.

Any ideas? Thanks!

Thank you HallsofIvy,

The amount op pollution is in m3/min at constant rate that comes into the oven. The requirement is that amount of pollution doesn't exceed a certain vol% inside the oven. So if you know what volume of polution is as function of time than you can easly calculate %vol of pollution relative to the total volume of the oven which is constant. Then %vol of pollution = V(polution)/V(oven)*100%
So I am looking for differential equation that discribes V(pollution) in the oven as function of time. Note: not only pollution is flushed out, also the air in the oven is being flushed out.
At t=0 you have 100% air in the oven. Then for example at t=1 min you have V(pollution) in m3 and V(air) in m3. Both amounts are being flushed at flush rate Vf [m3/min). I expect that at certain time the same amount of pollution is flushed out as it comes into the oven (equilibrium). Does your equation describe this situation? What is the outcome of your equation if you differentiate dt? In your equation V means oven volume?
I called V(polution) = Vg. Your said "Vg cubic meters per minute of pollution goes out each minute so the dV/dt= Vg- Vf(P/V)." But Vg actually goes into the oven not OUT.
Thank you very much.

Originally Posted by adosana

Thank you HallsofIvy,

The amount op pollution is in m3/min at constant rate that comes into the oven. The requirement is that amount of pollution doesn't exceed a certain vol% inside the oven. So if you know what volume of polution is as function of time than you can easly calculate %vol of pollution relative to the total volume of the oven which is constant. Then %vol of pollution = V(polution)/V(oven)*100%
So I am looking for differential equation that discribes V(pollution) in the oven as function of time. Note: not only pollution is flushed out, also the air in the oven is being flushed out.
At t=0 you have 100% air in the oven. Then for example at t=1 min you have V(pollution) in m3 and V(air) in m3. Both amounts are being flushed at flush rate Vf [m3/min). I expect that at certain time the same amount of pollution is flushed out as it comes into the oven (equilibrium). Does your equation describe this situation? What is the outcome of your equation if you differentiate dt? In your equation V means oven volume?

Yes, beause you said "A furnace has volume V(oven) [m3]."

I called V(polution) = Vg

No, you specifically said "During thermical treatment a polution comes into the furnace with a volume rate Vg [m3/min]." So Vg is a rate not a volume.

Your said "Vg cubic meters per minute of pollution goes out each minute so the dV/dt= Vg- Vf(P/V)." But Vg actually goes into the oven not OUT.
Thank you very much.

Yes, that last was an error. I intended to say "Vg cubic meters per minute of pollution goes in each minute". That's why, in the differential equation, Vg is positive while Vf(P/V) is negative.

That's correct Vg and Vf are rates. But I don't understand you. If P/V = kg/m3 and Vf=m3/min then you get for Vf(P/V)=m3/min*kg/m3 then the unit of Vf(P/V) becomes kg/min.
How do you combine kg/min with the rest of your equation (Vg) which is in m3/min?
Can you solve your equation please? Thanks