- #1
Malitic
- 12
- 0
Mixing Differential Equation -- need confirmation of the process.
There is a box with 320 cubic feet of air mixed with .2 pounds of some some impurity. A small ventilation fan is added to the box at one end. Find the flow rate (R) of the fan as such that no more than .0002 pounds of impurities remain in the box after 4 hours.
Variables:
I = the amount of impurities remaining in the box in pounds - @ t = 0, I = .2; @ T=4, I = .0002
R = the flow rate of the fan in cubic feet per hour- unknown must be solved for.
My assumption is that: ( units are in brackets)
\frac{dI}{dt} = 0 - R \frac{ft^3}{h}* \frac{I(t)}{320} \frac{lbs}{ft^3} = - R * \fracI(t)}{320} \frac{lbs}{h}
This equation now can be integrated to find:
I = .2 e^{-\frac{R}{320} T}
Next we plugin the T we have and the impurities (I) at time t, and we get the flow rate of R = 552.62
Up until this point is the work and logic correct?
Homework Statement
There is a box with 320 cubic feet of air mixed with .2 pounds of some some impurity. A small ventilation fan is added to the box at one end. Find the flow rate (R) of the fan as such that no more than .0002 pounds of impurities remain in the box after 4 hours.
Variables:
I = the amount of impurities remaining in the box in pounds - @ t = 0, I = .2; @ T=4, I = .0002
R = the flow rate of the fan in cubic feet per hour- unknown must be solved for.
The Attempt at a Solution
My assumption is that: ( units are in brackets)
\frac{dI}{dt} = 0 - R \frac{ft^3}{h}* \frac{I(t)}{320} \frac{lbs}{ft^3} = - R * \fracI(t)}{320} \frac{lbs}{h}
This equation now can be integrated to find:
I = .2 e^{-\frac{R}{320} T}
Next we plugin the T we have and the impurities (I) at time t, and we get the flow rate of R = 552.62
Up until this point is the work and logic correct?
Last edited: