
#1
Feb1713, 07:30 PM

P: 13

Sam is seeping into a room from the basement at a rate of
$$\frac{1}{6} \times 10^6 \frac{\mathrm{pCi}}{\mathrm{hr}}$$ (pCi=picocuries). The room contains $$10^6$$ liters of air. (The rate was chosen so that the room reaches the EPA action level of $$4 \frac{pCi}{liter}$$ after $$24$$ hours.) Air in the room is being exchanged with the outside air at a rate of $$R$$ liters/hr. The outside air has a concentration of $$0.5$$ pCi/liter. Set up an equation for the total radiation $x$ in the room in picocuries, assuming instantaneous uniform mixing, that is, the indoor concentration of radioactivity is $$\frac{x}{10^{6}}$$ pCi/liter. Explain your equation using the notations $$\Delta x$$ pCi, $$\Delta t$$ hr, and display the units of every variable explicitly to show that they match (e. g., (pCi/liter)(liter/hr) = pCi/hr). I wanted to try to express the differential equation under the form $$\frac{dx}{dt}=abx$$ such that x is the value of the radiation, but I am not sure how to use the remaining information. 



#2
Feb1713, 11:04 PM

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The form of your target equation says radiation is entering at rate a and leaving at a rate proportional to the amount in the room. So far so good.
What are the sources of radiation entering the room. What is the rate for each source? If the concentration in the room is x, at what rate is it leaving the room? 



#3
Feb1813, 05:44 AM

P: 13

So we have a=1/6*10^6. I still don't understand what is b ..




#4
Feb1813, 05:51 AM

P: 13

Radiation and differential equations
I think I found. Is it: $$\frac{\Delta x}{\Delta t}=\frac{1}{6}\times 10^6+0.5RR\frac{x}{10^6}$$




#5
Feb1813, 06:18 AM

P: 13

I am having a problem now with the final question:
Find the rate R at which the equilibrium is 1.5 pCi/liter (well below the EPA action level). At this value for R, how many times per day is the total volume of air of the room exchanged? I tried to solve the above differential equation and obtained: $$x(t)=\frac{1}{6} \times \frac{10^{12}}{R}+5 \times 10^5+ \lambda \exp(\frac{R}{10^6}t)$$ Perhaps it is the wrong procedure, but I don't really know how to proceed with this question. 



#6
Feb1813, 06:55 AM

P: 204

I don't think you need to solve for x(t) at all. Equilibrium occurs when dx/dt = 0. So just plug that in and solve for R given that you want x/(10^6) = 1.5
The rest should be pretty trivial once you have R. 



#7
Feb1813, 07:08 AM

P: 13

We find R=$$\frac{1}{6} \times 10^6$$ but I don't understand the rest of the question.




#8
Feb1813, 07:19 AM

P: 204

Well, in the simplest terms, that means that onesixth of the air is exchanged every hour. So... how many times is the entire volume exchanged over 24 hours?




#9
Feb1813, 07:21 AM

P: 13

4 times




#10
Feb1813, 07:23 AM

P: 204

Yup.




#11
Feb1813, 07:24 AM

P: 13

Ok Thank you very much for all your help. Have a nice day.




#12
Feb1813, 07:36 AM

P: 204

No problem, and the same to you.




#13
Feb1813, 04:06 PM

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P: 9,143




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