vee043324
White Claw Pusher
Update: You have to solve using equation 21-6 from C&A. Have yet to do it but will solve during lunch today or sometime tonight.
October 2018 Exam Study Progress
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vee043324
White Claw Pusher
vee043324
White Claw Pusher
I feel like part of what makes this problem so "hard" is that I'm so not used to these units.
OOC, what are the possible answers?
edit: And if it's not multiple choice, what do they provide for the answer? I think I've solved it with unit analysis and some knowledge of HP, but I have no actual experience or training in this sort of problem.
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In/PE/Out
Strong coffee drinking!
There aren’t answers provided for the end-of-chapter problems. I’m not sure this problem would even be representative of what would actually be on the exam, based from what I’ve seen in various study materials.
My guess would be to try to apply mass balance on the room in question. This probably isn’t even close to being right, but...
Radon in= 9.5 pCi/(m^2/s) * 2500ft^2 * 0.093 (m^2/ft^2) * 3600s/hr * 0.9 = 7.16 * 10^6 pCi/hr
or 1.72 * 10^8 pCi/day
[edit: Deleted the rest of my work because it was embarrassingly bad.]
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Hockey Eng
Well-known member
There's a solutions manual available for Cooper & Alley. It's around on the internet but I didn't bother to try and download it, and I'm not sure of the legitimacy of any of the sites that claim to have it. There's also a site called Chegg.com that has the solutions online. They have some of the end of chapter problems completely worked out, the rest you have to pay for. It's $20 so might be worth it down the road but not (for me) 2 days ahead of the exam. Anyway, for this particular problem they do show the first step in solving it, which is calculating the emission rate into the building.
S = 90% x (radon flux) x (floor surface area) = 7,151,487 pCi/hr
Go here, then in the solution box you can pick the chapter and problem # for the solution. Some have them worked out all the way others just the first step. Still pretty convenient even if only some of the answers are available.
https://www.chegg.com/homework-help/air-pollution-control-4th-edition-solutions-9781577666783
Disclaimer - I'n not affiliated with this website or company in any way.
S = 90% x (radon flux) x (floor surface area) = 7,151,487 pCi/hr
Go here, then in the solution box you can pick the chapter and problem # for the solution. Some have them worked out all the way others just the first step. Still pretty convenient even if only some of the answers are available.
https://www.chegg.com/homework-help/air-pollution-control-4th-edition-solutions-9781577666783
Disclaimer - I'n not affiliated with this website or company in any way.
Okay here's my take, I'll identify the one part that I'm not 100% on in italics.
First put radon emanation rate in consistent working units:
9.5 pCi/m^2*s * 1 m^2/10.7639 sf * 3600 s/hr = 3177 pCi/sf*hr
Next determine how much enters house (steady state)
3177 pCi/sf*hr * 2500 sf * 0.9 = 7.15e6 pCi/hr
Determine removal of radon via radioactive decay
A(t)=A(0)*exp(-lamda*t) --> A(t)=7.15e6 pCi/hr * exp(-7.6e-3 hr^-1 * 1 hr) = 7.10e6 pCi/hr
Determine removal rate from house via (HVAC?). This is where I'm not sure of the method, I'm a NukeE, and while I have degrees in EnvE and ME too, I don't have any coursework or experience here.
7.10e6 pCi/hr * 1 hr/1.5[air change] = 4.73e6 pCi
Determine concentration in house. Assuming uniform distribution in the structures volume (something I know won't be true, but whatever). I am certain on this step.
4.73e6 pCi/20000cf = 236 pCi/cf
Optional: convert to correct regulatory unit
236 pCi/cf * 1 cf/28.322 L = 8.35 pCi/L
Which is more than 4x the regulatory limit
First put radon emanation rate in consistent working units:
9.5 pCi/m^2*s * 1 m^2/10.7639 sf * 3600 s/hr = 3177 pCi/sf*hr
Next determine how much enters house (steady state)
3177 pCi/sf*hr * 2500 sf * 0.9 = 7.15e6 pCi/hr
Determine removal of radon via radioactive decay
A(t)=A(0)*exp(-lamda*t) --> A(t)=7.15e6 pCi/hr * exp(-7.6e-3 hr^-1 * 1 hr) = 7.10e6 pCi/hr
Determine removal rate from house via (HVAC?). This is where I'm not sure of the method, I'm a NukeE, and while I have degrees in EnvE and ME too, I don't have any coursework or experience here.
7.10e6 pCi/hr * 1 hr/1.5[air change] = 4.73e6 pCi
Determine concentration in house. Assuming uniform distribution in the structures volume (something I know won't be true, but whatever). I am certain on this step.
4.73e6 pCi/20000cf = 236 pCi/cf
Optional: convert to correct regulatory unit
236 pCi/cf * 1 cf/28.322 L = 8.35 pCi/L
Which is more than 4x the regulatory limit
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The nuclear terminology they use is the problem is a bit weird. I've never heard of a "decay coefficient". I'm assuming "decay coefficient" is actually the "decay constant". I confirmed that the decay constant of Rn-222 is ~0.007553585 /hr, which is close enough to the 7.6e-3 given in the problem.
Hockey Eng
Well-known member
I believe the "decay coefficient" in this instance is actually the removal reaction rate constant, used in steady state box model for calculating indoor air concentrations. You'd have to have the book in order to get the context - the problems in the book generally follow along with the topics as they are presented, and this particular problem matches up well with the discussion of steady state concentration calculations. They typically don't introduce concepts such a radioactive decay into a problem without at least brushing over it in the text. Not saying you're wrong, in fact it kind of makes sense. Just that when it comes to solving the practice problems they tend to keep them relatively in-line with the material presented.
Edit - they do discuss the radioactivity of radon, but from what I'm gathering from the text, for the purposes of indoor air quality calculations it's treated as a gas rather than a radionuclide (even though the units are in pCi, which is what makes it so hard to work with, for me anyway).
And hey, since you're here, do you know of any way to convert pCi to units of concentration (micrograms/cubic meter) or similar? When I was trying to solve this earlier that's where I got hung up, since the formula I was using solves for concentration (and which the problem is asking for).
Edit - they do discuss the radioactivity of radon, but from what I'm gathering from the text, for the purposes of indoor air quality calculations it's treated as a gas rather than a radionuclide (even though the units are in pCi, which is what makes it so hard to work with, for me anyway).
And hey, since you're here, do you know of any way to convert pCi to units of concentration (micrograms/cubic meter) or similar? When I was trying to solve this earlier that's where I got hung up, since the formula I was using solves for concentration (and which the problem is asking for).
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I follow your logic. OOC, when the book discusses "decay coefficient" what formulation does it use? Something like C * exp( - [decay coeff] * time) or something else?
So when describing radioactive concentrations, the appropriate units are curies (Ci) (imperial) or bequerel (Bq) (SI) divided by a volume (liters, cf, cc, cubic meters, etc). I used pCi/L in the solution earlier because the regulations are written in pCi/L.
You can convert that to a mass per volume, but its cumbersome and almost no one uses that metric. It's not difficult per se, it happens early in the "201" class, but it can be confusing at first because the units are often all over the place in practice.
Activity = number atoms * decay constant
mass = number atoms/ NA (6.02e23) * molar mass
So to use my solution to above, leaving aside the L^-1
Activity = 8.35 pCi but need to convert to decay/sec (aka Bq)
8.35 pCi = 8.35e-12 Ci -> 8.35 Ci * 3.7e10 Bq/Ci = 0.31 Bq = decay/sec
decay constant = lambda = ln(2)/t(1/2). I'll solve it for the above two ways
given in problem 7.6e-3 hr^1 -> 7.6e-3 hr^-1 * 1 hr /3600 s = 2.1e-6 s^-1
or t(1/2)Rn-222= 3.8235 d = 330350 s
lambda = ln(2)/330350 s = 2.1e-6 s^-1
So Number atoms = activity /decay constant
# atoms = 0.31 s^-1 /2.1e-6 s^-1 = 147744 atoms = 1.5e5 atoms
mass = (1.5e5 atoms / 6.02e23 atoms/mol ) * 222 grams/mol = 5e-17 grams
Throw back in the /L and you get 5e-17 g/L. Converting to another concentration metric is on you.
Hockey Eng
Well-known member
For that problem, like I said they use 'removal rate constant' which is just a first order coefficient (hour to -1 power), which matches what they have in the word problem. So you can see why there might be some confusion, lack of consistent terminology. There's no formula (not in that section of the book anyway).
I wish you hadn't gone to all that trouble for the conversion, looks like a lot of work. But appreciated!
Maybe we should all just chip in a couple bucks each to purchase the solutions manual, get it figured out once and for all :reading:
I wish you hadn't gone to all that trouble for the conversion, looks like a lot of work. But appreciated!
Maybe we should all just chip in a couple bucks each to purchase the solutions manual, get it figured out once and for all :reading:
Hi all,
Good luck to all taking the EnvPE! Was fortunate to pass the first time, and can appreciate all the last minute study efforts).
I took a stab at the problem, using a simple steady state material balance approach and thought I would share (I don't know if I'm applying the decay coefficient correctly, but it seems consistent with what I could find in literature). I apologize in advance for the text issues
Chouse = radon concentration in house at time t
pCiin = rate of radon coming into house = 7,152,530 pCi/hr
V = volume of house = 20,000 ft3
Qhouse = air flow out of house = 30,000 ft3/hr
K = decay coefficient = 7.6 x 10-3/hr
Using a material balance approach,
ChouseV =pCiint - kChousetV- ChouseQhouset
dChouse /dt = (pCiin - ChouseQhouse)/V - kChouse
sinceChouse is steady state (i.e., constant), then dChouse /dt = 0
which gives 0 = (pCiin - ChouseQhouse)/V - kChouse
0 = pCiin/V - ChouseQhouse/V - kChouse
ChouseQhouse/V - kChouse = pCiin/V
Chouse(Qhouse/V - kChouse) = pCiin/V
Chouse = pCiin/V [1/(Qhouse/V - kChouse)]
This gives Chouse = 237 pCi/ft3
Good luck to all taking the EnvPE! Was fortunate to pass the first time, and can appreciate all the last minute study efforts).
I took a stab at the problem, using a simple steady state material balance approach and thought I would share (I don't know if I'm applying the decay coefficient correctly, but it seems consistent with what I could find in literature). I apologize in advance for the text issues
Chouse = radon concentration in house at time t
pCiin = rate of radon coming into house = 7,152,530 pCi/hr
V = volume of house = 20,000 ft3
Qhouse = air flow out of house = 30,000 ft3/hr
K = decay coefficient = 7.6 x 10-3/hr
Using a material balance approach,
ChouseV =pCiint - kChousetV- ChouseQhouset
dChouse /dt = (pCiin - ChouseQhouse)/V - kChouse
sinceChouse is steady state (i.e., constant), then dChouse /dt = 0
which gives 0 = (pCiin - ChouseQhouse)/V - kChouse
0 = pCiin/V - ChouseQhouse/V - kChouse
ChouseQhouse/V - kChouse = pCiin/V
Chouse(Qhouse/V - kChouse) = pCiin/V
Chouse = pCiin/V [1/(Qhouse/V - kChouse)]
This gives Chouse = 237 pCi/ft3
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Hockey Eng
Well-known member
This problem has become our White Whale...
vee043324
White Claw Pusher
I take 24 hours away from EB and come back to this. Y’all are amazing. It seems like 237 pci / cf is the majority answer. I also got that.
i confirmed with my ppi professor that there is no need to convert pci back to ug!
i confirmed with my ppi professor that there is no need to convert pci back to ug!
Call me Ishmael.
I think we got it and @joeenv probably used the correct EnvE formulation to solve the problem.
p.s. I still think the whole decay coefficient thing is weird, and I don't know how transferable it is to other isotopes and time-frames.
Ok, so @In/PE/Out posted his good luck cat...I am posting a pic of my pupper. He was my study buddy my first two attempts, but then couldn't understand why I had to study at the dining room table during this exam cycle. He would try to get me to play with him while I was studying, and act naughty to get my attention...so cute. I took a pic of him yesterday while tabbing some notes. He said he's concerned about how hard we've all been working, and wants us to relax today.
...they’re the best !
