Swimming pool dilution problem
November 4th, 2013 at 8:36:54 AM
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You are given:
Please put answers and solutions in spoiler tags.
- A swimming pool has a capacity of 7 million cubic inches.
- Clean water flows into the pool at a rate of 30 cubic inches per second.
- Pool water exits the pool at the same rate.
- A kid pees in the pool. The volume of pee is 8 cubic inches, which displaces the same volume of clean water.
- As fresh water pours in the pool, it is immediately diluted. In other words, the concentration of pee is the same throughout the pool.
- You deem the pool safe to swim in if the total volume of pee is less than or equal to one cubic inch.
- From the time the kid pees, how long do you have to wait to swim?
Please put answers and solutions in spoiler tags.
Have a nice day.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
November 4th, 2013 at 9:15:09 AM
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I get 5.6 days.
I heart Crystal Math.
November 4th, 2013 at 9:44:24 AM
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Any more? I'll wait a little longer before commenting on CM's answer.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
November 4th, 2013 at 10:12:33 AM
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Well, still fiddlin' with it, but I got a different answer than CM, which generally means I'm wrong.
2.36 days
If the House lost every hand, they wouldn't deal the game.
November 4th, 2013 at 10:13:40 AM
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Quote: beachbumbabsWell, still fiddlin' with it, but I got a different answer than CM, which generally means I'm wrong.
:) Thanks.
I heart Crystal Math.
November 4th, 2013 at 10:13:53 AM
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Here's my guess:
485,203 seconds, which is 5.62 days
November 4th, 2013 at 10:15:22 AM
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Quote: WizardYou are given:
- A swimming pool has a capacity of 7 million cubic inches.
- Clean water flows into the pool at a rate of 30 cubic inches per second.
- Pool water exits the pool at the same rate.
- A kid pees in the pool. The volume of pee is 8 cubic inches, which displaces the same volume of clean water.
- As fresh water pours in the pool, it is immediately diluted. In other words, the concentration of pee is the same throughout the pool.
- You deem the pool safe to swim in if the total volume of pee is less than or equal to one cubic inch.
- From the time the kid pees, how long do you have to wait to swim?
Please put answers and solutions in spoiler tags.Have a nice day.
In other news, nothin' about a little kid pee the chlorine won't fix....
WC Fields: "Little girl, little girl, don't drink that water!!!"
Child: "But sir, whyever not?" (Kids talked like that 80 years ago)
WC Fields: "Fish {bleep} in it!!"
If the House lost every hand, they wouldn't deal the game.
November 4th, 2013 at 10:18:42 AM
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I get the same answer as CrystalMath (Edit: and ChesterDog, who posted while I was typing), so I'll post my full solution behind a spoiler.
Let C = current concentration of pee (fraction) and Vpee = current total volume of pee in the pool
V=7x106 cu. in. (given)
Flow = F = 30 cu. in. per second (given)
Vpee=C • V
dVpee/dt = -F • C = V dC/dt
dC/dt = -(F/V) • C
C = A • e-(F/V)•t
For t=0, C = 8/V --> A = 8/V
For "clean enough", C = 1/V.
C = 1/V = (8/V) • e-(F/V)•t
1/8 = e-(F/V)•t
ln(1/8) = -(F/V) • t
t = -(V/F) • ln(1/8) = (-(7x106)/30) • ln(1/8) = 485,203 seconds = 134.78 hours = 5.6157 days
V=7x106 cu. in. (given)
Flow = F = 30 cu. in. per second (given)
Vpee=C • V
dVpee/dt = -F • C = V dC/dt
dC/dt = -(F/V) • C
C = A • e-(F/V)•t
For t=0, C = 8/V --> A = 8/V
For "clean enough", C = 1/V.
C = 1/V = (8/V) • e-(F/V)•t
1/8 = e-(F/V)•t
ln(1/8) = -(F/V) • t
t = -(V/F) • ln(1/8) = (-(7x106)/30) • ln(1/8) = 485,203 seconds = 134.78 hours = 5.6157 days
November 4th, 2013 at 10:20:28 AM
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Quote: WizardYou are given:
- A swimming pool has a capacity of 7 million cubic inches.
- Clean water flows into the pool at a rate of 30 cubic inches per second.
- Pool water exits the pool at the same rate.
- A kid pees in the pool. The volume of pee is 8 cubic inches, which displaces the same volume of clean water.
- As fresh water pours in the pool, it is immediately diluted. In other words, the concentration of pee is the same throughout the pool.
- You deem the pool safe to swim in if the total volume of pee is less than or equal to one cubic inch.
- From the time the kid pees, how long do you have to wait to swim?
Please put answers and solutions in spoiler tags.
I figured it reduces urine by 37.028% per 24 hours. For a total of 109 hours and 17 min roughly
I'm not good at math, but that was my attempt!!!
November 4th, 2013 at 10:21:14 AM
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Quote: DocI get the same answer as CM, so I'll post my solution behind a spoiler.
Let C = current concentration of pee (fraction) and Vpee = current total volume of pee in the pool
V=7x106 cu. in. (given)
Flow = F = 30 cu. in. per second (given)
Vpee=C • V
dVpee/dt = -F • C = V dC/dt
dC/dt = -(F/V) • C
C = A • e-(F/V)•t
For t=0, C = 8/V --> A = 8/V
For "clean enough", C = 1/V.
C = 1/V = (8/V) • e-(F/V)•t
1/8 = e-(F/V)•t
ln(1/8) = -(F/V) • t
t = -(V/F) • ln(1/8) = (-(7x106)/30) • ln(1/8) = 485,203 seconds = 134.78 hours = 5.6157 days
The combined talents of CM, Chester, and Doc have intimidated me into requesting partial credit for the class clown maneuver. I can't possibly be correct, but at least I can be retained for comic relief.
If the House lost every hand, they wouldn't deal the game.
November 4th, 2013 at 10:26:23 AM
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Quote: beachbumbabsQuote: DocI get the same answer as CM, so I'll post my solution behind a spoiler.
Let C = current concentration of pee (fraction) and Vpee = current total volume of pee in the pool
V=7x106 cu. in. (given)
Flow = F = 30 cu. in. per second (given)
Vpee=C • V
dVpee/dt = -F • C = V dC/dt
dC/dt = -(F/V) • C
C = A • e-(F/V)•t
For t=0, C = 8/V --> A = 8/V
For "clean enough", C = 1/V.
C = 1/V = (8/V) • e-(F/V)•t
1/8 = e-(F/V)•t
ln(1/8) = -(F/V) • t
t = -(V/F) • ln(1/8) = (-(7x106)/30) • ln(1/8) = 485,203 seconds = 134.78 hours = 5.6157 days
The combined talents of CM, Chester, and Doc have intimidated me into requesting partial credit for the class clown maneuver. I can't possibly be correct, but at least I can be retained for comic relief.
I know how you feel. I did this on an adding machine and a 3" x 5" Phone Call Message piece of paper with writing already all over it. LOL
November 4th, 2013 at 10:29:48 AM
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Quote: beerseasonI did this on an adding machine and a 3" x 5" Phone Call Message piece of paper ...
I would have used one of the several slide rules in my desk drawer, but I wouldn't have been able to list the unreasonable number of allegedly significant digits as I showed. Instead, I played 21st century and used my cell phone.
November 4th, 2013 at 10:31:20 AM
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Quote: beachbumbabsThe combined talents of CM, Chester, and Doc have intimidated me into requesting partial credit for the class clown maneuver. I can't possibly be correct, but at least I can be retained for comic relief.
I agree that the three of them got the correct answer. You, as the class clown, may wear the dunce cap.
Let p = volume of pee in the pool, in cubic inches.
Let t = time since the kid peed in the pool, in seconds.
dp/dt = (-30/7000000)×p
dp = (-30/7000000)×p dt
dp (7000000/30p)= dt
Taking the integral of each side...
(1) -(7000000/30)×ln(p)= t + c, where c is the constant of integration.
We're given that when t=0, v=8. Plug that into the equation to find c.
-(7000000/30)×ln(8)= 0 + c
c = -(7000000/30)×ln(8)
Putting that value of c in equation (1) ...
-(7000000/30)×ln(p)= t -(7000000/30)×ln(8)
t = -(7000000/30)×ln(p) + (7000000/30)×ln(8)
t = (7000000/30)×(ln(8)-ln(p))
t = (7000000/30)×ln(8/p)
The question at hand is what is t when p = 1
t = (7000000/30)×ln(8/1)
t = 485,203 seconds = 8,087 minutes = 134.8 hours = 5.6158 days.
Let t = time since the kid peed in the pool, in seconds.
dp/dt = (-30/7000000)×p
dp = (-30/7000000)×p dt
dp (7000000/30p)= dt
Taking the integral of each side...
(1) -(7000000/30)×ln(p)= t + c, where c is the constant of integration.
We're given that when t=0, v=8. Plug that into the equation to find c.
-(7000000/30)×ln(8)= 0 + c
c = -(7000000/30)×ln(8)
Putting that value of c in equation (1) ...
-(7000000/30)×ln(p)= t -(7000000/30)×ln(8)
t = -(7000000/30)×ln(p) + (7000000/30)×ln(8)
t = (7000000/30)×(ln(8)-ln(p))
t = (7000000/30)×ln(8/p)
The question at hand is what is t when p = 1
t = (7000000/30)×ln(8/1)
t = 485,203 seconds = 8,087 minutes = 134.8 hours = 5.6158 days.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
November 4th, 2013 at 10:35:46 AM
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p = 7000000/30 ln (1/8) = 485,203 seconds?
I was running through the calculations just before my project manager made me take him out for lunch!!!
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You want the truth! You can't handle the truth!
November 4th, 2013 at 10:44:37 AM
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Quote: WizardI agree that the three of them got the correct answer. You, as the class clown, may wear the dunce cap.
Let p = volume of pee in the pool, in cubic inches.
Let t = time since the kid peed in the pool, in seconds.
dp/dt = (-30/7000000)×p
dp = (-30/7000000)×p dt
dp (7000000/30p)= dt
Taking the integral of each side...
(1) -(7000000/30)×ln(p)= t + c, where c is the constant of integration.
We're given that when t=0, v=8. Plug that into the equation to find c.
-(7000000/30)×ln(8)= 0 + c
c = -(7000000/30)×ln(8)
Putting that value of c in equation (1) ...
-(7000000/30)×ln(p)= t -(7000000/30)×ln(8)
t = -(7000000/30)×ln(p) + (7000000/30)×ln(8)
t = (7000000/30)×(ln(8)-ln(p))
t = (7000000/30)×ln(8/p)
The question at hand is what is t when p = 1
t = (7000000/30)×ln(8/1)
t = 485,203 seconds = 8,087 minutes = 134.8 hours = 5.6158 days.
I was no where close to using the right formula lol. I just took 8 cubic inches and reduced it by 37% per 24 hrs. I assume why it took longer is because it was constantly being diluted whereas I calculated it on a per day basis?
November 4th, 2013 at 10:50:10 AM
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It's a common calculus problem. I originally thought you would just divide out 7000000/30 x 7/8. When I realized it was a concentration problem, I realized that calculus was involved. I didn't know it first hand but had to cheat (aka, take another solved problem and apply it to this one) to figure it out.
My kid's taking Grade 12 Functions. I wonder if she can figure it out.
My kid's taking Grade 12 Functions. I wonder if she can figure it out.
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You want the truth! You can't handle the truth!
November 4th, 2013 at 11:01:13 AM
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I took calculus 3 years ago, to graduate. Didn't understand it one bit, and don't ever care to i suppose. It was one of those thing I just didn't "get."
November 4th, 2013 at 12:02:32 PM
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A common undergrad chemical engineering problem because reactors need to often be "continuously stirred", which this problem assumes. I'm excited!!! Let's see if I can do it without google's help and not screw it up! It's been 12 years since the relevant class...
:-\
:-\
Mass balance on pee!!
accumulation = input - output + generation
m = mass of pee
V_pool = 7,000,000
V_exit = 30
dm/dt = 0 - m_exit + 0
Out of convenience, I am turning the generation term into the initial concentration condition.
Since continuously "stirred" and pool volume is constant, turn mass into concentrations.
C = pee concentration
dC/dt*V_pool = 0 - C*V_exit
C_initial = 8/7,000,000
C_final = 1/7,000,000
dC/C = - V_exit/V_pool*dt
Integrate both sides:
ln(C_final/C_initial) = -V_exit/V_pool*(t-0)
solve for t:
t = V_pool/V_exit*ln(C_initial/C_final)
Note: C_initial/C_final = 8
Plug-in:
t = 7,000,000/30*ln(8) = 485203 seconds = 5.62 days.
In reality, I dunno how good of an assumption "continuously stirred" is here considering how pools filter, but I digress.
accumulation = input - output + generation
m = mass of pee
V_pool = 7,000,000
V_exit = 30
dm/dt = 0 - m_exit + 0
Out of convenience, I am turning the generation term into the initial concentration condition.
Since continuously "stirred" and pool volume is constant, turn mass into concentrations.
C = pee concentration
dC/dt*V_pool = 0 - C*V_exit
C_initial = 8/7,000,000
C_final = 1/7,000,000
dC/C = - V_exit/V_pool*dt
Integrate both sides:
ln(C_final/C_initial) = -V_exit/V_pool*(t-0)
solve for t:
t = V_pool/V_exit*ln(C_initial/C_final)
Note: C_initial/C_final = 8
Plug-in:
t = 7,000,000/30*ln(8) = 485203 seconds = 5.62 days.
In reality, I dunno how good of an assumption "continuously stirred" is here considering how pools filter, but I digress.
November 4th, 2013 at 1:12:18 PM
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is ln(8)* refresh rate = 485,203 seconds or about 5 days 14 hours 46 minutes 43 seconds after pee
refresh rate =7E6/30
ln(8) derived from conc. initial/conc. OK
lol I thought this was a crossed claissden rxn :P
Some people need to reimagine their thinking.
November 4th, 2013 at 1:38:45 PM
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Quote: WizardI agree that the three of them got the correct answer. You, as the class clown, may wear the dunce cap.
Let p = volume of pee in the pool, in cubic inches.
Let t = time since the kid peed in the pool, in seconds.
dp/dt = (-30/7000000)×p
dp = (-30/7000000)×p dt
dp (7000000/30p)= dt
Taking the integral of each side...
(1) -(7000000/30)×ln(p)= t + c, where c is the constant of integration.
We're given that when t=0, v=8. Plug that into the equation to find c.
-(7000000/30)×ln(8)= 0 + c
c = -(7000000/30)×ln(8)
Putting that value of c in equation (1) ...
-(7000000/30)×ln(p)= t -(7000000/30)×ln(8)
t = -(7000000/30)×ln(p) + (7000000/30)×ln(8)
t = (7000000/30)×(ln(8)-ln(p))
t = (7000000/30)×ln(8/p)
The question at hand is what is t when p = 1
t = (7000000/30)×ln(8/1)
t = 485,203 seconds = 8,087 minutes = 134.8 hours = 5.6158 days.
(In corner, on thinking chair, wearing dunce cap, muttering to myself softly...."we don't swim in your toilet, please don't pee in our pool..." over and over)
If the House lost every hand, they wouldn't deal the game.
November 4th, 2013 at 1:49:27 PM
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Quote: beachbumbabs(In corner, on thinking chair, wearing dunce cap, muttering to myself softly...."we don't swim in your toilet, please don't pee in our pool..." over and over)
Very funny! For the rest of you not in on the inside joke, the "thinking chair" is in my house, where one must sit after being punished to reflect on his or her antisocial (but its always a "her") behavior.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
November 4th, 2013 at 1:56:31 PM
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All this is why I swim in the ocean.
Beware of all enterprises that require new clothes - Henry David Thoreau. Like Dealers' uniforms - Dan.
November 4th, 2013 at 1:56:49 PM
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Quote: WizardVery funny! For the rest of you not in on the inside joke, the "thinking chair" is in my house, where one must sit after being punished to reflect on his or her antisocial (but its always a "her") behavior.
Proud to be associated with your daughters in any capacity. I am Woman, hear me roar...or mutter...or whisper. WhatEVER, dad.
If the House lost every hand, they wouldn't deal the game.
November 4th, 2013 at 4:11:05 PM
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Volume of pool V = 7M.
Rate R = 30/sec.
Say, p(t) is the volume of pee at time t in the ppol (in cubic inches).
concentration is c(t) = p(t) / V.
At time t+dt, The volume of R*dt (at the concentration c(t)) has been replaced by fresh water, i.e. the amount of pee is reduced by c(t)*R*dt.
Hence p(t+dt) = p(t) - c(t)* R*dt
or dp/dt = - c(t)*R = - p(t) * R / V.
Integration gives
p(t) = p(0) * exp(-R/V*t), with p(0) = 8.
Solving p(t)<=1: t >= ln(8) * V / R = 135 hours.
November 4th, 2013 at 5:15:21 PM
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I have a feeling that the wizard's punishment for his antisocial behaviour involve purchases of expensive goods and/or services. ;)
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You want the truth! You can't handle the truth!
November 4th, 2013 at 5:28:52 PM
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Quote: PaigowdanAll this is why I swim in the ocean.
But where do you think all those fish take care of business?
All animals are equal, but some are more equal than others
November 4th, 2013 at 5:33:34 PM
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deleted
DUHHIIIIIIIII HEARD THAT!
November 5th, 2013 at 9:22:48 AM
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Quote: WizardYou are given:
- A swimming pool has a capacity of 7 million cubic inches.
- Clean water flows into the pool at a rate of 30 cubic inches per second.
- Pool water exits the pool at the same rate.
- A kid pees in the pool. The volume of pee is 8 cubic inches, which displaces the same volume of clean water.
- As fresh water pours in the pool, it is immediately diluted. In other words, the concentration of pee is the same throughout the pool.
- You deem the pool safe to swim in if the total volume of pee is less than or equal to one cubic inch.
- From the time the kid pees, how long do you have to wait to swim?
About 5 minutes: stand up, walk past the family pool toward the adults-only area, and go for a nice swim in a pool that is only diluted with tequila and rum.
Also, the science nerd in me wants to know how a liquid that instantaneously dilutes over a volume of 7m cubic inches (see #5) would interact with the human body. Something tells me the "water" would dissolve you as soon as you got in, but I never studied p-chem.
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice."
-- Girolamo Cardano, 1563
November 5th, 2013 at 9:37:12 AM
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Quote: MathExtremistAlso, the science nerd in me wants to know how a liquid that instantaneously dilutes over a volume of 7m cubic inches (see #5) would interact with the human body. Something tells me the "water" would dissolve you as soon as you got in, but I never studied p-chem.
Maybe there are lots of other kids playing in the pool, who aren't aware of the pee, or don't care, splashing about and evenly distributing the pee that way.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
November 5th, 2013 at 3:01:37 PM
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I suspect that for a pool with people in it, and perhaps with circulating pumps operating, a fully-mixed-water assumption is better than trying to solve the problem with Fick's law (or some really complex combination of approaches). Besides, who has good reference data for the diffusion coefficient of pee in chlorinated water?
