anonymous
  • anonymous
A 100-gal tank is initially half full of pure water. Then water containing 0.1 lb/gal of salt is added at a rate of 4 gal/min. The tank contens are well mixed. water flows out of the tank at a rate of 2 gal/min untul the tank is full. Once the tank is full it overflows. Find the amount and concentration of salt in the tank as function of t for the first 2 hr.
Differential Equations
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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abb0t
  • abb0t
Is this for differential equations?
anonymous
  • anonymous
yes..., this is one of applications of differential equations
abb0t
  • abb0t
Yeah, it's a mixing problem. Hmm..it's been a while since I've done one of these. Give me a minute to see if I can remember. lol

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anonymous
  • anonymous
ok..., :)
zepdrix
  • zepdrix
|dw:1356232472158:dw|\[\huge A'=R_{\text{In}}-R_{\text{Out}}\]Hmm let's see if we can figure this out..
anonymous
  • anonymous
actually i just wanna ask " if the volume remains constant, is the outflow rate of water is 4 gal/min again???"
zepdrix
  • zepdrix
Yes that sounds right. For the water to remain at half-full over the entire time span, it would have to have the same RateOut as RateIn. That doesn't appear to be what we have here though. :D
anonymous
  • anonymous
ok thank u :) .., do you have another problem like this? i'll try to solve it
zepdrix
  • zepdrix
\[\large R_{\text{In}}=\left(4\frac{gal}{\min}\right)\left(0.1\frac{lbs}{gal}\right)=\frac{4}{10}\left(\frac{lbs}{\min}\right)\] The Rate In is fairly easy to solve. I'm trying to remember how to setup the rate out when the water is changing. Do you mean an easier example? Having trouble with this one? Or did you already solve it? :o
anonymous
  • anonymous
i got Q (t) = 10 + C exp(-t/25) is this right?
anonymous
  • anonymous
it is for > 25 minutes.., for 0 >= 25 i got Q (t) = (0.4t^2 + 20t)/(2t+50)
anonymous
  • anonymous
finally i got Q(t) = 10 - 2.5e[exp(t/25)] - 2.5e
anonymous
  • anonymous
let me see.., \[\frac{ dV }{ dt } = 4-2= 2\] \[\int\limits dV = 2 \int\limits dt\] \[V = 2t + C\] because of at t=0 V of water = 50, C = 50, then \[V = 2t +50\] \[\frac{ dQ }{ dt }= 0.1 (4) - 2\left( \frac{ Q }{ 2t+50 } \right)\] \(\frac{ dQ }{ dt } +2\left( \frac{ Q }{ 2t+50 } \right) = 0.1 (4) \) .......(1) i got p(t) = 2/(2t+50), then for integration factor, i have: \[\mu(t) = \exp \left( \int\limits_{0}^{t} \frac{ 2 }{ 2t+50 } dt\right) \] \[\mu(t) = \exp(\ln |2t+50|) = 2t+50\] then multiplying both side of (1) equations by (2t+50), then: \[(2t+50)\frac{ dQ }{ dt }+ \frac{ 2 }{ 2t+50 } (2t+50) Q = 0.4 (2t+50)\] \[\int\limits d \left( (2t+50) Q \right) = 0.4 \int\limits (2t+50) dt\] \[(2t+50) Q = 0.4(t^{2} + 50t) + C\] \[Q = \frac{ 0.4t^{2}+ 20t }{ 2t+50 } + C\] because at t = 0 Q = 0, then i have C = 0, then i have \(Q = \frac{ 0.4t^{2}+ 20t }{ 2t+50 }\) .......... (2) (note: it is from t = 0 until t = 25 minutes, and further the volume remains constan at 100 gal), FOR 25 minutes to 2 hour (120 minutes) \[\frac{ dQ }{ dt } = 0.1(4) - 4\frac{ Q }{ 100 } \] \(\frac{ dQ }{ dt } + 4\frac{ Q }{ 100 } = 0.1(4) \) .......(3) i have p(t), is -4/100, then i have new integration factor: \[\mu(t) = \exp \left( \int\limits_{0}^{t} \frac{ 4 }{ 100 } dt \right) = \exp \left( \frac{ 4t }{ 100 } \right) = e^{\frac{ 4t }{ 100 }}\] multiplying both sides of (3) equations by exp (4t/100) \[e^{\frac{ 4t }{ 100 }}\frac{ dQ }{ dt } + 4 e^{\frac{ 4t }{ 100 }}\frac{ Q }{ 100 } = 0.4 e^{\frac{ 4t }{ 100 }}\] \[\int\limits d \left( e^{\frac{ 4t }{ 100 }} Q\right) = 0.4 \int\limits \exp \left( \frac{ 4t }{ 100 }\right) dt \] \[\exp \left( \frac{ 4t }{ 100 } \right) Q = 10 e^{\frac{ 4t }{ 100 }} + C\] \(Q (t) = 10 + C \exp \left( \frac{ -4t }{ 100 } \right)\) ......(4) OK.., back to (2) equations.., for t = 25, i have : \[Q(25) = \frac{ 0.4(25)^{2} + 20 (25) }{ 2(25) + 50} = \frac{ 0.4(625)+ 500 }{ 100 } = \frac{ 750 }{ 100 } = 7.5\] So, at t = 25 minutes, Q is equal to 7.5 lb, substitute this to (4) \[Q (25) = 10 + C \exp \left( \frac{ -4(25) }{ 100 } \right) \] \[7.5 = 10 + C \exp \left( \frac{- 100 }{ 100 } \right) \] \[C = -2.5 e\] finally i got the amount of salt formula as functions of t at the volume remains constant at 100 gal. \[Q(t) = 10 -(2.5e) \exp \left( \frac{-t }{ 25 } \right) \] SO, at t = 2 hours (120 minutes), the value of the amount of salt is: \[Q(120) = 10 -(2.5e) \exp \left( \frac{-(120) }{ 25 } \right) = 10 - 0.05593 \approx 9.944 lb\]
anonymous
  • anonymous
is this okay??
anonymous
  • anonymous
How do you do it???
anonymous
  • anonymous
oh yeah, because i'm in a hurry
anonymous
  • anonymous
i'm little bit confused.., What can't I take \[\frac{ dQ }{ dt } + \frac{ 2 }{ 50+2t } Q = 0.4\] and \[\mu(t) = \exp \left( \int\limits_{0}^{t} \frac{ 2 }{ 50+2t } dt \right)\] ????? what's wrong if i take it??
anonymous
  • anonymous
let me see.., if i have an integrations factor as: t + 25 and take integral both side as: ( t + 25 ) y = 0.4 ∫ ( t + 25 ) dt \[(t+25) y = 0.4 \int\limits\limits (t+25) dt = 0.4 \left( \int\limits\limits t dt + \int\limits\limits 25dt \right)\] \[(t+25)y = 0.4 (\frac{ t^{2} }{ 2 }) + 25 t = 0.2 t^{2} + 10t+C\] \[y = \frac{ 0.2t^{2}+10t }{ t+25 } + C\] on my answer review i got \[Q (t) = \frac{ 20t+0.4t^{2} }{ 50+2t }+C\] it seems like: \[Q(t) = \frac{ 2 }{ 2 } \left( \frac{ 10t+0.2t^{2} }{ 25+t } \right) +C\] isn't that the same?
anonymous
  • anonymous
are u still here @Chlorophyll ?
anonymous
  • anonymous
welcome again @zepdrix :)
zepdrix
  • zepdrix
lol hey :D sorry i lost interest in this one XDDD these darn tank problems!!
anonymous
  • anonymous
u can check my answer :)
anonymous
  • anonymous
thank u., :). But, there is no incorrect integrations factor here, just the way that is done is different, but the result remains the same.

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