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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
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Is this for differential equations?
yes..., this is one of applications of differential equations
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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ok..., :)
|dw:1356232472158:dw|\[\huge A'=R_{\text{In}}-R_{\text{Out}}\]Hmm let's see if we can figure this out..
actually i just wanna ask " if the volume remains constant, is the outflow rate of water is 4 gal/min again???"
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
ok thank u :) .., do you have another problem like this? i'll try to solve it
\[\large R_{\text{In}}=\left(4\frac{gal}{\min}\right)\left(0.1\frac{lbs}{gal}\right)=\frac{4}{10}\left(\frac{lbs}{\min}\right)\] 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
i got Q (t) = 10 + C exp(-t/25) is this right?
it is for > 25 minutes.., for 0 >= 25 i got Q (t) = (0.4t^2 + 20t)/(2t+50)
finally i got Q(t) = 10 - 2.5e[exp(t/25)] - 2.5e
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: \[Q(120) = 10 -(2.5e) \exp \left( \frac{-(120) }{ 25 } \right) = 10 - 0.05593 \approx 9.944 lb\]
is this okay??
How do you do it???
oh yeah, because i'm in a hurry
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??
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?
are u still here @Chlorophyll ?
welcome again @zepdrix :)
lol hey :D sorry i lost interest in this one XDDD these darn tank problems!!
u can check my answer :)
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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