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A 100gal 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.
 one year ago
 one year ago
A 100gal 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.
 one year ago
 one year ago

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abb0tBest ResponseYou've already chosen the best response.0
Is this for differential equations?
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.0
yes..., this is one of applications of differential equations
 one year ago

abb0tBest ResponseYou've already chosen the best response.0
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
 one year ago

zepdrixBest ResponseYou've already chosen the best response.2
dw:1356232472158:dw\[\huge A'=R_{\text{In}}R_{\text{Out}}\]Hmm let's see if we can figure this out..
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.0
actually i just wanna ask " if the volume remains constant, is the outflow rate of water is 4 gal/min again???"
 one year ago

zepdrixBest ResponseYou've already chosen the best response.2
Yes that sounds right. For the water to remain at halffull 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
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.0
ok thank u :) .., do you have another problem like this? i'll try to solve it
 one year ago

zepdrixBest ResponseYou've already chosen the best response.2
\[\large R_{\text{In}}=\left(4\frac{gal}{\min}\right)\left(0.1\frac{lbs}{gal}\right)=\frac{4}{10}\left(\frac{lbs}{\min}\right)\] 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
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.0
i got Q (t) = 10 + C exp(t/25) is this right?
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.0
it is for > 25 minutes.., for 0 >= 25 i got Q (t) = (0.4t^2 + 20t)/(2t+50)
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.0
finally i got Q(t) = 10  2.5e[exp(t/25)]  2.5e
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.0
let me see.., \[\frac{ dV }{ dt } = 42= 2\] \[\int\limits dV = 2 \int\limits dt\] \[V = 2t + C\] 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: \[Q(120) = 10 (2.5e) \exp \left( \frac{(120) }{ 25 } \right) = 10  0.05593 \approx 9.944 lb\]
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.0
How do you do it???
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.0
oh yeah, because i'm in a hurry
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.0
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??
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.0
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?
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.0
are u still here @Chlorophyll ?
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.0
welcome again @zepdrix :)
 one year ago

zepdrixBest ResponseYou've already chosen the best response.2
lol hey :D sorry i lost interest in this one XDDD these darn tank problems!!
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.0
u can check my answer :)
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.0
thank u., :). But, there is no incorrect integrations factor here, just the way that is done is different, but the result remains the same.
 one year ago
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