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 2 years 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.
 2 years 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.

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abb0t
 2 years ago
Best ResponseYou've already chosen the best response.0Is this for differential equations?

gerryliyana
 2 years ago
Best ResponseYou've already chosen the best response.0yes..., this is one of applications of differential equations

abb0t
 2 years ago
Best ResponseYou've already chosen the best response.0Yeah, 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

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

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

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.2Yes 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

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

zepdrix
 2 years ago
Best 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

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

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

gerryliyana
 2 years ago
Best ResponseYou've already chosen the best response.0finally i got Q(t) = 10  2.5e[exp(t/25)]  2.5e

gerryliyana
 2 years ago
Best ResponseYou've already chosen the best response.0let 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: \[Q(120) = 10 (2.5e) \exp \left( \frac{(120) }{ 25 } \right) = 10  0.05593 \approx 9.944 lb\]

gerryliyana
 2 years ago
Best ResponseYou've already chosen the best response.0How do you do it???

gerryliyana
 2 years ago
Best ResponseYou've already chosen the best response.0oh yeah, because i'm in a hurry

gerryliyana
 2 years ago
Best ResponseYou've already chosen the best response.0i'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??

gerryliyana
 2 years ago
Best ResponseYou've already chosen the best response.0let 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?

gerryliyana
 2 years ago
Best ResponseYou've already chosen the best response.0are u still here @Chlorophyll ?

gerryliyana
 2 years ago
Best ResponseYou've already chosen the best response.0welcome again @zepdrix :)

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

gerryliyana
 2 years ago
Best ResponseYou've already chosen the best response.0u can check my answer :)

gerryliyana
 2 years ago
Best ResponseYou've already chosen the best response.0thank 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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