stueck9356
New member
- Joined
- Sep 24, 2017
- Messages
- 2
I've been racking my brain and Can't figure part of this question out...
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A tank initially contains 100 gallons of water with 20 lbs of salt dissolved in int. A solution containing 2 lbs of salt per gallon of water is pumped into the tank at a rate of 1 gallon per minute and the well-mixed solution is pumped out of the tank at a rate of 2 gallons per minute.
A) How much salt will be in the tank after 25 mins?
B) When will there be 55 lbs of salt in the tank? [Note: There are two answers]
C) How much salt will be in the tank after 100 minutes?
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I started out with my initial equation
dS/dt = rate in - rate out
dS/dt = (2)(1) - (s/100)(2)
This was rewritten into a linear equation form
S' + (1/50)S = 2
and plugged in as
S = (∫e∫1/50 dt * 2 dt) / (e∫1/50 dt)
Some more magic gets me to
S = 100+Ce-t/50
and solving for C using information given in the problem gets me
S=100-80e-t/50
for A its just plug and chug - t=25 and I get S=51.47 lbs.
for B its more plug and chug - S=55 so I get T=28.76 Mins. However I cant find the second answer.
for C its pretty easy - the tank empties at T=100 so it's 0.
I am really stuck on part B. I know there are two answers, because the salt in the tank fills up to a maximum, then the flow of water out begins to overcome the saltwater solution going in, so my salt level in the tank decreases - the whole thing looks like an inverted parabola shape. However my equation doesn't give me this.
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A tank initially contains 100 gallons of water with 20 lbs of salt dissolved in int. A solution containing 2 lbs of salt per gallon of water is pumped into the tank at a rate of 1 gallon per minute and the well-mixed solution is pumped out of the tank at a rate of 2 gallons per minute.
A) How much salt will be in the tank after 25 mins?
B) When will there be 55 lbs of salt in the tank? [Note: There are two answers]
C) How much salt will be in the tank after 100 minutes?
------
I started out with my initial equation
dS/dt = rate in - rate out
dS/dt = (2)(1) - (s/100)(2)
This was rewritten into a linear equation form
S' + (1/50)S = 2
and plugged in as
S = (∫e∫1/50 dt * 2 dt) / (e∫1/50 dt)
Some more magic gets me to
S = 100+Ce-t/50
and solving for C using information given in the problem gets me
S=100-80e-t/50
for A its just plug and chug - t=25 and I get S=51.47 lbs.
for B its more plug and chug - S=55 so I get T=28.76 Mins. However I cant find the second answer.
for C its pretty easy - the tank empties at T=100 so it's 0.
I am really stuck on part B. I know there are two answers, because the salt in the tank fills up to a maximum, then the flow of water out begins to overcome the saltwater solution going in, so my salt level in the tank decreases - the whole thing looks like an inverted parabola shape. However my equation doesn't give me this.