How Can Calculus Be Used to Solve a Tank Draining/Filling Problem?

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In summary: Sorry, it seems like you are still having trouble with this problem. Can you please provide the full question or provide any additional information that might be helpful?
  • #1
the7joker7
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Homework Statement



A tank contains 90 kg of salt and 2000 L of water. Pure water enters a tank at the rate 12 L/min. The solution is mixed and drains from the tank at the rate 6 L/min.

Find the amount of salt in the tank after 1.5 hours.

Homework Equations



Assorted Calculus stuff.

The Attempt at a Solution



rate in = (.045kg per L)(12L per min) = .54kg/min

rate out = (y(t)/2000 kg per L)(6L per min) = (y(t))/(333.33)

dy/dt = .54 - ((y(t))/333.33) = (180 - y(t))/333.33

int(dy/(180 - y)) = int(dt/333.33)

-ln(180 - y) = t/333.33 + C

y(0) = 90

So -ln(90) = C

-ln(180 - y) = t/333.33 - ln(90)

180 - y = 90e^(-t/333.33)

y(t) = 180 - 90e^(t/333.33)

y(90) = 62.103

Actual Answer:

70.8661417323
 
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  • #2
the7joker7 said:

Homework Statement



A tank contains 90 kg of salt and 2000 L of water. Pure water enters a tank at the rate 12 L/min. The solution is mixed and drains from the tank at the rate 6 L/min.

Find the amount of salt in the tank after 1.5 hours.

rate in Ri=12 L/min

rate out Ro=6L/min
concentration in Ci=0
Vo(initial volume)=2000
amount of salt in at t=0 x(o)=90 kg
concentraton out Co= x(t)*Ro/(Vo+(Ri-Ro)t) so

dx/dt=RiCi-x(t)Ro/(Vo+(Ri-Ro)t now solve this one
 
  • #3
the7joker7 said:

Homework Statement



A tank contains 90 kg of salt and 2000 L of water. Pure water enters a tank at the rate 12 L/min. The solution is mixed and drains from the tank at the rate 6 L/min.

Find the amount of salt in the tank after 1.5 hours.

Homework Equations



Assorted Calculus stuff.

The Attempt at a Solution



rate in = (.045kg per L)(12L per min) = .54kg/min
Where did you get the ".045 kg per L"? That is the initial concentration of salt in the tank but has nothing to do with salt entering the tank. You are told that Pure water enters the tank at the rate of 12 L/min. There is 0 kg/min of salt entering the tank- none at all.

rate out = (y(t)/2000 kg per L)(6L per min) = (y(t))/(333.33)
Almost correct. If the volume of water in the tank were always 2000 Liters, then the concentration would be y/2000. But you have 12 L of water entering the tank every minute and only 6 L of water leaving. The amount of water in the tank is increasing by 12- 6= 6 L every minute. After t minutes, there will be 2000+ 6t Liters of water in the tank and that should be your denominator.

dy/dt = .54 - ((y(t))/333.33) = (180 - y(t))/333.33

int(dy/(180 - y)) = int(dt/333.33)

-ln(180 - y) = t/333.33 + C

y(0) = 90

So -ln(90) = C

-ln(180 - y) = t/333.33 - ln(90)

180 - y = 90e^(-t/333.33)

y(t) = 180 - 90e^(t/333.33)

y(90) = 62.103

Actual Answer:

70.8661417323
 
  • #4
This time I tried...

rate in = 0 kg/min

rate out = y(t)/(2000 + 3t)(3 L/min) = y(t)/(666.66 + t)

dy/dt = -y/(666.66+t)

int(dy/-y) = int(dt/666.66+t)

-ln(-y) = t/(666.66+t) + C

y(0) = 80

So -ln(80) = C

-ln(-y) = (t/(666.66+t)) - ln(80)

y(t) = -80e^(t/333.33)

No luck this way either. (this is a new problem of the same type.)
 
  • #5
the7joker7 said:
This time I tried...

rate in = 0 kg/min

rate out = y(t)/(2000 + 3t)(3 L/min) = y(t)/(666.66 + t)

dy/dt = -y/(666.66+t)

int(dy/-y) = int(dt/666.66+t)

-ln(-y) = t/(666.66+t) + C

y(0) = 80

So -ln(80) = C

-ln(-y) = (t/(666.66+t)) - ln(80)

y(t) = -80e^(t/333.33)

No luck this way either. (this is a new problem of the same type.)

have a closer look at my post #2, i set up all the necessary things you need for this problem,all you need to do is solve the diff. eq i set up!
 
  • #6
the7joker7 said:
This time I tried...

rate in = 0 kg/min

rate out = y(t)/(2000 + 3t)(3 L/min) = y(t)/(666.66 + t)

dy/dt = -y/(666.66+t)

int(dy/-y) = int(dt/666.66+t)

-ln(-y) = t/(666.66+t) + C

y(0) = 80

So -ln(80) = C

-ln(-y) = (t/(666.66+t)) - ln(80)

y(t) = -80e^(t/333.33)

No luck this way either. (this is a new problem of the same type.)

Oh, sorry! It seems like you have another problem now, right? so is this a new problem? What does it originally say?
 
  • #7
the7joker7 said:
rate out = y(t)/(2000 + 3t)(3 L/min) = y(t)/(666.66 + t)

.)
this defenitely cannot be right? You keep doing the same mistake as in your first problem!
 
Last edited:
  • #8
Okay, I regened the problem again.

A tank contains `70` kg of salt and `1000` L of water. Pure water enters a tank at the rate `8` L/min. The solution is mixed and drains from the tank at the rate `4` L/min.

Find the amount of salt in the tank after 2.5 hours.

It's the same concept as the two above, so what was I doing wrong there?
 
  • #9
the7joker7 said:
Okay, I regened the problem again.

A tank contains `70` kg of salt and `1000` L of water. Pure water enters a tank at the rate `8` L/min. The solution is mixed and drains from the tank at the rate `4` L/min.

Find the amount of salt in the tank after 2.5 hours.

It's the same concept as the two above, so what was I doing wrong there?

initial volume Vo=1000 L
Rate in Ri=8 L/min
Concentration in Ci=0
Rate out Ro=4 L/min
Concentration out = x(t)*Ro/(Vo+(Ri-Ro)t)

so

dx/dt= Ri*Ci- x(t)*Ro/(Vo+(Ri-Ro)t)

remember Co depends on the amount of the salt on the tank, and also on the volume of the tank, so

Co= x(t)/V, but the volume on the tank V, depends on the initial volume, and also on how the volume is changing over time as we pipe in and out water so, V=Vo+(Ri-Ro)t

Do you see your flaws now?
 
  • #10
Okay...before I try to punch in the problem, does this look right to everyone?

dx/dt = -(x(t)*4)/(1000 + (4)t)

Co = x(t)/(1000+(4)t)\

int(dx/-x) = int(dt/(1000+4t))

-ln(-x) = t/(1000+4t) + C

y(0) = 70

so -ln(70) = C

-ln(-x) = (t/(1000+4t) - ln(70)

x(t) = -70e^(t/(1000+4t))
 
  • #11
the7joker7 said:
This time I tried...

rate in = 0 kg/min

rate out = y(t)/(2000 + 3t)(3 L/min) = y(t)/(666.66 + t)

dy/dt = -y/(666.66+t)

int(dy/-y) = int(dt/666.66+t)

-ln(-y) = t/(666.66+t) + C
The integration is wrong. You can't just ignore the "t" in the denominator!
Do a substitution: Let u= 2000/3+ t (I hate using approximations to fractions!) so that du= dt and the integral on the right becomes [itex]\int du/u[/itex].

Also the integral of 1/(-y) is -ln|y|. Since y is the amount of salt in the tank, which can't be negative, you should have -ln(y). I think it would be simpler to leave the negative sign outside the integral.

y(0) = 80

So -ln(80) = C

-ln(-y) = (t/(666.66+t)) - ln(80)

y(t) = -80e^(t/333.33)

No luck this way either. (this is a new problem of the same type.)
 
Last edited by a moderator:
  • #12
the7joker7 said:
Okay...before I try to punch in the problem, does this look right to everyone?

dx/dt = -(x(t)*4)/(1000 + (4)t)

Co = x(t)/(1000+(4)t)\

int(dx/-x) = int(dt/(1000+4t))

-ln(-x) = t/(1000+4t) + C

y(0) = 70

so -ln(70) = C

-ln(-x) = (t/(1000+4t) - ln(70)

x(t) = -70e^(t/(1000+4t))

you did not integrate correctly!
 
  • #13
I'm a little confused as to the integration...do I just take out the negative to get...

x(t) = 70e^(t/(1000+4t))

Here are my integration steps, what exactly did I do wrong?

int(dx/x) = int(dt/(1000+4t))

-ln(x) = t/(1000+4t) + C
 
  • #14
your right hand side is not correctly integrated, you are doing the same mistake there!

[tex]\int \frac{dt}{1000+4t}[/tex] let [tex]u=1000+4t, \ \ du=4dt=>dt=\frac{1}{4}du[/tex] so

[tex]\frac{1}{4}\int\frac{du}{u}=\frac{1}{4}ln(u)+C[/tex] now go back and substitute for u, what do you get?
 
Last edited:
  • #15
So is this...

x(t) = 70e^(.25ln(1000 + 4t)))

The formula I want?
 
  • #16
x(0)=70, right?

[tex]x(t)=e^{C-.25ln(1000+4t)}=Ae^{-ln(1000+4t)^{\frac{1}{4}}}=A(1000+4t)^{-(\frac{1}{4})}[/tex] now [tex]x(0)=70=A(1000+4*0)^\frac{-1}{4}[/tex], what do u get for A?

EDIT: i forgot a minus sign on the power, i edited it!
 
Last edited:
  • #17
12.45?
 
  • #18
So are you saying the x(t) function is

12.45(1000 + 4x)^(1/4)

Because that function increases over time, I need one that decreases.
 
  • #19
the7joker7 said:
So are you saying the x(t) function is

12.45(1000 + 4x)^(1/4)

Because that function increases over time, I need one that decreases.


I edited my post #16, i had forgotten a minus sign on the power, have another look at it!
 

Related to How Can Calculus Be Used to Solve a Tank Draining/Filling Problem?

1. How does the height of the water in the tank affect the draining/filling rate?

The height of the water in the tank does not directly affect the draining/filling rate. However, it can indirectly affect it by changing the pressure at the bottom of the tank, which can impact the rate of flow.

2. Can the shape or size of the tank affect the draining/filling rate?

Yes, the shape and size of the tank can affect the draining/filling rate. For example, a taller and narrower tank will have a slower draining/filling rate compared to a shorter and wider tank, as there is less surface area for the water to flow through.

3. How does the viscosity of the liquid being drained/filled affect the process?

The viscosity of the liquid being drained/filled can significantly impact the process. A more viscous liquid will have a slower draining/filling rate compared to a less viscous liquid, as it will experience more resistance and require more energy to flow.

4. Is the draining/filling rate affected by the presence of air in the tank?

Yes, the presence of air in the tank can affect the draining/filling rate. As the water level decreases, air will take up more space in the tank and create pressure imbalances, which can slow down the rate of flow.

5. How can I calculate the time it will take to drain or fill a tank?

The time it takes to drain or fill a tank can be calculated using the volume of the tank, the rate of flow, and the initial and final water levels. The formula would be: time = (volume of tank) / (rate of flow * (initial water level - final water level)).

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