Solving Differential Equations for Tank of Water and Salt

In summary, the conversation discusses a problem involving a tank containing water with salt concentration, as well as water flowing in and out of the tank. The differential equations for the concentration of salt and volume of water are provided and can be solved for particular solutions. The concentration of salt in the tank when it first overflows can be calculated by finding the time when the tank overflows and plugging it into the concentration equation.
  • #1
sam_0017
22
0
can you help me ??

A 200 liter tank initially contains 100 liters of water with a salt concentration of 0.1 grams per liter.
Water with a salt concentration of 0.5 grams per liter flows into the tank at a rate of 20 liters per
minute. Assume that the fluid is mixed instantaneously and that this well-mixed fluid is pumped out
at a rate of 10 liters per minute. Let c (t) and
v(t), be the concentration of salt and the volume of
water in the tank at time t (in minutes), respectively. Then,
v`(t)=10
v(t) c`(t) +20c(t)=10

a) Solve these differential equations to find the particular solutions for v(t) and c(t).
b) What is the concentration of salt in the tank when the tank first overflows?
 
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  • #2


you can first sole the top one and then solve the lower one.

for what v(t) is the derivative 10. This will lead to an answer up to an additive constant (since the derivative of a constant is 0) but you know v(0)=100 so you can solve for that constant.

then write c'(t) out in terms of c(t) and solve that differential equation. and again you know the concentration for t=0 (this time you will actually find a multiplicative constant.

If it is the actual differential equations you're having trouble with I suggest you find a good introductory physics book or of course an introductory differential equations book.
 
  • #3


i tray to solve it and i find this resolute, can anyone tell me are my way right ??

v(0)= 100 L , c(0) = 0.1 gm/L , c(t)=0.5 gm/L

v`= 10 ==> v (t) = 10t + k
==> v(0) = 0 + k =100 ==> k=100
so : v(t) = 10 t + 100
(mass) m= c(t) * v(t)
m`=c`(t)v(t) + c(t) v`(t)
10 = c`(t)(10t+100)+ c(t) *10
c`(t) + c(t) (10/(10t+100)) = 10/(10t+100)
by solving this DE :
e^∫(1/ t+10) dx = t+10
Multiplying through by both sides gives:
(t+10)c`(t)+c(t)=1
∂/∂t{c(t) (t+10)} =1

==> by integral
c(t) ( t+ 10) = t
==> c(t) = t/(t+10) +k
c(0) = 0.1
==>
0.1= 0 + k
so c(t) = t/(t+10) +0.1
 
  • #4


any one know how to solve b
 
  • #5


c(t) = 0.5 does not make any sense since the concentration of salt does not remain the same for all time. To solve b simply calculate when the tank overflows (you know the volume of the tank). Then plug that time into c(t)
 

1. How do you set up a differential equation for a tank of water and salt?

To set up a differential equation for a tank of water and salt, you need to consider the rate of change of salt concentration in the tank. This can be determined by the flow rate of salt into the tank and the rate of mixing within the tank. The resulting equation will be in the form of a first-order linear differential equation.

2. What are the initial conditions needed to solve the differential equation for a tank of water and salt?

The initial conditions required to solve the differential equation for a tank of water and salt include the initial salt concentration in the tank, the initial water volume in the tank, and the initial flow rate of salt into the tank. These values will help determine the constant of integration in the solution.

3. Can the differential equation for a tank of water and salt be solved analytically?

Yes, the differential equation for a tank of water and salt can be solved analytically using various methods such as separation of variables, integrating factors, and substitution. However, for more complex systems, numerical methods may be required to find an approximate solution.

4. How do you incorporate changes in flow rate or volume into the differential equation for a tank of water and salt?

If there are changes in the flow rate or volume of the tank, the differential equation will need to be modified accordingly. This can be done by adding or subtracting terms to the original equation that represent the changes in flow rate and volume.

5. How does temperature affect the solution of the differential equation for a tank of water and salt?

Temperature can affect the solution of the differential equation for a tank of water and salt if it is a factor in the rate of mixing or the solubility of salt in water. In this case, the equation will need to be modified to include temperature as a variable, and the initial conditions will also need to be adjusted accordingly.

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