Solving Differential Equations for Tank of Water and Salt

Click For Summary
SUMMARY

This discussion focuses on solving differential equations related to a tank containing water and salt. The initial conditions are a 200-liter tank with 100 liters of water at a salt concentration of 0.1 grams per liter, with water flowing in at 20 liters per minute and out at 10 liters per minute. The solutions for the volume function v(t) and concentration function c(t) are derived as v(t) = 10t + 100 and c(t) = t/(t + 10) + 0.1, respectively. The concentration of salt when the tank overflows can be determined by calculating the time at which the tank reaches its full capacity and substituting that time into the concentration function c(t).

PREREQUISITES
  • Understanding of basic differential equations
  • Familiarity with initial value problems
  • Knowledge of saltwater mixing dynamics
  • Ability to perform integration and solve for constants
NEXT STEPS
  • Study the method of solving first-order linear differential equations
  • Learn about initial value problems in differential equations
  • Explore applications of differential equations in fluid dynamics
  • Investigate the concept of mixing problems in chemistry
USEFUL FOR

Students and professionals in mathematics, engineering, and physics who are working with differential equations, particularly in applications involving fluid dynamics and mixing processes.

sam_0017
Messages
17
Reaction score
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?
 
Physics news on Phys.org


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.
 


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
 


any one know how to solve b
 


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)
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
3K
Replies
1
Views
2K
  • · Replies 23 ·
Replies
23
Views
4K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
3
Views
1K
  • · Replies 1 ·
Replies
1
Views
6K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
50
Views
7K