Solving Differential Equations for Tank of Water and Salt

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)
 
Thread 'Direction Fields and Isoclines'
I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...
Back
Top