a tank contain 1000 liters water. Brine that contain 0.05kg of salt per liter of water enters the tank at a rate of 8 liter per minute. Brine that contain 0.04kg of salt per liter of water enters the tank at a rate of 5 liter per minute. The well-mixed solution drains from the tank at a rate of 13 liters per minute.
1. write differential equation that the amount of salt, y(t) must satisfy.
2. find the amount of salt in the tank at any time, t.
3. how much salt is present after 20 minutes?
4. what is the concentration of solution in the tank after a very long time?

Question

a tank contain 1000 liters water. Brine that contain 0.05kg of salt per liter of water enters the tank at a rate of 8 liter per minute. Brine that contain 0.04kg of salt per liter of water enters the tank at a rate of 5 liter per minute. The well-mixed solution drains from the tank at a rate of 13 liters per minute.
1. write differential equation that the amount of salt, y(t) must satisfy.
2. find the amount of salt in the tank at any time, t.
3. how much salt is present after 20 minutes?
4. what is the concentration of solution in the tank after a very long time?

paxpolaris · Answer

1.   $dy/dt=0.05∗8+0.04∗5−13y/1000$

anonymous · Answer

thanks, can you also solve the rest?

paxpolaris · Answer

no, i'm probably wrong

experimentx · Answer

Ah ... y is salt right??

paxpolaris · Answer

mass of salt in tank

experimentx · Answer

i think it is right ///

experimentx · Answer

these two are inputs 0.05∗8+0.04∗5
and this y13/1000 is decay constant

experimentx · Answer

13/1000 is decay constant

experimentx · Answer

I'm pretty sure this must be decay equation ... that must converge to input

experimentx · Answer

to find the amount of salt at any time t, solve equation 1)
y(t) = some function of t + some constant ----- (2

you will need an initial condition, assuming that there is no salt present initially,
y(0) = 0 <--- using this find the value of 'some constant'

then put the value of 'some constant' ... in above ... which is your answer of Q.2

Q.3 put t=20 in y(20) = ...

experimentx · Answer

I am guessing that the Q.4 should be 0.046

paxpolaris · Answer

yes

experimentx · Answer

Oh ... gotta sleep, good night!

anonymous · Answer

Could you please show me all the formula?

paxpolaris · Answer

simplifying differential equation for Q1.:
$$\large y\ '(t )= 0.6-0.013y$$

paxpolaris · Answer

$$\large {dy \over dt} = {0.6 -0.013y }$$$$\implies \large {dy \over 0.6-0.013y} = dt$$
Integrate both sides, and get rid of the constant using $\large y(0)=0$

paxpolaris · Answer

that will give you the equation for Q2

anonymous · Answer

ln(0.6-0.013y)=t+c.    ?

paxpolaris · Answer

http://www.wolframalpha.com/input/pod.jsp?id=MSP33211a1i0dghh09b64b4000050b9ah0hhb5g60ei&s=14&button=1

paxpolaris · Answer

$$\Large -\frac 1 {0.013}\ln(0.6-0.013y)=t+C_1$$

anonymous · Answer

Q3=10.5668kg?

paxpolaris · Answer

yes. http://www.wolframalpha.com/input/?i=-600%2F13*e%5E%28-0.013t%29%2B600%2F13+where+t%3D20

anonymous · Answer

How about Q4? Not 46?

paxpolaris · Answer

46.15 kg in 1000 liters

paxpolaris · Answer

so, concentration will approach 0.04615 kg/l

anonymous · Answer

Thanks

anonymous · Answer

Can you solve my another closed problem?

paxpolaris · Answer

post link

anonymous · Answer

dy/dx=y^3-2y^2-15y
What are the constant solutions?
For what values of y is y(x) increasing?  And what decreasing?
Use the information groom above to sketch the direction field for the given differential equation

paxpolaris · Answer

http://openstudy.com/users/jen-kai#/updates/4fd23c1fe4b057e7d2214a27