# A Mixing Problem....

• Jan 27th 2011, 07:01 PM
slapmaxwell1
A Mixing Problem....
ok here is a mixing problem that i tried to do, it seems pretty straight forward. Could someone take a look and let me know where i went wrong? thanks in advance...

Suppose that a large mixing tank initailly holds 500 liters of water, in which 20 kg of salt have been dissolved. another brine solution containing 1 kg/liter is pumped into the tank at a rate of 3 liters/min and the solution is well stirred. solution is then pumped out of the tank at a rate of 4 liters/min. I am supposed to write a differential equation for the amount of salt in the tank at a particular time.

so V(water in tank) = 500 liters
Initial mass(salt) = 20 kg
rate of brine in = 1 kg/liter @ 3 liters/min
rate of brine out = A kg/liters @ 4 liters/min

dA/dt = rate in - rate out

so dA/dt = [(1 kg/liters) (3 liters/min)] - [(A kg/500 liters) (4 liters/min)]

so dA/dt = (375-A)/125 ; A(0)= 20 kg.

it seemed really straight forward....
• Jan 27th 2011, 07:05 PM
dwsmith
Quote:

Originally Posted by slapmaxwell1
ok here is a mixing problem that i tried to do, it seems pretty straight forward. Could someone take a look and let me know where i went wrong? thanks in advance...

Suppose that a large mixing tank initailly holds 500 liters of water, in which 20 kg of salt have been dissolved. another brine solution containing 1 kg/liter is pumped into the tank at a rate of 3 liters/min and the solution is well stirred. solution is then pumped out of the tank at a rate of 4 liters/min. I am supposed to write a differential equation for the amount of salt in the tank at a particular time.

so V(water in tank) = 500 liters
Initial mass(salt) = 20 kg
rate of brine in = 1 kg/liter @ 3 liters/min
rate of brine out = A kg/liters @ 4 liters/min

dA/dt = rate in - rate out

so dA/dt = [(1 kg/liters) (3 liters/min)] - [(A kg/500 liters) (4 liters/min)]

so dA/dt = (375-A)/125 ; A(0)= 20 kg.

it seemed really straight forward....

In 3 out 4

$t(3-4) = -t$

$\displaystyle\frac{dA}{dt}=3-\frac{4A}{500-t}\Rightarrow A'+\frac{4}{500-t}A=3$

$\displaystyle P(t)=\frac{4}{500-t} \ \ Q(t)=3$

$\displaystyle \exp{\left(\int P(t)dt\right)}t=\int P(t)Q(t)dt$
• Jan 27th 2011, 07:12 PM
slapmaxwell1
I just got this message from a classmate...

"Keep in mind that the concentration of salt at any time is the amount of salt divided by the volume of water. Since there is initially 500 liters of water, and solution is being pumped in at a rate of 3L/min and solution is pumped out at a rate of 4L/min, the amount of water in the tank is increasing by 1L/min."

so after reading his message im thinking what am i screwing up on?
• Jan 27th 2011, 07:13 PM
slapmaxwell1
ok i think i see it, you have 500 - t, i just have 500.
• Jan 27th 2011, 07:14 PM
dwsmith
Quote:

Originally Posted by slapmaxwell1
I just got this message from a classmate...

"Keep in mind that the concentration of salt at any time is the amount of salt divided by the volume of water. Since there is initially 500 liters of water, and solution is being pumped in at a rate of 3L/min and solution is pumped out at a rate of 4L/min, the amount of water in the tank is increasing by 1L/min."

so after reading his message im thinking what am i screwing up on?

3-4 = -1 the tank is losing water.

Look over post 2. I showed you how to handle that.