So here's the problem that I'm working on at the moment:
Tank 1 initially contains 50 gals of water with 10 oz of salt in it, while Tank 2 initially
contains 20 gals of water with 15 oz of salt in it. Water containing 2 oz/gal of salt flows
into Tank 1 at a rate of 5 gal/min and the well-stirred mixture flows from Tank 1 into
Tank 2 at the same rate of 5 gal/min. The solution in Tank 2 flows out to the ground at
a rate of 5 gal/min. If x_1_(t) and x_2_(t) represent the number of ounces of salt in Tank 1 and Tank 2 respectively,
Set up but do not solve an initial value problem describing this system.
I know that this is a situation where we have:
x_1_(t)' = x_2_(t)
and then there's an expression for x_2_(t)
Here's what I've got:
dx1/dt = (2 oz/gal)(5 gal/min) – (5 gal/min)(x_1)/(50 gal)
dx1/dt = (10 oz/min) – ((x_1)/10)oz/min
dx2/dt = (x_1)/2 – (x_2)/4
I was not sure about initial conditions, but I believe that they should be x_1_(0) = 10 oz. and x_2_(0) = 15 oz.
Can anyone help me figure out where I have gone wrong? I think that my expression for the first tank is correct, but I'm struggling with the second tank. Any suggestions?
Thanks!
Best Answer
Your first expression is indeed correct: $$ \frac{dx_1}{dt} = \left(10 - \frac{1}{10}x_1 \right) \qquad\text{oz/min} $$ For the second tank, we use the exact same reasoning. Since the rate of water entering tank 2 equals the rate of water leaving tank 2, the volume of salt water in tank 2 is constant. Thus, the concentration of salt water leaving tank 2 is $(x_2/20)$ oz/gal. After multiplying this rate by $5$ gal/min so that we obtain the correct units of oz/min, we obtain: $$ \frac{dx_2}{dt} = \left(\frac{1}{10}x_1 - \frac{1}{4}x_2 \right) \qquad\text{oz/min} $$