Differential Equations - Mixture in an overflowing tank

[proctortom]

HOW IS THE FOLLOWING QUESTION DONE?


Milk chocolate is being produced in a 300 litre tank, which initially contains 100 litres of milk. The following things then occur simultaneously:

1. Liquid cocoa (made up of equal parts cocoa solids and cocoa butter, both in liquid form) is added at a rate of 6 litres per minute.

2. Milk is added at a rate of 3 litres per minute.  The well-stirred mixture leaves the tank via a tube, at a rate of 6 litres per minute.


Let x(t) be the amount of cocoa solids in the mixture


The differential equation is dx/dt + 6x / (100 + 3t) = 3
...Therefore...
x(t) = (9t^3 + 900t^2 + 30,000t) / (3t + 100)^2


After the tank is full, the process continues as above. However, in addition to the well-stirred mixture leaving via the tube, it also flows over the edges of the tank and is collected by overflow tubing which takes it to the cooling process. Let y(t) be the number of litres of cocoa solids present in the tank t minutes after it is full.

Find the differential equation satisfied by y(t)

 

[haruspex]

Not sure about the = 3 in your first eqn. Shouldn't it be 6?
The soln doesn't look quite right to me. Did you check it satisfied the differential eqn?
Please show your attempt at the last part.

 

[sad_song009]

It is 3 because proctortom is asking for cocoa solids. dx/dt = rate of cocoa solids in - rate of cocoa solids out. Which is dx/dt = 3 [Liquid cocoa (made up of equal parts cocoa solids and cocoa butter, both in liquid form) is added at a rate of 6 litres per minute] - (x/100+3t)*6 = dx/dt + 6x/100+3t = 3

I also get stuck in finding the differential equation satisfied by y(t). The eqn should be dy/dt (3/100)*(100-y). But I don't know how to get to this !

 

[nishcha7]

dy/dt = rate in - rate out

rate in = 3L/m (cocoa solids going into mixture)

rate out = concentration x flow rate out
= amount/volume x flow rate out

where the flow rate out must equal the flow rate in as the tank is remaining full! (6L/m via tube and 3L/m via overflow tubing)

= y/300 * 9
= 3y/100

so dy/dt = 3 - 3y/100
= 3/100 (100 - y)

 

[blue_raver22]

To find the differential equation satisfied by y(t), we need to consider the rate of change of cocoa solids in the tank after it is full. This is determined by the rate at which cocoa solids are added, the rate at which they leave through the tube, and the rate at which they leave through the overflow tubing. We can express this mathematically as:

dy/dt = 6 - 6x(t) - 6y(t)

Where x(t) is the amount of cocoa solids in the tank at time t, and y(t) is the amount of cocoa solids collected by the overflow tubing at time t. This equation takes into account the addition of cocoa solids at a rate of 6 litres per minute, the removal of cocoa solids through the tube at a rate of 6 litres per minute, and the additional removal of cocoa solids through the overflow tubing at a rate of 6 litres per minute. This differential equation can then be solved to find the function y(t), which represents the amount of cocoa solids collected by the overflow tubing at any given time t.