Answer to Question #238273 in Calculus for Anuj

Question #238273

A tank having a capacity of 1000 liters, initially contains 400 liters of sugar water having a concen-

tration of 0.2 Kg of sugar for each liter of water. At time zero, sugar water with a concentration of

50 gm of sugar per liter begins pumped into the tank at a rate of 2 liter per minute. Simultaneously,

a drain is opened at the bottom of the tank so that the volume of the sugar-water solution in the

tank reduces 1 liter per minute. Determine the following:


1
Expert's answer
2021-09-17T03:45:52-0400

Let s(t) = amount, in kg of sugar at time t. Then we have

"\\frac{dt}{ds}" = (rate of salt into tank) - (rate of salt out of tank)

"= (0.05 \\frac{kg}{L} \u00b7 5 \\frac{L}{min}) + (0.04 \\frac{kg}{Min} \u00b7 10 \\frac{L}{min}) \u2212 \\frac{s kg}{1000 L} \u00b7 15 \\frac{L}{min}\\\\\n= (0.25 \\frac{kg}{Min}) + (0.04 \\frac{kg}{Min} ) \u2212 \\frac{15s kg}{1000 min} \\\\"

So we get the differential equation

"\\frac{ds}{\ndt} = 0.65 \u2212\n\\frac{15s}{\n1000}\\\\\n\\frac{ds}{\ndt} =\\frac{\n130 \u2212 3s}{\n200}"

We separate s and t to get

"\\frac{1}{\n130 \u2212 3}\nds = \\frac{1}{\n200}\ndt"

Integrate

"\\int \\frac{1}{130-30s}ds=-\\frac{1}{30}\\ln \\left|130-30s\\right|+C\\\\\n\\frac{1}{200}t+C\\\\\n\\implies s =\n\\frac{130 \u2212 C_4e^{\u22123t\/200}}{\n3}"

Since we begin with pure water, we have s(0) = 0. Substituting,

"C_4 = 130"

"s =\n\\frac{130 \u2212 130e^{\u22123t\/200}}{\n3}\\\\\ns(60) =\n\\frac{130 \u2212 130e^{\u22123t\/200}}{\n3} = 25.7153"

Thus, after one hour there is about 25.72 kg of sugar in the tank.


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