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:
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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