A tank contains 2380 L of pure water Solution that contains

A tank contains 2380 L of pure water. Solution that contains 0.06 kg of sugar per liter enters the tank at the rate 7 L/min, and is thoroughly mixed into it. The new solution drains out of the tank at the same rate. How much sugar is in the tank at the beginning? Find the amount of sugar after t minutes. As t becomes large, what value is y(t) approaching ? In other words, calculate the following limit. lim_t rightarrow infinity y(t) = (kg)

Solution

Solution :

Let y(t) = amount of sugar in tank (in kg) at time t (in min) .

( a )

y(0) = 0 (since tank initially contains pure water)

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( b )

Rate of sugar flowing into tank:
0.06 kg/L * 7 L/min = 0.42 kg/min

Rate of sugar flowing out of tank = concentration of sugar in tank * 7 L/min

= y kg / 2380 L * 7L/min

= ( 7y / 2380 ) kg/min

dy/dt = rate in rate out

dy/dt = 0.42 ( 7y / 2380 )

dy/dt = ( 7 / 2380) (0.6 * 238 y)

dy/(142.8 y) =( 7 / 2380 ) dt

Integrate both sides

dy/(142.8 y) = (7 / 2380) dt

ln|142.8 y| = 7 / 2380 t + C

ln|142.8 y| = 7t / 2380 + C

142.8 y = Ce^{( 7t/2380)}

y = 142.8 - Ce^{( 7t/2380)}

y(0) = 0

142.8 Ce⁰ = 0

142.8 C = 0

C = 142.8

y(t) = 142.8 142.8e^{( 7t/2380)}

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( c )

As \' t \' becomes large, the value that is approaching infinity is \' t \' itself .

im[t] y (t)

im[t] 142.8 142.8e^{( 7t/2380)}

= 142.8 142.8e^{( )}
= 142.8 142.8(0)
= 142.8 kg