A chemical manufacturing company has a 1000 kl
holding tank which it uses to control the release of pollutants into a sewage system. Initially the tank has 360 kl of water containing 2kg of pollutant per kl. Water containing 3 kg pollutant per kl enters the tank at the rate of 80 kl per hour and is uniformly mixed with water already in the tank. Simultaneously, water is released from the tank at the rate of 40 kl per hour. Determine the rate (in kg/kl) at which pollutant is being released after 10 hours.
Text Solution
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Ans. 2109
Sol. Let P(t) be the amount of pollutant (in kg) in the tank at time t. The rate of change of pollutant in the tank is given by 
.
= (rate in) – (rate out)
The pollutant is entering the tank at the rate of
3 × 80 kg/hr (= rate in).
Water is entering the tank at the rate of 80 kl/hr. and is leaving at 40 kl/hr.
∴ Amount of water at time t is (360 + 40t)kl
Hence amount of pollutant leaving the tank is
= 40 
= 
∴ we have 
= 240 – 
= 
+ 
= 240,
which is a linear differential equation whose
I.F. = 
= 
= 9 + t
Thus: 
(P(9 + t)) = 240(9 + t)
⇒ P (9 + t) = 
(9 + t) 2 + C
⇒ P = 120(9 + t) + 
Since amount of pollutant at t = 0 is 2 × 360 kg
We have P(0) = 120 × 9 + 
= 720
∴ C = – 3240
Thus P(t) = 120(9 + t) – 
(kg)
Blow P(10) = 120(19) – 
= 
= 2109.47
i.e. 2109 kg (approximate)
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