In a tank of l00 liters of brine containing dissolved salt.

agreseza 2021-12-30 Answered
In a tank of l00 liters of brine containing dissolved salt. Pure water is alloved to run into the tank at the rate of 3 li/min . Brine runs out of the tank at rate of 2 liters minute. The instantaneous concentration in the tank is kept uniform by stirting. How much salt is in the tank at the end of 1 Hr?
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Orlando Paz
Answered 2021-12-31 Author has 42 answers
Let Q= amount of salt in the mixture
100+(32)t=100+t= volume of mixture at any time t
q100+t= concentration of salt
dqdt= rate of gain-rate of loss
dqdt=02(q100+t)
dqdt=2dt100+t+c
lnq=2ln(100+t)+c
lnq=22ln(100+t)2+c
When t=0:q=50kg
ln50=ln(100+1)2+c
c=13.12236
When t=60min,Q=?
lnq=ln(100+60)2+13.12236
lnq=2.972
q=19.53kg
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veiga34
Answered 2022-01-01 Author has 32 answers

Suppose that the amount of salt in the mixture be Q.
The volume of mixture at time t will be as follows:
100+(32)t
Recall the fact that the concentration of salt will be given as follows:
Concentration of salt=Amount of salt in mixtureVolume of mixture at any time t
=Q100+(32)t
=Q100+t
Also, the rate of gain minus rate of loss will be dQdt, where Q is the amount of salt in the mixture.
Here rate of gain is 0 and rate of loss is 2(Q100+t). That is,
dQdt=02(Q100+t)
dQQ=2(dt100+t)
dQQ=2(dt100+t)
lnQ=2ln(100+t)+C
lnQ=2ln(100+t)2+C
Initially the amount of salt in the mixture is 55 kg. That is, Q=55,t=0.
Substitute Q=55,t=0  lnQ=ln(100+t)2+C:
ln55=ln(100+0)2+C
C=13.218
By the end one 1 hours, t will be 60 minutes.
Substitute t=60  lnQ=ln(100+t)2+13.218:
lnQ=ln(100+60)2+13.218
lnQ=3.06765
elnQ=e3.06765
Q=21.4913
Thus, the amount of salt in tank after one hour is 21.4913 Kg.

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Vasquez
Answered 2022-01-09 Author has 460 answers

100+(32)t=100+t= volume of mixture at any time t
q100+t=dqdt=dqdt=02(q100+t)dqdt=2dt100+t+clnq=2ln(100+t)+clnq=22ln(100+t)2+cWhen t=0:q=50kgln50=ln(100+1)2+cc=13.12236When t=60min,Q=?lnq=ln(100+60)2+13.12236lnq=2.972q=19.53kg

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