A tank contains 400 liters of brine. Twelve liters of brine, each containing 2.5 N of dissolved salt, enter the tank per minute, and the mixture, assumed uniform leaves at the rate of 8 liters per min. If the concentration is to be 2 N/litre at the end of one hour, how many newtons of salt should there be present in the tank originally? Solution:
Consider the following illustration
d S d t = ( d S d t ) e n − ( d S d t ) e s \frac{dS}{dt}=\left(\frac{dS}{dt}\right)_{en}-\left(\frac{dS}{dt}\right)_{es} d t d S = ( d t d S ) e n − ( d t d S ) es Using,
V b r i n e + ( r a t e o f b r i n e o u t ) t d S d t = 8 L M ( S 400 + 4 t ) = 8 S ( 400 + 4 t ) d S d t = 2 S ( 100 + t ) V_{brine}+\left(rate\:of\:brine\:out\right)t\\\frac{dS}{dt}=\frac{8L}{M}\left(\frac{S}{400+4t}\right)\\=\frac{8S}{\left(400+4t\right)}\\\frac{dS}{dt}=\frac{2S}{\left(100+t\right)} V b r in e + ( r a t e o f b r in e o u t ) t d t d S = M 8 L ( 400 + 4 t S ) = ( 400 + 4 t ) 8 S d t d S = ( 100 + t ) 2 S Using the general solution:
d S d t = 30 − 2 S ( 100 + t ) d S d t + 2 S ( 100 + t ) = 30 \frac{dS}{dt}=30-\frac{2S}{\left(100+t\right)}\\\frac{dS}{dt}+\frac{2S}{\left(100+t\right)}=30 d t d S = 30 − ( 100 + t ) 2 S d t d S + ( 100 + t ) 2 S = 30 To solve we will use First Order Linear Differential Equation (FOLDE) where:
P ( t ) = 2 ( 100 + t ) , Q ( t ) = 30 P_{\left(t\right)}=\frac{2}{\left(100+t\right)}\:,\:Q_{\left(t\right)}=30 P ( t ) = ( 100 + t ) 2 , Q ( t ) = 30 Solve for the integrating factor using the formula:
σ = e ∫ P ( t ) d t \sigma =e^{\int \:P_{\left(t\right)}dt} σ = e ∫ P ( t ) d t Apply,
σ = e ∫ 2 100 + t d t σ = e 2 l n ( 100 + t ) σ = e l n ( 100 + t ) 2 σ = ( 100 + t ) 2 \sigma =e^{\int \:\frac{2}{100+t}dt}\\\sigma =e^{2ln\left(100+t\right)}\\\sigma \:=e^{ln\left(100+t\right)^2}\\\sigma \:=\left(100+t\right)^2 σ = e ∫ 100 + t 2 d t σ = e 2 l n ( 100 + t ) σ = e l n ( 100 + t ) 2 σ = ( 100 + t ) 2 Substitute the given value to the formula:
S σ = ∫ σ Q ( t ) d t + C S\sigma =\int \:\sigma Q\left(t\right)dt+C S σ = ∫ σ Q ( t ) d t + C Apply,
S ( 100 + t ) 2 = ∫ ( 100 + t ) 2 30 d t + C S ( 100 + t ) 2 = 30 ∫ ( 100 + t ) 2 d t + C S ( 100 + t ) 2 = 30 ( 100 + t ) 3 3 d t + C S ( 100 + t ) 2 = 10 ( 100 + t ) 3 + C → e q n . 1 S\left(100+t\right)^2=\int \:\left(100+t\right)^230dt+C\\S\left(100+t\right)^2=30\int \:\left(100+t\right)^2dt+C\\S\left(100+t\right)^2=30\:\frac{\left(100+t\right)^{^3}}{3}dt+C\\S\left(100+t\right)^2=10\left(100+t\right)^{^3}+C\rightarrow eqn.1 S ( 100 + t ) 2 = ∫ ( 100 + t ) 2 30 d t + C S ( 100 + t ) 2 = 30 ∫ ( 100 + t ) 2 d t + C S ( 100 + t ) 2 = 30 3 ( 100 + t ) 3 d t + C S ( 100 + t ) 2 = 10 ( 100 + t ) 3 + C → e q n .1 Evaluate C; @t=1hr
Convert 1hr to minutes, where 1hr is simply 60 minutes.
S ( 100 + 60 ) 2 = 10 ( 100 + 60 ) 3 C = 2 N L ; C = S ( 400 + 4 t ) S\left(100+60\right)\:2\:=10\left(100+60\right)\:^3\\C=\frac{2N}{L}\:;\:C=\frac{S}{\left(400+4t\right)} S ( 100 + 60 ) 2 = 10 ( 100 + 60 ) 3 C = L 2 N ; C = ( 400 + 4 t ) S Get the value of S using the equation:
C = S ( 400 + 4 t ) \:C=\frac{S}{\left(400+4t\right)} C = ( 400 + 4 t ) S Isolate S,
S = C ( 400 + 4 t ) ; C = 2 , t = 60 S = 2 ( 400 + 4 ( 60 ) ) S = 1280 N S=C\left(400+4t\right);\:C=2,\:t=60\\S=2\left(400+4\left(60\right)\right)\\S=1280N S = C ( 400 + 4 t ) ; C = 2 , t = 60 S = 2 ( 400 + 4 ( 60 ) ) S = 1280 N Get the value of C using Eqn.1
S ( 100 + t ) 2 = 10 ( 100 + t ) 3 + C ; S = 1280 , t = 60 1280 ( 100 + 60 ) 2 = 10 ( 100 + 60 ) 3 + C 32768000 = 40960000 + C 32768000 − 40960000 = C C = − 8192000 S\left(100+t\right)^2=10\left(100+t\right)^{^3}+C; S=1280 , t=60\\
1280\left(100+60\right)^2=10\left(100+60\right)^{^3}+C\\32768000=40960000+C\\32768000-40960000=C\\ C=-8192000 S ( 100 + t ) 2 = 10 ( 100 + t ) 3 + C ; S = 1280 , t = 60 1280 ( 100 + 60 ) 2 = 10 ( 100 + 60 ) 3 + C 32768000 = 40960000 + C 32768000 − 40960000 = C C = − 8192000 With the presence of the value of C we will now have our working equation:
S ( 100 + t ) 2 = 10 ( 100 + t ) 3 − 8192000 S\left(100+t\right)^2=10\left(100+t\right)^{^3}-8192000 S ( 100 + t ) 2 = 10 ( 100 + t ) 3 − 8192000 Using the given working equation, solve for the value of S @ t=0
S ( 100 + t ) 2 = 10 ( 100 + t ) 3 − 8192000 ; t = 0 S ( 100 + 0 ) 2 = 10 ( 100 + 0 ) 3 − 8192000 S ( 1000 ) 2 1000 = 1808000 1000 S = 180.8 N S\left(100+t\right)^2=10\left(100+t\right)^{^3}-8192000; t=0\\S\left(100+0\right)^2=10\left(100+0\right)^{^3}-8192000\\\frac{S\left(1000\right)^2}{1000}=\frac{1808000}{1000}\\S=180.8 N S ( 100 + t ) 2 = 10 ( 100 + t ) 3 − 8192000 ; t = 0 S ( 100 + 0 ) 2 = 10 ( 100 + 0 ) 3 − 8192000 1000 S ( 1000 ) 2 = 1000 1808000 S = 180.8 N
