Respuesta :
Answer:
The mass of salt in the tank after t minutes is
[tex]y(t) = 5-4.5e^{-\frac{2t}{25}}[/tex]
The concentration of salt in the tank reach 0.02 kg/L when [tex]t=-\frac{25\ln \left(\frac{83}{75}\right)}{2} \approx -1.2669[/tex]
Step-by-step explanation:
Let y(t) be the mass of salt (in kg) that is in the tank at any time, t (in minutes).
The main equation that we will be using to model this mixing process is:
Rate of change of [tex]\frac{dy}{dt}[/tex] = Rate of salt in - Rate of salt out
We need to determine the rate at which salts enters the tank. From the information given we know:
- The brine flows into the tank at a rate of [tex]8\:\frac{L}{min}[/tex]
- The concentration of salt in the brine entering the tank is [tex]0.05\:\frac{kg}{L}[/tex]
The Rate of salt in = (flow rate of liquid entering) x (concentration of salt in liquid entering)
[tex](8\:\frac{L}{min}) \cdot (0.05\:\frac{kg}{L})=0.4 \:\frac{kg}{min}[/tex]
Next, we need to determine the output rate of salt from the tank.
The Rate of salt out = (flow rate of liquid exiting) x (concentration of salt in liquid exiting)
The concentration of salt in any part of the tank at time t is just y(t) divided by the volume. From the information given we know:
The tank initially contains 100 L and the rate of flow into the tank is the same as the rate of flow out.
[tex](8\:\frac{L}{min}) \cdot (\frac{y(t)}{100} \:\frac{kg}{L})= \frac{2y(t)}{25} \:\frac{kg}{min}[/tex]
At time t = 0, there is 0.5 kg of salt, so the initial condition is y(0) = 0.5. And the mathematical model for the mixing process is
[tex]\frac{dy}{dt}=0.4-\frac{2y(t)}{25}, \quad{y(0)=0.5}[/tex]
[tex]\frac{dy}{dt}=0.4-\frac{2y(t)}{25}\\\\\mathrm{Rewrite\:in\:the\:form\:of\:a\:first\:order\:separable\:ODE}\\\\ \frac{1}{2}\frac{dy}{5-y}=\frac{1}{25}dt \\\\\frac{1}{2}\int \frac{dy}{5-y}=\int \frac{1}{25}dt\\\\\frac{1}{2}\left(-\ln \left|5-y\right|+C\right)=\frac{1}{25}t+C\\\\-\frac{1}{2}\ln \left|5-y\right|+C_1\right)=\frac{1}{25}t+C\\\\-\frac{1}{2}\ln \left|5-y\right|=\frac{1}{25}t+C_2\\\\5-y=C_3e^{-\frac{2t}{25} }\\\\y(t) =5-C_3e^{-\frac{2t}{25} }[/tex]
Using the initial condition y(0)=0.5
[tex]y(t) =5-C_3e^{-\frac{2t}{25} }\\y(0)=0.5=5-C_3e^{-\frac{2(0)}{25}} \\C_3=4.5[/tex]
The mass of salt in the tank after t minutes is
[tex]y(t) = 5-4.5e^{-\frac{2t}{25}}[/tex]
To determine when the concentration of salt is 0.02 kg/L, we solve for t
[tex]y(t) = 5-4.5e^{-\frac{2t}{25}}\\\\0.02=5-4.5e^{-\frac{2t}{25}}\\\\5\cdot \:100-4.5e^{-\frac{2t}{25}}\cdot \:100=0.02\cdot \:100\\\\500-450e^{-\frac{2t}{25}}=2\\\\500-450e^{-\frac{2t}{25}}-500=2-500\\\\-450e^{-\frac{2t}{25}}=-498\\\\e^{-\frac{2t}{25}}=\frac{83}{75}\\\\\ln \left(e^{-\frac{2t}{25}}\right)=\ln \left(\frac{83}{75}\right)\\\\\frac{2t}{25}=\ln \left(\frac{83}{75}\right)\\\\t=-\frac{25\ln \left(\frac{83}{75}\right)}{2} \approx -1.2669[/tex]
The concentration of salt in the tank reach 0.02 kg/L when [tex]t = \frac{-25ln\frac{83}{75} }{2} = 1.26[/tex]
Given that,
A brine solution of salt flows at a constant rate of 8L/min into a large tank that initially held 100L of brine solution in which was dissolved 0.5kg of salt.
If the concentration of salt in the brine entering the tank is 0.05kg/L,
We have to determine,
The mass of salt in the tank after t minutes. When will the concentration of salt in the tank reach 0.02kg/L.
According to the question,
Taking the overall balance, since the total Volume of the setup is constant, then flowrate in = flow rate out.
Let y(t) be the mass of salt (in kg) that is in the tank at any time, t (in minutes).
The main equation that we will be using to model this mixing process is:
Rate of change of [tex]\frac{dy}{dx}[/tex] = Rate of salt in - Rate of salt out
Then the component balance for the salt,
Rate of accumulation = rate of flow into the tank - rate of flow out of the tank.
The Rate of salt in = (flow rate of liquid entering) x (concentration of salt in liquid entering),
[tex](8\frac{l}{min)} .(0.05\frac{kg}{l} ) = 0.4\frac{min}{kg}[/tex]
Then,
The Rate of salt out = (flow rate of liquid exiting) x (concentration of salt in liquid exiting)
The concentration of salt in any part of the tank at time t is just y(t) divided by the volume.
The tank initially contains 100 L and the rate of flow into the tank is the same as the rate of flow out.
[tex](8\frac{l}{min)} .(\frac{y(t)}{100} \frac{kg}{l} ) = \frac{2y(t)}{25} \frac{kg}{min}[/tex]
At time t = 0, there is 0.5 kg of salt, so the initial condition is y(0) = 0.5. And the mathematical model for the mixing process is,
[tex]\frac{dy}{dx} = 0.4 - \frac{2y(t)}{25} , y(0) = 0.5\\\\\frac{dy}{dt} = 0.4-\frac{2y(t)}{25}[/tex]
[tex]= \frac{1}{2} \frac{dy}{5-y} = \frac{1dt}{25} \\\\= \int\frac{1}{2} \frac{dy}{5-y} = \int\frac{1dt}{25} \\\\= \frac{1}{2} (-ln|5-y|+ C) = \frac{1 t}{25} + c\\\\=\frac{-1}{2} ln|5-y| +C_1 = \frac{1 t}{25} + c[/tex]
[tex]\frac{-1}{2} ln|5-y|= \frac{1t}{25}+ C_2[/tex]
[tex]5 - y = C_3e^{\frac{-2t}{25} } \\y(t) = 5 - y = C_3e^{\frac{-2t}{25} } \\[/tex]
Using the initial condition y(0)=0.5,
[tex]y(t) = 5 - y = C_3e^{\frac{-2t}{25} } \\y(0)= 5 - y = C_3e^{\frac{-2t}{25} } \\c_3 = 4.5[/tex]
The mass of salt in the tank after t minutes is,
[tex]y(t)= 5 - y = 4.5e^{\frac{-2t}{25} } \\[/tex]
To determine when the concentration of salt is 0.02 kg/L, we solve for t,
[tex]y(t) = 5-4.5e^{\frac{-2t}{25} } \\0.02 = 5-4.5e^{\frac{-2t}{25} } \\ (5)(100)-4.5e^{\frac{-2t}{25} } . (100) = 0.02-100\\500- 450e^{\frac{-2t}{25} }= 2\\= ln(e^{\frac{-2t}{y25} }) = ln\frac{83}{75}[/tex]
[tex]t = \frac{-25ln\frac{83}{75} }{2} = 1.26[/tex]
Hence, The concentration of salt in the tank reach 0.02 kg/L when [tex]t = \frac{-25ln\frac{83}{75} }{2} = 1.26[/tex].
For more information about Change in rate click the link given below.
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