A leaf filter is operated at 1 atm (gauge). The volume of filtrate collected \(V\) (in \(m^3\)) is related with the volumetric flow rate of the filtrate \(q\) (in \(m^3/s\)) as: \[ \frac{1}{q} = \frac{1}{\frac{dV}{dt}} = 50V + 100 \] The volumetric flow rate of the filtrate at 1 hour is ___________ \( \times 10^{-3} \, m^3/s\) (rounded off to 2 decimal places).
A leaf filter is operated at 1 atm (gauge). The volume of filtrate collected \(V\) (in \(m^3\)) is related with the volumetric flow rate of the filtrate \(q\) (in \(m^3/s\)) as: \[ \frac{1}{q} = \frac{1}{\frac{dV}{dt}} = 50V + 100 \] The volumetric flow rate of the filtrate at 1 hour is ___________ \( \times 10^{-3} \, m^3/s\) (rounded off to 2 decimal places).
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Solution and Explanation
Step 1: Rearranging the Equation. The given equation relates the inverse of the volumetric flow rate \(q\) with the volume \(V\): \[ q = \frac{1}{50V + 100} \] Step 2: Expressing the Rate of Change of Volume. The rate of change of volume is: \[ \frac{dV}{dt} = 50V + 100 \] Step 3: Integrating the Equation. We integrate both sides of the equation: \[ \int \frac{1}{50V + 100} \, dV = \int dt \] After integrating, we get: \[ \frac{1}{50} \ln(50V + 100) = t + C \] Step 4: Finding the Constant of Integration. Using the initial condition \(V = 0.002 \, {m}^3\) at \(t = 0\): \[ C = \frac{1}{50} \ln(100.1) \] Step 5: Finding \(V\) at \(t = 3600\) Seconds. Substitute \(t = 3600\) seconds into the equation and solve for \(V\).
Step 6: Calculate the Volumetric Flow Rate. Finally, we substitute the obtained volume \(V\) at \(t = 3600\) seconds into the equation: \[ q = \frac{1}{50V + 100} \] The final volumetric flow rate is \(q = 1.61 \times 10^{-3} \, {m}^3/{s}\).
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