Consider A Tank Used In Certain Hydrodynamic Experiments
Consider A Tank Used In Certain Hydrodynamic Experiments After One Ex
Consider a tank used in certain hydrodynamic experiments. After one experiment, the tank contains liters of a dye solution with a concentration of g/liter. To prepare for the next experiment, the tank is to be rinsed with fresh water flowing in at a rate of liters/min, with the well-stirred solution flowing out at the same rate. Find the time that will elapse before the concentration of dye in the tank reaches a specified fraction of its original value. Round your answer to two decimal places.
Paper For Above instruction
The problem involves modeling the dye concentration in a tank undergoing a process of dilution with incoming fresh water. This scenario is classic in the study of differential equations related to mixing problems, often referred to as "tank problems," which are key to understanding fluid dynamics and chemical process engineering.
The fundamental assumptions are that the tank is perfectly mixed at all times (well-stirred), the inflow and outflow rates are constant and equal, and the initial concentration of dye is known. The goal is to determine the time it takes for the dye concentration to reach a certain fraction of its initial value.
Initial Conditions and Variables:
- Let \( V \) denote the volume of the solution in the tank at the start. Although the exact volume is not specified, the problem typically assumes a constant volume, or that the initial volume is known.
- Let \( C(t) \) be the concentration of dye in the tank at time \( t \).
- The initial concentration is \( C_0 \) (g/liter).
- The inflow rate of water is \( Q \) (liters/min), and the outflow rate is also \( Q \) (liters/min), maintaining a constant volume \( V \).
Mathematical Modeling:
Given the well-mixed assumption, the rate of change of the amount of dye \( A(t) \) in the tank is described by the differential equation:
\[
\frac{dA}{dt} = \text{(Rate in)} - \text{(Rate out)}.
\]
Since incoming water is dye-free (assuming pure water inflow), the rate in is zero:
\[
\text{Rate in} = 0.
\]
The rate out depends on the current concentration:
\[
\text{Rate out} = \frac{A(t)}{V} \times Q,
\]
because the amount of dye leaves at the concentration \( C(t) = \frac{A(t)}{V} \).
The volume \( V \) remains constant because inflow equals outflow:
\[
\frac{dA}{dt} = - \frac{Q}{V} A(t).
\]
This is a standard first-order linear differential equation:
\[
\frac{dA}{dt} + \frac{Q}{V} A = 0.
\]
Solution of the Differential Equation:
The general solution is:
\[
A(t) = A_0 e^{-\frac{Q}{V} t},
\]
where \( A_0 = C_0 V \) (initial total dye amount).
The concentration at time \( t \) then becomes:
\[
C(t) = \frac{A(t)}{V} = C_0 e^{-\frac{Q}{V} t}.
\]
Determining the Time:
Suppose we are asked to find the time \( t \) when the concentration reaches a specified fraction \( f \) of the initial concentration \( C_0 \):
\[
C(t) = f C_0.
\]
Substituting into the exponential decay:
\[
f C_0 = C_0 e^{-\frac{Q}{V} t}.
\]
Dividing both sides by \( C_0 \):
\[
f = e^{-\frac{Q}{V} t}.
\]
Taking natural logarithm:
\[
\ln(f) = -\frac{Q}{V} t.
\]
Solving for \( t \):
\[
t = -\frac{V}{Q} \ln(f).
\]
Conclusion:
The time required for the dye concentration to reach a fraction \( f \) of its initial value is:
\[
t = -\frac{V}{Q} \ln(f).
\]
Application:
Assuming the problem specifies particular values for \( V \), \( Q \), \( C_0 \), and the target fraction \( f \), these can be substituted into the formula to compute \( t \). For example, if the initial volume \( V \) is 100 liters, inflow rate \( Q \) is 10 liters/min, and the target concentration is 10% of the original (\( f=0.1 \)), then:
\[
t = -\frac{100}{10} \ln(0.1) = -10 \times (-2.3026) = 23.03 \text{ minutes}.
\]
Summary:
In general, the key formula for this mixing problem is:
\[
\boxed{t = -\frac{V}{Q} \ln(f).}
\]
The specific numerical answer depends on the second piece of data provided—initial volume, inflow rate, and desired fraction— which should be plugged into this formula. Rounding to two decimal places will give the final answer.
References
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