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Seawater with a concentration of \(8000 \mathrm{g} / \mathrm{m}^{3}\) is pumped into a well-mixed tank at a rate of \(0.6 \mathrm{m}^{3} / \mathrm{hr}\). Because of faulty design work, water is evaporating from the tank at a rate of \(0.025 \mathrm{m}^{3} / \mathrm{hr}\) The salt solution leaves the tank at a rate of \(0.6 \mathrm{m}^{3} / \mathrm{hr}\) (a) If the tank originally contains \(1 \mathrm{m}^{3}\) of the inlet solution, how long after the outlet pump is turned on will the tank run dry? (b) Use numerical methods to determine the salt concentration in the tank as a function of time.

Short Answer

Expert verified
In summary, to find the time when the tank runs dry, we first model the rate of change of volume and salt in the tank. Then, we write the differential equation for the salt concentration and solve it to find the time when the tank is empty, which is 40 hours after the outlet pump is turned on. To determine the salt concentration as a function of time, we use numerical methods such as Euler's method or Runge-Kutta method to solve the differential equation for the salt mass, and thereby computing the salt concentration.

Step by step solution

01

Model the rate of change of volume and salt in the tank

The rate of change of volume in the tank can be expressed as follows: Rate of change of volume \(= (\text{Inflow rate of water}) - (\text{Evaporation rate}) - (\text{Outflow rate of water})\) Let V denote the volume of the tank. We have: \(\frac{dV}{dt} = 0.6 - 0.025 - 0.6 = -0.025 \: \mathrm{m}^3/\mathrm{hr}\) Now, we need to consider the rate of change of salt mass in the tank. Let S be the mass of the salt in the tank at any time t. The rate of change of mass of salt in the tank can be expressed as: Rate of change of salt mass \(= (\text{Mass of salt entering the tank}) - (\text{Mass of salt leaving the tank})\) The mass of salt entering the tank per hour is given as 8000 g/m³ and the inflow rate is 0.6 m³/hr, so the rate of change of mass of salt entering the tank is 8000 * 0.6 g/hr. Let C represent the concentration of salt in the tank, then the mass of salt leaving the tank per unit volume is given as C g/m³ and the outflow rate is 0.6 m³/hr. So, the rate of change of mass of salt leaving the tank is C * 0.6 g/hr. Thus, we have: \(\frac{dS}{dt} = 8000 \times 0.6 - C \times 0.6\)
02

Write the differential equation for the salt concentration

Now, we'll express the salt concentration, C, as a function of the mass of the salt (S) and the volume of the tank (V). The salt concentration is simply the mass of salt divided by the volume of the tank: C = S / V Using the chain rule and combining the equations obtained in Step 1, we get the following differential equation for the salt concentration: \(\frac{dS}{dt} = 8000 \times 0.6 - \frac{S}{V} \times 0.6\) We also know the volume of the tank at any time t: \(\frac{dV}{dt} = -0.025 \: \mathrm{m}^3/\mathrm{hr}\) Integrating this equation, we find: \(V(t) = -0.025t + V_0\) where V0 is the initial volume of the tank, which is given as 1 m³.
03

Solve the differential equation and compute time for the tank to run dry

We need to solve the differential equation for S(t) to find the time when the tank runs dry. This is equivalent to finding the time when V(t) = 0: \(-0.025t + 1 = 0\) Solving for t, we get: \(t = 40 hours\) The tank will run dry in 40 hours after the outlet pump is turned on.
04

Determine the salt concentration as a function of time using numerical methods

To find the salt concentration as a function of time, we will use numerical methods like Euler's method or Runge-Kutta method to solve the differential equation for S(t): \(\frac{dS}{dt} = 8000 \times 0.6 - \frac{S}{V} \times 0.6\) with initial condition S(0) = 8000 g (since the initial tank has 1 m³ of the inlet solution with a concentration of 8000 g/m³). Using software, libraries, or programming languages like Python, MATLAB, or Wolfram Mathematica, implement the numerical method of your choice to obtain the salt concentration, C(t), as a function of time. Make sure to take a sufficiently small step size and use the solution for S(t) up until the 40-hour mark since the tank runs dry after that point.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differential Equations
Differential equations are mathematical equations that relate a function to its derivatives. They are essential in describing various dynamic systems and processes. In this exercise, differential equations help us understand how both the volume and salt concentration in the tank change over time.
The rate of change of these variables with respect to time gives us insight into the process dynamics. For instance, the rate of volume change is \(\frac{dV}{dt} = -0.025\ m³/hr\), as calculated from the inflow rate, outflow rate, and evaporation rate.Similarly, the rate of change of salt is modeled by another differential equation: \(\frac{dS}{dt} = 8000 \times 0.6 - C \times 0.6\). These differential equations form the basis for further analysis using numerical methods, allowing us to predict future behaviors of the system, such as when the tank will run dry.
Concentration Calculations
Concentration calculations are crucial for understanding the composition of solutions. For this exercise, we're dealing with a salt solution whose concentration changes over time. Concentration is defined as the mass of solute (salt, in this case) divided by the volume of solvent (water).We represent concentration (C) as:\(C = \frac{S}{V}\)Where \(S\) is the mass of salt, and \(V\) is the volume of the mix in the tank. As water evaporates and more salt solution leaves the tank, both \(S\) and \(V\) change, affecting the concentrations. Calculating how each of these contribute to changes in concentration helps us predict the behavior of the tank's contents over time. These calculations are then used to set up and solve the differential equations.
Evaporation Rate
The evaporation rate is the speed at which water evaporates from the tank. Here, it is given as \(0.025\ m³/hr\). This constant rate directly affects the volume of water remaining in the tank since it reduces the water content without removing any salt.Understanding this rate is crucial because it continuously decreases the tank's water volume, influencing the time it will take for the tank to run dry. This evaporation is subtracted from the volume inflow in the differential equation. It ultimately leads to a net outflow, as both evaporation and outflow take place, impacting the salt concentration because only the water, not the salt, is lost to evaporation.
Euler's Method
Euler's Method is a simple numerical procedure for solving ordinary differential equations (ODEs) with a given initial value. Because the equations governing our tank's salt concentration are too complex to solve analytically, we use this method to approximate solutions.The method involves the following steps:
  • Choose a small step size for time \(dt\).
  • Calculate increments of \(S\) (salt amount) using the previously computed rates of change \(\frac{dS}{dt}\).
  • Iterate over time until the desired endpoint (40 hours, in this case).
Euler's Method is advantageous due to its simplicity, but it is less accurate for problems requiring more precision. It's an excellent starting point for those new to numerical methods.
Runge-Kutta Method
The Runge-Kutta Method is another numerical method for solving ODEs, known for its higher accuracy over simpler methods like Euler's. It does this by calculating several intermediate values, which provide a better estimate at each step.For our tank's salt concentration problem:
  • Runge-Kutta utilizes multiple evaluations of \(\frac{dS}{dt}\) at different points within each time interval.
  • It averages these evaluations to produce a more refined slope estimate.
  • This results in a more accurate approximation of \(S(t)\).
While this method is more computationally intense, the increase in accuracy is especially beneficial for tracking more gradual changes in concentration over the 40 hours until the tank empties.

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Most popular questions from this chapter

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