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Chapter 4: Problem 60

At time \(t=0\) the valve on an initially empty (perfect vacuum, \(\rho=0\) ) tank is opened and air rushes in. If the tank has a volume of \(V_{0}\) and the density of air within the tank increases as \(\rho=\rho_{\infty}\left(1-e^{-b t}\right),\) where \(b\) is a constant, determine the time rate of change of mass within the tank.

Short Answer

Expert verified
The time rate of change of mass in the tank is given by \(\frac{dm}{dt} = V_{0}\rho_{infty}b e^{-bt}\)

Step by step solution

01

Understand the Given Function

The density increases as \(\rho = \rho_{\infty}(1 - e^{-bt})\), where \(\rho_{\infty}\) and \(b\) are constants. At \(t=0\), the tank is empty which means \(\rho=0\) and as \(t\) goes to infinity, \(\rho =\rho_{\infty}\). The density function gives an idea of how the density of air increases with time when air rushes into an empty vacuum tank.
02

Find the Mass Function

The mass of the air in the tank can be calculated by multiplying the volume of the tank by the density of the air. Since the volume \(V_{0}\) is constant, the mass \(m\) at time \(t\) is given by the formula \(m = V_{0} \rho\). Substituting \(\rho = \rho_{\infty}(1 - e^{-bt})\) into the equation, we have \(m=V_{0}\rho_{\infty}(1-e^{-bt})\).
03

Determine the Time Rate of Change of Mass

To find the time rate of change of mass, take the derivative of the mass function with respect to time \(t\). The derivative of \(V_{0}\rho_{\infty}(1 - e^{-bt})\) with respect to \(t\) can be calculated as follows: \(\frac{dm}{dt} = V_{0}\rho_{\infty}b e^{-bt}\). This is the time rate of change of mass in the tank.

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Most popular questions from this chapter

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