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GwtBc
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- Homework Statement
- Due to the global outbreak of COVID-19, in order to limit the spread of the virus, the government has placed limits on the number of people that can be inside a building at any given time. Due to this initiative, a queue often forms at the entrance to the Woolworths on Lygon Street; assume this queue is one long, straight, line.
The government regulations also mandate that there must be a minimum of 1.5m between each customer—this determines the maximum density of customers in the queue. While waiting for their morning coffee, a research scientist has observed that, if the queue is empty, a customer entering the queue will walk at 1.4m/s—this is the maximum velocity of customers in the queue.
They have also determined, again by observation, that there is a linear relationship between the density of customers and their velocity, which is defined by the bounds given above. The first 10 metres of the queue is sectioned off and so people are only able to enter if they are already in the queue. However, behind this point, people become confused about where to enter the queue and so they enter at a random position, at a rate of α customers per metre per second; model this random entry as a constant (per unit time, per unit length).
During the lunch time peak hour, the supermarket is at its maximum government mandated capacity, however it is functioning efficiently and so customers in the queue are moving at maximum flux. At time t = 0, a malfunction in the checkout machines results in customers unable to exit the supermarket, and so customers at the beginning of the queue cannot enter the building. Solve the governing equations using the method of characteristics and construct the space-time diagram. In so doing, identify the presence of any shocks and fans (if either exist).
- Relevant Equations
- ##\frac{\partial\rho}{\partial t} + \frac{\partial J}{\partial x} = f(x,t)##
##J = \rho v##
Hello all, this question really has me and some friends stomped so advice would be appreciated.
Ok so, the relevant (dimensionless) continuity equation I have found to be
$$\frac{\partial\rho}{\partial t} + (1-2\rho)\frac{\partial \rho}{\partial x} = \begin{cases} \beta, \hspace{3mm} x < 0 \\ 0, \hspace{3mm} x \geq 0 \end{cases}$$
Where ##\beta## is the rate at which people enter the queue behind ##x = 0##, and we are using length scale of ##10## s.t. ##10m \rightarrow x=1##. Now on a characteristic given by $$ \dot{x} = 1-2\rho$$, we find $$ \dot{\rho} = \begin{cases} \beta, \hspace{3mm} x < 0 \\ 0, \hspace{3mm} x \geq 0 \end{cases}$$
From the statement of the question, the relevant initial conditions are ##x(0)=s##, and ##\rho(s,0)= \begin{cases} 1/2 ,\hspace{3mm} x < 1 \\ 1, \hspace{3mm} x \geq \end{cases}##
We also need to ensure that at every point ##\rho(x,t) \leq 1## since this gives the maximum allowed density in the queue (since ρ=ρreal/ρmaxin dimensionless form).
Now until now in my attempts to solve this question I have naively written $$ \dot{\rho} = \begin{cases} \beta, \hspace{3mm} x < 0 \\ 0, \hspace{3mm} x \geq 0 \end{cases} \Rightarrow \rho = \begin{cases} \beta t, \hspace{3mm} s < 0 \\ 0, \hspace{3mm} s \geq 0 \end{cases} + \rho(s,0)$$ In other words, I have somewhat ignored the x dependence of the discontinuity in ##\dot{\rho}## by saying that it only depends on the initial condition. I don't like ignoring the x-dependence of the ODE but I'm not sure how to deal with it otherwise. Additionally, this approach implies that for##s < 0## the density goes to infinity as ##t \rightarrow \infty##, which does not make sense. Is this the right approach? If so, then how are we to ensure that ##\rho## does not exceed ##1##?
edit:fixed formatting
Ok so, the relevant (dimensionless) continuity equation I have found to be
$$\frac{\partial\rho}{\partial t} + (1-2\rho)\frac{\partial \rho}{\partial x} = \begin{cases} \beta, \hspace{3mm} x < 0 \\ 0, \hspace{3mm} x \geq 0 \end{cases}$$
Where ##\beta## is the rate at which people enter the queue behind ##x = 0##, and we are using length scale of ##10## s.t. ##10m \rightarrow x=1##. Now on a characteristic given by $$ \dot{x} = 1-2\rho$$, we find $$ \dot{\rho} = \begin{cases} \beta, \hspace{3mm} x < 0 \\ 0, \hspace{3mm} x \geq 0 \end{cases}$$
From the statement of the question, the relevant initial conditions are ##x(0)=s##, and ##\rho(s,0)= \begin{cases} 1/2 ,\hspace{3mm} x < 1 \\ 1, \hspace{3mm} x \geq \end{cases}##
We also need to ensure that at every point ##\rho(x,t) \leq 1## since this gives the maximum allowed density in the queue (since ρ=ρreal/ρmaxin dimensionless form).
Now until now in my attempts to solve this question I have naively written $$ \dot{\rho} = \begin{cases} \beta, \hspace{3mm} x < 0 \\ 0, \hspace{3mm} x \geq 0 \end{cases} \Rightarrow \rho = \begin{cases} \beta t, \hspace{3mm} s < 0 \\ 0, \hspace{3mm} s \geq 0 \end{cases} + \rho(s,0)$$ In other words, I have somewhat ignored the x dependence of the discontinuity in ##\dot{\rho}## by saying that it only depends on the initial condition. I don't like ignoring the x-dependence of the ODE but I'm not sure how to deal with it otherwise. Additionally, this approach implies that for##s < 0## the density goes to infinity as ##t \rightarrow \infty##, which does not make sense. Is this the right approach? If so, then how are we to ensure that ##\rho## does not exceed ##1##?
edit:fixed formatting
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