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Mathematical model of continuous and batch (discrete) system combined

by Warren Holden
Tags: batch, combined, continuous, discrete, mathematical, model
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Warren Holden
#1
Dec11-12, 08:09 AM
P: 4
I'm having difficulties trying to establish the best approach to create a mathematical model of a process that has a combined continuous and discrete (batch) element to it. I explain as follows:
The system is a hopper (vessel), open to atmosphere, with dry granular material being fed in by conveyor (i.e. continous), the outlet at bottom of hopper has a valve that opens to discharge granular material initially discharging a lot of material quickly into a container below then a more slow gradual but constant outlet flow as the container is moved forward and filled. The hopper discharge valve then closes and waits a delay time until the next empty container is ready below (meanwhile the hopper is being filled), and the process repeats again.
I have tried Integral Mass Balance approach but I don't think this is right. Ideally I need a Laplace transform expression for the system. I know the time values for the valve open discharge and rates of mass flow as well as material in-flow to hopper and hopper volume.
If anyone has any ideas on the best way to approach this problem it would be gratefully received.
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haruspex
#2
Dec28-12, 06:40 PM
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What is the problem to be solved?
Warren Holden
#3
Dec31-12, 12:57 AM
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Hi Haruspex,

I thought that I'd explained the problem previously? Anyway I'll explain again....I have a container (hopper) with a material infeed that I know (i.e. kg/s). The output from the hopper is controlled by a slide valve opening & closing. The discharge from the hopper goes into a moving container below that travels forwards slowly thus filling the container. Initially, when the slide valve opens there will be a large volume of material discharged from the hopper, then as the container below the hopper moves forwards there will be a constant out flow from the hopper (much less than the initial discharge). I am working on the values of the discharge mass flow flow rates from the hopper.
My question is...if I know the hopper in-feed and discharge mass flow rates over time can I construct a Laplace transform to describe the system? and how do I go about this?

pasmith
#4
Dec31-12, 05:25 PM
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Mathematical model of continuous and batch (discrete) system combined

I think the simplest model one could reasonably look at is this: Let [itex]V(t)[/itex] be the amount of material in the hopper at time [itex]t[/itex]. Suppose that material falls in to the hopper at a constant rate [itex]k_0 > 0[/itex]. Suppose that the rate at which material is discharged is, when the exit valve is open, [itex]k_0 + k_1V[/itex] with [itex]k_1 > 0[/itex] (chosen on the assumption that, if the hopper is empty when the valve is open, material falls straight through from the conveyor).

Suppose that the valve is closed for [itex]0 < t < T_1[/itex] and open for [itex]T_1 < t < T_2[/itex]. Then the amount of material in the hopper over a single cycle can be modelled by the ordinary differential equation
[tex]
\frac{\mathrm{d}V}{\mathrm{d}t} = \left\{\begin{array}{r@{\quad}l}
k_0 & 0 < t < T_1 \\
-k_1V & T_1 < t < T_2\end{array}\right.
[/tex]
subject to [itex]V(0) = V_0[/itex].
This equation can be solved analytically to give
[tex]
V(t) = \left\{\begin{array}{r@{\quad}l}
V_0 + k_0t & 0 \leq t \leq T_1 \\
(V_0 + k_0T_1)\exp(-k_1(t - T_1)) & T_1 < t \leq T_2\end{array}\right.
[/tex]
A drawback of the linear model is that the hopper never completely empties: [itex](V_0 + k_0T_1)\exp(-k_1(t - T_1)) > 0[/itex] for all [itex]T_1 < t \leq T_2[/itex] and all choices of the parameters.

However, it is possible to show that the amount of material in the hopper at the end of a cycle tends to a finite limit: Let [itex]V_n = V(nT_2)[/itex] for positive integer [itex]n[/itex]. Then
[tex]
V_{n+1} = (V_n + k_0T_1)\exp(-k_1(T_2 - T_1))
[/tex]
This is a linear recurrence relation, which can be solved to give
[tex]
V_n = (V_0 - B)a^n + B
[/tex]
where
[tex]a = \exp(-k_1(T_2 - T_1)) \\
B = \frac{k_0T_1\exp(-k_1(T_2 - T_1))}{1 - \exp(-k_1(T_2 - T_1))}[/tex]
Since [itex]0 < a < 1[/itex], [itex]V_n \to B[/itex] as [itex]n \to \infty[/itex].

EDIT: one can tune the parameters so that [itex]V(0) = V(T_2) = V_0 \neq 0[/itex]: one must have
[tex]
k_1(T_2 - T_1) = \ln \left(1 + \frac{k_0T_1}{V_0}\right).
[/tex]
Warren Holden
#5
Dec31-12, 09:43 PM
P: 4
Hi pasmith,

Many thanks for your explanation. I can't say that I fully understand it, particularly the solving of the 'linear recurrence relation', and I cannot see a way that I can use this information in practice.
Looking at the problem from another perspective, if I know the rate of material entering the hopper and also if I know the rate of material discharging from the hopper (initially when the slide valve opens there will be a sudden large discharge rate, then a steady discharge rate which is much less than the initial surge). This information will provide three (3) cases. If I have this data in the form of a graph of hopper material balance (i.e. hopper material vs time) over a cycle of these three cases, can I produce a mathematical model from this data?
haruspex
#6
Jan4-13, 12:15 AM
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Quote Quote by Warren Holden View Post
My question is... can I construct a Laplace transform to describe the system? and how do I go about this?
Ah, ok. In the OP I read your 'ideally' as saying a Laplace transform would be a means to an end, the end being unstated.
haruspex
#7
Jan4-13, 12:22 AM
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Quote Quote by pasmith View Post
I think the simplest model one could reasonably look at is this: Let [itex]V(t)[/itex] be the amount of material in the hopper at time [itex]t[/itex]. Suppose that material falls in to the hopper at a constant rate [itex]k_0 > 0[/itex]. Suppose that the rate at which material is discharged is, when the exit valve is open, [itex]k_0 + k_1V[/itex] with [itex]k_1 > 0[/itex] (chosen on the assumption that, if the hopper is empty when the valve is open, material falls straight through from the conveyor).

Suppose that the valve is closed for [itex]0 < t < T_1[/itex] and open for [itex]T_1 < t < T_2[/itex]. Then the amount of material in the hopper over a single cycle can be modelled by the ordinary differential equation
[tex]
\frac{\mathrm{d}V}{\mathrm{d}t} = \left\{\begin{array}{r@{\quad}l}
k_0 & 0 < t < T_1 \\
-k_1V & T_1 < t < T_2\end{array}\right.
[/tex]
That's not how I read the OP. If the valve is open long enough, the discharge rate must become the input rate. Since only two rates are described, that must be the second of them. So dV/dt=0 in [T1,T2]. This means there's a minimum occupancy, which maybe we can subtract out from V, leaving V=0 in [T1, T2].


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