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Differential Equation- Height of water in tank

  • Thread starter Abide
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  • #1
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Homework Statement


A water tank is filled by an inflow x(t), the tank is emptied by the outflow y(t)

The outflow if controlled by a resistance R

The water depth in the tank is represented by d(t)

The surface area of the water is A, independent of depth

The tank is 1.5m high with a 1m diameter and a valve resistance of 10s/m^2

We are being asked to write a differential equation in terms of tank dimensions and valve resistance


Homework Equations



The outflow is related by y(t) = d(t)/R

d/dt(Volume of tank) = x(t)-y(t)

dh/dt = -a*e^(10t) --> I pulled this from another example online

The Attempt at a Solution



I don't understand how to express this without using the terms in regards to outflow and inflow. From the given equation we can calculate...
d(t) = y(t)R --> From here I assume the following
d(t ) = x(t) - y(t)R , can we assume that x(t) is the initial volume of the tank?
I honestly don't know where to go from here, the only times I have seen a problem like this has been in regards to inflow and outflow, not tank dimensions and valve resistance.

d(t) = pi(.5^2)(1.5) - y(t)*10

I feel like I am overlooking something very simple here, I haven't slept in about a day and a half so that might be it :p
 

Answers and Replies

  • #2
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The net volumetric flow (m^3/sec) which is expressed as

dv/dt = x(t) - y(t) = x(t) - h/R where h is the height of water in the tank

But dv/dt = A * dh/dt where A is cross sectional area.

So write your ODE. The independent variable is time, and the dependent variable is h.
 
  • #3
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So then after some algebraic manipulation I have

dh/dt= x(t)/A - h/RA

This still uses output flow as a parameter, which isnt what the question wants. I'm sorry that I'm so hung up on this problem
 
  • #4
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So then after some algebraic manipulation I have

dh/dt= x(t)/A - h/RA

This still uses output flow as a parameter, which isnt what the question wants. I'm sorry that I'm so hung up on this problem
The output flow is y(t). It's not in the ODE. The input flow is present.
 
  • #5
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Right but that isn't in the terms requested right? We can't have it in terms of output flow..or does y(t) somehow relate to valve resistance?
 
  • #6
1,198
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dh/dt= x(t)/A - h/RA

A is the area and it is based on the tank dimension diameter. R is the resistance. The overall height of the tank is 1.5 m but that is extraneous information. You could change the A to pi*D^2/4.
 

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