Differential equations to determine water level

In summary, the problem is to find the time it takes for a cylindrical water reservoir with an average level of 300,000,000 m3 to drop to 90% of the average level during a drought, considering factors such as stream inflow, rainfall, evaporation, and community outflow. The model for this problem is dh/dt = K1-(K2/S)+(Bh0/S)-(BK/S)h + (2K0/S)(1/(t+1)), where h is the height of the water in the reservoir and t is time. To solve for t, separation of variables can be used.
  • #1
aquameatwad
5
0
Hi, the problem i have is this:
How long will it take a water reservoir with an Average level of 300,000,000 m3 to drop to 90% of the average level, if there is a drought, taking into account average rainfall, evaporation and amount of water taken in and out?

Assumption are:
Reservoir is cylindrical.
Drought consists of zero rainfall
the only factors that affect reservoir level is stream inflow, rainfall, community outflow and evaporation.


the variables are:
Avg initial volume, V0=300,000,000 m3
Surface area of reservoir, S=10,000,000 m3
Avg in flow, K0=6,000,000 m3/Day
Avg evaporation rate, K1=.0012 m/sec
Avg out flow, K2=6,000,000 m3/Day
Height of water in reservoir, h0=30m
constant B=1/h0
constant a= 1day/m2
time, t=?
dependent variable, h=?

So my model is this dv/dt=IN-OUT
IN is a(K0/(t+1)) ,(stream inflow) since it will go down since there is a drought
OUT is K1(S), (evaporation) & K2-(B(h0-h)K2)K2 ,(Community outflow)

so the model is this
dv/dt=S(dh/dt)=[a(K0/(t+1))]-[K1(S)-(K2-(B(h0-h)K2)K2)]

which turns into
dh/dt=[K1-(K2/S)+(Bh0/S)] - [(BK/S)H] + [(2K0/S)(1/(t+1))]

so i guess in turn looks like the format:
dh/dt = C +(-Dh) + E/(t+1)

Now... is this even correct? cause i can't figure out how to solve E/(t+1). i know i could use variation of parameters if it was a 2nd order and find the Aux Eqn etc. But I am kinda stuck?Help, please.
 
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  • #2
Yes, your model is correct. To solve for E/(t+1), you can use separation of variables. This involves separating the variables dh/dt and t and then integrating to find h as a function of t. After that, you can solve for t given the condition that h = 90%*V0.
 

FAQ: Differential equations to determine water level

What are differential equations?

Differential equations are mathematical equations that describe how a quantity changes over time. They involve derivatives, which represent the rate of change of the quantity with respect to another variable.

How are differential equations used to determine water level?

Differential equations can be used to model the behavior of a system, such as the amount of water in a tank. By setting up an equation that describes the rate of change of water level over time, we can solve for the water level at any given time using known initial conditions.

What factors affect the water level in a tank according to a differential equation?

The factors that affect water level in a tank according to a differential equation include the rate at which water is added or removed from the tank, the shape and size of the tank, and any external forces acting on the water (such as gravity or pressure).

Are differential equations the only way to determine water level in a tank?

No, there are other mathematical methods that can be used to determine water level in a tank, such as algebraic equations or numerical simulations. However, differential equations are often the most accurate and efficient way to model the behavior of a system over time.

Can differential equations be used to predict future water levels?

Yes, if the initial conditions and factors affecting the system are known, differential equations can be used to predict future water levels. However, any small changes in these conditions or factors can affect the accuracy of the prediction.

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