The problem is this = a tank is fed by 2 streams A & B - yes A = 120 litres per min and B = 30 litres per min - tank height = 6m, sectional area = 4m^2
and outlet flow = 5x10^-4
basically assuming that the system is at a steady state with the inlet flows ive given ( A & B )
How do i demonstrate that if the system is to attain steady state when inlet flow A is stopped flowing, basically showing 5 iterations of Euler method using a step size of 20 mins
I am busy right now, but in the meanwhile, a couple of additional questions. "outlet flow = 5x10^-4": in what units? Is there also a hole through which the water flows out? Is it in the bottom or in the wall? "sectional area = 4m^2": of the tank or of the outlet flow?
I don't see "then" in the statement you need to show.How do i demonstrate that if the system is to attain steady state when inlet flow A is stopped flowing
In general, you would do well if you provide exhaustive details the first time.
ok its getting to complicated now - Sectional area is the tank and yes there is a hole in tank at bottom which is proportional to the liquid in the tank
5x10^-4 = m^2s^-1
basically this is question i have been given - which has 5 parts to it- i have managed to do 3 parts which is calculate the time for tank to fill = (6x4) - 0.15= 23.85 mins
and managed to write mass balance equation and rate equation. Im just stuck on eulers method calculations - not familiar
but ignore that from my first post im not so familar with eulers method so i ask on here. thanks
if i knew how to answer this - at a steady state I have 2 inlets in a tank ( A & B )
A= 120litres , B = 30litres If inlet A was stopped, How do i demonstrate that if the system is to attain steady state when inlet flow A is stopped flowing,
basically Question = what would the 5 iterations of euler's method be when i use a step size of 20 minutes.
http://www.4shared.com/photo/L7TAy0J-/eulermethod.html
or picture uploaded onto beginning of post
OK, for the outlet we have the coefficient $\displaystyle 5\times10^-4\, \frac{m^2}{s}=30\,\frac{l}{m\cdot min}$. This means that it has to be multiplied by the height of the liquid in the tank to get the rate of leaking. To find the height h at the steady state, we have the equation 30h = 120 + 30, so the steady height is 5 m.
Suppose the inlet A = 120 l/min is closed. Given volume V at some point, the height is V/4, so the net rate of flow is 30 - 30 * V/4 = 30 - 7.5V liters/min. At first, when V = 20, the rate is 30 - 150 = -120 liters/min. Assuming the rate is constant for 20 min (that's how Euler's method works), after 20 min the new volume is 20 - 120 * 20 / 1000 = 17.6 m^3. At that point the new rate is 30 - 7.5 * 17.6 = 102 liters/min. After 20 min, the new volume is 17.6 - 102 * 20 / 1000 = 15.56 m^3. Continuing in this way, we get the following data.
$\displaystyle \begin{array}{rllllll}
\mbox{Time} & 0 & 20 & 40 & 60 & 80 & 100\\
\mbox{Rate} & 120.00 & 102.00 & 86.70 & 73.70 & 62.64 & 53,24\\
\mbox{Volume} & 20.00 & 17.60 & 15.56 & 13.83 & 12.35 & 11.10
\end{array}$
The new equilibrium height is 1 m.