- #1
jzcrouse
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<< Mentor Note -- thread moved from the schoolwork forums to ME for better views >>
1. Homework Statement
Not really homework as this is for work. We are washing a cylindrical container (15.5 cm) with only one (1.8" dia) hole at the bottom the rest is sealed. Our machine sticks a .5" pipe with a sprayer up through the hole with a 12.5 GPM output. The customer wants to reduce the time it takes to wash and drain the container. My main question is when the container begins filling with water (faster than the exit of water through the same hole), I need to find the rate the water level height increases. Need to know the height at 10 second increments up to 90 seconds. Also if I put a hollow tube spanning the length of the pipe sprayer so it is inside and outside the tank. Can I assume ATM pressure at the top of the container for the most part? Height of the container is 23 inches. All dimensions were converted to be in inches and seconds. Area of tank is ##A_{t}##, Area of hole is ##A_{h}## The flow in will go for 90 seconds (the container should be filled by then) We would prefer not to have it fill but understand that may not feasible if flow rate in doesn't change. The Volume accumulation rate in the tank should more than coming out from actual results
##A_{h}V_{1}=A_{t}V_{2}##
##v=\sqrt{2g(h_{1}-h_{0})}## assuming atm at top of container. If not true finding pressure above water would be difficult i assume since it would change when the tank begins filling and ATM is pushing up while you only have hydrostatic + whatever pressure is there via ideal gas too much math since were running a simulation soon.
Rate in - Rate out = Rate Accumulation
##12.5 GPM- v*A_{h}= \frac{dh*A_{t}}{dt}##
##\frac{12.5 gpm - \sqrt{2g(h)}*A_{h}}{A_{t}}=\frac{dh}{dt}##
##\int \, dt=\int\frac{ A_{t}}{12.5 gpm - \sqrt{2g(h)}*A_{h}}\, dh##
integrate and I end up with ##t=-\frac{2*188.69*(\sqrt{h}* 65.26 + 48.125* \log(\sqrt{h} *65.26 - 48.125))}{65.26^2}##
which doesn't work because Ill end up with a negative inside my log function. Perhaps I am just screwing up units but i tripled checked those. I was hoping to just input various times and solve for h.
1. Homework Statement
Not really homework as this is for work. We are washing a cylindrical container (15.5 cm) with only one (1.8" dia) hole at the bottom the rest is sealed. Our machine sticks a .5" pipe with a sprayer up through the hole with a 12.5 GPM output. The customer wants to reduce the time it takes to wash and drain the container. My main question is when the container begins filling with water (faster than the exit of water through the same hole), I need to find the rate the water level height increases. Need to know the height at 10 second increments up to 90 seconds. Also if I put a hollow tube spanning the length of the pipe sprayer so it is inside and outside the tank. Can I assume ATM pressure at the top of the container for the most part? Height of the container is 23 inches. All dimensions were converted to be in inches and seconds. Area of tank is ##A_{t}##, Area of hole is ##A_{h}## The flow in will go for 90 seconds (the container should be filled by then) We would prefer not to have it fill but understand that may not feasible if flow rate in doesn't change. The Volume accumulation rate in the tank should more than coming out from actual results
Homework Equations
##A_{h}V_{1}=A_{t}V_{2}##
##v=\sqrt{2g(h_{1}-h_{0})}## assuming atm at top of container. If not true finding pressure above water would be difficult i assume since it would change when the tank begins filling and ATM is pushing up while you only have hydrostatic + whatever pressure is there via ideal gas too much math since were running a simulation soon.
The Attempt at a Solution
Rate in - Rate out = Rate Accumulation
##12.5 GPM- v*A_{h}= \frac{dh*A_{t}}{dt}##
##\frac{12.5 gpm - \sqrt{2g(h)}*A_{h}}{A_{t}}=\frac{dh}{dt}##
##\int \, dt=\int\frac{ A_{t}}{12.5 gpm - \sqrt{2g(h)}*A_{h}}\, dh##
integrate and I end up with ##t=-\frac{2*188.69*(\sqrt{h}* 65.26 + 48.125* \log(\sqrt{h} *65.26 - 48.125))}{65.26^2}##
which doesn't work because Ill end up with a negative inside my log function. Perhaps I am just screwing up units but i tripled checked those. I was hoping to just input various times and solve for h.
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