Depends on the amount of DEI in the construction and maintenance. If DEI constructed the bottom will probably fall out. If DEI maintained I doubt you could get the valve open all the way.
Nope. Tank X is wider at the top, so it's height will initially lower more slowly than Y. Once tank Y gets lower, it's lower pressure will cause it to empty more slowly.
Pressure decreases as the water flows out. The one on the right will have more water at lower pressure/lower water height. I'd expect the one on the left to empty slightly faster.
Don't forget the Coriolis effect. If both are stable, and still at the start, they should be equal. However, any circular momentum will improve the flow. Presuming the bell shaped container is flat bottomed, the liquid rotation will be impeded. Additionaly, the conical bottom of the funnel shaped container can impose, and amplify rotation. As the quantity of liquid dispenses, the volume of the remaining quantity also decreases. A funnel shape, therefore, maintains a taller column during dispensing, whilst the bell shape increasingly decreases the column height. So, I would utilize the funnel shaped one. YMMV.
Gentlemen - one of us is a retard! Check me, if you will:
The formula for vessel pressure is
Pressure = fluid column height x gravity x density
Doesn’t matter what the shape of the vessel is; if you have a totally enclosed vessel with a 1” vent at the top open to atmosphere and it has the same fluid, with the same density at the same atmospheric pressure … it will have the same pressure at the bottom of the tank.
So I guess it is a trick question? Unless we know the height of the fluid, and the fluid density… we can’t mathematically prove anything one way or another. 🤬
Gentlemen - we must preserve our self esteem at any cost in the world of vicious retard-haters. I’m going to give us all part marks.
Head decreases quicker on y, which slows the flow and thus x empties quicker. I used to be able to write and integrate the equations, but am to old and dumb now.
The height of the water or pressure head is the kicker. Correct At the start the pressure hence the flow rate is the same. However tank y will lower more for the same volume removed. This will lower the pressure and the flow of tank y faster than tank x. It temains to be seen if the flow of the lower parts will exactly balance the difference of the higher parts.
all things equal, given the atmospheric pressure is the same they will both empty at the same time, given the outlets are equal in flow rate. but tank X has a tapered bottom verses tank Y giving it a better volumetric efficiency. i'm thinking that tank tank X should empty slightly faster than tank Y. that's my 3rd grade education showing.
Yes but fluid pressure is based on height of water column above it. In tank x water pressure has higher head dur to shape of tank all the way to bottom. head in tank y falls off as height of column actually decreases as water is drained out( ie wider not deeper as tank drains)
From what I remember the distance of the visible horizon is the square root of 3 multiplied by the height (in feet) divided by 2 with the answer in miles. I hope this helps
Grok says: Tank X will empty first.From the image, Tank X is the one shaped like a funnel pointing down — wide at the top, narrow at the bottom. Tank Y is the opposite — narrow at the top, wide at the bottom (like an upright cone or bell shape).Both hold the same volume (1000 liters) and have identical taps/outlets at the bottom, so the outflow starts with the same speed (by Torricelli's law, exit velocity depends only on the initial water height above the outlet, which appears similar here).The key difference is how the water level (height) drops as volume leaves:In Tank X (wide top → narrow bottom), most of the water is stored up high in the wider part. When water drains, the level drops relatively slowly at first (because you're removing volume from a large cross-sectional area). This keeps the height (and therefore pressure/head at the outlet) higher for longer during most of the draining process. Higher average head → higher average flow rate over time → empties faster overall. In Tank Y (narrow top → wide bottom), most of the water is in the lower, wider part. As soon as draining starts, the level drops quickly at first (small cross-sectional area up top means small volume loss causes big height change). This quickly reduces the head/pressure, so the flow rate slows down earlier and stays lower for more of the process.
Mathematically, the time to empty is proportional to the integral of A(h) / √h dh over the height (where A(h) is cross-sectional area at height h). The shape where A(h) is smaller when h is larger (Tank X) results in less total time.Many physics discussions, simulations (e.g., some using tools like LS-DYNA), and calculations confirm Tank X empties first — often by a noticeable margin (up to ~2x faster in extreme cone cases, though less here since they're frustums).So the answer is Tank X (the one on the left in your image, wide opening at top). Tank Y will still have water left when X runs dry.
Depends on the amount of DEI in the construction and maintenance. If DEI constructed the bottom will probably fall out. If DEI maintained I doubt you could get the valve open all the way.
ReplyDeleteBasic physics, X will empty sooner.
ReplyDeleteFitty
Nope. Tank X is wider at the top, so it's height will initially lower more slowly than Y. Once tank Y gets lower, it's lower pressure will cause it to empty more slowly.
ReplyDeletePressure decreases as the water flows out. The one on the right will have more water at lower pressure/lower water height. I'd expect the one on the left to empty slightly faster.
ReplyDeleteWater will evaporate out of tank X faster so it will empty first.
ReplyDeleteDon't forget the Coriolis effect. If both are stable, and still at the start, they should be equal. However, any circular momentum will improve the flow. Presuming the bell shaped container is flat bottomed, the liquid rotation will be impeded. Additionaly, the conical bottom of the funnel shaped container can impose, and amplify rotation.
ReplyDeleteAs the quantity of liquid dispenses, the volume of the remaining quantity also decreases. A funnel shape, therefore, maintains a taller column during dispensing, whilst the bell shape increasingly decreases the column height. So, I would utilize the funnel shaped one. YMMV.
😡
ReplyDeleteGentlemen - one of us is a retard! Check me, if you will:
The formula for vessel pressure is
Pressure = fluid column height x gravity x density
Doesn’t matter what the shape of the vessel is; if you have a totally enclosed vessel with a 1” vent at the top open to atmosphere and it has the same fluid, with the same density at the same atmospheric pressure … it will have the same pressure at the bottom of the tank.
So I guess it is a trick question? Unless we know the height of the fluid, and the fluid density… we can’t mathematically prove anything one way or another.
🤬
Gentlemen - we must preserve our self esteem at any cost in the world of vicious retard-haters. I’m going to give us all part marks.
Head decreases quicker on y, which slows the flow and thus x empties quicker. I used to be able to write and integrate the equations, but am to old and dumb now.
ReplyDeleteThe height of the water or pressure head is the kicker.
ReplyDeleteCorrect
At the start the pressure hence the flow rate is the same. However tank y will lower more for the same volume removed. This will lower the pressure and the flow of tank y faster than tank x.
It temains to be seen if the flow of the lower parts will exactly balance the difference of the higher parts.
It's why water towers are big at the top instead of a big lower tank with a 100ft riser.
ReplyDeleteall things equal, given the atmospheric pressure is the same they will both empty at the same time, given the outlets are equal in flow rate.
ReplyDeletebut tank X has a tapered bottom verses tank Y giving it a better volumetric efficiency. i'm thinking that tank tank X should empty slightly faster than tank Y. that's my 3rd grade education showing.
Yes but fluid pressure is based on height of water column above it. In tank x water pressure has higher head dur to shape of tank all the way to bottom. head in tank y falls off as height of column actually decreases as water is drained out( ie wider not deeper as tank drains)
ReplyDeleteFast is slow, slow is fast.
ReplyDeleteChutes Magoo
From what I remember the distance of the visible horizon is the square root of 3 multiplied by the height (in feet) divided by 2 with the answer in miles. I hope this helps
ReplyDeleteI think what you mean to say is the heat of the meat times the crack of the ass equals the angle of the...well, you know the rest...
DeleteComments sound like Vizzini explaining the poison switcheroo in The Princess Bride.
ReplyDeleteThe one filled with beer.
ReplyDeleteGrok says:
ReplyDeleteTank X will empty first.From the image, Tank X is the one shaped like a funnel pointing down — wide at the top, narrow at the bottom. Tank Y is the opposite — narrow at the top, wide at the bottom (like an upright cone or bell shape).Both hold the same volume (1000 liters) and have identical taps/outlets at the bottom, so the outflow starts with the same speed (by Torricelli's law, exit velocity depends only on the initial water height above the outlet, which appears similar here).The key difference is how the water level (height) drops as volume leaves:In Tank X (wide top → narrow bottom), most of the water is stored up high in the wider part. When water drains, the level drops relatively slowly at first (because you're removing volume from a large cross-sectional area). This keeps the height (and therefore pressure/head at the outlet) higher for longer during most of the draining process. Higher average head → higher average flow rate over time → empties faster overall.
In Tank Y (narrow top → wide bottom), most of the water is in the lower, wider part. As soon as draining starts, the level drops quickly at first (small cross-sectional area up top means small volume loss causes big height change). This quickly reduces the head/pressure, so the flow rate slows down earlier and stays lower for more of the process.
Mathematically, the time to empty is proportional to the integral of A(h) / √h dh over the height (where A(h) is cross-sectional area at height h). The shape where A(h) is smaller when h is larger (Tank X) results in less total time.Many physics discussions, simulations (e.g., some using tools like LS-DYNA), and calculations confirm Tank X empties first — often by a noticeable margin (up to ~2x faster in extreme cone cases, though less here since they're frustums).So the answer is Tank X (the one on the left in your image, wide opening at top). Tank Y will still have water left when X runs dry.
As I stated.
ReplyDelete