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Measuring pressure of water tank

by PA3040
Tags: measuring, pressure, tank, water
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PA3040
#1
May11-13, 05:32 AM
P: 10
Dear All

My water tank fixed on roof top 5M from the ground

Height of tank 1m
width of tank .75m ( Cylinder type water tank )
if water out from the tank by .5 inch hose

how can I measure the pressure at 1m height from the ground

Please advice
Thanks in advance
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HallsofIvy
#2
May11-13, 06:05 AM
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Quote Quote by PA3040 View Post
Dear All

My water tank fixed on roof top 5M from the ground

Height of tank 1m
width of tank .75m ( Cylinder type water tank )
if water out from the tank by .5 inch house
You mean ".5 inch hose", right? That puzzled me for a while.

The pressure on a surface is the weight on that surface divided by the area. Since any fluid, like water, exerts the same pressure in all directions, to find the pressure at the end of a hose .5 inches in diameter, of height h, you could calculate the volume, [itex]\pi(.5)^2h[/itex], then multiply by the density (the density of water is 1000 kg per cubic m or 62.3 pounds per cubic foot) and finally divide by the area of the end of the hose, [itex]\pi (.5)^2[/itex]. Of course the "[itex]\pi (.5)^2[/itex]" terms will just cancel out, leaving "density times height".

It seems odd that you give the diameter of the pipe in "inches" and the height in meters but, since only the height (5- 1= 4 meters) is relevant, use 1000 kg per cubic meter (approximately) as the density of water.
how can I measure the pressure at 1m height from the ground

Please advice
Thanks in advance
PA3040
#3
May11-13, 06:48 AM
P: 10
Thanks for the reply
how can I measure the pressure at 1m height from the ground
I need to measure the pressure of .5inch hose at 1m height from the ground. the hose is connected to the water tank,

Please also advice how measure the pressure when the thank water level going down

Thanks in advance

Integral
#4
May11-13, 11:31 AM
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Measuring pressure of water tank

Probably the best way is with a gauge.

It is hard to know the loses due to the hose. You know that the final pressure will be less then the ideal pressure given by ρgh. ρ is the density of water, g the acceleration due to gravity and h=4m.
PA3040
#5
May13-13, 03:50 AM
P: 10
Quote Quote by HallsofIvy View Post
You mean ".5 inch hose", right? That puzzled me for a while.

The pressure on a surface is the weight on that surface divided by the area. Since any fluid, like water, exerts the same pressure in all directions, to find the pressure at the end of a hose .5 inches in diameter, of height h, you could calculate the volume, [itex]\pi(.5)^2h[/itex], then multiply by the density (the density of water is 1000 kg per cubic m or 62.3 pounds per cubic foot) and finally divide by the area of the end of the hose, [itex]\pi (.5)^2[/itex]. Of course the "[itex]\pi (.5)^2[/itex]" terms will just cancel out, leaving "density times height".

It seems odd that you give the diameter of the pipe in "inches" and the height in meters but, since only the height (5- 1= 4 meters) is relevant, use 1000 kg per cubic meter (approximately) as the density of water.
Dear HallsofIvy

Could you please advice me according to my attached picture
How may I calculate the pressure location of Letter "A"

It could be much appreciated if you can sow me step by step

Thanks
Attached Thumbnails
WATER TANK1.JPG  
HallsofIvy
#6
May13-13, 08:27 AM
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The pressure is, as I said before, the total weight of the water above that point, divided by the area it is pressing against. Here, you have cylinder of water, 1 m in diameter, so .5 m in radius and 1 m high. That has volume [itex]\pi (.5)^2(1)= \pi/4= 0.7854 cubic meters. Water, at standard temperature and pressure, has a density of 1000 kg per cubic meter (that's essentially the definition of "kg") so that water has a mass of 785.4 kg and so a weight of (785.4)(9.81)= 7705 Newtons.

Your pipe is relatively so small the water in it adds only a very tiny amount to the weight of the water and can be ignored. However, the pipe has radius .5 inch so 1.25 cm= .0125 m and area [itex]\pi (.0125)^2[/itex]= 0.0004909[/itex] square meters. The pressure at A is 7705/.0004909= 15696497 Newtons per square meter.
PA3040
#7
May13-13, 11:24 AM
P: 10
Quote Quote by HallsofIvy View Post
The pressure is, as I said before, the total weight of the water above that point, divided by the area it is pressing against. Here, you have cylinder of water, 1 m in diameter, so .5 m in radius and 1 m high. That has volume [itex]\pi (.5)^2(1)= \pi/4= 0.7854 cubic meters. Water, at standard temperature and pressure, has a density of 1000 kg per cubic meter (that's essentially the definition of "kg") so that water has a mass of 785.4 kg and so a weight of (785.4)(9.81)= 7705 Newtons.

Your pipe is relatively so small the water in it adds only a very tiny amount to the weight of the water and can be ignored. However, the pipe has radius .5 inch so 1.25 cm= .0125 m and area [itex]\pi (.0125)^2[/itex]= 0.0004909[/itex] square meters. The pressure at A is 7705/.0004909= 15696497 Newtons per square meter.
Dear HallsofIvy
Thank you so much for details reply
Can you please tell me where did you get the value ( 9.81)
What is the meaning of (itex)


Please advice
PA3040
#8
May24-13, 09:40 PM
P: 10
Quote Quote by HallsofIvy View Post
The pressure is, as I said before, the total weight of the water above that point, divided by the area it is pressing against. Here, you have cylinder of water, 1 m in diameter, so .5 m in radius and 1 m high. That has volume [itex]\pi (.5)^2(1)= \pi/4= 0.7854 cubic meters. Water, at standard temperature and pressure, has a density of 1000 kg per cubic meter (that's essentially the definition of "kg") so that water has a mass of 785.4 kg and so a weight of (785.4)(9.81)= 7705 Newtons.

Your pipe is relatively so small the water in it adds only a very tiny amount to the weight of the water and can be ignored. However, the pipe has radius .5 inch so 1.25 cm= .0125 m and area [itex]\pi (.0125)^2[/itex]= 0.0004909[/itex] square meters. The pressure at A is 7705/.0004909= 15696497 Newtons per square meter.
Hi

I got 9.81 which Earth gravity and [itex] too.

I should think you not consider water tank fixed from 5m from the ground in above calculation right?
SteamKing
#9
May25-13, 12:43 AM
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I'm sorry, but the calculations for pressure at point A have taken a weird turn. According to Halls of Ivy, the pressure at point A is some 15.7 MPa, which is in excess of 2200 psi, or more than 150 atmospheres, all produced by a column of water 4 meters tall.

According to Pascal's law, the static pressure in the hose at point A is the fluid head or 4 meters of water. This is equivalent to a pressure of:

rho*g*h = 1000 kg/m^3 * 9.81 m/s^2 * 4 m = 39,240 N/m^2 = 39.24 kPa = 5.69 psi

which a hose or pipe should be able to withstand without blowing up and killing everyone standing nearby.

Remember, all of the water in the tank is not pressing on the hose at point A, just the column of water immediately above the point A in the hose and the tank itself.
PA3040
#10
May25-13, 03:27 AM
P: 10
Quote Quote by SteamKing View Post
I'm sorry, but the calculations for pressure at point A have taken a weird turn. According to Halls of Ivy, the pressure at point A is some 15.7 MPa, which is in excess of 2200 psi, or more than 150 atmospheres, all produced by a column of water 4 meters tall.

According to Pascal's law, the static pressure in the hose at point A is the fluid head or 4 meters of water. This is equivalent to a pressure of:

rho*g*h = 1000 kg/m^3 * 9.81 m/s^2 * 4 m = 39,240 N/m^2 = 39.24 kPa = 5.69 psi

which a hose or pipe should be able to withstand without blowing up and killing everyone standing nearby.

Remember, all of the water in the tank is not pressing on the hose at point A, just the column of water immediately above the point A in the hose and the tank itself.
Dear SteamKing Thanks for the reply

Actually my pressure measuring sensor measuring range is 1 to 100 kPa

http://www.freescale.com/files/senso...et/MPX5100.pdf

As per SteamKing I can implement the my project which is displaying water level in the LCD display

Please advice
zgozvrm
#11
Jun1-13, 08:43 AM
P: 754
The easy answer is:
231' of water (measured vertically) creates 100 psi, no matter what the shape of the hose or tank.
(Or, more simply, 2.31' of water equals 1 psi).
So, if your tank is half full (filled to the 0.5m level), that would mean that at 1m from the ground, you'd have a vertical
measurement of 4.5m of water (4 meters up to the bottom of the tank, plus 0.5m to the surface of the water).
4.5m is approximately 177.165 inches.
177.165 divided by 2.31 equals roughly 76.69 psi.

So, simply measure the height of the water in the tank in meters (0m to 1m), add 4 meters, and
divide by 2.31 to get the psi at a point 1m above the ground.

Of course if the tank is empty, you would simply measure the vertical height of the water in the hose alone (0m to 4m).


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