Measuring pressure of water tank

by PA3040
Tags: measuring, pressure, tank, water
 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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P: 39,491
 Quote by PA3040 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, $\pi(.5)^2h$, 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, $\pi (.5)^2$. Of course the "$\pi (.5)^2$" 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
P: 10
 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

 Mentor P: 7,318 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.
P: 10
 Quote by HallsofIvy 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, $\pi(.5)^2h$, 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, $\pi (.5)^2$. Of course the "$\pi (.5)^2$" 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

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

 Math Emeritus Sci Advisor Thanks PF Gold P: 39,491 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 $\pi (.5)^2(1)= \pi/4= 0.7854 cubic meters. Water, at standard temperature and pressure, has a density of 1000 kg per cubic meter (that's essentially the definition of "kg") so that water has a mass of 785.4 kg and so a weight of (785.4)(9.81)= 7705 Newtons. Your pipe is relatively so small the water in it adds only a very tiny amount to the weight of the water and can be ignored. However, the pipe has radius .5 inch so 1.25 cm= .0125 m and area [itex]\pi (.0125)^2$= 0.0004909[/itex] square meters. The pressure at A is 7705/.0004909= 15696497 Newtons per square meter.
P: 10
 Quote by HallsofIvy 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 $\pi (.5)^2(1)= \pi/4= 0.7854 cubic meters. Water, at standard temperature and pressure, has a density of 1000 kg per cubic meter (that's essentially the definition of "kg") so that water has a mass of 785.4 kg and so a weight of (785.4)(9.81)= 7705 Newtons. Your pipe is relatively so small the water in it adds only a very tiny amount to the weight of the water and can be ignored. However, the pipe has radius .5 inch so 1.25 cm= .0125 m and area [itex]\pi (.0125)^2$= 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)

P: 10
 Quote by HallsofIvy 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 $\pi (.5)^2(1)= \pi/4= 0.7854 cubic meters. Water, at standard temperature and pressure, has a density of 1000 kg per cubic meter (that's essentially the definition of "kg") so that water has a mass of 785.4 kg and so a weight of (785.4)(9.81)= 7705 Newtons. Your pipe is relatively so small the water in it adds only a very tiny amount to the weight of the water and can be ignored. However, the pipe has radius .5 inch so 1.25 cm= .0125 m and area [itex]\pi (.0125)^2$= 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?
 Emeritus Sci Advisor HW Helper Thanks PF Gold P: 6,471 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.
P: 10
 Quote by SteamKing 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