In the previous section, we saw the pressures in the horizontal direction. In this section, we will see hydrostatic paradox. Later in this section we will see gauge pressure and absolute pressure
Hydrostatic paradox can be described in 5 steps:
1. Consider the four vessels A, B, C and D in fig.10.12 below:
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| Fig.10.12 |
• They have different shapes
• All four are connected by a horizontal pipe at the bottom
2. Water (or any other liquid) poured in any one of the vessels, will reach all the other three vessels
• This is due to the interconnection through the horizontal pipe at the bottom
3. When the poured water settle down, we will see that:
• The level of water is same in all the vessels
• This is known a hydrostatic paradox
4. If some more water is poured, a new common level will be attained
5. The word ‘paradox’ is used because, different vessels hold different quantities of water. Even then, the levels are the same
A basic explanation can be written in 3 steps
1. We have Eq.10.1: Pa + 𝛒gz
• This equation gives the pressure at any depth z
2. We derived this equation using the three cylinders in fig.10.7 in a previous section
• The red, yellow and green cylinders have the same base area A
• But this ‘A’ is not coming in the final result
• What ever be the area, we will get the same Eq.10.1
• That means, the size or shape have no role in determining the pressure
3. Let us apply this equation to the vessels in fig.10.12 above
• We want the pressure at the bottom of each vessel
♦ The height of water is the same 'h' in all vessels
♦ So for all the vessels, z will be equal to h
♦ 𝛒 will be the same because, we are using the same liquid in all the vessels
• So at the bottom, pressure will be the same for all the vessels
In the above fig.10.12, different vessels hold different quantities of liquid. Yet, the pressure is same at the bottom. Upon seeing this situation, some questions arise in our minds. Those questions and their answers are given in the form of the solved example 10.6 below:
Solved example 10.6
Two vessels have the same base area but different shapes. The first vessel takes twice the volume of water that the second vessel requires to fill up to a particular common height. Is the force exerted by the water on the base of the vessel the same in the two cases ? If so, why do the vessels filled with water to that same height give different readings on a weighing scale ?
Solution:
• The solution has two parts: The analysis part and the actual solution part
• The analysis part helps to get a good understanding about the question. It can be written in 4 steps:
1. There are two vessels. Those two vessels satisfy three conditions:
(i) Both have the same base area
(ii) Both are filled water to the same height h
(iii) Volume of water in the first vessel = 2 × Volume of water in the second vessel
2. Let us try to make two vessels satisfying the above three conditions. It can be written in 6 steps:
(i) Consider fig.10.13(a) below:
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| Fig.10.13 |
• A vessel is filled with water upto a height h
♦ The vessel has a rectangular shape
♦ So the water also has a rectangular shape
✰ The rectangle representing water is shown in blue color
(ii) This blue rectangle is divided into two equal parts. This is shown in fig.b
• The division is done diagonally. So we get two equal triangles
♦ They are shown in yellow and magenta colors
(iii) The yellow triangle is flipped both vertically and horizontally
• Then it is attached to the left side of the blue rectangle
(iv) The magenta triangle is flipped horizontally
• Then it is attached to the right side of the blue triangle
(v) So the blue rectangle now has to triangles attached to it's sides
• The final shape is shown in fig.c
(vi) Now take the two vessels:
• The vessel in fig.a and the vessel in fig.c
■ When taken together, they will satisfy all the three conditions written in (1)
♦ They have the same water height h
♦ They have the same base area A
♦ Volume in one is twice than the other
3. We are asked this question:
Is the force exerted by the water on the base of the vessel the same in the two cases ?
• Analysis of this question can be written in 4 steps:
(i) Which is the force and where is it applied?
♦ The force is exerted by water
♦ The force is exerted at the base
(ii) Calculation of the force in first vessel:
• We have: Force = Pressure × Area
♦ So Force = 𝛒gz × A = 𝛒ghA
(iii) Calculation of the force in second vessel:
• We have: Force = Pressure × Area
♦ So Force = 𝛒gz × A = 𝛒ghA
(iv) So we conclude that, force is same in both the cases
4. We are asked this question:
If so, why do the vessels filled with water to that same height give different readings on a weighing scale ?
• Analysis of this question can be written in 5 steps:
(i) We are asked to compare the weights. So we put them on a weighing scale
This is shown in fig.10.14 below:
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| Fig.10.14 |
(ii) It is clear that, reading in fig.a will be greater
(iii) The weighing apparatus records the force exerted on it by the vessels
• The force is exerted through the base of the vessel
(iv). So a confusion arises:
♦ In (3), we see that: Forces at the base are the same
♦ But now we see that, weights are different
(v) We are asked to give an explanation for such a difference
■ The analysis part is complete. Now we write the solution part. It can be written in 5 steps:
1. Consider fig.10.15(a) below:
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| Fig.10.15 |
• The rectangular portion is in the middle
♦ It's weight is denoted as FR
• The triangular portions are on the sides
♦ Each of them has a weight denoted as FT
2. FR acts directly on the base of the vessel
• But FT acts on the sides of the vessel
♦ FT does not act directly on the base
3. We know that, the fluid pressure '𝛒gh' actually arises due to the weight of the fluid
• But here, the 'weight FT due to the triangular portions' does not act on the base
4. FT is first transmitted to the sides
• From the sides, they are transmitted to the base
• So the weighing apparatus will experience the total force: (FT + FR + FT)
5. In the case of the ordinary rectangular vessel in fig.10.13(a):
♦ the reading in the weighing apparatus
♦ will be same as
♦ the force due to hydrostatic pressure
We have solved the above problem. At this stage, it is better to see the basics of another problem also, which is closely related. This can be written in steps:
1. Consider the ordinary rectangular vessel that we saw earlier in fig.10.13(a)
• In the above solved example, two triangles were added to it’s sides
2. Now we remove two triangles
• This is shown in fig.10.15 (b) above
• The result is a conical vessel with same base area
• This is shown in fig.c
3. As before, we will be mislead by the resulting forces
♦ The force at the base of the rectangular vessel = 𝛒ghA
♦ The force at the base of the conical vessel = 𝛒ghA
• We get the same force because, base area and heights are same
4. The weighing apparatus will of course show different readings
• An explanation can be written in 3 steps:
(i) Consider the conical vessel in fig.10.15(c)
• The water will exert pressure in a direction perpendicular to the walls
• The walls are sloping inwards
• So the pressure will not be horizontal. It will be sloping. This is denoted by the two green arrows
(ii) So there will be a net upward force on the walls
• Due to the upward force, the weighing apparatus will not experience the full force (Accurate mathematical steps can be written using calculus. We will see them in higher classes)
(iii) We can write:
• For ordinary rectangular vessel:
♦ Reading in the weighing apparatus = 𝛒ghA
• For vessel with sides sloping outwards:
♦ Reading in the weighing apparatus = 𝛒ghA + weights of the two triangular portions
• For vessel with sides sloping inwards:
♦ Reading in the weighing apparatus = 𝛒ghA - weights of the two triangular portions
Let us write a summary of what we discussed so far in this section. It can be written in two steps:
1. In fig.10.12, we saw that, inter connected vessels will always have the same height of liquid
2. Using figs.10.13, 10.14 and 10.15, we gave a satisfactory explanation of how ‘same height’ is maintained even when shapes are different
Interconnected vessels have many practical applications. Here we will see one such application. It can be written in 20 steps:
1. Consider the water tank in fig.10.16(a) below
• The tank is shown in yellow color
• It is kept at an elevated position
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| Fig.10.16 |
2. Water is taken out from the bottom of that tank
• For taking the water out, a pipe of medium diameter is used
• This pipe is shown in red color
3. This red pipe goes down to below ground level
• A pipe of small diameter is taken out from the red pipe
• This small diameter pipe is shown in magenta color
4. If proper support is available, this magenta pipe can be taken vertically upwards to a large height
• Water will rise in the magenta pipe, up to the same height as in the tank
♦ This 'same height' is indicated by the white dotted line
♦ All houses which are below the white dotted line will get water supply
♦ But houses above the white dotted line will not get water supply
5. Now consider fig.b
• A tap is taken out from an intermediate point along the magenta pipe
♦ The tap is at a vertical distance of h from the water surface in the tank
• Now we recall a common phenomenon that we experience at our homes
♦ If the tap is in the first floor of the building, h will be less
✰ The water from the tap will be flowing at a low speed
✰ This indicates low pressure
♦ If the tap is in the ground floor of the building, h will be high
✰ The water from the tap will be flowing at a high speed
✰ This indicates high pressure
6. We want to know the exact pressure available at the tap
• We have Eq.10.1: Pa + 𝛒gz
• In our present case, z = h
• So the pressure available at the tap is equal to Pa + 𝛒gh
7. So we want the value of h
♦ The water tank may be far away from the tap
♦ So it may not be easy to measure h directly
■ In such situations, we use a device known as: The open tube manometer or U-tube manometer
• Let us see how it works. The following steps from (8) to (20) explains the working of a manometer
8. The manometer consists of a U-tube
• The left limb of the U-tube is attached to the tap. Then the tap is opened. This is shown in fig.10.17(a) below:
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| Fig.10.17 |
• The U-tube is shown in white color
9. The right limb of the U-tube do not have much height
• So when the tap is opened, water will flow out through the right limb
• In order to avoid such a ‘flow out’, we fill a high density fluid (such as mercury) on the right limb
10. This is shown in fig,b
♦ Fig.b shows the 'portion of the tap and the manometer' separately
• We see that, the column of mercury can effectively suppress the outward pressure from the tap
11. Now we mark the important points:
• Point A indicates the bottom tip of the water column in the left limb of the manometer
• We draw a green dotted line through A
• This dotted line intersects the mercury column in the right limb at B
• This dotted line is the datum
12. Next we measure the heights:
• Height of water (above datum) in the left limb = h1
• Height of mercury (above datum) in the right limb = h2
13. Next we write the pressures:
• The total pressure acting at the datum on the left side
= Atmospheric pressure + Pressure due to water of height h + Pressure due to water of height h1
= Pa + 𝛒wgh + 𝛒wgh1
• The total pressure acting at the datum on the right side
= Atmospheric pressure + Pressure due to mercury of height h2
= Pa + 𝛒mgh2
• Where:
♦ 𝛒w is the density of water
♦ 𝛒m is the density of mercury
14. At equilibrium, the two pressures will be equal. So we get:
Pa + 𝛒wgh + 𝛒wgh1 = Pa + 𝛒mgh2
• Pa is to be included in the left side because:
♦ The atmosphere is pushing down on the water in the tank
♦ The effect of that push will be felt at the tap
• Pa is included in the right side because:
♦ The atmosphere is pushing down at the top of the mercury column
♦ That push also contributes to suppress the water in the left limb
• Since Pa is present on both sides, they cancel out
Thus we get: 𝛒wgh + 𝛒wgh1 = 𝛒mgh2
⇒ 𝛒wgh = 𝛒mgh2 - 𝛒wgh1
⇒ 𝛒wh = 𝛒mh2 - 𝛒wh1
15. In the above equation, h is the only unknown. So it can be easily calculated
• Note that, h1 and h2 are the only readings that we have to take
• An example can be written in 4 steps:
(i) Let h1 = 0.3 m and h2 = 0.65 m
(ii) Substituting the values, we get:
1 × 103 × h = 13.6 × 103 × 0.65 - 1 × 103 × 0.3
⇒ h = 8.54 m
(iii) So the pressure available at the tap = Pa + 𝛒wgh = 1.01 × 105 + 1 × 103 × 9.8 × 8.54 = 184692 Pa
(iv) The height of a two storeyed building is approximately equal to 8.54 m
• So we can write:
The tap is at the ground floor and the tank is placed on top of the first floor
16. Using the equation in (14), we get h
• Once we get h, we can find 𝛒gh
• That means:
♦ Using just the two readings h1 and h2 from the manometer, we can calculate h
♦ And using that h, we can calculate 𝛒gh
17. Now consider Eq.10.1: Pa + 𝛒gh
• We see that 𝛒gh is one of the two components in Eq.10.1 (the other component being Pa)
• The component '𝛒gh' is easily obtained from the readings in the manometer
• The 'manometer' is a type of 'pressure gauge'
• So the 'manometer pressure gauge' readings are used to calculate 𝛒gh
■ So '𝛒gh' is called gauge pressure
• In Eq.10.1, we are actually adding this gauge pressure to the atmospheric pressure Pa
■ The sum thus obtained (Pa + 𝛒gh) is called absolute pressure
18. In the above discussion, we calculated the pressure in a system
• It is called a 'system' because, it consists of various components:
A water tank, proper supports to keep the tank at an elevated position, pipes of various diameters, tap etc.,
• These components work together for the proper functioning of the system
• Boilers, Distillation plants etc., are some other examples of system
♦ Many components work together harmoniously in such systems
• For simple problems involving manometers, we do not have to show the details of the system
• This can be explained in 3 steps:
(i) Consider the red circle in fig.10.18(a) below:
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| Fig.10.18 |
• This 'red circle and the blue liquid inside it' together denotes the system
(ii) Point A is the center of the circle
• We will be asked to find the pressure at A
(iii) Thus we represent a large system by a simple circle filled with fluid
19. Pressure at A can be calculated in 4 simple steps:
(i) Let P be the pressure at A
(ii) This pressure pushes the mercury down in the left limb
• So mercury rises in the right limb
(iii) The mercury column in the right limb, together with the atmospheric pressure, balances the pressures on the left side
(iv) From the fig., it is clear that:
P + 𝛒gh1 = Pa + 𝛒mgh2
⇒ P - Pa = 𝛒mgh2- 𝛒gh1
Where:
♦ 𝛒 is the density of the liquid in the system
♦ 𝛒m is the density of mercury
20. Consider the result in (19)
• On the left side, Pa is subtracted from the total pressure P
♦ We know that, when Pa is subtracted, the remaining pressure is the gauge pressure
✰ We saw this in step (17)
• For computing the right side, h1 and h2 are the only information we need
♦ We know that, h1 and h2 are simple manometer readings
• That means:
♦ The left side gives the gauge pressure
♦ The right side is calculated using manometer readings
■ Thus it is proved again that:
The manometer readings give gauge pressure
Now we will see a solved example:
Solved example 10.7
The right limb of a U-tube manometer containing mercury is open to the atmosphere. The left limb is connected to a system filled with a liquid of relative density 0.9. The level of mercury in the right limb is 12 cm above the center of the system. The difference in levels of mercury in the two limbs is 20 cm. Calculate the pressure at the center of the system
Solution:
1. The arrangement is shown in fig.10.18(b) above
• It is clear that h1 = 8 cm and h2 = 20 cm
2. The absolute pressure at the center point A will be given by: P = Pa + 𝛒mgh2- 𝛒gh1
• Substituting the values, we get:
P = 1.01 × 105 + 13.6 × 103 × 9.8 × 0.20 - 0.9 × 103 × 9.8 × 0.08 = 126950.4 Pa
In the next section, we will see some solved examples. We will also see barometer
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