Fluid
Mass of unit volume, Called density
Density at a point of liquid described by
density is a positive scalar quantity
SI unit = Kg/m3
CGS unit = gm/cm3
Dimension = [ML-3]
It is the ratio of density of the given liquid to the density of pure water at 4°C
Relative density or specific gravity is unitless, dimensionless. It is a positive scalar physical Quantity.
Value of R.D. is same in SI and CGS system due to dimensionless/unitless
It is the ratio of the weight of given liquid to the weight of pure water at 4°C
Specific Gravity = = = Relative density of liquid i.e. than the specific gravity of a liquid is approximately equal to the relative density. For calculation, they can be interchanged
It is the pressure of the earth's atmosphere. This changes with weather and elevation. Normal atmospheric pressure at sea level (an average value) is 1.013 × 105 Pa. Thus
1 atm = 1.013 × 105 Pa
Note: Fluid pressure acts perpendicular to any surface in the fluid no matter how that surface is oriented. Hence, pressure has no intrinsic direction of its own, its a scalar. By contrast, force is a vector with a definite direction.
Following is the formula of Pascal law: F = PA
Where,
F is the force applied,
P is the pressure transmitted, and
A is the cross-sectional area.
Let's explore how Pascal's law works with an example.
Pascal's Law Derivation
Let ad, bd, and cd represent the area of the faces ABFE, ABDC, and CDFE, respectively.
Let P1, P2, and P3 denote the pressure acting on the faces ABFE, ABDC, and CDFE.
Pressure creates a force that acts perpendicular to the surface.
Let P1 create a force F1 on the surface ABFE.
Let P2 create a force F2 on the surface ABDC.
Let P3 create a force F3 on the surface CDFE.
Therefore, the forces can be expressed as:
Also,
Additionally, the total force on the prism will be zero because the prism is stable.
Thus, pressure increases linearly with depth, if r and g are uniform, A graph between P and h is shown below.
Further, the pressure is the same at any two points at the same level in the fluid. The shape of the container does not matter.
It is a device used to measure atmospheric pressure.
In principle, any liquid can be used to fill the barometer, but mercury is the substance of choice because its great density makes possible an instrument of reasonable size.
P1 = P2
Here, P1 = atompsheric pressure (P0)
and P2 = 0 + pgh = pgh
Here, p = density of mercury
p0 = pgh
Thus, the mercury barometer reads the atmosphereic pressure (P0) directly from the height of the mercury column.
For example if the height of mercury in a barometer is 760 mm, then atmospheric pressure will be,
P0 = rgh = (13.6 × 103) (9.8) (0.760)= 1.01 × 105 N/m2
Barometer
Force on the sidewall of the vessel can not be directly determined as at different depths pressures are different. To find this we consider a strip of width dx at a depth x from the surface of the liquid as shown in figure,
and on this strip the force due to the liquid is given as:
dF = xrg × bdx
This force is acting in the direction normal to the sidewall.
Net force can be evaluated by integrating equation
...(2.4)
The absolute pressure on the sidewall cannot be evaluated because at different depths on this wall pressure is different. The average pressure on the wall can be given as :
= ...(2.5)
Equation (2.5) shows that the average pressure on side vertical wall is half of the net pressure at the bottom of the vessel.
As shown in figure, due to the force dF, the sidewall experiences a torque about the bottom edge of the side which is given as
= xrgb dx (h -x)
This net torque is
It is a device used to measure the pressure of a gas inside a container.
Manometer
The U-shaped tube often contains mercury.
P1 = P2
Here, P1 = pressure of the gas in the container (P)
and P2 = atmospheric pressure (P0) + rgh
P = P0 hrg
This can also be written as
P -P0 = gauge pressure = hrg
Here, r is the density of the liquid used in U - tube
Thus by measuring h we can find absolute (or gauge) pressure in the vessel.
Example 1. Two liquid which do not react chemically are placed in a bent tube as shown in figure. Find out the displacement of the liquid in equilibrium position.
Sol. The pressure at the interface must be same, calculated via either tube. Since both tube all open to the atmosphere, we must have.
Example 2. Three liquid which do not react chemically are placed in a bent tube as shown in figure (initially) then fluid out the displacement of the liquid in equilibrium position.
Sol. Let us assume that level of liquid having density 3r displaced below by x as shown in figure below.
⇒
We've already discussed that when a liquid is filled in a container, generally its free surface remains horizontal as shown in figure (a) as for its equilibrium its free surface must be normal to gravity i.e. horizontal. Due to the same reason we said that pressure at every point of a liquid layer parallel to its free surface remains constant. Similar situation exist when liquid is in an accelerated frame as shown in figure (b). Due to acceleration of container, liquid filled in it experiences a pseudo force relative to container and due to this the free surface of liquid which normal to the gravity now is filled as:
...(2.22)
Now from equilibrium of liquid we can state that pressure at every point in a liquid layer parallel to the free surface (which is not horizontal), remains same for example if we find pressure at a point A in the accelerated container as shown in figure (a) is given as
Where h is the depth of the point A below the free surface of liquid along effective gravity and P0 is the atmospheric pressure acting on free surface of the liquid.
The pressure at point A can also obtained in an another way as shown in figure (b). If l1 and l2 are the vertical and horizontal distances of point A from the surface of liquid then pressure at point A can also be given as
Here l1rg is the pressure at A due to the vertical height of liquid above A and according to Pascal's Law pressure at A is given as
...(2.25)
Here we can write l1 as
or from equation (2.25)
Similarly if we consider the horizontal distance of point A from free surface of liquid, which is l2 then due to pseudo acceleration of container the pressure at point A is given as
...(2.26)
Here l2 is given as
From equation (2.24), we have
Here students should note that while evaluating pressure at point A from vertical direction we haven't mentioned any thing about pseudo acceleration as along vertical length l1, due to pseudo acceleration at every point pressure must be constant similarly in horizontal direction at every point due to gravity pressure remains constant.
Example 3. Figure shows a tube in which liquid is filled at the level. It is now rotated at an angular frequency w about an axis passing through arm A find out pressure difference at the liquid interfaces.
Sol. To solve the problem we take a small mass dm from the axis at `a' distance x in displaced condition.
Net inward force = (P +dP) A -PA
= AdP
This force is balanced by centripetal force in equilibrium
Example 4. A liquid of density r is in a bucket that spins with angular velocity w as shown in figure.
Show that the pressure at a radial distance r from the axis is
where P0 is the atmospheric pressure.
Sol. Consider a fluid particle P of mass m at coordinates
(x, y). From a non-inertial rotating frame of reference two forces are acting on it,
(i) pseudo force (mxw2)
(ii) weight (mg)
in the directions shown in figure.
The net force on it should be perpendicular to the free surface (in equilibrium). Hence.
or
This is the equation of the free surface of the liquid, which is a parabola.
97 videos|382 docs|103 tests
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1. What is atmospheric pressure and how is it measured? |
2. How does pressure vary with depth in a fluid? |
3. What is a manometer and how is it used to measure pressure? |
4. What is the force exerted on the side walls of a vessel containing fluid? |
5. How does pressure distribution behave in an accelerated frame? |
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