hydrostatic pressure and pressure at depth equation
In fluid statics or hydrostatics we study fluids at rest and we also study about pressure fluids exerts on a body immersed in fluids. From ordinary day to day observations we can see that an effort is required to immerse the hand in water. Again if you try to immerse a piece of wood in water, water offers a resistance to the immersion. This resistance arises due to the fluid pressure or hydrostatic pressure that acts on the surface of wood (or body) immersed in water.
When a fluid rests in contact with a rigid body, a mutual force of resistance is called into action at every point of the common surface of contact. Direction of force is normal to that surface of contact. [from Section I, Elementary Hydrostatics by J.B. Phear]
You can even try these activities at home. So it is a matter of ordinary observation that fluids are capable of exerting presure.
The pressure of a liquid on any surface with which it is in contact is perpandicular to the surface.[from Chapter 1, Elementary hydrostatics by W.H. Besant]
Above statement is also a fundamental property of fluids.
Fluid or hydrostatic pressure
When a fluid (either liquid or in gas) is at rest or in equilibrium, it exerts a force to any surface in contact with it, such as walls of a container. While fluid as a whole is at rest, the molecules that make up the fluid are in motion; the force exerted by the fluid is due to molecules colliding with their surroundings. To feel that liquid exerts pressure try to stop the liquid flowing out of a vessel by pressing finger again the hole. Your finger will experience an outward force due to liquid in the vessel. This clearly shows that liquid exerts pressure on the walls and base of the containing vessel.
What is thrust ?
Total force exerted by a liquid on any surface in contact with it is called thrust of liquid or fluid. It is because of this thrust that a liquid flows out through the holes of the containing vessel. Since thrust is a force, its SI unit is Newton (N).
Let us now define Hydrostatic pressure or fluid pressure.
Pressure at a point on a surface is the thrust acting normally per unit area around that point. If \(F\) is the thrust exerted by liquid on surface of area \(A\), then pressure is given by relation,
\[
\bbox[aqua,5px,border:2px solid red]{
P=\frac{F}{A}
}
\]
Again, if force is not distributed uniformly over the given surface, then pressure would be different at different points. If \( \Delta P\) is the force acting normally on a small area \( \Delta A \) surrounding a given point, then pressure at that point will be
\[
\bbox[aqua,5px,border:2px solid red]{
P=\mathop{\lim}_{\Delta A\rightarrow 0}\,\frac{\Delta F}{\Delta A}=\frac{dF}{dA}
}
\]
Hydrostatic pressure unit
Unit of pressure is \(dyne \, cm^{-2}\) in cgs system and \(Nm^{-2}\) in SI system.
A pressure of \(1\, Nm^{-2}\) is also called \(1 \, Pascal \,(Pa) \).
Dimension formula of pressure is \([ML^{-1}T^{-2}]\)
Pressure is a scholar quantity, because fluid pressure at a particular point in fluid has same magnitude in all direction. So definite direction is not associated with fluid pressure.
Other units of pressure are
Centimeter of mercury:- It is pressure exerted by \(1 cm\) of mercury column of any surface.
\( 1cm\, Hg = 1.33\times 10^3 \, N/m^2\)
Weather forecast in usually demonstrated in bar and milibar
\[1\, bar = 10^5 \, N/m^2\]
\[1\, milibar=10^2 \, N/m^2 \]\
\( \text{76 cm of Hg } = 1.013\, bar = 1013\, milibar =1\, atm\)
Or,
\(1atm=1.013 \times 10^5 \, Pa\)
Force applied per unit area on the surface in contact with atmosphere is known as atmospheric pressure.
\[\text{1 mm of Hg} = 1\, torr =\frac{1}{760}\, atm\]
\[
\rho_{mercury}=13.6 \times 10^3 \, kg/m^3 \]
\[ \rho_{\text{sea water}} = 1.03 \times 10^3 \, kg/m^3
\]
Practical applications of pressure
Paper pins and nails one made to have pointed ends due to which they have small area. When me apply force over the head of a pin, it will exert a large pressure on the surface and will penetrate easily into the surface.
Sharp knife cuts better than a blunt one. The area of a sharp edge is much less than the area of a blunt edge. Form the some force the effective force per unit area (or pressure) is more for the sharp edge than the blunt edge. Hence a sharp knife cuts better.
It is painful to walk on a road covered with pebbles having sharp edges. The weight of our body is supported on the pebbles under our feet. In turn, the pebbles given equal reaction to the feet. As the pebbles have very small area at their sharp edges, the presser exerted by them on the feet is very large and they prove painful.
Fluids in equilibrium
Consider the figure given below where a liquid is contained in a vessel in the equilibrium state at rest.
Suppose the liquid exerts a force \(F\) on the bottom surface in an inclined direction \( OP\). The surface then exerts an equal reaction \(R\) to water along \(OQ\).
Figure: Liquid in vessel is in equilibrium.
This reaction \(R\) along \(OQ\) has two rectangular components (i) Tangential component, \(OT=R \cos \theta\) (ii) Normal component, \(OS=R \sin \theta\).
As mentioned earlier liquids yields to any tangential force, so the liquid near \(O\) should begin to flow along \(OT\) .But the liquid is at rest, the force along \(OT\) must be zero.
Therefore, \(R \cos \theta = 0\)
\[
as \,\, R\ne 0 \,\, so, \,\, \cos \theta =0 \,\, or, \,\, \theta =90^{\circ}
\]
Hence a liquid always exerts Force \(\bot ^{r} \) to the surface of container at every point.
Fluid pressure equation or pressure at depth equation
When we go down in a lake or ocean pressure in lake or ocean increases. Similarly when we go up at greater attitudes atmospheric pressure decreases with increase in attitude. This variation of pressure in fluids with depth or height can be derived easily.
Consider a large vessel filled with liquid such as water. Consider two points \(R\) and \(Q\) separated by a small vertical height \(\Delta h\) where \(h\) is the depth measured from the top of the liquid.
Figure: Variation of fluid pressure with depth.
Again imagine horizontal area \(\Delta \text{A}\) containing points \(R\) and \(Q\) shown in the figure. Join the boundaries of the areas enclosing \(R\) and \(Q\) and consider it as a slab of liquid.
Let \(P_1\) and \(P_2\) be the pressure at points \(R\) and \(Q\) respectively. Now vertical forces acting on this fluid are
(a) \(P_1 \Delta A \rightarrow\) is the force acting downward by fluid above the slab. (b) \(P_2 \Delta A \rightarrow\) is the force exerted on the slab in upward direction by fluid below the slab. (c) Weight \(W\) of the slab in the vertically downward direction.
Since the imaginary slab in the liquid is in equilibrium, the net force on it must be zero. Thus,
\begin{equation}
P_1\Delta A + W = P_2\Delta A
\end{equation}
Where, \(W\) is the weight of the slab. If \(m\) is the mass of the slab then \(W=mg\). If \(rho\) is the density of the fluid then
\[
m=\rho \Delta A\Delta h\\
\Rightarrow \text{weight W}=\left( \rho \Delta A\Delta h \right) g\\
\]
Putting it in equation (1) we get
\[
\left( P_2-P_1 \right) \Delta A=\rho g\left( \Delta h \right) \Delta A
\]
or,
\begin{equation}
\Delta P=\rho g\Delta h
\end{equation}
Equation (2) gives pressure difference corresponding to depth difference \(\Delta h\) between points \(R\) and \(Q\) in the fluid. From equation (2) we have
\[\frac{\Delta P}{\Delta h}=\rho g\]
If we take limit \(\Delta h\rightarrow 0\) then \(\frac{\Delta P}{\Delta h}\) becomes derivative \(\frac{dP}{dh}\). Thus,
\begin{equation}
\frac{dP}{dh}=\rho g
\end{equation}
And this equation holds for any fluid.
If the liquid under consideration is incompressible i.e., density of liquid is approximately independent of pressure which is the case with most of the liquids. For ex- water, alcohol etc. then,
\[\frac{dP}{dh}=\rho g=constant\]
or,
\[ dP\propto dh \]
This proportionality shows that pressure various linearly with \(h\). integrating above equation we have,
\[\int_{P_0}^P{dP=\rho g\int_0^h{dh}}\]
Where \(P_0\) is pressure at height \(h=0\). On integrating we have,
\begin{equation}
P=P_0+\rho gh
\end{equation}
From this equation (4) we conclude that pressure at a point in a fluid in static equilibrium depends
on the vertical depth of the point and on the density of fluid under consideration. It does not depend on the shape and size of the container.
Again if we shift we point \(R\) in discussion to top of the fluid surface which is open to the atmosphere then in equation (4) \(P_0\) may be replaced by \(P_a\) , which is atmospheric pressure. Thus,
\[ P=P_a + \rho gh \]
Thus pressure \(P\) at a depth h below the surface of liquid open to atmosphere is greater than atmospheric pressure by an amount \(\rho gh\). This excess pressure at depth \(h\) is called gauge pressure at that point. Note:- Results derived in this section is not valid in cases of gases when \(h\) is large. The density of gases decreases with height. Theis result has been derived on the assumption that density is constant. This equation is an reasonable approximation for small \(h\). Densities of gases are low and for small h pressure variation with depth is also small.
Characteristics of fluid pressure or hydrostatic pressure
(i) The fluid or hydrostatic pressure is the same at all points at the same horizontal level or at same depth when the liquid is at rest. (ii) Liquids at rest exerts lateral pressure, which increases with depth. (iii) Pressure acts normally on any area in whatever orientation the area may be held. (iv) Free surface of a liquid at rest remains horizontal.
