pressure, depth, and pascals Law



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If the weight of the fluid can be ignored, the pressure in a fluid is the same throughout its volume. We used that approximation in our discussion of bulk stress and strain in Section 11.4. But often the fluid’s weight is not negligible, and pressure variations are important. Atmospheric pressure is less at high altitude than at sea level, which is why airliner cabins have to be pressurized. When you dive into deep water, you can feel the increased pressure on your ears.

We can derive a relationship between the pressure p at any point in a fluid at rest and the elevation y of the point. We’ll assume that the density r has the same value throughout the fluid (that is, the density is uniform), as does the acceleration due to gravity g. If the fluid is in equilibrium, any thin element of the fluid with thickness dy is also in equilibrium (Fig. 12.4a). The bottom and top surfaces each have area A, and they are at elevations y and y+dy above some reference level where y=0. The fluid element has volume dV=A dy, mass dm=r dV=rA dy, and weight dw=dm g=rgA dy.

pressure, depth, and pascals Law

 

 

 

 

 

 

 

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What are the other forces on this fluid element (Fig 12.4b)? Let’s call the pressure at the bottom surface p; then the total y-component of upward force on this surface is pA. The pressure at the top surface is p+dp, and the total y-component of (downward) force on the top surface is -(p+dp2). The fluid element is in equilibrium, so the total y-component of force, including the weight and the forces at the bottom and top surfaces, must be zero:

ΣFy=0  so  pA+(p+dp)A-rgA dy=0

When we divide out the area A and rearrange, we get

pressure, depth, and pascals Law

 

 

 

 

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This equation shows that when y increases, p decreases; that is, as we move upward in the fluid, pressure decreases, as we expect. If p1 and p2 are the pressures at elevations y1 and y2 , respectively, and if r and g are constant, then

pressure, depth, and pascals Law

 

 

 

It’s often convenient to express Eq. (12.5) in terms of the depth below the surface of a fluid (Fig. 12.5). Take point 1 at any level in the fluid and let p rep-resent the pressure at this point. Take point 2 at the surface of the fluid, where the pressure is p0 (subscript zero for zero depth). The depth of point 1 below the surface is h=y2-y1 , 

  and Eq. (12.5) becomes 

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  p0-p=-rg1y2-y12=-rgh  or

 

pressure, depth, and pascals Law

 

 

 

The pressure p at a depth h is greater than the pressure p0 at the surface by an amount rgh. Note that the pressure is the same at any two points at the same level in the fluid. The shape of the container does not matter (Fig. 12.6).Equation (12.6) shows that if we increase the pressure p0 at the top surface, possibly by using a piston that fits tightly inside the container to push down on the fluid surface, the pressure p at any depth increases by exactly the same amount. This observation is called Pascal’s law.

pressure, depth, and pascals Law

 

 

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pressure, depth, and pascals Law

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