__Fluid Statics__

__Fluid Statics__

In *ﬂuid statics *there is no relative motion between ﬂuid particles so there are no shear stresses present (a shear results from a velocity gradient). This does not mean that the ﬂuid particles are not moving, only that they are not moving relative to one another. If they are moving, as in a can of water rotating about its axis, they move as a solid body. The only stress involved in ﬂuid statics is the normal stress, the pressure. It is the pressure acting over an area that gives rise to the force. Three types of problems are presented in this chapter: ﬂuids at rest, as in the design of a dam; ﬂu- ids undergoing linear acceleration, as in a rocket; and ﬂuids that are rotating about an axis.

__2.1 Pressure Variation__

__2.1 Pressure Variation__

Pressure is a quantity that acts at a point. But, does it have the same magnitude in all directions at the point? To answer this question, consider Fig. 2.1. A pressure *p *is assumed to act on the hypotenuse and different pressures *p *and *p *on the other two sides of the inﬁnitesimal element which has a uniform depth *dz *into the paper.

**F****i****g****u****re 2.1 **Pressure acting on an inﬁnitesimal element.

The ﬂuid particle occupying the ﬂuid element could be accelerating so we apply Newton’s second law in both the *x- *and *y*-directions:

Here we see that the quantities on the right-hand sides of the equations are inﬁnitesimal (multiplied by *dx *and *dy*), and can be neglected2 so that Since the angle *b *is arbitrary, this holds for all angles. We could have selected dimensions *dx *and *dz *and arrived at *p**x *= *p**z *= *p*. So, in our applications to ﬂuid statics, the pressure is a scalar function that acts equally in all directions at a point.

In the preceding discussion, pressure at a point was considered. But, how does pressure vary from point to point? The ﬂuid element of depth *dy *in Fig. 2.2 is assumed to be accelerating. Newton’s second law provides

If the element were shown in the *y*-direction also, the *y*-component equation would be

**F****i****g****u****re 2.2 **Forces acting on an element of ﬂuid.

Finally, the pressure differential can be written as

This can be integrated to give the desired difference in pressure between speciﬁed points in a ﬂuid.

In a ﬂuid at rest, there is no acceleration so the pressure variation from Eq. (2.8) is

This implies that as the elevation *z *increases, the pressure decreases, a fact that we are aware of in nature; the pressure increases with depth in the ocean and decreases with height in the atmosphere.

If *g *is constant, Eq. (2.9) allows us to write

where Δ*p *is the pressure change over the elevation change Δ*z*. If we desire an expression for the pressure, a distance *h *below a free surface (the pressure is zero on a free surface), it would be

*p *= γ *h *2.11)

where *h *= −Δ*z*. Equation (2.11) is used to convert pressure to an equivalent height of a liquid; atmospheric pressure is often expressed as millimeters of mercury (the pressure at the bottom of a 30-in column of mercury is the same as the pressure at the earth’s surface due to the entire atmosphere).

If the pressure variation in the atmosphere is desired, Eq. (2.9) would be used with the ideal-gas law *p *= ρ*R**T *to give

where *p _{0 }*is the pressure at

*z*= 0. If the temperature could be assumed constant over the elevation change, this could be integrated to obtain

In the troposphere (between the earth’s surface and an elevation of about 10 km), the temperature in kelvins is *T *= 288 − 0.0065*z*; Eq. (2.12) can be integrated to give the pressure variation.

__EXAMPLE 2.1__

Convert 230 kPa to millimeters of mercury, inches of mercury, and feet of water.

__Solution__

Equation (2.11) is applied using the speciﬁc weight of mercury as 13.6γ _{water}:

We could have converted meters of mercury to feet of mercury and then multi- plied by 13.6 to obtain feet of water.

__2.2 Manometer____s__

__2.2 Manometer__

__s__

A *manometer *is an instrument that uses a column of a liquid to measure pressure. A typical U-tube manometer containing mercury is attached to a water pipe, as shown in Fig. 2.3. There are several ways to analyze a manometer. One way is to identify two points that have the same pressure, i.e., they are at the same elevation in the same liquid, such as points 2 and 3. Then we write

Since point 4 is shown to be open to the atmosphere, *p _{4 }*= 0. Thus, the manometer would measure the pressure

*p*in the pipe to be

_{1}**F****i****g****u****re 2.3 **A U-tube manometer using water and mercury.

Some manometers will have several ﬂuids with several interfaces. Each interface should be located with a point when analyzing a manometer.

**EXAMPLE 2.2**

A manometer connects an oil pipeline and a water pipeline, as shown. Determine the difference in pressure between the two pipelines using the readings on the manometer. Use *S*_{oil} = 0.86 and *S*_{Hg} = 13.6.

The points of interest have been positioned on the manometer shown. The pres- sure at point 2 is equal to the pressure at point 3:

Observe that the heights must be in meters. The pressure at point 4 is essentially the same as that at point 5 since the speciﬁc weight of air is negligible compared to that of the oil. So,