Category: Fluid Mechanics

The cross section of a pipe is most frequently circular, but other shapes may be encountered. For example, the rectangular cross section of many domestic hotair heating ducts should be apparent to most people living in the United States. The situation for a horizontal duct is illustrated in Fig. 3.14; the cross-sectional shape is quite arbitrary—it doesn’t have to be rectangular as shown—as long as it is uniform at all locations. There, A is the cross-sectional area and P is the wetted perimeter—defined as the length of wall that is actually in contact with the fluid. For the flow of a gas, P will always be the length of the complete periphery of the duct; for liquids, however, it will be somewhat less than the periphery if the liquid has a free surface and incompletely fills the total cross section.

Fig. 3.14 Flow in a duct of noncircular cross section.

If τ_w is the wall shear stress and there is a pressure drop –Δp over a length L, a momentum balance in the direction of flow yields:

The pressure drop is therefore:

Thus, the equation for the pressure drop is identical with that of Eqn. (3.33) for a circular pipe provided that D is replaced by the hydraulic mean diameter D_e, defined by:

The reader may wish to check that D_e = D for a circular duct. Following similar lines as those used previously, the frictional dissipation per unit mass can be deduced as:

and this expression can then be employed for inclined ducts of noncircular cross section.

Steady flow in open channels. A similar treatment follows for a liquid flowing steadily down a channel inclined at an angle θ to the horizontal, such as a river or irrigation ditch, shown in Fig. 3.15. Again, as long as the cross section is uniform along the channel, it can be quite arbitrary in shape, not necessarily rectangular. The driving force is now gravity, there being no variation of pressure because the free surface is uniformly exposed to the atmosphere.

Fig. 3.15 Flow in an open channel.

If the wetted perimeter is again P and the cross-sectional area occupied by the liquid is A, a steady-state momentum balance in the direction of flow gives:

Noting that:

division of Eqn. (3.56) by –ρA gives:

in which the second term can be rearranged as:

Comparison with the overall energy balance:

gives the frictional dissipation per unit mass as:

which has exactly the same form as Eqns. (3.54) and (3.55).

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Example 3.6—Flow in an Irrigation Ditch

The irrigation ditch shown in Fig. E3.6 has a cross section that is 6 ft wide × 6 ft deep. It conveys water from location 1 to location 2, between which there is a certain drop in elevation.

Fig. E3.6 Cross section of irrigation ditch.

With a flow rate of Q = 72 ft³/s of water, the ditch is filled to a depth of 4 ft. If the same ditch, transporting water between the same two locations, were completely filled to a depth of 6 ft, by what percentage would the flow rate increase? Start by applying the overall energy balance between points 1 and 2, and assume that the friction factor remains constant.

Solution

Apply the energy equation between two points separated by a distance L:

or,

Whether the ditch is filled to 4 ft or 6 ft, all quantities in Eqn. (E3.6.1) remain the same except u_m and D_e, so that:

in which c is a factor that incorporates everything that remains constant.

Case 1. When the ditch is filled to a depth of only 4 ft, the hydraulic mean diameter is:

But the mean velocity is:

so that the value of the constant in Eqn. (E3.6.2) is:

Case 2. When the ditch is filled to a depth of 6 ft, the hydraulic mean diameter is:

From Eqns. (E3.6.2) and (E3.6.5), the mean velocity is:

so that the flow rate is now:

The percentage increase in flow rate is therefore:

Thus, the increase in flow rate is somewhat more than the 50% increase in the depth of water. The reason is that the increased area for flow is accompanied by a somewhat lower increase in the length of the wetted perimeter.

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Pressure drop across pipe fittings. A variety of auxiliary hardware such as valves and elbows is associated with most piping installations. These fittings invariably cause the flow to deviate from its normal straight course and hence induce additional turbulence and frictional dissipation. Indeed, the resulting additional pressure drop is sometimes comparable to that in the pipeline itself.

The basic procedure is to recognize that the fitting causes an additional pressure drop that would be produced by a certain length of pipe into which the fitting is introduced. Therefore, we substitute for the fitting an extra contribution to the length of the pipe, based on the equivalent length (L/D)_e of the fitting. For example, referring to Table 3.4, three standard 90° elbows in a 6–in.-diameter line cause a pressure drop that is equivalent to an extra 45 ft of pipe.

Table 3.4 Equivalent Lengths of Pipe Fittings^9,10

⁹ See for example, G.G. Brown et al., Unit Operations, Wiley & Sons, New York, 1950, p. 140.

¹⁰ For sudden expansions and contractions, the diameter ratio is given in Table 3.4; also, the equivalent length in these cases is based on the smaller diameter.

The gate valve uses a retractable circular plate that normally has one of two extreme positions: (a) complete obstruction of the flow, or (b) essentially no obstruction. The gate valve cannot be used for fine control of the flow rate, for which the globe or needle valve, with an adjustable plug or needle partly obstructing a smaller orifice, is more effective.

Laminar and turbulent velocity profiles. The parabolic velocity profile already encountered in laminar flow in a pipe is again illustrated on the left of Fig. 3.16. On the right, we see for the first time the general shape of the velocity profile for turbulent flow. Chapter 9 shows that although in turbulent flow the velocities exhibit random fluctuations, it is still possible to work in terms of a time-averaged axial velocity. For simplicity at this stage, we shall still use u to denote such a quantity, although in Chapter 9 it will be replaced with a symbol such as .

Fig. 3.16 Laminar and turbulent velocity profiles.

In addition to the generally higher velocities, note that the overall turbulent velocity profile consists of two profoundly different zones:

1. A very thin region, known as the laminar sublayer, in which turbulent effects are essentially absent, the shear stress is virtually constant, and there is an extremely steep velocity gradient. The following equation is derived in Section 9.7 for the thickness δ of the laminar sublayer relative to the pipe diameter D as a function of the Reynolds number in the pipe:

Observe that the laminar sublayer becomes thinner as the Reynolds number increases, because the greater intensity of turbulence extends closer to the wall.

2. A turbulent core, which extends over nearly the whole cross section of the pipe. Here, the velocity profile is relatively flat because rapid turbulent radial momentum transfer tends to “iron out” any differences in velocity. A representative equation for the ratio of the velocity u at a distance y from the wall to the centerline velocity u_c is:

in which n in the exponent varies somewhat with the Reynolds number, as in Table 3.5. For n = 1/7, Eqn. (3.63) is plotted in Fig. 9.11.

Table 3.5 Exponent n for Equation (3.63)¹¹

¹¹ H. Schlichting, Boundary-Layer Theory, McGraw-Hill, New York, 1955, p. 403.

Newton’s law of viscosity relating the shear stress to the velocity gradient has a ready interpretation based on momentum transport resulting from molecular diffusion. As an introduction, consider first the situation in Fig. 3.7, which shows a plan of two trolleys on frictionless tracks.

Fig. 3.7 Lateral transport of momentum between two trolleys on frictionless tracks.

A person of mass M jumps from trolley A, which is moving to the right with velocity u, onto trolley B, which is hitherto stationary. The person clearly transports an amount of momentum M = Mu from A to B, with the result that B accelerates; note that the momentum is in the x direction, but that it is transported in the transverse or y direction. If the transfer were continuous, with several people jumping successively—assuming that the trolley is large enough to accommodate them—the net effect would be a steady force in the x direction exerted by A on B, analogous to a shear force.

Momentum transport in laminar flow. A similar phenomenon occurs for laminar flow, except that the lateral transport is now due to random molecular movement, known as Brownian motion.³

³ G. Stix, Scientific American, Vol. 291, No. 3, September 2004, p. 46, says: “[In 1905] Albert Einstein published a paper in Annalen, ‘On the motion of small particles suspended in liquids at rest required by the molecular-kinetic theory of heat,’ that supplied a prediction of the number and mass of molecules in a given volume of liquid—and how these molecules would flit around. The erratic movements were known as Brownian motion, after the observation by botanist Robert Brown in 1827 of the irregular zigs and zags of particles inside pollen grains in water. Einstein suggested that the movements of the water molecules would be so great that they would jostle suspended particles, a dance that could be witnessed under a microscope.”

Fig. 3.8 shows part of the velocity profile for flow in the axial or z direction of a pipe. Consider a plane C–C at any radial location. Because of Brownian motion, molecules such as those at A and B will constantly be crossing C–C from below and above, at a mass rate m per unit area per unit time. A molecule such as B, whose z component of velocity is u_B, will bring down with it a slightly larger amount of z momentum than that taken upward by a molecule such as A, whose z component of velocity is only u_A. For a velocity profile as shown, there is a net positive rate of transfer of z momentum downward across C–C, and this is manifest as a shear stress given by:

Fig. 3.8 Molecular diffusion model for shear stress in laminar flow.

On the average, the molecules will travel a small distance δ, known as their mean free path, before they collide with other molecules and surrender their momentum. The velocity discrepancy u_B – u_A is therefore the product of δ and the local velocity gradient, yielding Newton’s law of viscosity:

where μ = mδ is the viscosity of the fluid. Since both m and δ are independent of the flow rate in the z direction, the viscosity is constant, independent of the bulk motion. Note that we now have a physical basis for the similar result of (3.4), in which the minus sign merely reflects an alternative viewpoint for the direction of the shear stress.

Newton, Sir Isaac, born 1642 in Lincolnshire, England; died 1727 in Kensington, buried in Westminster Abbey, London. He was an unsur-passed scientific genius. After a brief interruption of his education to help with the family farm, he entered Trinity College Cambridge as a student in 1661, obtained his B.A. degree in mathematics in 1665, and was elected a fellow of his college in 1667. Newton’s research in optics embraced reflection, refraction, and polarization of light, and he invented a reflecting telescope. About 1666, he deduced from Kepler’s laws of planetary motion that the gravitational attraction between the sun and a planet must vary inversely as the square of the distance between them. Newton was appointed Lucasian professor of mathematics in 1669, and elected a fellow of the Royal Society in 1671. His magnum opus was Philosophiæ Naturalis Principia Mathematica; published in three volumes in 1687, “Principia” included the laws of mechanics, celestial motions, hydrodynamics, wave motion, and tides. At this stage, despite all of his accomplishments, Newton had neither been rewarded monetarily nor with a position of national prominence. However, matters improved when he was appointed Master of the Royal Mint in London in 1699, a position he held with distinction until his death.

Source: The Encyclopædia Britannica, 11th ed., Cambridge University Press (1910–1911).

For gases, the mean free path can be predicted from kinetic theory, leading to the following theoretical expression for the viscosity, which is found to agree well in most cases with experiment, provided the pressure is less than approximately 10 atm:

Here, r is the effective radius of a gas molecule and m is its mass; k is Boltzmann’s constant (1.380 × 10 ^–16 ergs/molecule K) and T is the absolute temperature. Note the predictions that: (a) μ is independent of pressure, and (b) μ rises with increasing temperature.

For liquids, there is no simple corresponding expression for their viscosities. In general, however—and in contrast to gases—the viscosity of liquids falls with increasing temperature. Note also the empirical correlations of Eqns. (1.17) and (1.18) for the viscosities of liquids and gases, which are based on experimental observations.

Momentum transport in turbulent flow. An analogous situation holds for turbulent flow. However, the random molecular motion is now substantially augmented by a turbulent eddy motion, which is on a much larger scale. Turbulent flow can be described by the concept of the eddy kinematic viscosity, ν_T, giving:

it being noted—from experiment—that ν_T ≫ ν. The eddy kinematic viscosity ν_T is not constant, but is found experimentally to be approximately directly proportional to u_mD, where D is the pipe diameter; this result is also plausible on qualitative grounds—an increase in u_m causes more turbulence and a larger D permits eddies to travel further. The momentum transport is also proportional to the fluid density. Since du/dr is also roughly proportional to u_m (a doubling of u_m also approximately doubles the velocity gradient at any position), and inversely proportional to D, it follows that:

in which c is a constant, as yet unknown, and which depends on the particular radial location we are examining. A more detailed discussion of the eddy viscosity

3.1. Introduction

In chemical engineering process operations, fluids are typically conveyed through pipelines, in which viscous action—with or without accompanying turbulence—leads to “friction” and a dissipation of useful work into heat. Such friction is normally overcome either by means of the pressure generated by a pump or by the fluid falling under gravity from a higher to a lower elevation. In both instances, it is usually necessary to know what flow rate and velocity can be expected for a given driving force. This topic will now be discussed.

Fig. 3.1 shows a pipe fitted with pressure gauges that record the pressures p₁ and p₂ at the beginning and end of a test section of length L. A horizontal pipe is intentionally chosen because the observations are not then complicated by the effect of gravity. In addition, for good results, it is desirable that a substantial length of straight pipe should precede the first pressure gauge, in order that the flow pattern is fully developed (no longer varying with distance along the pipe) at that location.

Fig. 3.1 Pressure drop in a horizontal pipe.

Reynolds, Osborne, born 1842 in Belfast, Ireland; died 1912 in Somerset, England. Born into an Anglican clerical family, Reynolds entered an apprenticeship in 1861 with Edward Hayes, a mechanical engineer, before obtaining a degree in mathematics at Cambridge in 1867. After brief employment as a civil engineer, he competed successfully the next year for a newly created professorship at Owens College, Manchester, holding this position for the next 37 years. He worked on a wide range of topics in engineering and physics. He demonstrated the significance of the dimensionless group that now bears his name in his paper, “An experimental investigation of the circumstances which determine whether the motion of water in parallel channels shall be direct or sinuous and of the law of resistance in parallel channels,” Philosophical Transactions of the Royal Society, Vol. 174, pp. 935–982 (1883). His analogy between heat and momentum transport (see Chapter 9) was published in his paper “On the extent and action of the heating of steam boilers,” Proceedings of the Manchester Literary and Philosophical Society, Vol. 14, p. 8 (1874–1875). In 1885, he attributed the name dilatancy to the ability of closely packed granules to increase the volume of their interstices (the void fraction) when disturbed. He was elected a Fellow of the Royal Society in 1877.

Source: Dictionary of Scientific Biography, Charles Scribner’s Sons, New York, 1975.

For a given flow rate, repetition of the experiment for different lengths demonstrates that the pressure drop (p₁ – p₂) is directly proportional to L. Hence, it is appropriate to plot the pressure drop per unit length (p₁ – p₂)/L (the negative of the pressure gradient dp/dz, where z denotes distance along the pipe) against the volumetric flow rate Q. There are three distinct flow regimes in the resulting graph:

1. For flow rates that are low (in a sense to be defined shortly), the pressure gradient is directly proportional to the flow rate.

2. For intermediate flow rates, the results are irreproducible, and alternate seemingly randomly between extensions of regimes 1 and 3.

3. For high flow rates, the pressure gradient is closely proportional to the square of the flow rate.

These regimes are known as the laminar, transition, and turbulent zones, respectively. The situation is further illuminated by the famous 1883 experiment of Sir Osborne Reynolds (see box above), similar to that illustrated in Fig. 3.2, where a liquid flows in a transparent tube. A fine steady filament of a dye is introduced by a hypodermic needle into the center of the flowing liquid stream, care being taken to ensure that there is no instability due to an imbalance of velocities. (For gases, the flow can be visualized by injecting a filament of smoke, such as kerosene vapor.) Again, three distinct flow regimes are found, which correspond exactly to those already encountered above:

1. For low flow rates, Fig. 3.2(a), the injected dye jet maintains its integrity as a long filament that travels along with the liquid. (The jet actually broadens gradually, due to diffusion.)

2. For intermediate flow rates, the results are irreproducible, and seem to alternate between extensions of regimes 1 and 3.

3. For high flow rates, Fig. 3.2(b), the jet of dye mixes very rapidly with the surrounding liquid and becomes highly diluted, so that it soon becomes invisible. The reason is that the liquid flow in the pipe is unstable, consisting of random turbulent motions superimposed on the bulk flow to the right.

Fig. 3.2 The Reynolds experiment.

Further experiments show that the three regimes do not depend solely on the flow rate, but on a dimensionless combination of the mean fluid velocity u_m, its density ρ and viscosity μ, and the diameter D of the pipe. The combination or dimensionless group is defined by:

and is called the Reynolds number, and indicates the relative importance of inertial effects (as measured by —see Eqn. (3.26), for example) to viscous effects (μu_m/D). Table 3.1 shows which regime can be expected for a given Reynolds number. The exponent on Q is 1.8 or 2, depending on whether the pipe is hydraulically smooth or rough, respectively, in a sense to be defined later. In Sections 3.2 and 3.3 we shall study flow in the laminar and turbulent regimes more closely.

Table 3.1 Dependence of Pipe Flow Regime on the Reynolds Number

3.2. Laminar Flow

In order to avoid the additional complication of gravity (which will be included later), consider flow in the horizontal cylindrical pipe of radius a shown in Fig. 3.3.

Fig. 3.3 Forces acting on a cylindrical fluid element.

Consider further a moving cylinder of fluid of radius r and length L. In this case, there is zero convective transport of momentum across the two circular ends of the cylinder, and the analysis is simplified.¹ Because of the retarding action of the pipe wall, there will be a shear stress τ exerted to the left on the curved surface of the cylinder by the fluid between it and the pipe wall.² The net pressure force acting on the circular area πr² of the two ends is exactly counterbalanced by the shear stress acting on the curved surface, of area 2πrL.

Momentum. The general conservation law also applies to momentum M, which for a mass M moving with a velocity u, as in Fig. 2.13(a), is defined by:

Fig. 2.13 (a) Momentum as a product of mass and velocity; (b) velocity components in the three coordinate directions.

Strangely, there is no universally accepted symbol for momentum. In this text and elsewhere, the symbols M and m (or ) frequently denote mass and mass flow rate, respectively. Therefore, we arbitrarily denote momentum and the rate of transfer of momentum due to flow by the symbols M (“script” M) and , respectively. Momentum is a vector quantity, and for the simple case shown has the direction of the velocity u. More generally, there may be velocity components u_x, u_y, and u_z (sometimes also written as υ_x, υ_y, and υ_z, or as u, υ, and w) in each of the three coordinate directions, illustrated for Cartesian coordinates in Fig. 2.13(b). In this case, the momentum of the mass M has components Mu_x, Mu_y, and Mu_z in the x, y, and z directions.

For problems in more than one dimension, conservation of momentum or a momentum balance applies in each of the coordinate directions. For example, for the basketball shown in Fig. 2.14, momentum Mu_x in the x direction remains almost constant (drag due to the air would reduce it slightly), whereas the upward momentum Mu_z is constantly diminished—and eventually reversed in sign—by the downward gravitational force.

Fig. 2.14 Momentum varies with time in the x and y directions.

If a system, such as a river, consists of several parts each moving with different velocities u (boldface denotes a vector quantity), the total momentum of the system is obtained by integrating over all of its mass:

Law of momentum conservation. Following the usual law, Eqn. (2.2), the net rate of transfer of momentum into a system equals the rate of increase of the momentum of the system. The question immediately arises: “How can momentum be transferred?” The answer is that there are two principal modes, by a force and by convection, as follows in more detail.

1. Momentum Transfer by a Force

A force is readily seen to be equivalent to a rate of transfer of momentum by examining its dimensions:

In fluid mechanics, the most frequently occurring forces are those due to pressure (which acts normal to a surface), shear stress (which acts tangentially to a surface), and gravity (which acts vertically downwards). Pressure and stress are examples of contact forces, since they occur over some region of contact with the surroundings of the system. Gravity is also known as a body force, since it acts throughout a system.

A momentum balance can be applied to a mass M falling with instantaneous velocity u under gravity in air that offers negligible resistance. Considering momentum as positive downwards, the rate of transfer of momentum to the system (the mass M) is the gravitational force Mg, and is equated to the rate of increase of downwards momentum of the mass, giving:

Note that M can be taken outside the derivative only if the mass is constant. The acceleration is therefore:

and, although this is a familiar result, it nevertheless follows directly from the principle of conservation of momentum.

Another example is provided by the steady flow of a fluid in a pipe of length L and diameter D, shown in Fig. 2.15. The upstream pressure p₁ exceeds the downstream pressure p₂ and thereby provides a driving force for flow from left to right. However, the shear stress τ_ω exerted by the wall on the fluid tends to retard the motion.

Fig. 2.15 Forces acting on fluid flowing in a pipe.

A steady-state momentum balance to the right (note that a direction must be specified) is next performed. As the system, choose for simplicity a cylinder that is moving with the fluid, since it avoids the necessity of considering flows entering and leaving the system. The result is:

Here, the first term is the rate of addition of momentum to the system resulting from the net pressure difference p₁– p₂, which acts on the circular area πD²/4. The second term is the rate of subtraction of momentum from the system by the wall shear stress, which acts to the left on the cylindrical area πDL. Since the flow is steady, there is no change of momentum of the system with time, and dM/dt = 0. Simplification gives:

which is an important equation from which the wall shear stress can be obtained from the pressure drop (which is easy to determine experimentally) independently of any constitutive equation relating the stress to a velocity gradient.

2. Momentum Transfer by Convection or Flow

The convective transfer of momentum by flow is more subtle, but can be appreciated with reference to Fig. 2.16, in which water from a hose of cross-sectional area A impinges with velocity u on the far side of a trolley of mass M with (for simplicity) frictionless wheels. The dotted box delineates a stationary system within which the momentum Mv is increasing to the right, because the trolley clearly tends to accelerate in that direction. The reason is that momentum is being transferred across the surface BC into the system by the convective action of the jet. The rate of transfer is the mass flow rate m = ρuA times the velocity u, namely:

Fig. 2.16 Convection of momentum.

which is again momentum per unit time.

The acceleration of the trolley will now be found by applying momentum balances in two different ways, depending on whether the control volume is stationary or moving. In each case, the water leaves the nozzle of cross-sectional area A with velocity u and the trolley has a velocity υ. Both velocities are relative to the nozzle. The reader should make a determined effort to understand both approaches, since momentum balances are conceptually more difficult than mass and energy balances. It is also essential to perform the momentum balance in an inertial frame of reference—one that is either stationary or moving with a uniform velocity, but not accelerating.

1. Control surface moving with trolley. As shown in Fig. 2.17, the control surface delineating the system is moving to the right at the same velocity υ as the trolley. The observer perceives water entering the system across BC not with velocity u but with a relative velocity (u – v), so that the rate m of convection of mass into the control volume is:

Fig. 2.17 Stationary observer—moving system.

A momentum balance (positive direction to the right) gives:

Note that to obtain the momentum flux, m is multiplied by the absolute velocity u [not the relative velocity (u – v), which has already been accounted for in the mass flux m]. Also, the mass of the system is not constant, but is increasing at a rate given by:

It follows from the last three equations that the acceleration a = dv/dt of the trolley to the right is:

2. Control volume fixed. In Fig. 2.18, the control surface is now fixed in space, so that the trolley is moving within it. Also—and not quite so obviously, that part of the jet of length L inside the control surface is lengthening at a rate dL/dt = υ and increasing its momentum ρALu, and this must of course be taken into account.

A mass balance first gives:

Rate of addition of mass = Rate of increase of mass,

Fig. 2.18 Stationary observer—fixed system.

Likewise, a momentum balance gives:

Rate of addition = Rate of increase.

By eliminating dM/dt between Eqns. (2.56) and (2.57) and rearranging, the acceleration becomes identical with that in (2.55), thus verifying the equivalence of the two approaches:

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Example 2.4—Impinging Jet of Water

Fig. E2.4 shows a plan of a jet of water impinging against a shield that is held stationary by a force F opposing the jet, which divides into several radially outward streams, each leaving at right angles to the jet. If the total water flow rate is Q = 1 ft³/s and its velocity is u = 100 ft/s, find F (lb_f).

Fig. E2.4 Jet impinging against shield.

Solution

Perform a momentum balance to the right on the system bounded by the dotted control surface. If the mass flow rate is m, the rate of transfer of momentum into the system by convection is mu. The exiting streams have no momentum to the right. Also, the opposing force amounts to a rate of addition of momentum F to the left. Hence, at steady state:

so that:

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Example 2.5—Velocity of Wave on Water

As shown in Fig. E2.5(a), a small disturbance in the form of a wave of slightly increased depth D + dD travels with velocity u along the surface of a layer of otherwise stagnant water of depth D. Note that no part of the water itself moves with velocity u—just the dividing line between the stagnant and disturbed regions; in fact, the water velocity just upstream of the front is a small quantity du, and downstream it is zero. If the viscosity is negligible, find the wave velocity in terms of the depth.

Fig. E2.5 (a) Moving wave front; (b) stationary wave front as seen by observer traveling with wave.

Estimate the warning times available to evacuate communities that are threatened by the following “avalanches” of water:

(a) A tidal wave (closely related to a tsunami²), generated by an earthquake 1,000 miles away, traveling across an ocean of average depth 2,000 ft.

² According to The New York Times of December 27, 2004, a tsunami is a series of waves generated by underwater seismic disturbances. In the disaster of the previous day, the Indian tectonic plate slipped under (and raised) the Burma plate off the coast of Sumatra. The article reported that the ocean floor could rise dozens of feet over a distance of hundreds of miles, displacing unbelievably enormous quantities of water. The waves reached Indonesia, Thailand, Sri Lanka, and elsewhere, causing immense destruction and the eventual loss of more than 225,000 lives.

(b) An onrush, caused by a failed dam 50 miles away, traveling down a river of depth 12 ft that is also flowing towards the community at 4 mph.

Solution

As shown in Fig. E2.5(a), the problem is a transient one, since the picture changes with time as the wave front moves to the right. The solution is facilitated by superimposing a velocity u to the left, as in Fig. E2.5(b)—that is, by taking the viewpoint of an observer traveling with the wave, who now “sees” water coming from the right with velocity u and leaving to the left with velocity u – du.

All forces, flow rates, and momentum fluxes will be based on unit width normal to the plane of the figure. Because the disturbance is small, second-order differentials such as (dD)² and du dD can be neglected. Referring to Fig. E2.5(b), the total downstream and upstream pressure forces are obtained by integration:

A mass balance on the indicated control volume relates the downstream and upstream velocities and their depths (remember that calculations are based on unit width, so D × 1 = D is really an area):

Since the viscosity is negligible, there is no shear stress exerted by the floor on the water above. Therefore, a steady-state momentum balance to the left on the control volume yields:

which simplifies to:

Substitution for dD from the mass balance yields the velocity u of the wave:

Note that the wave velocity increases in proportion to the square root of the depth of the water. Another viewpoint is that the Froude number, Fr = u²/gD, being the ratio of inertial (ρu²) to hydrostatic (ρgD) effects, is unity.

The calculations for the water “avalanches” now follow:

(a) Ocean wave:

so that the warning time is:

(a) River wave:

Since the river is itself flowing at 4 mph, the total wave velocity is 13.4 + 4 = 17.4 mph. Thus, the warning time is:

The velocity of a sinusoidally varying wave traveling on deep water is discussed in Section 7.10.

2.1. General Conservation Laws

The study of fluid mechanics is based, to a large extent, on the conservation laws of three extensive quantities:

1. Mass—usually total, but sometimes of one or more individual chemical species.

2. Total energy—the sum of internal, kinetic, potential, and pressure energy.

3. Momentum, both linear and angular.

For a system viewed as a whole, conservation means that there is no net gain nor loss of any of these three quantities, even though there may be some redistribution of them within a system. A general conservation law can be phrased relative to the general system shown in Fig. 2.1, in which can be identified:

1. The system V.

2. The surroundings S.

3. The boundary B, also known as the control surface, across which the system interacts in some manner with its surroundings.

Fig. 2.1 (a) System and its surroundings; (b) transfers to and from a system. For a chemical reaction, creation and destruction terms would also be included inside the system.

The interaction between system and surroundings is typically by one or more of the following mechanisms:

1. A flowing stream, either entering or leaving the system.

2. A “contact” force on the boundary, usually normal or tangential to it, and commonly called a stress.

3. A “body” force, due to an external field that acts throughout the system, of which gravity is the prime example.

4. Useful work, such as electrical energy entering a motor or shaft work leaving a turbine.

Let X denote mass, energy, or momentum. Over a finite time period, the general conservation law for X is:

Nonreacting system

For a mass balance on species i in a reacting system

The symbols are defined in Table 2.1. The understanding is that the creation and destruction terms, together with the superscript i, are needed only for mass balances on species i in chemical reactions, which will not be pursued further in this text.

Table 2.1 Meanings of Symbols in Equation (2.1)

It is very important to note that Eqn. (2.1a) cannot be applied indiscriminately, and is only observed in general for the three properties of mass, energy, and momentum. For example, it is not generally true if X is another extensive property such as volume, and is quite meaningless if X is an intensive property such as pressure or temperature.

In the majority of examples in this book, it is true that if X denotes mass and the density is constant, then Eqn. (2.1a) simplifies to the conservation of volume, but this is not the fundamental law. For example, if a gas cylinder is filled up by having nitrogen gas pumped into it, we would very much hope that the volume of the system (consisting of the cylinder and the gas it contains) does not increase by the volume of the (compressible) nitrogen pumped into it!

Equation (2.1a) can also be considered on a basis of unit time, in which case all quantities become rates; for example, ΔX_system becomes the rate, dX_system/dt, at which the X-content of the system is increasing, x_in (note the lower-case “x”) would be the rate of transfer of X into the system, and so on, as in Eqn. (2.2):

2.2. Mass Balances

The general conservation law is typically most useful when rates are considered. In that case, if X denotes mass M and x denotes a mass “rate”m (the symbol  can also be used) the transient mass balance (for a nonreacting system) is:

in which the symbols have the meanings given in Table 2.2.

Table 2.2 Meanings of Symbols in Equation (2.3)

The majority of the problems in this text will deal with steady-state situations, in which the system has the same appearance at all instants of time, as in the following examples:

1. A river, with a flow rate that is constant with time.

2. A tank that is draining through its base, but is also supplied with an identical flow rate of liquid through an inlet pipe, so that the liquid level in the tank remains constant with time.

Steady-state problems are generally easier to solve, because a time derivative, such as dM_system/dt, is zero, leading to an algebraic equation.

A few problems—such as that in Example 2.1—will deal with unsteady-state or transient situations, in which the appearance of the system changes with time, as in the following examples:

1. A river, whose level is being raised by a suddenly elevated dam gate downstream.

2. A tank that is draining through its base, but is not being supplied by an inlet stream, so that the liquid level in the tank falls with time.

Transient problems are generally harder to solve, because a time derivative, such as dM_system/dt, is retained, leading to a differential equation.

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Example 2.1—Mass Balance for Tank Evacuation

The tank shown in Fig. E2.1(a) has a volume V = 1 m³ and contains air that is maintained at a constant temperature by being in thermal equilibrium with its surroundings.

Fig. E2.1(a) Tank evacuation.

If the initial absolute pressure is p₀ = 1 bar, how long will it take for the pressure to fall to a final pressure of 0.0001 bar if the air is evacuated at a constant rate of Q = 0.001 m³/s, at the pressure prevailing inside the tank at any time?

Solution

First, and fairly obviously, choose the tank as the system, shown by the dashed rectangle. Note that there is no inlet to the system, and just one outlet from it. A mass balance on the air in the system (noting that a rate of loss is negative) gives:

Note that since the tank volume V is constant, dV/dt = 0. For an ideal gas:

so that:

Cancellation of M/RT gives the following ordinary differential equation, which governs the variation of pressure p with time t:

Separation of variables and integration between t = 0 (when the pressure is p₀) and a later time t (when the pressure is p) gives:

The resulting solution shows an exponential decay of the tank pressure with time, also illustrated in Fig. E2.1(b):

Fig. E2.1(b) Exponential decay of tank pressure.

Thus, the time t_f taken to evacuate the tank from its initial pressure of 1 bar to a final pressure of p_f = 0.0001 bar is:

Problem 2.1 contains a variation of the above, in which air is leaking slowly into the tank from the surrounding atmosphere.

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Steady-state mass balance for fluid flow. A particularly useful and simple mass balance—also known as the continuity equation—can be derived for the situation shown in Fig. 2.2, where the system resembles a wind sock at an airport. At station 1, fluid flows steadily with density ρ₁ and a uniform velocity u₁ normally across that part of the surface of the system represented by the area A₁. In steady flow, each fluid particle traces a path called a streamline. By considering a large number of particles crossing the closed curve C, we have an equally large number of streamlines that then form a surface known as a stream tube, across which there is clearly no flow. The fluid then leaves the system with uniform velocity u₂ and density ρ₂ at station 2, where the area normal to the direction of flow is A₂.

Fig. 2.2 Flow through a stream tube.

Referring to Eqn. (2.3), there is no accumulation of mass because the system is at steady state. Therefore, the only nonzero terms are m₁ (the rate of addition of mass) and m₂ (the rate of removal of mass), which are equal to ρ₁A₁u₁ and ρ₂A₂u₂, respectively, so that Eqn. (2.3) becomes:

which can be rewritten as:

where m (= m₁ = m₂) is the mass flow rate entering and leaving the system.

For the special but common case of an incompressible fluid, ρ₁ = ρ₂, so that the steady-state mass balance becomes:

in which Q is the volumetric flow rate.

Equations (2.4a/b) would also apply for nonuniform inlet and exit velocities, if the appropriate mean velocities u_m1 and u_m2 were substituted for u₁ and u₂. However, we shall postpone the concept of nonuniform velocity distributions to a more appropriate time, particularly to those chapters that deal with microscopic fluid mechanics.

2.3. Energy Balances

Equation (2.1) is next applied to the general system shown in Fig. 2.3, it being understood that property X is now energy. Observe that there is both flow into and from the system. Also note the quantities defined in Table 2.3.

Fig. 2.3 Energy balance on a system with flow in and out.

Table 2.3 Definitions of Symbols for Energy Balance

A differential energy balance results by applying Eqn. (2.1) over a short time period. Observe that there are two transfers into the system (incoming mass and heat) and two transfers out of the system (outgoing mass and work). Since the mass transfers also carry energy with them, there results:

in which each term has units of energy or work. In the above, the system is assumed for simplicity to be homogeneous, so that all parts of it have the same internal, potential, and kinetic energy per unit mass; if such were not the case, integration would be needed throughout the system. Also, multiple inlets and exits could be accommodated by means of additional terms.

Since the density ρ is the reciprocal of υ, the volume per unit mass, e + p/ρ = e + pv, which is recognized as the enthalpy per unit mass. The flow energy term p/ρ in Eqn. (2.6), also known as injection work or flow work, is readily explained by examining Fig. 2.4. Consider unit mass of fluid entering the stream tube under a pressure p₁. The volume of the unit mass is:

Fig. 2.4 Flow of unit mass to and from stream tube.

which is the product of the area A₁ and the distance 1/ρ₁A₁ through which the mass moves. (Here, the “1” has units of mass.) Hence, the work done on the system by p₁ in pushing the unit mass into the stream tube is the force p₁A₁ exerted by the pressure multiplied by the distance through which it travels:

Likewise, the work done by the system on the surroundings at the exit is:

Steady-state energy balance. In the following, all quantities are per unit mass flowing. Referring to the general system shown in Fig. 2.5, the energy entering with the inlet stream plus the heat supplied to the system must equal the energy leaving with the exit stream plus the work done by the system on its surroundings. Therefore, the right-hand side of Eqn. (2.6) is zero under steady-state conditions, and division by dM_in = dM_out gives:

Fig. 2.5 Steady-state energy balance.

in which each term represents an energy per unit mass flowing.

For an infinitesimally small system in which differential changes are occurring, Eqn. (2.10) may be rewritten as:

in which, for example, de is now a differential change, and υ = 1/ρ is the volume per unit mass.

Now examine the increase in internal energy de, which arises from frictional work dF dissipated into heat, heat addition dq from the surroundings, less work pdv done by the fluid. That is: de = dF + dq – pdv. Thus, eliminate the change de in the internal energy from Eqn. (2.11), and expand the term d(pv),

which simplifies to the differential form of the mechanical energy balance, in which heat terms are absent:

For a finite system, for flow from point 1 to point 2, Eqn. (2.13) integrates to:

in which a finite change is consistently the final minus the initial value, for example:

An energy balance for an incompressible fluid of constant density permits the integral to be evaluated easily, giving:

In the majority of cases, g will be virtually constant, in which case there is a further simplification to:

which is a generalized Bernoulli equation, augmented by two extra terms—the frictional dissipation, F, and the work w done by the system. Note that F can never be negative—it is impossible to convert heat entirely into useful work. The work term w will be positive if the fluid flows through a turbine and performs work on the environment; conversely, it will be negative if the fluid flows through a pump and has work done on it.

Power. The rate of expending energy in order to perform work is known as power, with dimensions of ML²/T³, typical units being W (J/s) and ft lb_f/s. The relations in Table 2.4 are available, depending on the particular context.

Table 2.4 Expressions for Power in Different Systems

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Example 2.2—Pumping n-Pentane

Fig. E2.2 shows an arrangement for pumping n-pentane (ρ = 39.3 lb_m/ft³) at 25 °C from one tank to another, through a vertical distance of 40 ft. All piping is 3-in. I.D. Assume that the overall frictional losses in the pipes are given (by methods to be described in Chapter 3) by:

Fig. E2.2 Pumping n-pentane.

For simplicity, however, you may ignore friction in the short length of pipe leading to the pump inlet. Also, the pump and its motor have a combined efficiency of 75%. If the mean velocity u_m is 25 ft/s, determine the following:

(a) The power required to drive the pump.

Finally, consider the shape of the free surface for the situation shown in Fig. 1.20(a), in which a cylindrical container, partly filled with liquid, is rotated with an angular velocity ω—that is, at N = ω/2π revolutions per unit time. The analysis has applications in fuel tanks of spinning rockets, centrifugal filters, and liquid mirrors.

Fig. 1.20 Pressure changes for rotating cylinder: (a) elevation, (b) plan.

Point O denotes the origin, where r = 0 and z = 0. After a sufficiently long time, the rotation of the container will be transmitted by viscous action to the liquid, whose rotation is called a forced vortex. In fact, the liquid spins as if it were a solid body, rotating with a uniform angular velocity ω, so that the velocity in the direction of rotation at a radial location r is given by υ_θ = rω. It is therefore appropriate to treat the situation similar to the hydrostatic investigations already made.

Suppose that the liquid element P is essentially a rectangular box with cross-sectional area dA and radial extent dr. (In reality, the element has slightly tapering sides, but a more elaborate treatment taking this into account will yield identical results to those derived here.) The pressure on the inner face is p, whereas that on the outer face is p + (∂p/∂r)dr. Also, for uniform rotation in a circular path of radius r, the acceleration toward the center O of the circle is rω². Newton’s second law of motion is then used for equating the net pressure force toward O to the mass of the element times its acceleration:

Note that the use of a partial derivative is essential, since the pressure now varies in both the horizontal (radial) and vertical directions. Simplification yields the variation of pressure in the radial direction:

so that pressure increases in the radially outward direction.

Observe that the gauge pressure at all points on the interface is zero; in particular, p_O = p_Q = 0. Integrating from points O to P (at constant z):

However, the pressure at P can also be obtained by considering the usual hydro-static increase in traversing the path QP:

Elimination of the intermediate pressure p_P between Eqns. (1.45) and (1.46) relates the elevation of the free surface to the radial location:

Thus, the free surface is parabolic in shape; observe also that the density is not a factor, having been canceled from the equations.

There is another type of vortex—the free vortex—that is also important, in cyclone dust collectors and tornadoes, for example, as discussed in Chapters 4 and 7. There, the velocity in the angular direction is given by υ_θ = c/r, where c is a constant, so that υ_θ is inversely proportional to the radial position.

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Example 1.8—Overflow from a Spinning Container

A cylindrical container of height H and radius a is initially half-filled with a liquid. The cylinder is then spun steadily around its vertical axis Z-Z, as shown in Fig. E1.8. At what value of the angular velocity ω will the liquid just start to spill over the top of the container? If H = 1 ft and a = 0.25 ft, how many rpm (revolutions per minute) would be needed?

Fig. E1.8 Geometry of a spinning container: (a) at rest, (b) on the point of overflowing.

Solution

From Eqn. (1.47), the shape of the free surface is a parabola. Therefore, the air inside the rotating cylinder forms a paraboloid of revolution, whose volume is known from calculus to be exactly one-half of the volume of the “circumscribing cylinder,” namely, the container.⁸ Hence, the liquid at the center reaches the bottom of the cylinder just as the liquid at the curved wall reaches the top of the cylinder. In Eqn. (1.47), therefore, set z = H and r = a, giving the required angular velocity:

⁸ Proof can be accomplished as follows. First, note for the parabolic surface in Fig. E1.8(b), r = a when z = H, so, from Eqn. (1.47), ω²/2g = H/a². Thus, Eqn. (1.47) can be rewritten as:

The volume of the paraboloid of air within the cylinder is therefore:

which is exactly one-half of the volume of the cylinder, πa²H. Since the container was initially just half filled, the liquid volume still accounts for the remaining half.

For the stated values:

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Problems for Chapter 1

1. Units conversion—E. How many cubic feet are there in an acre-foot? How many gallons? How many cubic meters? How many tonnes of water?

2. Units conversion—E. The viscosity μ of an oil is 10 cP and its specific gravity s is 0.8. Reexpress both of these (the latter as density ρ) in both the lb_m, ft, s system and in SI units.

3. Units conversion—E. Use conversion factors to express: (a) the gravitational acceleration of 32.174 ft/s² in SI units, and (b) a pressure of 14.7 lb_f/in² (one atmosphere) in both pascals and bars.

4. Meteorite density—E. The Barringer Crater in Arizona was formed 30,000 years ago by a spherical meteorite of diameter 60 m and mass 10⁶ t (tonnes), traveling at 15 km/s when it hit the ground.⁹ (Clearly, all figures are estimates.) What was the mean density of the meteorite? What was the predominant material in the meteorite? Why? If one tonne of the explosive TNT is equivalent to five billion joules, how many tonnes of TNT would have had the same impact as the meteorite?

⁹ Richard A.F. Grieve, “Impact cratering on the earth,” Scientific American, Vol. 262, No. 4, p. 68 (1990).

5. Reynolds number—E. What is the mean velocity u_m (ft/s) and the Reynolds number Re = ρu_mD/μ for 35 gpm (gallons per minute) of water flowing in a 1.05-in. I.D. pipe if its density is ρ = 62.3 lb_m/ft³ and its viscosity is μ = 1.2 cP? What are the units of the Reynolds number?

6. Pressure in bubble—E. Consider a soap-film bubble of diameter d. If the external air pressure is p_a, and the surface tension of the soap film is σ, derive an expression for the pressure p_b inside the bubble. Hint: Note that there are two air/liquid interfaces.

7. Reservoir waterflooding—E. Fig. P1.7(a) shows how water is pumped down one well, of depth H, into an oil-bearing stratum, so that the displaced oil then flows up through another well. Fig. P1.7(b) shows an enlargement of an idealized pore, of diameter d, at the water/oil interface.

Fig. P1.7 Waterflooding of an oil reservoir.

If the water and oil are just starting to move, what water inlet pressure p_w is needed if the oil exit pressure is to be p_o? Assume that the oil completely wets the pore (not always the case), that the water/oil interfacial tension is σ, and that the densities of the water and oil are ρ_ω and ρ_o, respectively.¹⁰

¹⁰ D.L. Katz et al., Handbook of Natural Gas Engineering, McGraw-Hill, New York, 1959, p. 57, indicates a wide range of wettability by water, varying greatly with the particular rock formation.

8. Barometer reading—M. In your house (elevation 950 ft above sea level) you have a barometer that registers inches of mercury. On an average day in January, you telephone the weather station (elevation 700 ft) and are told that the exact pressure there is 0.966 bar. What is the correct reading for your barometer, and to how many psia does this correspond? The specific gravity of mercury is 13.57.

9. Two-layer buoyancy—E. As shown in Fig. P1.9, a layer of an unknown liquid A (immiscible with water) floats on top of a layer of water W in a beaker. A completely submerged cylinder of specific gravity 0.9 adjusts itself so that its axis is vertical and two-thirds of its height projects above the A/W interface and one-third remains below. What is the specific gravity of A? Solve the problem two ways—first using Archimedes’ law, and then using a momentum or force balance.

Fig. P1.9 Cylinder immersed in water and liquid A.

10. Differential manometer—E. The U-tube shown in Fig. P1.10 has legs of unequal internal diameters d₁ and d₂, which are partly filled with immiscible liquids of densities ρ₁ and ρ₂, respectively, and are open to the atmosphere at the top.

Fig. P1.10 U-tube with immiscible liquids.

If an additional small volume υ₂ of the second liquid is added to the right-hand leg, derive an expression—in terms of ρ₁, ρ₂, υ₂, d₁, and d₂—for δ, the amount by which the level at B will fall. If ρ₁ is known, but ρ₂ is unknown, could the apparatus be used for determining the density of the second liquid?

Hints: The lengths h_A, h_B, and h_C have been included just to get started; they must not appear in the final result. After adding the second liquid, consider h_C to have increased by a length Δ—a quantity that must also eventually be eliminated.

11. Ascending bubble—E. As shown in Fig. P1.11, a hollow vertical cylinder with rigid walls and of height H is closed at both ends, and is filled with an incompressible oil of density ρ. A gauge registers the pressure at the top of the cylinder. When a small bubble of volume υ₀ initially adheres to point A at the bottom of the cylinder, the gauge registers a pressure p₀. The gas in the bubble is ideal, and has a molecular weight of M_w. The bubble is liberated by tapping on the cylinder and rises to point B at the top. The temperature T is constant throughout. Derive an expression in terms of any or all of the specified variables for the new pressure-gauge reading p₁ at the top of the cylinder.

Fig. P1.11 Bubble rising in a closed cylinder.

12. Ship passing through locks—M. A ship of mass M travels uphill through a series of identical rectangular locks, each of equal superficial (bird’s-eye view) area A and elevation change h. The steps involved in moving from one lock to the next (1 to 2, for example) are shown as A–B–C in Fig. P1.12. The lock at the top of the hill is supplied by a source of water. The initial depth in lock 1 is H, and the density of the water is ρ.

(a) Derive an expression for the increase in mass of water in lock 1 for the sequence shown in terms of some or all of the variables M, H, h, A, ρ, and g.

(b) If, after reaching the top of the hill, the ship descends through a similar series of locks to its original elevation, again derive an expression for the mass of water gained by a lock from the lock immediately above it.

(c) Does the mass of water to be supplied depend on the mass of the ship if: (i) it travels only uphill, (ii) it travels uphill, then downhill? Explain your answer.

Fig. P1.12 Ship and locks.

13. Furnace stack—E. Air (ρ_a = 0.08 lb_m/ft³) flows through a furnace where it is burned with fuel to produce a hot gas (ρ_g = 0.05 lb_m/ft³) that flows up the stack, as in Fig. P1.13. The pressures in the gas and the immediately surrounding air at the top of the stack at point A are equal.

Fig. P1.13 Furnace stack.

What is the difference Δh (in.) in levels of the water in the manometer connected between the base B of the stack and the outside air at point C? Which side rises? Except for the pressure drop across the furnace (which you need not worry about), treat the problem as one in hydrostatics. That is, ignore any frictional effects and kinetic energy changes in the stack. Also, neglect compressibility effects.

14. Hydrometer—E. When a hydrometer floats in water, its cylindrical stem is submerged so that a certain point X on the stem is level with the free surface of the water, as shown in Fig. P1.14. When the hydrometer is placed in another liquid L of specific gravity s, the stem rises so that point X is now a height z above the free surface of L.

Fig. P1.14 Hydrometer in water and test liquid L.

Derive an equation giving s in terms of z. If needed, the cross-sectional area of the stem is A, and when in water a total volume V (stem plus bulb) is submerged.

15. Three-liquid manometer—E. In the hydrostatic case shown in Fig. P1.15, a = 6 ft and c = 4 ft. The specific gravities of oil, mercury, and water are s_o = 0.8, s_m = 13.6, and s_ω = 1.0. Pressure variations in the air are negligible. What is the difference b in inches between the mercury levels, and which leg of the manometer has the higher mercury level? Note: In this latter respect, the diagram may or may not be correct.

Variation of pressure with elevation. Here, we investigate how the pressure in a stationary fluid varies with elevation z. The result is useful because it can answer questions such as “What is the pressure at the summit of Mt. Annapurna?” or “What forces are exerted on the walls of an oil storage tank?” Consider a hypothetical differential cylindrical element of fluid of cross-sectional area A, height dz, and volume A dz, which is also surrounded by the same fluid, as shown in Fig. 1.13. Its weight, being the downwards gravitational force on its mass, is dW = ρA dz g. Two completely equivalent approaches will be presented:

Fig. 1.13 Forces acting on a cylinder of fluid.

Method 1. Let p denote the pressure at the base of the cylinder; since p changes at a rate dp/dz with elevation, the pressure is found either from Taylor’s expansion or the definition of a derivative to be p + (dp/dz)dz at the top of the cylinder.⁷ (Note that we do not anticipate a reduction of pressure with elevation here; hence, the plus sign is used. If, indeed—as proves to be the case—pressure falls with increasing elevation, then the subsequent development will tell us that dp/dz is negative.) Hence, the fluid exerts an upward force of pA on the base of the cylinder, and a downward force of [p + (dp/dz)dz]A on the top of the cylinder.

⁷ Further details of this fundamental statement can be found in Appendix A and must be fully understood, because similar assertions appear repeatedly throughout the book.

Next, apply Newton’s second law of motion by equating the net upward force to the mass times the acceleration—which is zero, since the cylinder is stationary:

Cancellation of pA and division by A dz leads to the following differential equation, which governs the rate of change of pressure with elevation:

Method 2. Let p_z and p_z+dz denote the pressures at the base and top of the cylinder, where the elevations are z and z + dz, respectively. Hence, the fluid exerts an upward force of p_zA on the base of the cylinder, and a downward force of p_z+dzA on the top of the cylinder. Application of Newton’s second law of motion gives:

Isolation of the two pressure terms on the left-hand side and division by A dz gives:

As dz tends to zero, the left-hand side of Eqn. (1.32) becomes the derivative dp/dz, leading to the same result as previously:

The same conclusion can also be obtained by considering a cylinder of finite height Δz and then letting Δz approach zero.

Note that Eqn. (1.30) predicts a pressure decrease in the vertically upward direction at a rate that is proportional to the local density. Such pressure variations can readily be detected by the ear when traveling quickly in an elevator in a tall building, or when taking off in an airplane. The reader must thoroughly understand both the above approaches. For most of this book, we shall use Method 1, because it eliminates the steps of taking the limit of dz → 0 and invoking the definition of the derivative.

Pressure in a liquid with a free surface. In Fig. 1.14, the pressure is p_s at the free surface, and we wish to find the pressure p at a depth H below the free surface—of water in a swimming pool, for example.

Fig. 1.14 Pressure at a depth H.

Separation of variables in Eqn. (1.30) and integration between the free surface (z = H) and a depth H (z = 0) gives:

Assuming—quite reasonably—that ρ and g are constants in the liquid, these quantities may be taken outside the integral, yielding:

which predicts a linear increase of pressure with distance downward from the free surface. For large depths, such as those encountered by deep-sea divers, very substantial pressures will result.

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Example 1.3—Pressure in an Oil Storage Tank

What is the absolute pressure at the bottom of the cylindrical tank of Fig. E1.3, filled to a depth of H with crude oil, with its free surface exposed to the atmosphere? The specific gravity of the crude oil is 0.846. Give the answers for (a) H = 15.0 ft (pressure in lb_f/in²), and (b) H = 5.0 m (pressure in Pa and bar). What is the purpose of the surrounding dike?

Fig. E1.3 Crude oil storage tank.

Solution

(a) The pressure is that of the atmosphere, p_a, plus the increase due to a column of depth H = 15.0 ft. Thus, setting p_s = p_a, Eqn. (1.34) gives:

The reader should check the units, noting that the 32.2 in the numerator is g [=] ft/s², and that the 32.2 in the denominator is g_c [=] lb_m ft/lb_f s².

(b) For SI units, no conversion factors are needed. Noting that the density of water is 1,000 kg/m³, and that p_a  1.01 × 10⁵ Pa absolute:

p = 1.01 × 10⁵ + 0.846 × 1,000 × 9.81 × 5.0 = 1.42 × 10⁵ Pa = 1.42 bar.

In the event of a tank rupture, the dike contains the leaking oil and facilitates prevention of spreading fire and contamination of the environment.

Epilogue

When he arrived at work in an oil refinery one morning, the author saw first-hand the consequences of an inadequately vented oil-storage tank. Rain during the night had caused partial condensation of vapor inside the tank, whose pressure had become sufficiently lowered so that the external atmospheric pressure had crumpled the steel tank just as if it were a flimsy tin can. The refinery manager was not pleased.

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Example 1.4—Multiple Fluid Hydrostatics

The U-tube shown in Fig. E1.4 contains oil and water columns, between which there is a long trapped air bubble. For the indicated heights of the columns, find the specific gravity of the oil.

Fig. E1.4 Oil/air/water system.

Solution

The pressure p₂ at point 2 may be deduced by starting with the pressure p₁ at point 1 and adding or subtracting, as appropriate, the hydrostatic pressure changes due to the various columns of fluid. Note that the width of the U-tube (2.0 ft) is irrelevant, since there is no change in pressure in the horizontal leg. We obtain:

in which ρ_o, ρ_a, and ρ_w denote the densities of oil, air, and water, respectively. Since the density of the air is very small compared to that of oil or water, the term containing ρ_a can be neglected. Also, p₁ = p₂, because both are equal to atmospheric pressure. Equation (E1.4.1) can then be solved for the specific gravity s_o of the oil:

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Pressure variations in a gas. For a gas, the density is no longer constant, but is a function of pressure (and of temperature—although temperature variations are usually less significant than those of pressure), and there are two approaches:

1. For small changes in elevation, the assumption of constant density can still be made, and equations similar to Eqn. (1.34) are still approximately valid.

2. For moderate or large changes in elevation, the density in Eqn. (1.30) is given by Eqn. (1.7) or (1.8), ρ = M_wp/RT or ρ = M_wp/ZRT, depending on whether the gas is ideal or nonideal. It is understood that absolute pressure and temperature must always be used whenever the gas law is involved. A separation of variables can still be made, followed by integration, but the result will now be more complicated because the term dp/p occurs, leading—at the simplest (for an isothermal situation)—to a decreasing exponential variation of pressure with elevation.

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Example 1.5—Pressure Variations in a Gas

For a gas of molecular weight M_w (such as the earth’s atmosphere), investigate how the pressure p varies with elevation z if p = p₀ at z = 0. Assume that the temperature T is constant. What approximation may be made for small elevation increases? Explain how you would proceed for the nonisothermal case, in which T = T (z) is a known function of elevation.

Solution

Assuming ideal gas behavior, Eqns. (1.30) and (1.7) give:

Separation of variables and integration between appropriate limits yields:

since M_wg/RT is constant. Hence, there is an exponential decrease of pressure with elevation, as shown in Fig. E1.5:

Fig. E1.5 Variation of gas pressure with elevation.

Since a Taylor’s expansion gives e^–x = 1 – x + x²/2 – …, the pressure is approximated by:

For small values of M_wgz/RT, the last term is an insignificant second-order effect (compressibility effects are unimportant), and we obtain:

in which ρ₀ is the density at elevation z = 0; this approximation—essentially one of constant density—is shown as the dashed line in Fig. E1.5 and is clearly applicable only for a small change of elevation. Problem 1.19 investigates the upper limit on z for which this linear approximation is realistic. If there are significant elevation changes—as in Problems 1.16 and 1.30—the approximation of Eqn. (E1.5.5) cannot be used with any accuracy. Observe with caution that the Taylor’s expansion is only a vehicle for demonstrating what happens for small values of M_wgz/RT. Actual calculations for larger values of M_wgz/RT should be made using Eqn. (E1.5.3), not Eqn. (E1.5.4).

For the case in which the temperature is not constant, but is a known function T(z) of elevation (as might be deduced from observations made by a meteorological balloon), it must be included inside the integral:

Since T(z) is unlikely to be a simple function of z, a numerical method—such as Simpson’s rule in Appendix A—will probably have to be used to approximate the second integral of Eqn. (E1.5.6).

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Total force on a dam or lock gate. Fig. 1.15 shows the side and end elevations of a dam or lock gate of depth D and width W. An expression is needed for the total horizontal force F exerted by the liquid on the dam, so that the latter can be made of appropriate strength. Similar results would apply for liquids in storage tanks. Gauge pressures are used for simplicity, with p = 0 at the free surface and in the air outside the dam. Absolute pressures could also be employed, but would merely add a constant atmospheric pressure everywhere, and would eventually be canceled out. If the coordinate z is measured from the bottom of the liquid upward, the corresponding depth of a point below the free surface is D – z. Hence, from Eqn. (1.34), the differential horizontal force dF on an infinitesimally small rectangular strip of area dA = Wdz is:

Fig. 1.15 Horizontal thrust on a dam: (a) side elevation, (b) end elevation.

Integration from the bottom (z = 0) to the top (z = D) of the dam gives the total horizontal force:

Horizontal pressure force on an arbitrary plane vertical surface. The preceding analysis was for a regular shape. A more general case is illustrated in Fig. 1.16, which shows a plane vertical surface of arbitrary shape. Note that it is now slightly easier to work in terms of a downward coordinate h.

Fig. 1.16 Side view of a pool of liquid with a submerged vertical surface.

Again taking gauge pressures for simplicity (the gas law is not involved), with p = 0 at the free surface, the total horizontal force is:

But the depth h_c of the centroid of the surface is defined as:

Thus, from Eqns. (1.37) and (1.38), the total force is:

in which p_c is the pressure at the centroid.

The advantage of this approach is that the location of the centroid is already known for several geometries. For example, for a rectangle of depth D and width W:

in agreement with the earlier result of Eqn. (1.36). Similarly, for a vertical circle that is just submerged, the depth of the centroid equals its radius. And, for a vertical triangle with one edge coincident with the surface of the liquid, the depth of the centroid equals one-third of its altitude.

Horizontal pressure force on a curved surface. Fig. 1.17(a) shows the cross section of a submerged surface that is no longer plane. However, the shape is uniform normal to the plane of the diagram.

Fig. 1.17 Thrust on surface of uniform cross-sectional shape.

In general, as shown in Fig. 1.17(b), the local pressure force pdA on an element of surface area dA does not act horizontally; therefore, its horizontal component must be obtained by projection through an angle of (π/2 – θ), by multiplying by cos(π/2 – θ) = sin θ. The total horizontal force F is then:

in which dA* = dA sin θ is an element of the projection of A onto the hypothetical vertical plane A*. The integral of Eqn. (1.41) can be obtained readily, as illustrated in the following example.

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Example 1.6—Hydrostatic Force on a Curved Surface

A submarine, whose hull has a circular cross section of diameter D, is just submerged in water of density ρ, as shown in Fig. E1.6. Derive an equation that gives the total horizontal force F_x on the left half of the hull, for a distance W normal to the plane of the diagram. If D = 8 m, the circular cross section continues essentially for the total length W = 50 m of the submarine, and the density of sea water is ρ = 1,026 kg/m³, determine the total horizontal force on the left-hand half of the hull.

Fig. E1.6 Submarine just submerged in seawater.

Solution

The force is obtained by evaluating the integral of Eqn. (1.41), which is identical to that for the rectangle in Fig. 1.15:

Insertion of the numerical values gives:

Thus, the total force is considerable—about 3.62 × 10⁶ lb_f.

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Archimedes, ca. 287–212 B.C. Archimedes was a Greek mathematician and inventor. He was born in Syracuse, Italy, where he spent much of his life, apart from a period of study in Alexandria. He was much more interested in mathematical research than any of the ingenious inventions that made him famous. One invention was a “burning mirror,” which focused the sun’s rays to cause intense heat. Another was the rotating Archimedean screw, for raising a continuous stream of water. Presented with a crown supposedly of pure gold, Archimedes tested the possibility that it might be “diluted” by silver by separately immersing the crown and an equal weight of pure gold into his bath, and observed the difference in the overflow. Legend has it that he was so excited by the result that he ran home without his clothes, shouting , “I have found it, I have found it.” To dramatize the effect of a lever, he said, “Give me a place to stand, and I will move the earth.” He considered his most important intellectual contribution to be the determination of the ratio of the volume of a sphere to the volume of the cylinder that circumscribes it. [Now that calculus has been invented, the reader might like to derive this ratio!] Sadly, Archimedes was killed during the capture of Syracuse by the Romans.

Source: The Encyclopædia Britannica, 11th ed., Cambridge University Press (1910–1911).

Buoyancy forces. If an object is submerged in a fluid, it will experience a net upward or buoyant force exerted by the fluid. To find this force, first examine the buoyant force on a submerged circular cylinder of height H and cross-sectional area A, shown in Fig. 1.18.

Fig. 1.18 Pressure forces on a submerged cylinder.

The forces on the curved vertical surface act horizontally and may therefore be ignored. Hence, the net upward force due to the difference between the opposing pressures on the bottom and top faces is:

which is exactly the weight of the displaced liquid, thus verifying Archimedes’ law, (the buoyant force equals the weight of the fluid displaced) for the cylinder. The same result would clearly be obtained for a cylinder of any uniform cross section.

Fig. 1.19 shows a more general situation, with a body of arbitrary shape. However, Archimedes’ law still holds since the body can be decomposed into an infinitely large number of vertical rectangular parallelepipeds or “boxes” of infinitesimally small cross-sectional area dA. The effect for one box is then summed or “integrated” over all the boxes, and again gives the net upward buoyant force as the weight of the liquid displaced.

Fig. 1.19 Buoyancy force for an arbitrary shape.

Mass, weight, and force. The mass M of an object is a measure of the amount of matter it contains and will be constant, since it depends on the number of constituent molecules and their masses. On the other hand, the weight w of the object is the gravitational force on it, and is equal to Mg, where g is the local gravitational acceleration. Mostly, we shall be discussing phenomena occurring at the surface of the earth, where g is approximately 32.174 ft/s² = 9.807 m/s² = 980.7 cm/s². For much of this book, these values are simply taken as 32.2, 9.81, and 981, respectively.

Newton’s second law of motion states that a force F applied to a mass M will give it an acceleration a:

from which is apparent that force has dimensions ML/T². Table 1.7 gives the corresponding units of force in the SI (meter/kilogram/second), CGS (centimeter/gram/second), and FPS (foot/pound/second) systems.

Table 1.7 Representative Units of Force

The poundal is now an archaic unit, hardly ever used. Instead, the pound force, lb_f, is much more common in the English system; it is defined as the gravitational force on 1 lb_m, which, if left to fall freely, will do so with an acceleration of 32.2 ft/s². Hence:

When using lb_f in the ft, lb_m, s (FPS) system, the following conversion factor, commonly called “g_c,” will almost invariably be needed:

Some writers incorporate g_c into their equations, but this approach may be confusing since it virtually implies that one particular set of units is being used, and hence tends to rob the equations of their generality. Why not, for example, also incorporate the conversion factor of 144 in²/ft² into equations where pressure is expressed in lb_f/in²? We prefer to omit all conversion factors in equations, and introduce them only as needed in evaluating expressions numerically. If the reader is in any doubt, units should always be checked when performing calculations.

SI Units. The most systematically developed and universally accepted set of units occurs in the SI units or Système International d’Unités⁶; the subset we mainly need is shown in Table 1.8.

⁶ For an excellent discussion, on which Tables 1.8 and 1.9 are based, see Metrication in Scientific Journals, published by The Royal Society, London, 1968.

Table 1.8 SI Units

The basic units are again the meter, kilogram, and second (m, kg, and s); from these, certain derived units can also be obtained. Force (kg m/s²) has already been discussed; energy is the product of force and length; power amounts to energy per unit time; surface tension is energy per unit area or force per unit length, and so on. Some of the units have names, and these, together with their abbreviations, are also given in Table 1.8.

Tradition dies hard, and certain other “metric” units are so well established that they may be used as auxiliary units; these are shown in Table 1.9. The gram is the classic example. Note that the basic SI unit of mass (kg) is even represented in terms of the gram, and has not yet been given a name of its own!

Table 1.9 Auxiliary Units Allowed in Conjunction with SI Units

Table 1.10 shows some of the acceptable prefixes that can be used for accommodating both small and large quantities. For example, to avoid an excessive number of decimal places, 0.000001 s is normally better expressed as 1 μs (one microsecond). Note also, for example, that 1 μkg should be written as 1 mg—one prefix being better than two.

Table 1.10 Prefixes for Fractions and Multiples

Some of the more frequently used conversion factors are given in Table 1.11.

Table 1.11 Commonly Used Conversion Factors

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Example 1.1—Units Conversion

Part 1. Express 65 mph in (a) ft/s, and (b) m/s.

Solution

The solution is obtained by employing conversion factors taken from Table 1.11:

Part 2. The density of 35 °API crude oil is 53.1 lb_m/ft³ at 68 °F and its viscosity is 32.8 lb_m/ft hr. What are its density, viscosity, and kinematic viscosity in SI units?

Solution

Or, converting to SI units, noting that P is the symbol for poise, and evaluating ν:

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Example 1.2—Mass of Air in a Room

Estimate the mass of air in your classroom, which is 80 ft wide, 40 ft deep, and 12 ft high. The gas constant is R = 10.73 psia ft³/lb-mol °R.

Solution

The volume of the classroom, shown in Fig. E1.2, is:

V = 80 × 40 × 12 = 3.84 × 10⁴ ft³.

Fig. E1.2 Assumed dimensions of classroom.

If the air is approximately 20% oxygen and 80% nitrogen, its mean molecular weight is M_w = 0.8 × 28 + 0.2 × 32 = 28.8 lb_m/lb-mol. From the gas law, assuming an absolute pressure of p = 14.7 psia and a temperature of 70 °F = 530 °R, the density is:

Hence, the mass of air is:

M = ρV = 0.0744 (lb_m/ft³) × 3.84 × 10⁴ (ft³) = 2, 860 lb_m.

There are three physical properties of fluids that are particularly important: density, viscosity, and surface tension. Each of these will be defined and viewed briefly in terms of molecular concepts, and their dimensions will be examined in terms of mass, length, and time (M, L, and T). The physical properties depend primarily on the particular fluid. For liquids, viscosity also depends strongly on the temperature; for gases, viscosity is approximately proportional to the square root of the absolute temperature. The density of gases depends almost directly on the absolute pressure; for most other cases, the effect of pressure on physical properties can be disregarded.

Typical processes often run almost isothermally, and in these cases the effect of temperature can be ignored. Except in certain special cases, such as the flow of a compressible gas (in which the density is not constant) or a liquid under a very high shear rate (in which viscous dissipation can cause significant internal heating), or situations involving exothermic or endothermic reactions, we shall ignore any variation of physical properties with pressure and temperature.

Densities of liquids. Density depends on the mass of an individual molecule and the number of such molecules that occupy a unit of volume. For liquids, density depends primarily on the particular liquid and, to a much smaller extent, on its temperature. Representative densities of liquids are given in Table 1.1.² (See Eqns. (1.9)–(1.11) for an explanation of the specific gravity and coefficient of thermal expansion columns.) The accuracy of the values given in Tables 1.1–1.6 is adequate for the calculations needed in this text. However, if highly accurate values are needed, particularly at extreme conditions, then specialized information should be sought elsewhere.

² The values given in Tables 1.1, 1.3, 1.4, 1.5, and 1.6 are based on information given in J.H. Perry, ed., Chemical Engineers’ Handbook, 3rd ed., McGraw-Hill, New York, 1950.

Table 1.1 Specific Gravities, Densities, and Thermal Expansion Coefficients of Liquids at 20 °C

Density. The density ρ of a fluid is defined as its mass per unit volume, and indicates its inertia or resistance to an accelerating force. Thus:

in which the notation “[=]” is consistently used to indicate the dimensions of a quantity.³ It is usually understood in Eqn. (1.5) that the volume is chosen so that it is neither so small that it has no chance of containing a representative selection of molecules nor so large that (in the case of gases) changes of pressure cause significant changes of density throughout the volume. A medium characterized by a density is called a continuum, and follows the classical laws of mechanics—including Newton’s law of motion, as described in this book.

³ An early appearance of the notation “[=]” is in R.B. Bird, W.E. Stewart, and E.N. Lightfoot, Transport Phenomena, Wiley, New York, 1960.

Degrees A.P.I. (American Petroleum Institute) are related to specific gravity s by the formula:

Note that for water, °A.P.I. = 10, with correspondingly higher values for liquids that are less dense. Thus, for the crude oil listed in Table 1.1, Eqn. (1.6) indeed gives 141.5/0.851 – 131.5  35° A.P.I.

Densities of gases. For ideal gases, pV = nRT, where p is the absolute pressure, V is the volume of the gas, n is the number of moles (abbreviated as “mol” when used as a unit), R is the gas constant, and T is the absolute temperature. If M_w is the molecular weight of the gas, it follows that:

Thus, the density of an ideal gas depends on the molecular weight, absolute pressure, and absolute temperature. Values of the gas constant R are given in Table 1.2 for various systems of units. Note that degrees Kelvin, formerly represented by “°K,” is now more simply denoted as “K.”

Table 1.2 Values of the Gas Constant, R

For a nonideal gas, the compressibility factor Z (a function of p and T) is introduced into the denominator of Eqn. (1.7), giving:

Thus, the extent to which Z deviates from unity gives a measure of the nonideality of the gas.

The isothermal compressibility of a gas is defined as:

and equals—at constant temperature—the fractional decrease in volume caused by a unit increase in the pressure. For an ideal gas, β = 1/p, the reciprocal of the absolute pressure.

The coefficient of thermal expansion α of a material is its isobaric (constant pressure) fractional increase in volume per unit rise in temperature:

Since, for a given mass, density is inversely proportional to volume, it follows that for moderate temperature ranges (over which α is essentially constant) the density of most liquids is approximately a linear function of temperature:

where ρ₀ is the density at a reference temperature T₀. For an ideal gas, α = 1/T, the reciprocal of the absolute temperature.

The specific gravity s of a fluid is the ratio of the density ρ to the density ρSC of a reference fluid at some standard condition:

For liquids, ρSC is usually the density of water at 4 °C, which equals 1.000 g/ml or 1,000 kg/m³. For gases, ρSC is sometimes taken as the density of air at 60 °F and 14.7 psia, which is approximately 0.0759 lb_m/ft³, and sometimes at 0 °C and one atmosphere absolute; since there is no single standard for gases, care must obviously be taken when interpreting published values. For natural gas, consisting primarily of methane and other hydrocarbons, the gas gravity is defined as the ratio of the molecular weight of the gas to that of air (28.8 lb_m/lb-mol).

Values of the molecular weight M_w are listed in Table 1.3 for several commonly occurring gases, together with their densities at standard conditions of atmospheric pressure and 0 °C.

Table 1.3 Gas Molecular Weights and Densities (the Latter at Atmospheric Pressure and 0 °C)

Viscosity. The viscosity of a fluid measures its resistance to flow under an applied shear stress, as shown in Fig. 1.8(a). There, the fluid is ideally supposed to be confined in a relatively small gap of thickness h between one plate that is stationary and another plate that is moving steadily at a velocity V relative to the first plate.

Fig. 1.8 (a) Fluid in shear between parallel plates; (b) the ensuing linear velocity profile.

In practice, the situation would essentially be realized by a fluid occupying the space between two concentric cylinders of large radii rotating relative to each other, as in Fig. 1.1. A steady force F to the right is applied to the upper plate (and, to preserve equilibrium, to the left on the lower plate) in order to maintain a constant motion and to overcome the viscous friction caused by layers of molecules sliding over one another.

Under these circumstances, the velocity u of the fluid to the right is found experimentally to vary linearly from zero at the lower plate (y = 0) to V itself at the upper plate, as in Fig. 1.8(b), corresponding to no-slip conditions at each plate. At any intermediate distance y from the lower plate, the velocity is simply:

Recall that the shear stress τ is the tangential applied force F per unit area:

in which A is the area of each plate. Experimentally, for a large class of materials, called Newtonian fluids, the shear stress is directly proportional to the velocity gradient:

The proportionality constant μ is called the viscosity of the fluid; its dimensions can be found by substituting those for F (ML/T²), A (L²), and du/dy (T^–1), giving:

Representative units for viscosity are g/cm s (also known as poise, designated by P), kg/m s, and lb_m/ft hr. The centipoise (cP), one hundredth of a poise, is also a convenient unit, since the viscosity of water at room temperature is approximately 0.01 P or 1.0 cP. Table 1.11 gives viscosity conversion factors.

The viscosity of a fluid may be determined by observing the pressure drop when it flows at a known rate in a tube, as analyzed in Section 3.2. More sophisticated methods for determining the rheological or flow properties of fluids—including viscosity—are also discussed in Chapter 11; such methods often involve containing the fluid in a small gap between two surfaces, moving one of the surfaces, and measuring the force needed to maintain the other surface stationary.

The kinematic viscosity ν is the ratio of the viscosity to the density:

and is important in cases in which significant viscous and gravitational forces coexist. The reader can check that the dimensions of ν are L²/T, which are identical to those for the diffusion coefficient D in mass transfer and for the thermal diffusivity α = k/ρc_p in heat transfer. There is a definite analogy among the three quantities—indeed, as seen later, the value of the kinematic viscosity governs the rate of “diffusion” of momentum in the laminar and turbulent flow of fluids.

Viscosities of liquids. The viscosities μ of liquids generally vary approximately with absolute temperature T according to:

and—to a good approximation—are independent of pressure. Assuming that μ is measured in centipoise and that T is either in degrees Kelvin or Rankine, appropriate parameters a and b are given in Table 1.4 for several representative liquids. The resulting values for viscosity are approximate, suitable for a first design only.

Table 1.4 Viscosity Parameters for Liquids

Viscosities of gases. The viscosity μ of many gases is approximated by the formula:

in which T is the absolute temperature (Kelvin or Rankine), μ₀ is the viscosity at an absolute reference temperature T₀, and n is an empirical exponent that best fits the experimental data. The values of the parameters μ₀ and n for atmospheric pressure are given in Table 1.5; recall that to a first approximation, the viscosity of a gas is independent of pressure. The values μ₀ are given in centipoise and correspond to a reference temperature of T₀  273 K  492 °R.

Table 1.5 Viscosity Parameters for Gases

Surface tension.⁴ Surface tension is the tendency of the surface of a liquid to behave like a stretched elastic membrane. There is a natural tendency for liquids to minimize their surface area. The obvious case is that of a liquid droplet on a horizontal surface that is not wetted by the liquid—mercury on glass, or water on a surface that also has a thin oil film on it. For small droplets, such as those on the left of Fig. 1.9, the droplet adopts a shape that is almost perfectly spherical, because in this configuration there is the least surface area for a given volume.

1.1. Fluid Mechanics in Chemical Engineering

Aknowledge of fluid mechanics is essential for the chemical engineer because the majority of chemical-processing operations are conducted either partly or totally in the fluid phase. Examples of such operations abound in the biochemical, chemical, energy, fermentation, materials, mining, petroleum, pharmaceuticals, polymer, and waste-processing industries.

There are two principal reasons for placing such an emphasis on fluids. First, at typical operating conditions, an enormous number of materials normally exist as gases or liquids, or can be transformed into such phases. Second, it is usually more efficient and cost-effective to work with fluids in contrast to solids. Even some operations with solids can be conducted in a quasi-fluidlike manner; examples are the fluidized-bed catalytic refining of hydrocarbons, and the long-distance pipelining of coal particles using water as the agitating and transporting medium.

Although there is inevitably a significant amount of theoretical development, almost all the material in this book has some application to chemical processing and other important practical situations. Throughout, we shall endeavor to present an understanding of the physical behavior involved; only then is it really possible to comprehend the accompanying theory and equations.

1.2. General Concepts of a Fluid

We must begin by responding to the question, “What is a fluid?” Broadly speaking, a fluid is a substance that will deform continuously when it is subjected to a tangential or shear force, much as a similar type of force is exerted when a water-skier skims over the surface of a lake or butter is spread on a slice of bread. The rate at which the fluid deforms continuously depends not only on the magnitude of the applied force but also on a property of the fluid called its viscosity or resistance to deformation and flow. Solids will also deform when sheared, but a position of equilibrium is soon reached in which elastic forces induced by the deformation of the solid exactly counterbalance the applied shear force, and further deformation ceases.

A simple apparatus for shearing a fluid is shown in Fig. 1.1. The fluid is contained between two concentric cylinders; the outer cylinder is stationary, and the inner one (of radius R) is rotated steadily with an angular velocity ω. This shearing motion of a fluid can continue indefinitely, provided that a source of energy—supplied by means of a torque here—is available for rotating the inner cylinder. The diagram also shows the resulting velocity profile; note that the velocity in the direction of rotation varies from the peripheral velocity Rω of the inner cylinder down to zero at the outer stationary cylinder, these representing typical no-slip conditions at both locations. However, if the intervening space is filled with a solid—even one with obvious elasticity, such as rubber—only a limited rotation will be possible before a position of equilibrium is reached, unless, of course, the torque is so high that slip occurs between the rubber and the cylinder.

Fig. 1.1 Shearing of a fluid.

There are various classes of fluids. Those that behave according to nice and obvious simple laws, such as water, oil, and air, are generally called Newtonian fluids. These fluids exhibit constant viscosity but, under typical processing conditions, virtually no elasticity. Fortunately, a very large number of fluids of interest to the chemical engineer exhibit Newtonian behavior, which will be assumed throughout the book, except in Chapter 11, which is devoted to the study of non-Newtonian fluids.

A fluid whose viscosity is not constant (but depends, for example, on the intensity to which it is being sheared), or which exhibits significant elasticity, is termed non-Newtonian. For example, several polymeric materials subject to deformation can “remember” their recent molecular configurations, and in attempting to recover their recent states, they will exhibit elasticity in addition to viscosity. Other fluids, such as drilling mud and toothpaste, behave essentially as solids and will not flow when subject to small shear forces, but will flow readily under the influence of high shear forces.

Fluids can also be broadly classified into two main categories—liquids and gases. Liquids are characterized by relatively high densities and viscosities, with molecules close together; their volumes tend to remain constant, roughly independent of pressure, temperature, or the size of the vessels containing them. Gases, on the other hand, have relatively low densities and viscosities, with molecules far apart; generally, they will rapidly tend to fill the container in which they are placed. However, these two states—liquid and gaseous—represent but the two extreme ends of a continuous spectrum of possibilities.

The situation is readily illustrated by considering a fluid that is initially a gas at point G on the pressure/temperature diagram shown in Fig. 1.2. By increasing the pressure, and perhaps lowering the temperature, the vapor-pressure curve is soon reached and crossed, and the fluid condenses and apparently becomes a liquid at point L. By continuously adjusting the pressure and temperature so that the clockwise path is followed, and circumnavigating the critical point C in the process, the fluid is returned to G, where it is presumably once more a gas. But where does the transition from liquid at L to gas at G occur? The answer is at no single point, but rather that the change is a continuous and gradual one, through a whole spectrum of intermediate states.

Fig. 1.2 When does a liquid become a gas?

1.3. Stresses, Pressure, Velocity, and the Basic Laws

Stresses. The concept of a force should be readily apparent. In fluid mechanics, a force per unit area, called a stress, is usually found to be a more convenient and versatile quantity than the force itself. Further, when considering a specific surface, there are two types of stresses that are particularly important.

1. The first type of stress, shown in Fig. 1.3(a), acts perpendicularly to the surface and is therefore called a normal stress; it will be tensile or compressive, depending on whether it tends to stretch or to compress the fluid on which it acts. The normal stress equals F/A, where F is the normal force and A is the area of the surface on which it acts. The dotted outlines show the volume changes caused by deformation. In fluid mechanics, pressure is usually the most important type of compressive stress, and will shortly be discussed in more detail.

Fig. 1.3(a) Tensile and compressive normal stresses F/A, acting on a cylinder, causing elongation and shrinkage, respectively.

2. The second type of stress, shown in Fig. 1.3(b), acts tangentially to the surface; it is called a shear stress τ, and equals F/A, where F is the tangential force and A is the area on which it acts. Shear stress is transmitted through a fluid by interaction of the molecules with one another. A knowledge of the shear stress is very important when studying the flow of viscous Newtonian fluids. For a given rate of deformation, measured by the time derivative dγ/dt of the small angle of deformation γ, the shear stress τ is directly proportional to the viscosity of the fluid (see Fig. 1.3(b)).

Fig. 1.3(b) Shear stress τ = F/A, acting on a rectangular parallelepiped, shown in cross section, causing a deformation measured by the angle γ (whose magnitude is exaggerated here).

Pressure. In virtually all hydrostatic situations—those involving fluids at rest—the fluid molecules are in a state of compression. For example, for the swimming pool whose cross section is depicted in Fig. 1.4, this compression at a typical point P is caused by the downwards gravitational weight of the water above point P. The degree of compression is measured by a scalar, p—the pressure.

Fig. 1.4 (a) Balloon submerged in a swimming pool; (b) enlarged view of the compressed balloon, with pressure forces acting on it.

A small inflated spherical balloon pulled down from the surface and tethered at the bottom by a weight will still retain its spherical shape (apart from a small distortion at the point of the tether), but will be diminished in size, as in Fig. 1.4(a). It is apparent that there must be forces acting normally inward on the surface of the balloon, and that these must essentially be uniform for the shape to remain spherical, as in Fig. 1.4(b).

Although the pressure p is a scalar, it typically appears in tandem with an area A (assumed small enough so that the pressure is uniform over it). By definition of pressure, the surface experiences a normal compressive force F = pA. Thus, pressure has units of a force per unit area—the same as a stress.

The value of the pressure at a point is independent of the orientation of any area associated with it, as can be deduced with reference to a differentially small wedge-shaped element of the fluid, shown in Fig. 1.5.

Fig. 1.5 Equilibrium of a wedge of fluid.

Due to the pressure there are three forces, p_AdA, p_BdB, and p_CdC, that act on the three rectangular faces of areas dA, dB, and dC. Since the wedge is not moving, equate the two forces acting on it in the horizontal or x direction, noting that p_AdA must be resolved through an angle (π/2 – θ) by multiplying it by cos(π/2 – θ) = sin θ:

The vertical force p_BdB acting on the bottom surface is omitted from Eqn. (1.1) because it has no component in the x direction. The horizontal pressure forces acting in the y direction on the two triangular faces of the wedge are also omitted, since again these forces have no effect in the x direction. From geometrical considerations, areas dA and dC are related by:

These last two equations yield:

verifying that the pressure is independent of the orientation of the surface being considered. A force balance in the z direction leads to a similar result, p_A = p_B.¹

¹ Actually, a force balance in the z direction demands that the gravitational weight of the wedge be considered, which is proportional to the volume of the wedge. However, the pressure forces are proportional to the areas of the faces. It can readily be shown that the volume-to-area effect becomes vanishingly small as the wedge becomes infinitesimally small, so that the gravitational weight is inconsequential.

For moving fluids, the normal stresses include both a pressure and extra stresses caused by the motion of the fluid, as discussed in detail in Section 5.6.

The amount by which a certain pressure exceeds that of the atmosphere is termed the gauge pressure, the reason being that many common pressure gauges are really differential instruments, reading the difference between a required pressure and that of the surrounding atmosphere. Absolute pressure equals the gauge pressure plus the atmospheric pressure.

Velocity. Many problems in fluid mechanics deal with the velocity of the fluid at a point, equal to the rate of change of the position of a fluid particle with time, thus having both a magnitude and a direction. In some situations, particularly those treated from the macroscopic viewpoint, as in Chapters 2, 3, and 4, it sometimes suffices to ignore variations of the velocity with position. In other cases—particularly those treated from the microscopic viewpoint, as in Chapter 6 and later—it is invariably essential to consider variations of velocity with position.

Velocity is not only important in its own right, but leads immediately to three fluxes or flow rates. Specifically, if u denotes a uniform velocity (not varying with position):

1. If the fluid passes through a plane of area A normal to the direction of the velocity, as shown in Fig. 1.6, the corresponding volumetric flow rate of fluid through the plane is Q = uA.

Fig. 1.6 Fluid passing through an area A: (a) Uniform velocity, (b) varying velocity.

2. The corresponding mass flow rate is m = ρQ = ρuA, where ρ is the (constant) fluid density. The alternative notation with an overdot, , is also used.

3. When velocity is multiplied by mass it gives momentum, a quantity of prime importance in fluid mechanics. The corresponding momentum flow rate passing through the area A is  = mu = ρu²A.

If u and/or ρ should vary with position, as in Fig. 1.6(b), the corresponding expressions will be seen later to involve integrals over the area A: 

Basic laws. In principle, the laws of fluid mechanics can be stated simply, and—in the absence of relativistic effects—amount to conservation of mass, energy, and momentum. When applying these laws, the procedure is first to identify a system, its boundary, and its surroundings; and second, to identify how the system interacts with its surroundings. Refer to Fig. 1.7 and let the quantity X represent either mass, energy, or momentum. Also recognize that X may be added from the surroundings and transported into the system by an amount X_in across the boundary, and may likewise be removed or transported out of the system to the surroundings by an amount X_out.

Fig. 1.7 A system and transports to and from it.

The general conservation law gives the increase ΔX_system in the X-content of the system as:

Although this basic law may appear intuitively obvious, it applies only to a very restricted selection of properties X. For example, it is not generally true if X is another extensive property such as volume, and is quite meaningless if X is an intensive property such as pressure or temperature.

In certain cases, where Xⁱ is the mass of a definite chemical species i, we may also have an amount of creation  or destruction  due to chemical reaction, in which case the general law becomes:

The conservation law will be discussed further in Section 2.1, and is of such fundamental importance that in various guises it will find numerous applications throughout all of this text.

To solve a physical problem, the following information concerning the fluid is also usually needed:

1. The physical properties of the fluid involved, as discussed in Section 1.4.

2. For situations involving fluid flow, a constitutive equation for the fluid, which relates the various stresses to the flow pattern.

1.4. Physical Properties—Density, Viscosity, and Surface Tension

There are three physical properties of fluids that are particularly important: density, viscosity, and surface tension. Each of these will be defined and viewed briefly in terms of molecular concepts, and their dimensions will be examined in terms of mass, length, and time (M, L, and T). The physical properties depend primarily on the particular fluid. For liquids, viscosity also depends strongly on the temperature; for gases, viscosity is approximately proportional to the square root of the absolute temperature. The density of gases depends almost directly on the absolute pressure; for most other cases, the effect of pressure on physical properties can be disregarded.

Typical processes often run almost isothermally, and in these cases the effect of temperature can be ignored. Except in certain special cases, such as the flow of a compressible gas (in which the density is not constant) or a liquid under a very high shear rate (in which viscous dissipation can cause significant internal heating), or situations involving exothermic or endothermic reactions, we shall ignore any variation of physical properties with pressure and temperature.

Densities of liquids. Density depends on the mass of an individual molecule and the number of such molecules that occupy a unit of volume. For liquids, density depends primarily on the particular liquid and, to a much smaller extent, on its temperature. Representative densities of liquids are given in Table 1.1.² (See Eqns. (1.9)–(1.11) for an explanation of the specific gravity and coefficient of thermal expansion columns.) The accuracy of the values given in Tables 1.1–1.6 is adequate for the calculations needed in this text. However, if highly accurate values are needed, particularly at extreme conditions, then specialized information should be sought elsewhere.