Chapter 3Chapter 3

The Equations of ChangeThe Equations of Change

for Isothermal Systemsfor Isothermal Systems

In Chapter 2, velocity distributions were determined for several simple flow systems by the shell momentum balance method. The resulting velocity distributions were then used to get other quantities, such as the average velocity and drag force.

For more complex problems we need a general mass balance and a general momentum balance that can be applied to any problem, including problems with nonrectilinear motion. That is the main point of this chapter.

The two equations that we derive are called the equation of continuity (for the mass balance) and the equation of motion (for the momentum balance). These equations can be used as the starting point for studying all problems involving the isothermal flow of a pure fluid.

Those equations are called as equations of change because they describe the change of velocity due to the change of time and position in the fluid system.

§3.1. §3.1. THE EQUATION OF CONTINUITY THE EQUATION OF CONTINUITY (MASS BALANCE)(MASS BALANCE)

This equation is developed by writing a mass balance over a volume element x.y.z , fixed in space, through which a fluid is flowing (see Fig. 3.1-1):

The rate of mass entering the volume element through the shaded face at x is (vx)|x y.z, and the rate of mass leaving through the shaded face at x + x is (vx)|x+x y.z.

Similar expressions can be written for the other two pairs of faces. The rate of increase of mass within the volume element is x.y.z(∂/∂t). The mass balance then becomes

By dividing the entire equation by x.y.z and taking the limit as x, y, and z go to zero, and then using the definitions of the partial derivatives, we get

x x x x x y y y y y

z z z z z

x y z y z v v x z v vt

x y v v

(3.1-2)

. (3.1-3)

This is the equation of continuity, which describes the time rate of change of the fluid density at a fixed point in space. This equation can be written more concisely by using vector notation as follows:.

.v = "divergence of v"

x y zv v vt x y z

The vector v is the mass flux, and its divergence has a simple meaning: it is the net rate of mass efflux per unit volume. A very important special form of the equation of continuity is that for a fluid of constant density, for which Eq. 3.1-4 assumes the particularly simple form (incompressible fluid)

.

Of course, no fluid is truly incompressible, but frequently in engineering and biological applications, the assumption of constant density results in considerable simplification and very little error.

Example 3.1-1. Normal Stresses at Solid Example 3.1-1. Normal Stresses at Solid Surfaces for Incompressible Newtonian Surfaces for Incompressible Newtonian FluidsFluidsShow that for any kind of flow pattern, the normal stresses are

zero at fluid-solid boundaries, for Newtonian fluids with constant density. This is an important result that we shall use often.

SOLUTIONWe visualize the flow of a fluid near some solid surface, which

may or may not be flat. The flow may be quite general, with all three velocity components being functions of all three coordinates and time.

At some point P on the surface we erect a Cartesian coordinate system with the origin at P.

We now ask what the normal stress zz is at P.According to Table B.l or Eq. 1.2-6, zz = -2(dvz/dz), because

for incompressible fluids. Then at point P on the surface of the solid 0 .v

.

First we replaced the derivative dvz/dz by using Eq. 3.1-3 with constant. However, on the solid surface at z = 0, the velocity vx is zero at any position of x by the no-slip condition (see §2.1), and therefore the derivative dvx/dx on the surface = 0. The same is true of dvy/dy on the surface. Therefore zz is zero.

It is also true that xx and yy are zero at the surface because of the vanishing of the derivatives at z = 0. (Note: The vanishing of the normal stresses on solid surfaces does not apply to polymeric fluids, which are viscoelastic).

For compressible fluids, the normal stresses at solid surfaces are zero if the density is not changing with time (see Problem 3C.2.)

According to mass balance for incompressible flow

z

x

y

0

§3.2. §3.2. THE EQUATION OF MOTION THE EQUATION OF MOTION (MOMENTUM BALANCE) (MOMENTUM BALANCE) To get the equation of motion we write a momentum balance

over the volume element x.y.z in Fig. 3.2-1 of the form.

Note that Eq. 3.2-1 is an extension of Eq. 2.1-1 to unsteady-state problems.

The fluid is allowed to move through all six faces of the volume element.

Remember that Eq. 3.2-1 is a vector equation with components in each of the 3 coordinate directions x, y or z containing shear stresses, normal stresses and convective momentum fluxes.

We develop the x-component of each term in Eq. 3.2-1.The y- and z-components may be treated analogously.First, we consider the rates of flow of the x-component

of momentum into and out of the volume element shown in Fig. 3.2-1. (second subscript x: directions of shear and normal stresses for molecular transport and direction of convective momentum flux for convective transport)

Momentum enters and leaves x.y.z by two mechanisms: molecular transport (see §1.2) and convective transport (see §1.7).

First subscripts x, y, and z (cause): directions of momentum transfers due to the change of velocity for molecular transport or due to convection represented by velocity in x, y and z directions for convective transport. Cause can be from all directions.Second subscripts x (effect): directions of shear stress or normal stress or direction of convective momentum flux represented by direction of mass flux (in x direction). Effect is only in one direction in a momentum balance

Second subscripts of all components are the same, i.e. x (direction of effects)

The rate at which the x-component of momentum enters across the shaded face at x by all mechanisms-both convective and molecular-is xx|x yz and the rate at which it leaves the shaded face at x + x is xx|x+x yz.

The rates at which x-momentum enters and leaves through the faces at y and y + y are yx|y zx and yx|y+y zx respectively.

Similarly, the rates at which x-momentum enters and leaves through the faces at z and z + z are zx|z xy and zx|z+z xy

When these contributions are added we get for the net rate of addition of x-momentum across all three pairs of faces.

.

Next there is the external force (typically the gravitational force) acting on the fluid in the volume element. The x-component of this force is

Equations 3.2-2 and 3.2-3 give the x-components of the three terms on the right side of Eq. 3.2-1.

The sum of these terms must then be equated to the rate of increase of x-momentum within the volume element:

x.y.z ∂(vx)/∂t. When this is done, we have the x-component of the momentum balance. When this

equation is divided by x.y.z and the limit is taken as

x, y and z → zero, the following equation results:

x-direction

Here we have made use of the definitions of the partial derivatives. Similar equations can be developed for the y- and z-components of the momentum balance:

.

By using vector-tensor notation, these three equations can be written as follows:

.

This is a vector equation (vector dot tensor = vector)

That is, by letting i be successively x, y, and z, Eqs. 3.2-4,5, and 6 can be reproduced. The quantities vi are the Cartesian components of the vector v, which is the momentum per unit volume at a point in the fluid.

Similarly, the quantities gi are the components of the vector g, which is the external force per unit volume. The term -[.]i is the ith component of the vector -[.].

When the magnitude of ith component of Eq. 3.2-7 is multiplied by the unit vector in the ith direction and the three components are added together vectorially, we get

which is the differential statement of the law of conservation of momentum. It is the translation of Eq. 3.2-1 into mathematical symbols.

In Eq. 1.7-1 it was shown that the combined momentum flux tensor is the sum of the convective momentum flux tensor vv and the molecular momentum flux tensor , and that the latter can be written as p + . When we insert = p + vv + into Eq. 3.2-8, we get the following equation of motion

Vector equation combining 3 directions of 2nd subscripts

In this equation p is a vector (=vector times scalar) called the "gradient of (the scalar) p" sometimes written as "grad p ". The symbol [.] is a vector (=vector dot tensor) called the "divergence of (the tensor) " and [.vv] is a vector (=vector dot tensor) called the "divergence of vv.“

In the next two sections we give some formal results that are based on the equation of motion. The equations of change for mechanical energy and angular momentum are not used for problem solving in this chapter, but will be referred to in Chapter 7 (this chapter is excluded from the lecture material!).

§3.5. §3.5. THE EQUATIONS OF CHANGE IN THE EQUATIONS OF CHANGE IN TERMS OF THE SUBSTANTIAL TERMS OF THE SUBSTANTIAL DERIVATIVEDERIVATIVE

The Partial Time Derivative ∂/∂t (derivative against one variable)Suppose we stand on a bridge and observe the concentration of

fish just below us as a function of time. We can then record the time rate of change of the fish concentration at a fixed location. The result is (∂c/∂t)|x,y,z the partial derivative of c with respect to t, at constant x, y, and z.

The Total Time Derivative d/dt (derivative against all variables)Now suppose that we jump into a motor boat and speed around

on the river, sometimes going upstream, sometimes downstream, and sometimes across the current as we wish.

All the time we are observing fish concentration. At any instant, the time rate of change of the observed fish concentration is

. (3.5-1)

in which dx/dt, dy/dt, and dz/dt are the components of the velocity of the boat.

The Substantial Time Derivative D/DtNext we climb into a canoe and we just float along with the

current to observe the fish concentration.

x,y ,z y ,x,t x,y ,tx,z ,t

dc c dx c dy c dz c

dt t dt x dt y dt z

In this situation the velocity of the observer = the velocity v of the stream, which has components vx, vy, and vz.

If at any instant we report the time rate of change of fish concentration, we are then giving

The special operator D/Dt = ∂/∂t + v. is called the substantial derivative (meaning that the time rate of change is reported as one moves with the "substance"). The terms material derivative, hydrodynamic derivative, and derivative following the motion are also used.

Now we need to know how to convert equations expressed in terms of ∂/∂t into equations written with D/Dt. For any scalar function f(x,y,z,t) we can do the following manipulations:

The quantity in the second parentheses in the second line = 0 according to the equation of continuity.

=0

According to mass balance

Consequently Eq. 3.5-3 can be written in vector form asSimilarly, for any vector function f(x,y,z,t),

These equations can be used to rewrite the equations of change given in §§3.1 to 3.4 in terms of the substantial derivative as shown in Table 3.5-1.

D/Dt = ∂/∂t + v.

temporal change

Spatial/positional change

: in Chapter 7, excluded in our lecture

Equation A in Table 3.5-1 tells how the density is decreasing or increasing as one moves along with the fluid, because of the compression [(.v) < 0] or expansion of the fluid [(.v) > 0].

Equation B can be interpreted as (mass) x (acceleration) = the sum of the pressure forces, viscous forces, and the external force. In other words, Eq. 3.2-9 is equivalent to Newton's second law of motion

(density x acceleration = summation of all forces /volume)

Three most common simplifications of the equation of motion:For constant and , insertion of the Newtonian expression

for from Eq. 1.2-7 into the equation of motion leads to the very famous Navier-Stokes equation, first developed from molecular arguments by Navier, a French engineer, and from continuum arguments by Stokes, an English mathematician:.

When the acceleration terms in Navier-Stokes equation are neglected-that is, when (Dv/Dt) = 0 -we get

which is called the Stokes flow equation. It is sometimes called the creeping flow equation, because the term (v.v] 0 when the flow is extremely slow and can be approached as steady flow.

When viscous forces in Navier-Stokes equation are neglected - that is, . = 2v = 0 - the equation of motion becomes (normal and shear stresses occur due to viscosity)

which is known as the Euler equation for "inviscid" fluid in unsteady flow. Of course, there are no truly "inviscid" fluids, but there are many flows in which the viscous forces are relatively unimportant (far from solid surfaces or very high velocity). Examples are the flow around airplane wings (except near the solid boundary), flow of rivers around the upstream surfaces of bridge supports, some problems in compressible gas dynamics, and flow of ocean current.

Example 3.5-1. The Bernoulli EquationExample 3.5-1. The Bernoulli Equationfor the Steady Flow of Inviscid Fluidsfor the Steady Flow of Inviscid Fluids

The Bernoulli equation for steady flow of inviscid, incompressible fluids (conditions in Stokes’ flow and Euler eq’s) is one of the most famous equations in classical fluid dynamics. Show how it is obtained from the Euler equation of motion.

SOLUTIONInviscid Fluids omit the time-derivative term in Eq. 3.5-

9, and then use the vector identity [.vvl = [v.vl = ½ (v.v) - [v x [ x v]] (Eq. A.4-23) to rewrite the Navier Stokes equation as

Next we divide Eq. 3.5-10 by and then form the dot product with the unit vector s = v/|v| in the flow direction. When the fluid is inviscid, then there is no vorticity ( x v = 0) and consequently v x ( x v) = 0, and (s.) can be replaced by d/ds, where s is the distance along a streamline. Thus we get

When this is integrated along a streamline from point 1 to point 2, we get

which is called the Bernoulli equation. It relates the velocity, pressure, and elevation of two points along a streamline in a fluid in steady-state flow of inviscid fluid.

§3.6. §3.6. USE OF THE EQUATIONS OF CHANGEUSE OF THE EQUATIONS OF CHANGETO SOLVE FLOW PROBLEMSTO SOLVE FLOW PROBLEMS

To describe the flow of a Newtonian fluid at constant temperature, we need in general The equation of continuity Eq. 3.1-4 The equation of motion Eq. 3.2-9 The components of Eq. 1.2-6 The equation of state = (p) The equations for the viscosities = (p, T)

These equations, along with the necessary boundary (related to positions) and initial (related to time) conditions, determine completely the pressure, density, and velocity distributions in the fluid.

They are seldom used in their complete form to solve fluid dynamics problems. Usually restricted forms are used for convenience, as in this chapter.

If it is appropriate to assume constant density and viscosity, then we use The equation of continuity Eq. 3.1-4 and Table B.4 The Navier-Stokes equation Eq. 3.5-6 and Tables B.5,

6, 7 along with initial and boundary conditions. From these one determines the pressure and velocity

distributions.

Example 3.6-1. Steady Flow in a Long Example 3.6-1. Steady Flow in a Long Circular TubeCircular Tube

Rework the tube-flow problem of Example 2.3-1 using the equations of continuity and motion. This illustrates the use of the tabulated equations for constant viscosity and density in cylindrical coordinates, given in Appendix B.5.

SOLUTIONWe postulate that v = zvz(r, z). This postulate implies that

there is no radial flow (vr = 0) and no tangential flow (v = 0), and that vz ≠ f ().

We assume that there is no change of velocity profile in z direction.

Consequently, we can discard many terms from the tabulated equations of change, leaving

Datum plane

The postulate and Eq. 3.6-1 indicates that vz depends only on r; hence the partial derivatives in the second term on the right side of Eq. 3.6-4 can be replaced by ordinary derivatives.

By using the modified pressure P = p - gh (where h is the height below some arbitrary datum plane and g is a constant), we avoid the necessity of calculating the components of g in r and coordinates, and we obtain a solution valid for any orientation of the axis of the tube.

Equations 3.6-2 and 3.6-3 show that P is a function of z alone, and the partial derivative in the first term of Eq. 3.6-4 may be replaced by an ordinary derivative.

For constant change of P against z, by introducing a constant C0, Eq. 3.6-4 reduces to

The P equation can be integrated at once. The vz-equation

can be integrated one operation after another on the left side (do not "work out" the compound derivative there). This gives

.

The 4 constants of integration can be found from the boundary conditions:

.

ln 0 = indefinite, so C2 must not be existence or mathematically C2= 0 to obtain definite vz.. The resulting solutions are:

.

As pointed out in Example 2.3-1, Eq. 3.6-13 is valid only in the laminar-flow regime, and at locations not too near the tube entrance and exit. For Re > about 2100, a turbulent-flow regime exists downstream of the entrance region, and Eq. 3.6-13 is no longer valid.

Example 3.6-2. Falling FilmExample 3.6-2. Falling Film

Set up the problem in Example 2.2-2 by using the equations of Appendix B.5. This illustrates the use of the equation of motion in terms of .

SOLUTIONAs in Example 2.2-2 we postulate a steady-state flow

with constant density. We postulate, as before, that the x- and y-components of

the velocity are zero (vx and vy = 0) and that vz = vz(x). With these postulates, the equation of continuity is zero.

According to Table B.l, the only nonzero components of are xz = zx = -(dvz/dx). The components of the equation of motion in terms of are, from Table B.5,

.

Integration of Eq. 3.6-14 gives.in which f(y, z) is an arbitrary function. Equation 3.6-15

shows that f cannot be a function of y.

We next recognize that the pressure in the gas phase is very nearly constant at the prevailing atmospheric pressure patm. Therefore, at the gas-liquid interface x = 0, the pressure is also constant at the value patm. Consequently, f can be set equal to patm, and we obtain finally from 3.6-14.

.(p is function of x only). Equation 3.5-16 then becomes.which is the same as Eq. 2.2-10. The remainder of the

solution is the same as in §2.2.

Example 3.6-3. Operation of a Couette Example 3.6-3. Operation of a Couette ViscometerViscometer

The viscosity may also be determined by measuring the torque required to turn a solid object in contact with a fluid. The forerunner of all rotational viscometers is the Couette instrument, which is sketched in Fig. 3.6-1.

Determine velocity distribution and shear stress for the laminar, tangential flow of an incompressible fluid between 2 co-axial vertical cylinders. Outer cylinder rotates with angular velocity o (see Figure 3.6-1). End-effects is negligible.

SolutionIn steady-state laminar flow, fluid moves in circular

direction with velocity components vr = 0 and vz = 0. There is no pressure gradient in direction (p = p(r,z)). It is expected that p depends on z due to gravity and on r due to centrifugal force.

For these postulates all the terms in the equation of continuity are zero, and the components of the equation of motion simplify to

The first equation tells how the centrifugal force affects the pressure.

The second equation gives the velocity distribution. The third equation gives the effect of gravity on the

pressure (the hydrostatic effect)For the problem at hand we need only the -component of

the equation of motion for velocity distribution

0 zp

gz

Integration of Eq. 3.6-21 results in.

The boundary conditions are that the fluid does not slip at the two cylindrical surfaces:

.

These boundary conditions can be used to get the constants of integration, which are then inserted in Eq. 3.6-26. This gives

.

From the velocity distribution we can find the momentum flux by using Table B.1:

The torque acting on the inner cylinder is then given by the product of the inward momentum flux (-r), the surface of the cylinder, and the lever arm, as follows:

.

Therefore, measurement of the angular velocity of the cup makes it possible to determine the viscosity. The same kind of analysis is available for other rotational viscometers.

For any viscometer it is essential to know when turbulence will occur. The critical Reynolds number (oR2/), above which the system becomes turbulent, is shown in Fig. 3.6-2 as a function of the radius ratio .

One might ask what happens if we hold the outer cylinder fixed and cause the inner cylinder to rotate with an angular velocity i (the subscript "i" stands for inner). Then the velocity distribution is

This is obtained by making the same postulates (see before Eq. 3.6-20) and solving the same differential equation (Eq. 3.6-21), but with a different set of boundary conditions.