CONSERVATION OF MOMENTUM

EXERCISES

4.4. CONSERVATION OF MOMENTUM

In this section, the momentum-conservation equivalent of (4.5) is developed from Newton’s second law, the fundamental principle governing fluid momentum. When applied to a material volumeV(t) with surface areaA(t), Newton’s second law can be stated directly as:

d dt

Z

VðtÞ

rðx,tÞuðx,tÞdV ¼ Z

VðtÞ

rðx,tÞgdVþ Z

AðtÞ

fðn,x,tÞdA, (4.13) whereruis the momentum per unit volume of the flowing fluid,gis the body force per unit mass acting on the fluid withinV(t),fis the surface force per unit area acting onA(t), andnis the outward normal onA(t). The implications of (4.13) are better displayed when the time derivative is expanded using Reynolds transport theorem (3.35) withF¼ruandb¼u:

Z

VðtÞ

v

vtðrðx,tÞuðx,tÞÞdVþ Z

AðtÞ

rðx,tÞuðx,tÞðuðx,tÞ,nÞdA

¼ Z

VðtÞ

rðx,tÞgdVþ Z

AðtÞ

fðn,x,tÞdA: (4.14)

This is a momentum-balance statement between integrated momentum changes withinV(t), integrated momentum contributions from the motion of A(t), and integrated volume and surface forces. It is the momentum conservation equivalent of (4.2).

To develop an integral equation that represents momentum conservation for an arbitrarily moving control volumeV*(t) with surfaceA*(t), (4.14) must be modified to involve integra- tions overV*(t) and A*(t). The steps in this process are entirely analogous to those taken between (4.2) and (4.5) for conservation of mass. First setF¼ru in (3.35) and rearrange it to obtain:

Z

VðtÞ

v

vtðrðx,tÞuðx,tÞÞdV ¼ d dt

Z

VðtÞ

rðx,tÞuðx,tÞdV Z

AðtÞ

rðx,tÞuðx,tÞb,ndA ¼ 0: (4.15)

Then chooseV*(t) to be instantaneously coincident withV(t) so that at the moment of interest:

Z

VðtÞ

v

vtðrðx,tÞuðx,tÞÞdV ¼ Z

VðtÞ

v

vtðrðx,tÞuðx,tÞÞdV, Z

AðtÞ

rðx,tÞuðx,tÞðuðx,tÞ,nÞdA ¼ Z

AðtÞ

rðx,tÞuðx,tÞðuðx,tÞ,nÞdA, Z

VðtÞ

rðx,tÞgdV ¼ Z

VðtÞ

rðx,tÞgdV, and Z

AðtÞ

fðn,x,tÞdA ¼ Z

AðtÞ

fðn,x,tÞdA:

(4.16a, 4.16b, 4.16c, 4.16d)

4.4. CONSERVATION OF MOMENTUM 101

Now substitute (4.16a) into (4.15) and use this result plus (4.16b, 4.16c, 4.16d) to convert (4.14) to:

d dt

Z

VðtÞ

rðx,tÞuðx,tÞdVþ Z

AðtÞ

rðx,tÞuðx,tÞðuðx,tÞ bÞ,ndA

¼ Z

VðtÞ

rðx,tÞgdVþ Z

AðtÞ

fðn,x,tÞdA: (4.17)

This is the general integral statement of momentum conservation for an arbitrarily moving control volume. Just like (4.5), it can be specialized to stationary, steadily moving, acceler- ating, or deforming control volumes by appropriate choice ofb. For example, whenb¼u, the arbitrary control volume becomes a material volume and (4.17) reduces to (4.13).

At this point, the forces in (4.13), (4.14), and (4.17) merit some additional description that facilitates the derivation of the differential equation representing momentum conservation and allows its simplification under certain circumstances.

The body force,rgdV, acting on the fluid element dVdoes so without physical contact.

Body forces commonly arise from gravitational, magnetic, electrostatic, or electromagnetic force fields. In addition, in accelerating or rotating frames of reference,fictitiousbody forces arise from the frame’s noninertial motion (see Section 4.7). By definition body forces are distributed through the fluid and are proportional to mass (or electric charge, electric current, etc.). In this book, body forces are specified per unit mass and carry the units of acceleration.

Body forces may be conservative or nonconservative.Conservative body forcesare those that can be expressed as the gradient of a potential function:

g ¼ VF or gj ¼ vF=vxj, (4.18)

whereF is called theforce potential; it has units of energy per unit mass. When thez-axis points vertically upward, the force potential for gravity isF¼gz, wheregis the acceleration of gravity, and (4.18) produces g ¼ egez. Forces satisfying (4.18) are called conservative because the work done by conservative forces is independent of the path, and the sum of fluid-particle kinetic and potential energies is conserved when friction is absent.

Surface forces, f, act on fluid elements through direct contact with the surface of the element. They are proportional to the contact area and carry units of stress (force per unit area). Surface forces are commonly resolved into components normal and tangential to the contact area. Consider an arbitrarily oriented element of areadAin a fluid (Figure 2.5).

If n is the surface normal with components ni, then from (2.15) the components fj of the surface force per unit areaf(n,x,t) on this element arefj¼nisij, wheresijis the stress tensor.

Thus, the normal component offisn,f¼n_if_i, while the tangential component is the vector fe(n,f)nwhich has componentsf_ke(n_if_i)n_k.

Other forces that influence fluid motion are surface- and interfacial-tension forces that act on lines or curves embedded within interfaces between liquids and gases or between immis- cible liquids (see Figure 1.4). Although these forces are commonly important in flows with such interfaces, they do not appear directly in the equations of motion, entering instead through the boundary conditions.

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Before proceeding to the differential equation representing momentum conservation, the use of (4.5) and (4.17) for stationary, moving, and accelerating control volumes having a variety of sizes and shapes is illustrated through a few examples. In all four examples, equations representing mass and momentum conservation must be solved simultaneously.

EXAMPLE 4.1

A long bar with constant cross section is held perpendicular to a uniform horizontal flow of speedUN, as shown inFigure 4.2. The flowing fluid has densityrand viscositym(both constant).

The bar’s cross section has characteristic transverse dimensiond, and the span of the bar islwith l[d. The average horizontal velocity profile measured downstream of the bar isU(y), which is less thanUNdue to the presence of the bar. Determine the required force per unit span,eFD/l, applied to the ends of the bar to hold it in place. Assume the flow is steady and two dimensional in the plane shown. Ignore body forces.

Solution

Before beginning, it is important to explain the sign convention for fluid dynamic drag forces.

The drag force on an inanimate object isthe force applied to the object by the fluid. Thus, for stationary objects, drag forces are positive in the downstream direction, the direction the object would accelerate if released. However, the control volume laws are written forforces applied to the contents of the volume. Thus, from Newton’s third law, a positive drag force on an object implies a negative force on the fluid. Therefore, theFDappearing inFigure 4.2is a positive number and this will be borne out by the final results. Here we also note that since the horizontal velocity downstream of the bar, the wake velocityU(y), is less thanU_N, the fluid has been decelerated inside the control volume and this is consistent with a force from the body opposing the motion of the fluid as shown.

The basic strategy is to select a stationary control volume, and then use (4.5) and (4.17) to determine the forceFDthat the body exerts on the fluid per unit span. The first quantitative step in the solution is to select a rectangular control volume with flat control surfaces aligned with the coordinate directions. The inlet, outlet, and top and bottom sides of such a control volume are shown inFigure 4.2. The vertical sides parallel to thex-yplane are not shown. However, the flow does not vary in the third direction and is everywhere parallel to these surfaces so these merely need be selected a comfortable distancelapart. The inlet control surface should be far enough upstream of the bar so that the inlet fluid velocity isUNex, the pressure ispN, and both are uniform.

H

d x

U y

U(y) U –FD

inlet outlet

top

bottom

FIGURE 4.2 Momentum and mass balance for flow past long bar of constant cross section placed perpendicular to the flow. The intersection of the recommended stationary control volume with thex-yplane is shown with dashed lines. The forceeFDholds the bar in place and slows the fluid that enters the control volume.

4.4. CONSERVATION OF MOMENTUM 103

The top and bottom control surfaces should be separated by a distanceHthat is large enough so that these boundaries are free from shear stresses, and the horizontal velocity and pressure are so close toUNand pN that any difference can be ignored. And finally, the outlet surface should be far enough downstream so that streamlines are nearly horizontal there and the pressure can again be treated as equal top_N.

For steady flow and the chosen stationary volume, the control surface velocity isb¼0 and the time derivative terms in (4.5) and (4.17) are both zero. In addition, the surface force integral contributeseFDe_xwhere the beam crosses the control volume’s vertical sides parallel to thex-y plane. The remainder of the surface force integral contains only pressure terms since the shear stress is zero on the control surface boundaries. After setting the pressure top_Non all control surfaces, (4.5) and (4.17) simplify to:

Z

AðtÞ

ruðxÞ,ndA ¼ 0, and Z

A

ruðxÞuðxÞ,ndA ¼ Z

A

pNndAFDex:

In this case the pressure integral may be evaluated immediately by using Gauss’ divergence theorem:

Z

A

pNndA ¼ Z

V

VpNdV ¼ 0,

with the final value (zero) occurring becausepNis a constant. After this simplification, denote the fluid velocity components by (u,v) ¼ u, and evaluate the mass andx-momentum conservation equations:

Z

inlet

rUNldyþ Z

top

rvldx Z

bottom

rvldxþ Z

outlet

rUðyÞldy ¼ 0, and

Z

inlet

rU²_Nldyþ Z

top

rUNvldx Z

bottom

rUNvldxþ Z

outlet

rU²ðyÞldy ¼ FD

whereu,ndAis:eU_Nldyon the inlet surface,þvldxon the top surface,evldxon the bottom surface, andþU(y)ldyon the outlet surface wherelis the span of the flow into the page. Dividing both equations byrl, and combining like integrals produces:

Z

top

vdx Z

bottom

vdx ¼

þH=2Z

H=2

ðU_NUðyÞÞdy, and

U_N 0 B@ Z

top

vdx Z

bottom

vdx 1 CAþ

þH=2Z

H=2

U² y

U_N²

dy ¼ FD=rl:

Eliminating the top and bottom control surface integrals between these two equations leads to:

FD=l ¼ r Z

þH=2

H=2

UðyÞðU_NUðyÞÞdy,

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which produces a positive value ofFDwhenU(y) is less thanUN. An essential feature of this anal- ysis is that there are nonzero mass fluxes through the top and bottom control surfaces. The final formula here is genuinely useful in experimental fluid mechanics since it allowsFD/lto be deter- mined from single-component velocity measurements made in the wake of an object.

EXAMPLE 4.2

Using a stream-tube control volume of differential length ds, derive the Bernoulli equation, (½)rU²þgzþp/r¼constant along a streamline, for steady, inviscid, constant density flow where Uis the local flow speed.

Solution

The basic strategy is to use a stationary stream-tube-element control volume, (4.5), and (4.17) to determine a simple differential relationship that can be integrated along a streamline. The geometry is shown inFigure 4.3. For steady inviscid flow and a stationary control volume, the control surface velocityb¼0, the surface friction forces are zero, and the time derivative terms in (4.5) and (4.17) are both zero. Thus, these two equations simplify to:

Z

AðtÞ

ruðxÞ,ndA ¼ 0 and Z

AðtÞ

ruðxÞuðxÞ,ndA ¼ Z

VðtÞ

rgdV Z

AðtÞ

pndA:

The geometry of the volume plays an important role here. The nearly conical curved surface is tangent to the velocity while the inlet and outlet areas are perpendicular to it. Thus,u,ndAis:eUdA on the inlet surface, zero on the nearly conical curved surface, andþ[Uþ(vU/vs)ds]dAon the outlet surface. Therefore, conservation of mass with constant density leads to

ds

ρ

p A

Uρ

θ

x

y z

stream tube

g p+

(

∂ ∂^s

)

^ds

A+

(

∂ ∂^s

)

^ds

U+

(

∂ ∂^s

)

^ds

extra pressure

force

p A

U

FIGURE 4.3 Momentum and mass balance for a short segment of a stream tube in steady inviscid constant density flow. Here, the inlet and outlet areas are perpendicular to the flow direction, and they are small enough so that only first-order corrections in the stream direction need to be considered. The alignment of gravity and stream tube leads to a vertical change of sinqds¼dzbetween its two ends. The area difference between the two ends of the stream tube leads to an extra pressure force.

4.4. CONSERVATION OF MOMENTUM 105

erUAþr

UþvU

vsds AþvA

vsds ¼ 0,

where first-order variations in U and A in the stream-wise direction are accounted for. Now consider the stream-wise component of the momentum equation recalling thatu ¼ Ueuand setting g ¼egez. For inviscid flow, the only surface force is pressure, so the simplified version of (4.17) becomes

rU²Aþr

UþvU vsds

2 AþvA

vsds

¼ rgsinq

AþvA vs

ds

2 dsþpAþ

pþvp vs

ds 2

vA vsds

pþvp

vsds AþvA vsds : Here, the middle pressure term comes from the extra pressure force on the nearly conical surface of the stream tube.

To reach the final equation, use the conservation of mass result to simplify the flux terms on the left side of the stream-wise momentum equation. Then, simplify the pressure contributions by canceling common terms, and note that sinqds¼dzto find

rU²AþrU

UþvU

vsds A¼ rUAvU vsds

¼ rg

AþvA vs

ds

2 dzþvp vs

vA vs

ðdsÞ² 2 Avp

vsdsvp vs

vA vsðdsÞ²:

Continue by dropping the second-order terms that containðdsÞ²ordsdz, and divide byrAto reach:

UvU

vsds ¼ gdz1 r

vp

vsds, or

dðU²=2Þ þgdzþ ð1=rÞdp ¼ 0

along a streamline:

Integrate the final differential expression along the streamline to find:

1

2U²þgzþp=r ¼ a constant along a streamline: (4.19)

EXAMPLE 4.3

Consider a small solitary wave that moves from right to left on the surface of a water channel of undisturbed depthh(Figure 4.4). Denote the acceleration of gravity byg. Assuming a small change in the surface elevation across the wave, derive an expression for its propagation speed,U, when the channel bed is flat and frictionless.

Solution

Before starting the control volume part of this problem, a little dimensional analysis goes along way toward determining the final solution. The statement of the problem has only three parameters,

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U, g, and h, and there are two independent units (length and time). Thus, there is only one dimensionless group,U²=gh, so it must be a constant. Therefore, the final answer must be in the form:U ¼ const, ffiffiffiffiffi

pgh

, so the value of the following control volume analysis lies merely in deter- mining the constant.

Choose the control volume shown and assume it is moving at speedb ¼ Uex. Here we assume that the upper and lower control surfaces coincide with the water surface and the channel’s fric- tionless bed. They are shown close to these boundaries inFigure 4.4for clarity. Apply the integral conservation laws for mass and momentum.

d dt

Z

VðtÞ

rdV þ Z

AðtÞ

rðubÞ,ndA ¼ 0, d dt

Z

VðtÞ

rudVþ Z

AðtÞ

ruðubÞ,ndA

¼ Z

VðtÞ

rgdVþ Z

AðtÞ

fdA:

With this choice of a moving control volume, its contents are constant so both thed/dtterms are zero; thus,

Z

A

rðuþUexÞ,ndA ¼ 0, and Z

AðtÞ

ruðuþUexÞ,ndA ¼ Z

VðtÞ

rgdVþ Z

AðtÞ

fdA:

Here, all velocities are referred to a stationary coordinate frame, so thatu ¼ 0on the inlet side of the control volume in the undisturbed fluid layer. In addition, label the inlet (left) and outlet (right) water depths ashin andhout, respectively, and save consideration of the simplifications that occur whenðhouthinÞ ðhoutþhinÞ=2for the end of the analysis. LetUoutbe the horizontal flow speed on the outlet side of the control volume and assume its profile is uniform. Therefore ðuþUexÞ,ndAiseUldyon the inlet surface, andþ(UoutþU)ldyon the outlet surface, wherelis (again) the width of the flow into the page. With these replacements, the conservation of mass equation becomes:

rUhinlþrðUoutþUÞhoutl¼ 0, or Uhin ¼ ðUoutþUÞhout, U

g

h

x y

FIGURE 4.4 Momentum and mass balance for a small amplitude water wave moving into quiescent water of depthh. The recommended moving control volume is shown with dashed lines. The wave is driven by the imbalance of static pressure forces on the vertical inlet (left) and outlet (right) control surfaces.

4.4. CONSERVATION OF MOMENTUM 107

and the horizontal momentum equation becomes:

rð0Þð0þUÞhinlþrUoutðUoutþUÞhoutl ¼ Z

inlet

pn,exdA Z

outlet

pn,exdA Z

top

pn,exdA:

Here, no friction terms are included, and the body force term does not appear because it has no hori- zontal component. First consider the pressure integral on the top of the control volume, and let y¼h(x) define the shape of the water surface:

po

Z

n,exdA ¼ po

Z ðffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffidh=dx, 1Þ 1þ ðdh=dxÞ²

q ,ð1, 0Þl

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1þ ðdh=dxÞ² q

dx

¼ po

Z dh

dx dx ¼ po

Z^hout

hin

dh ¼ poðhouthinÞ

where the various square-root factors arise from the surface geometry;pois the (constant) atmo- spheric pressure on the water surface. The pressure on the inlet and outlet sides of the control volume is hydrostatic. Using the coordinate system shown, integrating (1.8), and evaluating the constant on the water surface producesp¼poþrg(hey). Thus, the integrated inlet and outlet pressure forces are:

Z

inlet

pdA Z

outlet

pdA Z

top

pon,exdA

¼ Z^hin

0

poþrg

hiny ldy

Z^hout

0

poþrg

houty

ldyþpoðhouthinÞl

¼ Z^hin

0

rgðhinyÞldy Z^hout

0

rgðhoutyÞldy ¼ rg h²_in 2 h²_out

2

! l

where the signs of the inlet and outlet integrals have been determined by evaluating the dot prod- ucts and we again note that the constant reference pressurepodoes not contribute to the net pres- sure force. Substituting this pressure force result into the horizontal momentum equation produces:

r 0

0þU

hinlþrUout

UoutþU

houtl ¼ rg 2

h²_inh²_out l:

Dividing by the common factors ofrandl, Uout

UoutþU

hout ¼ g

2

h²_inh²_out

,

and eliminatingUoutvia the conservation of mass relationship,Uout ¼ ðhinhoutÞU=hout, leads to:

UðhinhoutÞ hout

UðhinhoutÞ

hout þU hout ¼ g 2

h²_inh²_out :

Dividing by the common factor of (hin e hout) and simplifying the left side of the equation produces:

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108

U²hin

hout ¼ g

2ðhinþhoutÞ, or U ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ghout

2hinðhinþhoutÞ s

z ffiffiffiffiffi pgh

,

where the final approximate equality holds when the inlet and outlet heights differ by only a small amount with both nearly equal toh.

EXAMPLE 4.4

Derive the differential equation for the vertical motion for a simple rocket having nozzle areaAe

that points downward, exhaust discharge speedVe, and exhaust densityr_e,without considering the internal flow within the rocket (Figure 4.5). Denote the mass of the rocket byM(t) and assume the discharge flow is uniform.

Solution

Select a control volume (not shown) that contains the rocket and travels with it. This will be an accelerating control volume and its velocityb ¼ b(t)ezwill be the rocket’s vertical velocity. In addition, the discharge velocity is specified with respect to the rocket, so in a stationary frame of reference, the absolute velocity of the rocket’s exhaust isu¼uzez¼(eVeþb)ez.

The conservation of mass and vertical-momentum equations are:

d dt

Z

VðtÞ

rdV þ Z

AðtÞ

rðubÞ,ndA ¼ 0, d dt

Z

VðtÞ

ruzdVþ Z

AðtÞ

ruzðubÞ,ndA

¼ g Z

VðtÞ

rdVþ Z

AðtÞ

fzdA:

Here we recognize the first term in each equation as the time derivative of the rocket’s massM, and the rocket’s vertical momentumMb, respectively. (The second of these identifications is altered when the rocket’s internal flows are considered; seeThompson, 1972, pp. 43e47.) For ordinary rocketry, the

Ae

ρe, V_e b(t)

g z

V_e M(t)

FIGURE 4.5 Geometry and parameters for a simple rocket having massM(t) that is moving vertically at speedb(t). The rocket’s exhaust area, density, and velocity (or specific impulse) areAe,re, andVe, respectively.

4.4. CONSERVATION OF MOMENTUM 109

rocket exhaust exit will be the only place that mass and momentum cross the control volume boundary and heren¼eez; thusðubÞ,ndA ¼ ðVeezÞ,ðezÞdA ¼ VedAover the nozzle exit. In addition, we will denote the integral of vertical surface stresses by FS, a force that includes the aerodynamic drag on the rocket and the pressure thrust produced when the rocket nozzle’s outlet pressure exceeds the local ambient pressure. With these replacements, the above equations become:

dM

dt þr_eVeAe ¼ 0, d

dtðMbÞ þr_eðVeþbÞVeAe ¼ MgþFS: Eliminatingr_eVeAebetween the two equations produces:

d

dtðMbÞ þ ðVeþbÞ

dM

dt ¼ MgþFS, which reduces to:

Md²zR

dt² ¼ VedM

dt MgþFS,

wherezRis the rocket’s vertical location anddzR/dt¼b. From this equation it is clear that negative dM/dt (mass loss) may produce upward acceleration of the rocket when its exhaust discharge velocity Veis high enough. In fact,Veis the crucial figure of merit in rocket propulsion and is commonly referred to as thespecific impulse, the thrust produced per unit rate of mass discharged.

Returning now to the development of the equations of motion, the differential equation that represents momentum conservation is obtained from (4.14) after collecting all four terms into the same volume integration. The first step is to convert the two surface integrals in (4.14) to volume integrals using Gauss’ theorem (2.30):

Z

AðtÞ

rðx,tÞuðx,tÞðuðx,tÞ,nÞdA ¼ Z

VðtÞ

V,ðrðx,tÞuðx,tÞuðx,tÞÞdV ¼ Z

VðtÞ

v vx_i

ruiuj

dV, and Z

AðtÞ

fðn,x,tÞdA ¼ Z

AðtÞ

n_isijdA ¼ Z

VðtÞ

v vx_i

sij dV,

(4.20a, 4.20b) where the explicit listing of the independent variables has been dropped upon moving to index notation. Substituting (4.20a, 4.20b) into (4.14) and collecting all the terms on one side of the equation into the same volume integration produces:

Z

VðtÞ

v vt

ru_j

þ v vxi

ru_iu_j

rg_j v vxi

sij

dV ¼ 0: (4.21)

Similarly to (4.6), the integral in (4.21) can only be zero for any material volume if the inte- grand vanishes at every point in space; thus (4.21) requires:

v vt

ruj

þ v vx_i

ruiuj

¼ rgjþ v vx_i

sij

: (4.22)

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