Problem Solving with the Momentum Equation in ControlVolume Form
In applying the integralmomentum equation to a given problem, there are two important things to remember:
1. In choosing a suitable control surface, some choices may lead to simpler solutions than others.
2. All of the surface – and bodyforces and momentumflux terms must be accounted for with special attention given to the sign of each term.
Again, recall that we are working with a vector equation, and it may be necessary to account for all three components in some situations.
Notice the analogy between the control volume as used here for fluidmechanics problems and the freebody diagram learned in mechanics problems. Whenever we work with forces and moments, it is necessary to use the methods of dynamics in relating the force system to the motion of the system of particles. When we define a control volume that fits the geometry of a given problem, it also is useful to visualize it as a freebody diagram. The forces on the boundaries of the control volume represent interaction with the outside world. It is important that all forces of interaction are represented. In Table 3.3, only the forces due to fluiddynamic interactions are displayed. It is possible that other forces may need to be accounted for; for example, suppose that the control volume is attached to a beam, strut, spring, or other structure. Such directly applied forces also must appear in the force balance. Incorporation of such mechanical constraints should be second nature from earlier studies of statics and dynamics. We review the freebodydiagram method as extended to fluid dynamics by means of the example problems in this section.
So as not to miss a term or a sign in assessing the control volume for a given problem, a fourstep procedure is suggested for working problems using the integral – momentum equation, as follows: 1
along a coordinate direction, it must be resolved into components in the coordinate system chosen.
2. Identify all forces, including those due to mechanical constraints; some may have unknown magnitude and direction. If the direction of a force component is known, apply the correct sign depending on whether it is in the positive or negative coordinate direction.
3. Identify the parts of the control surface across which there is a flow of momentum. Recognize that the momentumflux integrand contains two signed terms. The mass flow part, p(V« n)dS, is a scalar and therefore has no dependence on coordinate direction; the sign depends only on whether the flux is out of the control volume (positive sign) or into the surface of the control volume (negative sign). The V part of the integrand is the momentum vector (per unit mass) and thus carries a sign depending on how the velocity vector is pointing with respect to the coordinate axes. This sign must be consistent with the positive coordinate directions chosen for the force terms. If the momentum vector is not aligned with a coordinate direction, then a component in that direction must be included.
4. It may be necessary to use the continuity equation as well as the momentum equation to solve for all of the unknowns in the problem. It is generally easiest to do the mass balance (i. e., application of the continuity equation) first. If flow channel areas are initially unknown, they usually can be determined from continuity. Recall that unknown velocity components may be in either direction. If a positive sign is assumed initially for an undetermined velocity component, the solution indicates the proper sign. That is, if it is pointed in the direction opposite to the assumed one, the final solution for the scalar magnitude includes a minus sign indicating this result.
Before presenting example problems using the integralmomentum equation, it is necessary to recall the two reference values of pressure: gauge and absolute (see Chapter 2). A gauge pressure is defined as the pressure above (plus) or below (minus) the atmospheric or ambient pressure. Pressure levels with respect to zero reference (i. e., a perfect vacuum) are termed absolute pressures and are always positive. Thus, at standard sea level, the gauge pressure in quiescent air is 0 psig and the absolute pressure is 14.7 psia. Most thermodynamic relationships, such as the First and Second Laws of Thermodynamics and the Equation of State, require the use of absolute pressure. In the case of the integralmomentum equation, either gauge or absolute pressure can be used for a given problem and leads to the same answer if all surfaces are considered. The benefit of using gauge pressure is that on surfaces on which only quiescent fluid interacts, the pressure force is zero. Of course, it is necessary to use gauge pressure to evaluate all remaining surfaces. The use of absolute pressure may lead to a more complicated solution in some cases, as illustrated in Example 3.3.
example 3.3 Given: A twodimensional free jet of water (p = 62.4 lbm/ft3) impinges with a steady flow onto a stationary flat plate. The jet divides into two streams flowing in opposite directions along the plate. The entire flow is assumed to be frictionless (i. e., no flow losses and no frictional force along
the plate surface). Because there are no flow losses, the velocity and static pressure of the two jets leaving the plate are the same as in the jet. The static pressure in the flow is everywhere ambient pressure because the flow forms a free surface with respect to the ambient air, and there can be no pressure difference across a free surface. Assume all flow segments to be onedimensional. Certain velocity magnitudes and dimensions are known, as indicated in the following drawing:
Required: Find the jet heights “q” and “t2” and predict the magnitude and direction of the force exerted on the plate (per unit depth) by the impinging jet.
Approach: The flow is steady and incompressible. The momentum equation is used to predict the unknown force. Because t1 and t2 also are unknowns, another equation is needed—namely, the continuity equation. The control surface is arbitrary, so choose the rectangular shape ABCD as indicated by the dotted line. The problem is solved in terms of gauge pressure; later, the choice of absolute pressure is examined.
Solution: From continuity Eq. 3.5:
j] (V. n)dS
Hence, A1 + A2 = 2/12 ft2. Now, apply the integralmomentum equation (Momentum Theorem, Eq. 3.37) for steady flow:
jj pndS = jjpV(Vn)dS.
S S
With the problem expressed in terms of gauge pressure, the pressure forces on sides AB, BC, and CD are zero because the pressures on all of these faces
are ambient pressure, including Flow Areas 1, 2, and 3. The pressure force, FAD, on side AD of the control volume, is the integrated pressure force of the plate acting on the control volume. A force equal and opposite to this force is the (unknown) force of the control volume (i. e., of the jet) acting on the plate. Taking the component of Eq. 3.37 in the xdirection, the netforce term on the left side of the equation is zero because there are no gaugepressure forces present (recall that the inviscid assumption means that there are no shear forces along the plate surface AD).
Taking the component of Eq. 3.37 in the уdirection, the only term on the left side is the pressure force, FAD, acting in the positive уdirection. There are three momentumflux terms in the xcomponent equation but only one in the уcomponent equation. Thus,
xdirection:
0 = p(Vi)^^1) + p^2)^2A2) + P^3 cos40°)(V3 A3).
Examine this equation and carefully consider the signs associated with each term. Simplifying and recognizing that V1 = V2 = V3 = V,
V2(A2 – A1) – V2 A3 cos 40° = 0.
Eliminating V and writing A1 in terms of A2 from continuity, this relation becomes:
A2 – (2/12 – A2) = (2/12) cos 40°.
Solving, A2 = 0.195 ft2 (t2 = 0.195 ft) and A1 = 0.147 ft2 (t1 = 0.147 ft). уdirection:
FaD = (0) + (0) + ( – V3 sin 40)( – V3A3)
FAD = V32 A3 sin 40° = (62.4/32.2)(50)2(2/12)sin 40° = 519 lbf/ft.
The plus sign indicates that FAD acts in the positive уdirection. The force on the plate is equal and opposite to this—namely, a force of 519 pounds pushing the plate downward and to the right.
Note: Either of the following two controlvolume choices would lead to the same result (the student should verify this). Choice (a) would make the inflow – flux calculation easier because there would be no need to resolve the velocity components:
Appraisal: The predicted direction of the force on the plate agrees with physical intuition. If the problem were solved using absolute pressures, the result would
be the same but the solution would not have been as straightforward. Working with absolute pressures, there would have been a pressure force on sides AB and CD (which still would cancel each other) and a force on BC having a magnitude of (ambient pressure) x (length BC). This force would have a negative sign (negative уdirection) and, when subtracted in Eq. 3.37 would have the effect of increasing the magnitude of FAD to F’A_D.
(ambient pressure) x (length AD = length BC)
However, the force that the jet exerts on the plate is now partially resisted by the ambient pressure beneath the plate. Thus, the net force on the plate is the same as that calculated previously. The benefit of working this problem in gauge pressure is apparent.
example 3.4 Given: A steady flow of a liquid (p = 2.0 slugs/ft3) enters and leaves the pipe coupling as shown:
From measurement, p1 = 16.78 psia and
A1 = 4 ft2, V1 = 40 ft/s A2 = 0.5 ft2, V2 = 20 ft/s A3 = 1.5 ft2, V3 = 20 ft/s.
(Note that given any five of these areas or velocities, continuity would provide the sixth.) The flow exhausts into the atmosphere at Stations 2 and 3 as a free jet. Hence, the jetexhaust pressure p2 = p3 = ambient pressure (assumed to be 2,116 psfa.) Assume that the rubber bellows at Station 1 cannot withstand a horizontal force (i. e., it can support only hoop stresses).
Required: Predict the xcomponent, Fx, of force F on the support strut (i. e., magnitude and direction).
Approach: The continuity equation is not needed in this example because all of the required mass (volume) flow information is given in the problem statement. Use the integralmomentum equation to find the required unknown force. Choose a control surface that coincides with the inside of the pipe coupling. An externally applied mechanical force must be included in the momentum equation. (Note that cases displayed in Table 3.3 do include such external forces.)
Solution: For this steady flow, apply momentum Eq. 3.37. Adding the unknown external reaction force to Eq. 3.37, we must solve:
JJpV(Vn)dS = F – JJ pndS. (3.38)
S S
Elect to work in gauge pressure, lbf/ft2 (or psfa), as suggested in the discussion preceding Example 3.4. An appropriate control volume is sketched herein. Notice that this also is a freebody diagram and all forces must be displayed. The presence of the mechanical constraint due to the supporting strut is represented by the unknown reaction force, F, at the point where the strut connects to the pipe.
[1]
Consider the xcomponent of the vectormomentum equation. If gauge pressure is used, there is only one nonzero pressureforce term—namely the pressure force at Station 1 (the pressure forces are zero at Stations 2 and 3 because the pressure in the two jets has zero gauge value). There also is the horizontal component of the strut reaction, FX that must be calculated. Then, the net force (xcomponent) acting on the surface of the control volume is given by:
p1 A1 + FX = [(16.78)(144) – 2,116] (4) + FX = 1,200 + FX.
Because the direction of the force is unknown, it is represented as a positive unknown, FX. If the solution is a positive magnitude, then the choice of sign was correct; that is, the reaction of the strut force, FX, on the control surface acts to the right. If FX is negative in the solution, then the assumed direction was incorrect and FX actually acts to the left.
Three momentumflux terms must be evaluated:
net momentum flux = (E1)(E1 Ax) + (V2 cos 60)(V2 A2) + (V3)(V3 A3)
= (V2 A1 – A2 cos 60 + V32 A3) = 11,800 lbf.
Equating the xcomponents of the force and flux terms as indicated in Eq.
3.38: 1,200 + FX = 11,800, or FX = 13,000 pounds.
The negative sign indicates that the strut reaction force exerted on the control surface by the pipe is in the negative xdirection. Thus, the equal and opposite force of the control surface on the pipe (and, hence, on the strut) is to the right with a magnitude of 13,000 pounds. This is the force that the strut must resist if the pipe is to be immobilized.
Appraisal: The predicted force on the strut agrees with physical intuition.
example 3.5 Given: the drag coefficient of a twodimensional airfoil (i. e., a section of a wing of infinite span) may be measured experimentally by installing a wing model that spans the windtunneltest section. The drag of wind – tunnel models is usually measured with a force balance. However, in the twodimensional case, an alternate method for measuring the drag is to make a velocity survey in a vertical direction through the wake suitably far downstream of the model. The model reaches from one wall to the other and, if the ends are sealed, there is no flow around the tips and the model effectively has infinite span. Application of the integralmomentum equation yields an indirect measurement of the drag. This is the example treated herein.
Consider a wing section near midspan. There is a drag force on the airfoil section due to pressure and shear forces acting on the surface. A wake trails downstream, within which there is a velocity deficit that can be measured by suitable instrumentation. The wing section is assumed to be symmetrical and the wing is assumed to be set at zero angle of attack. Under these assumptions, the wake is symmetrical about the xaxis.
y
Two different choices of control volume are discussed. In both cases, it is assumed that the control volume is suitably far from the airfoil that the local pressure acting on the control surface is the undisturbed value of freestream static pressure. This requires that the wake survey be carried out several chord lengths downstream of the windtunnel model. The flow is assumed to be steady and incompressible. An inviscid flow in the wake is assumed, with the effects of viscosity represented by a force on the model and a resulting velocity deficit in the wake. For simplicity, the measured velocity profile is assumed to be linear. Thus, the velocity distribution within the wake is assumed to
acter (i. e., velocity minimum at the center of the wake) but is not necessarily physically realistic.
Required: Predict the drag coefficient of the airfoil under test if the wake profile is linear.
Approach: Select a rectangular control surface as a simple choice. Assume that it is suitably far from the airfoil that the static pressure and velocity (except in the wake) are essentially the undisturbed tunnelflow conditions:







Solution: For this profile, assume that it is known that h = 0.01c and arbitrarily assume H to be H = 10h, where c is the chord (width) of the model, and H is the halfheight of the assumed control volume. Choose the rectangular control surface shown here. At “D”, the control surface extends upstream to encircle the airfoil (EGH) and then returns on itself to “J”. Because the front and rear of the control surface are far from the airfoil and at freestream static pressure, this datum is assumed to be zero gauge pressure. The pressure forces on AM and BL are equal and opposite, whereas the pressure forces on AB and LM are at right angles to the xdirection. Any pressure or viscous forces along DE are equal and opposite to those on HJ. Thus, the only unbalanced force on the control surface in the xdirection is the force, F, of the airfoil acting on the control surface along EGH. This force is entered into the equations with a plus sign and the direction established later.
There clearly is a momentum flux in the left face, AM, and out the right face, BL. Closer inspection reveals that there also must be a mass flux out of AB and LM because the mass flux out through BL is less than the mass flux in through AM due to a mass (i. e., velocity) deficit along CK. The detailed outflow pattern along the top and bottom is not needed but continuity requires that:
+rmAB + mLM + mBL – mAM = 0
or
This mass flux of pVM h carries with it an xmomentum given by (pVM h)V„. Each fluid particle exiting the top and bottom of the control surface is traveling with a velocity component in the xdirection of nearly VM because the control surface is suitably far from the airfoil.
Thus, there are four fluxofmomentum terms needed for this control volume:
flux[AM] + flux[(A – B) + (M + L)] +
+ flux[(B – C) + (K – L)] + (A – B) + flux[C – K]
or
pV„(2 VoH ) + p VO h+p Vo [2 Vo (H – h)] –
Assembling the components of the momentum equation in the xdirection,
F = – pV2(h /3).
According to the sign, the force of the body on the control surface is to the left (i. e., upstream). Therefore, the force of the control surface on the airfoil is in the opposite direction (i. e., to the right, or downstream.) Then, the drag on the airfoil per unit span is D’ = pV2(h / 3) and the drag coefficient for the twodimensional airfoil is:
Notice that any control surface height H > h gives the same result because the mass and momentum flux across side AM through any height H > h simply is canceled by the flux out across the segments BC and KL.
Appraisal: The drag force on the airfoil has the correct direction (i. e., downstream) and the drag coefficient is dimensionless and has a reasonable magnitude. The magnitude is not conclusive because the wakevelocity profile was idealized. Physically, the drag on the airfoil is perceived as equal to the momentum loss or defect in the wake.
Alternate Approach: As mentioned previously, there is a more natural choice of control volume that yields the same result. Consider a control volume in which the upper and lower surfaces are streamlines, as follows:
The advantage of this controlsurface choice is that there is no flow across AB and LM because they are streamlines; hence, there is no momentum flux across these surfaces. However, because of the mass deficit in the wake, H2 ф H1. The link between these two heights is supplied by continuity. Thus,
Hi h pV v H2
2 pVKdv + 2J + 2jpVMdv = 0
0 0 h h
h h
Hi = 2 + (H – h) ^ H2 = Hi + ^.
With this relationship established, the net momentum flux through the upstream and downstream faces is calculated and set equal to the force of the airfoil acting on the control surface. The windtunnel staticpressure forces acting normal
to AB and LM have a component in the downstream direction, but it is balanced by the same pressure acting over the projected area 2(H2 – H). The final result for the predicted airfoil drag is the same as that obtained by using the rectangular control surface.
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