Momentum theorem

The consequences of this impulse momentum theorem are rather profound. If there are no external forces acting on an object, then the impulse (force times time) is zero. The change in momentum is also zero because it is equal to the force. Hence, if an object has no external forces acting on it, the momentum of the object can never change. This law is the law of conservation of momentum. There are no known exceptions to this fundamental law of physics. Like other conservation laws (such as conservation of energy), the law of conservation of momentum is a very powerful tool for understanding the universe. 

Hence Eqs. (213) and (214), [i.e., (204) and (208)] are constraints for translational invariance of expectation values. If the momentum theorem is satisfied, then, according to (214), the average angular momentum is independent of origin. 

This relation is referred to as the angular momentum theorem. The first term on the RHS represents the rate of change of angular momentum of the fluid contained in the control volume, and the second term represents the net flux of angular momentum leaving the control volume. The sum is equal to the total torque applied to the mass of fluid that is contained in the control volume at time t. 

Por such motions about a fixed centre the conservation of angular momentum (theorem of areas) holds as well as the conservation of energy. The former implies in the first place that the motion is all in one plane. We take this plane as the ajy-plane and transform to polar co-ordinates by means of the equations x = r coscf), y — r sin. The momentum theorem then gives... 

We can use the momentum theorem by itself to obtain useful results. In this example, we apply the momentum theorem to relate the force on a body (catalyst particle) to the properties of the flow that is some distance away from the particle. This technique is useful in wind tunnel tests and is the basis of several fundamental theorems that are related to the lift and the induced drag of wings or, in our... 

For such flows it is more convenient to use the momentum theorem... 

In the above equations s is the scaled streamwise coordinate, n the scaled normal coordinate, A=An/An with An denoting the normalized separation between streamlines constituting a streamtube, is the normalized radius of curvature of the streamlines,6 is the direction of flow velocity, and the rest of the variables are the same as those of the previous section. Moreover, for use in nearly frozen flows the momentum theorem. ... 

The conservation of momentum is more complex. In a steady state, considering a fluid domain, the momentum flux leaving the domain equals that entering the domain, if no force is exerted at the domain boundaries. The momentum theorem enables the evaluation of the forces exerted by a flow on a wall, or the estimation of the global head loss produced in a volume, depending on the circumstances. 

Figure 2.5. Momentum theorem. Definition of a closed volume, D, of its envelope having solid surfaces (solid lines) and throughfiow surfaces Sj (dashed lines), and of the normal oriented toward the outside of the domain...

To explain the momentum theorem, it is assiuned that the flow is steady and this is the only necessary assumptiom It is by no means necessaiy for the fluid to be ideal, or even for the fluid s rheology to be Newtonian. ... 

The momentum theorem is established by integrating the local equations of motion on domain D, given in Chapter 1, which are written so as to exhibit the stress deviator, I (the fluid is not necessarily Newtonian). We then obtain ... 

Exercise I in Chapter 7 uses the momentum theorem for a Bingham fluid. 

Forces R and K are the global pressure and friction forces, rcspectively, exerted by the flow on solid walls (a minus sign appears in equation [2.17] in front of K). The left-hand side of equation [2.18] represents the impulse flux. While applying the momentum theorem, it is important to specify whether the forces under consideration are the ones exerted by the flow on solid walls or the other way round. Sign errors frequently occur when there is lack of specificity in this respect. To apply the momentum theorem, the first step is to specify domain D, throughflow surfaces Sjand solid surfaces S. Care must then be taken to orientate the normals in the proper direction. 

The momentum theorem is used to answer two types of questions ... 

Consider the flow in a straight pipe, whose walls are parallel to the axis (Figure 2.6). Here, the cross-sectional geometry is arbitrary. In addition, the gravity forces are ignored and it is also assumed that the flow is unidirectional in the direction. The momentum theorem applied to the closed domain botmded by the walls and the fluid inlet and outlet sections is written as ... 

Figure 2.6. Application of the momentum theorem to the fluid contained in a cylindrical pipe...

Figure 2.7. Application of the momentum theorem on a virtual, circular cylindrical volume of radius r (shaded volume) contained within a pipe of radius R...

The momentum theorem is equally applicable to any circular cyUndrical volume of radius r internal to the pipe (i.e. rfriction stress exerted by the fluid contained inside the cylinder of radius r on the fluid contained outside that cylinder. Extrapolating equation [2.24] shows that this quantity varies linearly with the radius ... 

This remarkable result holds for a laminar flow as well as a turbulent flow. Such examples illustrate the power of momentum theorem local knowledge of the flow is not always necessary to determine wall friction. However, the pressure gradient enables its determination. 

Consider the flow in a sudden expansion (Figure 2.8(a)) between two straight pipes whose walls are parallel to the (9 , axis. In the expansion area, the flow is not parallel to the direction. The domain to which the momentum theorem is applied (shaded domain in Figure 2.8(a)) is taken between the expansion section and a downstream section S2 chosen sufficiently far away for the flow therein to become parallel again. Here, the gravity forces are ignored (the way gravity can be accounted for is discussed in section 2.7). Consequently, the pressure is uniform in sections Si and S2, and is denoted by Pi and P2, respectively. 

The momentum theorem, of which only the projection onto vector 1 is considered here, is written as ... 

Therefore, the momentum theorem makes it possible to calculate the pressure change between the inlet and outlet sections. The head change (equation [2.7]) between the inlet and outlet is then estimated (using [2.26] and [2.29]) based on the velocity in the inlet section only ... 

The examples presented in the last two sections show that Bemoulh s theorem and the momentum theorem carmot always be applied simultaneously. When the momentum theorem is applied assuming that the pressures on the throughflow surfaces surrounding the domain considered are known, the forces on the walls can be deduced therefrom (as in section 2.5). When the forces are assumed known, the momentum theorem makes it possible to calculate the pressure change between the... 

Here, we analyze what can be learnt from the application of the Bernoulli s and momentum theorems when they are appUed to the emptying of a tank in which the water level is H. 

It is assumed that the fluids are again completely mixed in the outlet section containing point D. Explain why the head cannot be conserved between B and D. By applying the momentum theorem on a domain to be specified, calculate the pressure difference, Pb - Pd, between points B and D. 

The notion of a hydrauhc diameter originates from the momentum theorem (Chapter 2, section 2.5). The factor 4 in [4.4] equates the hydraulic diameter with the physical diameter for a pipe of circular or square cross-section. For a rectangular pipe whose sides have lengths a and b, we have Dh = lab a+b). 

The hnear pressme drop coefficient is related in a simple way to a friction coefficient. The momentum theorem (Chapter 2, section 2.5) established that the average friction stress r exerted by the flow on the pipe is such that ... 

The operating characteristics of a centrifugal pump are determined by an application of the momentum theorem, which provides a connection between the force exerted by the rotor blades on the fluid and the pressures and velocities at the inlet and outlet of the rotor. Such design work is not treated herein. We mainly wish to provide elements of understanding for the selection of a pump, with an emphasis on the dimensional forms of the sizing laws for a centrifugal pump. 

By applying the momentum theorem, calculate the minimum pressure difference, AP, that should be applied between the inlet and outlet of the pipe to set the fluid in motion. By applying the momentum theorem on a domain to be specified, determine the variation of (r) with r and AP. Justify the fact that the momentum theorem can be applied for a Bingham fluid. 

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