Shear stress balance

Bubbles and drops tend to deform when subject to external fluid fields until normal and shear stresses balance at the fluid-fluid interface. When compared with the infinite number of shapes possible for solid particles, fluid particles at steady state are severely limited in the number of possibilities since such features as sharp corners or protuberances are precluded by the interfacial force balance. 

To this point, we have considered only the component of the stress balance (2-134), in the direction normal to the interface. There will be, in general, two tangential components of (2-134), which we obtain by taking the inner product with the two orthogonal unit tangent vectors that are normal to n. If we denote these unit vectors as t (with i = 1 or 2), the so-called shear-stress balances can be written symbolically in the form,... 

In this case, very tiny changes in the interface concentration produce a large contribution to the shear-stress balance, as can be seen from (7-268). As a result, because the shear stress is 0(1) in the dimensionless form of (7-268), it follows that T is always very close to Teq, and the leading-order approximation is that of a drop with no-slip boundary conditions on the interface. It is because of the latter condition that the incompressible limit is said to resemble the insoluble surfactant limit. However, in this case T = Teq, not 1 (i.e. T Too), and bulk-phase mass transport will play an important role in determining the departure from this limiting case. This case is formulated as Problem 7 21 at the end of this chapter. 

If we combine (7-323) with the zero-order solution, (7-305), we see, as expected, that the presence of surfactant on the interface retards the flow. This is consistent with our qualitative expectations based on the fact that the surfactant concentration increases as we move from the front to the back of the drop. However, one surprising feature of the solution (7-323) is that there is no dependence on the viscosity ratio. This flow is established as a consequence of the shear-stress balance, (7-320). Clearly, the shear-stress difference [the left-hand side of (7-320)] does depend on the viscosity ratio however we see from (3-322) that the Marangoni stress that drives the flow also depends on viscosity ratio in precisely the same form. 

At the liquid-liquid interface (y = 0), due to the presence of the surface charge, shear stress is not continuous at the interface. The matching conditions are the continuities of velocity and the shear stress balance, which jumps abruptly at the interface. 

At the interface between the drop and the suspending medium, the following normal stress and shear stress balances must be satisfied ... 

When a dislocation segment of length L is pinned at the ends under the influence of an applied shear stress t, a balance between the line tension and the applied stress produces a radius of curvature R given by [37]... 

When a shearing stress is imposed on a solid, deformation occurs, until a point is reached when the internal stresses produced balance the shearing stresses. Provided the elastic limit for the material is not exceeded the solid will return to its original shape when the load is removed. 

In this method [89], a single fiber is taken and partially embedded in a drop of uncured resin placed on a holder. The resin is then cured with the fiber held upright. The holder, with resin and fiber, is held in a grip attached to the crosshead and then pulled out from the resin. The force pulling the fiber out of the resin is balanced by shear stress at the resin-fiber interface holding the fiber in place. The maximum shear stress occurs as the embedded length tends to zero and is given by ... 

Consider a uniform cylindrical bar or tube to which some balanced torque T is applied (Figure 2-28). The bar will be subject to a torsional stress, or shear stress which increases with the radial position within the bar. 

The distribution of shear stress over the cross-section of a pipe is determined by a force balance and is independent of the nature of the fluid or the type of flow. 

It is assumed here that the fluid in contact with the surface is at rest and therefore h(j must be zero. Furthermore, all the fluid close to the surface is moving at very low velocity and therefore any changes in its momentum as it flows parallel to the surface must be extremely small. Consequently, the net shear force acting on any element of fluid near the surface is negligible, the retarding force at its lower boundary being balanced by the accelerating force at its upper boundary. Thus the shear stress Ro in the fluid near the surface must approach a constant value. 

The Martinelli correlations for void fraction and pressure drop are used because of their simplicity and wide range of applicability. France and Stein (6 ) discuss the method by which the Martinelli gradient for two-phase flow can be incorporated into a choked flow model. Because the Martinelli equation balances frictional shear stresses cuid pressure drop, it is important to provide a good viscosity model, especially for high viscosity and non-Newtonian fluids. 

As the name implies, the cup-and-bob viscometer consists of two concentric cylinders, the outer cup and the inner bob, with the test fluid in the annular gap (see Fig. 3-2). One cylinder (preferably the cup) is rotated at a fixed angular velocity ( 2). The force is transmitted to the sample, causing it to deform, and is then transferred by the fluid to the other cylinder (i.e., the bob). This force results in a torque (I) that can be measured by a torsion spring, for example. Thus, the known quantities are the radii of the inner bob (R ) and the outer cup (Ra), the length of surface in contact with the sample (L), and the measured angular velocity ( 2) and torque (I). From these quantities, we must determine the corresponding shear stress and shear rate to find the fluid viscosity. The shear stress is determined by a balance of moments on a cylindrical surface within the sample (at a distance r from the center), and the torsion spring ... 

As will be shown later, a momentum (force) balance on the fluid in the tube provides a relationships between the shear stress at the tube wall (rw) and the measured pressure drop ... 

This result can also be derived by equating the shear stress for a Newtonian fluid, Eq. (6-9), to the expression obtained from the momentum balance for tube flow, Eq. (6-4), and integrating to obtain the velocity profile ... 

This simple force balance has provided an extremely important result the wall shear stress for flow in a pipe can be determined from the frictional component of the pressure drop. In practice it is desirable to use the conditions in Example 1.7 so that the frictional component is the only component of the total pressure drop, which can be measured directly. 

The variation of the shear stress rrx with radial coordinate r can be determined by making a force balance similar to that in Example 1.7 but using an element extending from the centre-line to a general radial distance r. 

The reason for this variation of the shear stress is easily understood. For steady flow there must be a balance between the force due to the pressure difference and the shear force. As shown in equation 1.39, the pressure force is proportional to r2 but the shear force to r, so to maintain the... 

When analysing simple flow problems such as laminar flow in a pipe, where the form of the velocity profile and the directions in which the shear stresses act are already known, no formal sign convention for the stress components is required. In these cases, force balances can be written with the shear forces incorporated according to the directions in which the shear stresses physically act, as was done in Examples 1.7 and 1.8. However, in order to derive general equations for an arbitrary flow field it is necessary to adopt a formal sign convention for the stress components. 

As an illustration of the fact that the two sign conventions give the same results, the equivalents of equations 1.46 to 1.53 can be written for the positive sign convention. In the positive sign convention, the shear stress acting on the outer surface of the element is measured in the positive x-direction and that on the inner surface is measured in the negative x-direction. This is the opposite of the directions shown in Figure 1.17. The force balance, equivalent to equation 1.46 is now... 

Thus, only the normal Reynolds stresses (i = j) are directly dissipated in a high-Reynolds-number turbulent flow. The shear stresses (i / j), on the other hand, are dissipated indirectly, i.e., the pressure-rate-of-strain tensor first transfers their energy to the normal stresses, where it can be dissipated directly. Without this redistribution of energy, the shear stresses would grow unbounded in a simple shear flow due to the unbalanced production term Vu given by (2.108). This fact is just one illustration of the key role played by 7 ., -in the Reynolds stress balance equation. 

The average shear strength at the interface, t., whether bonded, debonded or if the surrounding matrix material is yielded, whichever occurs first, can be approximately estimated from a simple force balance equation for a constant interface shear stress (Kelly and Tyson, 1965) ... 

In the second approach shown in Fig 3.12(b), a force is applied continuously using a Vickers microhardness indenter to compress the fiber into the specimen surface (Marshall, 1984). For ceramic matrix composites where the bonding at the interface is typically mechanical in nature, the interface shear stress, Tf, against the constant frictional sliding is calculated based on simple force balance (Marshall, 1984) ... 

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