associated with irreversible effects. From the momentum balance, ef is also seen to be directly related to the stress between the fluid and the tube wall (rw), i.e., it can be interpreted as the work required to overcome the resistance to flow in the conduit. These interpretations are both correct and are equivalent.

Although the energy and momentum balances lead to equivalent results for this special case of one-dimensional fully developed flow in a straight uniform tube, this is an exception and not the rule. In general, the momentum balance gives additional information relative to the forces exerted on and/or by the fluid in the system through the boundaries, which is not given by the energy balance or Bernoulli equation. This will be illustrated shortly.

B. The Loss Coefficient

Looking at the Bernoulli equation, we see that the friction loss (ef) can be made dimensionless by dividing it by the kinetic energy per unit mass of fluid. The result is the dimensionless loss coefficient, Kf:

A loss coefficient can be defined for any element that offers resistance to flow (i.e., in which energy is dissipated), such as a length of conduit, a valve, a pipe fitting, a contraction, or an expansion. The total friction loss can thus be expressed in terms of the sum of the losses in each element, i.e., ef = i (KfjV}/2). This will be discussed further in Chapter 6.

As can be determined from Eqs. (5-47) and (5-49), the pipe wall stress can also be made dimensionless by dividing by the kinetic energy per unit volume of fluid. The result is known as the pipe Fanning friction factor, f:

Although pV2/2 represents kinetic energy per unit volume, pV2 is also the flux of momentum carried by the fluid along the conduit. The latter interpretation is more logical in Eq. (5-50), because rw is also a flux of momentum from the fluid to the tube wall. However, the conventional definition includes the (arbitrary) factor 2. Other definitions of the pipe friction factor are in use that are some multiple of the Fanning friction factor. For example, the Darcy friction factor, which is equal to 4f, is used frequently by mechanical and civil engineers. Thus, it is important to know which definition is implied when data for friction factors are used.

Because the friction loss and wall stress are related by Eq. (5-47), the loss coefficient for pipe flow is related to the pipe Fanning friction factor as follows:

4 fL

Example 5-6: Friction Loss in a Sudden Expansion. Figure 5-7 shows the flow in a sudden expansion from a small conduit to a larger one. We assume that the conditions upstream of the expansion (point j) are known, as well as the areas Aj and A2. We desire to find the velocity and pressure downstream of the expansion (V2 and P2) and the loss coefficient, Kf. As before, V2 is determined from the mass balance (continuity equation) applied to the system (the fluid in the shaded area). Assuming constant density,

For plug flow, the Bernoulli equation for this system is

which contains two unknowns, P2 and ef. Thus, we need another equation, the steady-state momentum balance:

where Vjx = Vj and V2x = V2, because all velocities are in the x direction. Accounting for all the forces that can act on the system through each section of the boundary, this becomes

Pj Aj + Pja(A2 — Aj) — P2A2 + Fwall = pVj Aj(V2 — Vj)

where Pja is the pressure on the left-hand boundary of the system (i.e., the "washer-shaped" surface), and Fwall is the force due to the drag of the wall on the fluid at the horizontal boundary of the system. The fluid pressure cannot change discontinuously, so Pja ' Pj. Also, because the contact area

Figure 5-7 Sudden expansion.

Figure 5-7 Sudden expansion.

with the wall bounding the system is relatively small, we can neglect Fwall with no serious consequences. The result is

This can be solved for (P2 — Pi), which, when inserted into the Bernoulli equation, allows us to solve for ef:

The loss coefficient is seen to be a function only of the geometry of the system (note that the assumption of plug flow implies that the flow is highly turbulent). For most systems (i.e., flow in valves, fittings, etc.), the loss coefficient cannot be determined accurately from simple theoretical concepts (as in this case) but must be determined empirically. For example, the friction loss in a sudden contraction cannot be calculated by this simple method due to the occurrence of the vena contracta just downstream of the contraction (see Table 7-5 in Chapter 7 and the discussion in Section IV of Chapter 10). For a sharp 90° contraction, the contraction loss coefficient is given by where ft is the ratio of the small to the large tube diameter.

Example 5-7: Flange Forces on a Pipe Bend. Consider an incompressible fluid flowing through a pipe bend, as illustrated in Fig. 5-8. We would like to determine the forces in the bolts in the flanges that hold the bend in the pipe, knowing the geometry of the bend, the flow rate through the bend, and the exit pressure (P2) from the bend. Taking the system to be the fluid within the pipe bend, a steady-state "x-momentum" balance is


Figure 5-8 Flange forces in a pipe bend.

Various factors contribute to the forces on the left-hand side of this equation:

X^^onsys = P1A1x + P2A2x — (Fx )on wall by fluid = P1A1 - P2A2 cos 0 — (Fx)on bolts

[The sign of the force resulting from the pressure acting on any area element is intuitive, because pressure acts on any system boundary from the outside, i.e., pressure on the left-hand boundary acts to the right on the system, and vice versa. This is also consistent with previous definitions, because the sign of a surface element corresponds to the direction of the normal vector that points outward from the bounded volume, and pressure is a compressive (negative) stress. Thus P1Ax1 is (+) because it is a negative stress acting on a negative area, and P2Ax2 is (—) because it is a negative stress acting on a positive area. These signs have been accounted for intuitively in the equation.]

The right-side of the momentum balance reduces to m(V2x — V1x) = m(V2cos 0 — V1) Equating these two expressions and solving for (Fx)onwall gives

(Fx)on wall = (Fx)on bolts = P1A1 — P2A2 cos 0 — m( V2 cos 0 — V1)

Similarly, the "y-momentum" balance is

(Fy)onwall = (Fy)on bolts = —P2A2 sin 0 — mV2 sin 0

This assumes that the xy plane is horizontal. If the y direction is vertical, the total weight of the bend, including the fluid inside, could be included as an additional (negative) force component due to gravity. The magnitude and direction of the net force are

where ' is the direction of the net force vector measured counterclockwise from the +x direction. Note that either P1 or P2 must be known, but the other is determined by the Bernoulli equation if the loss coefficient is known:

Methods for evaluating the loss coefficient Kf will be discussed in Chapter 6.

It should be noted that in evaluating the forces acting on the system, the effect of the external pressure transmitted through the boundaries to the system from the surrounding atmosphere was not included. Although this pressure does result in forces that act on the system, these forces all cancel out, so the pressure that appears in the momentum balance equation is the net pressure in excess of atmospheric, e.g., gage pressure.

C. Conservation of Angular Momentum

In addition to linear momentum, angular momentum (or the moment of momentum) may be conserved. For a fixed mass (m) moving in the x direction with a velocity Vx, the linear x-momentum (Mx) is mVx. Likewise, a mass m rotating counterclockwise about a center of rotation at an angular velocity ! = dQ/dt has an angular momentum (Le) equal to mVeR = mwR2, where R is the distance from the center of rotation to m. Note that the angular momentum has dimensions of "length times momentum,'' and is thus also referred to as the "moment of momentum.'' If the mass is not a point but a rigid distributed mass (M) rotating at a uniform angular velocity, the total angular momentum is given by

0 0

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