Pipe flow momentum balance

Pipe Flow For steady-state flow through a constant diameter duct, the mass flux G is constant and the governing steady-state momentum balance is ... 

HEM for Two-Phase Pipe Discharge With a pipe present, the backpressure experienced by the orifice is no longer qg, but rather an intermediate pressure ratio qi. Thus qi replaces T o iri ihe orifice solution for mass flux G. ri Eq. (26-95). Correspondingly, the momentum balance is integrated between qi and T o lo give the pipe flow solution for G,p. The solutions for orifice and pipe now must be solved simultaneously to make G. ri = G,p and to find qi and T o- This can be done explicitly for the simple case of incompressible single-phase (hquid) inclined or horizontal pipe flow The solution is implicit for compressible regimes. 

A momentum balance for the flow of a two-phase fluid through a horizontal pipe and an energy balance may be written in an expanded form of that applicable to single-phase fluid flow. These equations for two-phase flow cannot be used in practice since the individual phase velocities and local densities are not known. Some simplification is possible if it... 

The steady flow pressure drop in the pipe can be deduced from a momentum balance on a differential slice of the fluid-particle mixture in a... 

Momentum Balance in Dimensionless Variables For pipe flow, it is necessary to solve the momentum balance. The momentum balance is simplified by using the following dimensionless variables ... 

Setting 0 equal to py in Eq. (46) produces the macroscopic momentum balance. The term (j -n represents the reaction force of the wall of the pipe on the fluid arising from friction and changes in the direction of flow. The term R V represents the action of the body force pg on the total flow. Thus, Eq. (46) becomes... 

The shell balance method will be used to examine steady laminar flow of a fluid in a pipe. For the geometrical system illustrated in Figure 3B-1 and for steady laminar fully developed flow of a fluid, a shell momentum balance can be conducted (Bird et al., 1960 Geankoplis, 1983) using the cylindrical coordinates, r, 6, andz. The momentum balance is conducted on a control volume shell at a radius r with dimensions Ar and Az. 

EXAMPLE 2.4. Consider the nonlinear ordinary differential equation (ODE) for a gravity-flow tank, which is derived from a momentum balance around the exit pipe. 

Figure 7.19 shows a cylinder of fluid, radius r and length L, within a pipe of radius R. Consider a momentum balance in the axial (x) direction. Because the flow is fully developed = i 2x Pi = Pil hence from equation 7.4.10 the total force on the cylinder in the x direction is zero. That part of the force due to... 

EXAMPLE 2.8-4. Friction Loss in a Sudden Enlargement A mechanical-energy loss occurs when a fluid flows from a small pipe to a large pipe through an abrupt expansion, as shown in Fig. 2.8-4. Use the momentum balance and mechanical-energy balance to obtain an expression for the loss for a liquid. (Hint Assume thatPq = p, and Vq = Vj. Make a mechanical-energy balance between points 0 and 2 and a momentum balance between points 1 and 2. It will be assumed that pi and P2 are uniform over the cross-sectional area.)... 

Figure 2.9-2. Velocity and momentum flux profiles for laminar flow in a pipe. Sec. 2.9 Shell Momentum Balance and Velocity Profile in Laminar Flow...

The average velocity over the whole cross section of the pipe is precisely 0.5 times the maximum velocity at the center as given by the shell momentum balance in Eq. (2.9-13) for laminar flow. On the other hand, for turbulent flow, the curve is somewhat flattened in the center (see Fig. 2.10-1) and the average velocity is about 0.8 times the... 

Momentum Balance on Reducing Elbow and Friction Losses. Water at 20°C is flowing through a reducing bend, where a2 (see Fig. 2.8-3) is 120°. The inlet pipe diameter is 1.829 m, the outlet is 1.219 m, and the flow rate is8.50 m /s. The exit point Z2 is 3.05 m above the inlet and the inlet pressure is 276 kPa gage. Friction losses are estimated as 0.5vjl2 and the mass of water in the... 

An example of the use of momentum balance in pipe flow of a Newtonian fluid is... 

One typical example that can be addressed in this way is the start-up of a flow through a pipe. We can consider a one dimension problem for the profile in the radial direction. We evaluate now the unsteady fluid flow through a long pipe with constant fluid properties (p, p). The momentum balance is given by the following equation [5] ... 

Here, we will apply the momentum balance to determine the steady-state radial velocity profile for the flow in a pipe of a Newtonian fluid. Figure 6.3 describes the system. 

The fluid momentum balance applied to the case of laminar, fully developed flow of a power-law fluid in a horizontal pipe of diameter D yields the following expression for the relationship between the pressure drop, AP/L, and the average flow velocity, vav-... 

Calculates the velocity profile of a non-Newtonian fluid flowing in a circular pipe by solving the momentum balance equation (s hooting, m). 

A cross section of the annular die required to produce the pipe is shown in Figure 2.7. The approximate velocity and stress profiles for this flow are sketched here also. A thin cylindrical shell is now chosen of length L and thickness Ar as shown in Figure 2.8. The shell is selected so that the surface is parallel to the flow direction. A force (or momentum) balance is now performed on the shell. Because the flow is under steady-state conditions, the forces in the z direction must sum to zero as shown below ... 

For pipe-objects, on the basis of either a 5-equation or a 4-equation model, there is also the possibility to use method of integrated mass and momentum balances an option for fastrunning calculations. With the application of the method, the solution variables are now the object pressure, the mass flows at pipe inlet and outlet, and the local qualities and enthalpies (4-equation model) or temperatures (5-equation model). The local pressure and mass flow rates are obtained from algebraic equations as a function of solution variables. 

Isothermal Gas Flow in Pipes and Channels Isothermal compressible flow is often encountered in long transport lines, where there is sufficient heat transfer to maintain constant temperature. Velocities and Mach numbers are usually small, yet compressibihty effects are important when the total pressure drop is a large fraction of the absolute pressure. For an ideal gas with p = pM. JKT, integration of the differential form of the momentum or mechanical energy balance equations, assuming a constant fric tion factor/over a length L of a channel of constant cross section and hydraulic diameter D, yields,... 

While we laud the virtue of dynamic modeling, we will not duphcate the introduction of basic conservation equations. It is important to recognize that all of the processes that we want to control, e.g. bioieactor, distillation column, flow rate in a pipe, a drag delivery system, etc., are what we have learned in other engineering classes. The so-called model equations are conservation equations in heat, mass, and momentum. We need force balance in mechanical devices, and in electrical engineering, we consider circuits analysis. The difference between what we now use in control and what we are more accustomed to is that control problems are transient in nature. Accordingly, we include the time derivative (also called accumulation) term in our balance (model) equations. 

In two-phase flow, most investigations are carried out in one dimension in the steady state with constant flow rates. The system may or may not be isothermal, and heat and mass may be transferred either from liquid to gas, or vice versa. The assumption is commonly made that the pressure is constant at a given cross section of the pipe. Momentum and energy balances can then be written separately for each phase, and with the constraint that the static pressure drop, dP, is identical for both phases over the same increment of flow length dz, these balances can be added to give over-all expressions. However, it will be seen that the resulting over-all balances do not have the simple relationships to each other that exist for single-phase flow. 

Because, in the fully developed flow, there is bo change in the velocity profile with distance, z, along the pipe, there can be no change in fluid momentum through the control volume. The forces acting on the control must therefore balance, i.e. ... 

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