Incompressible fluid, energy balance

We assume the incompressible fluid energy balance is accurate for this liquid-phase reactor. If the heat removal is manipulated to maintain constant reactor temperature, the time derivative in Equation 6.15 vanishes leaving... 

Example 3 Venturi Flowmeter An incompressible fluid flows through the venturi flowmeter in Fig. 6-7. An equation is needed to relate the flow rate Q to the pressure drop measured by the manometer. This problem can he solved using the mechanical energy balance. In a well-made venturi, viscous losses are neghgihle, the pressure drop is entirely the result of acceleration into the throat, and the flow rate predicted neglecting losses is quite accurate. The inlet area is A and the throat area is a. 

The kinetic energy attributable to this velocity will be dissipated when the liquid enters the reservoir. The pressure drop may now be calculated from the energy balance equation and equation 3.19. For turbulent flow of an incompressible fluid ... 

Methods have been given for the calculation of the pressure drop for the flow of an incompressible fluid and for a compressible fluid which behaves as an ideal gas. If the fluid is compressible and deviations from the ideal gas law are appreciable, one of the approximate equations of state, such as van der Waals equation, may be used in place of the law PV = nRT to give the relation between temperature, pressure, and volume. Alternatively, if the enthalpy of the gas is known over a range of temperature and pressure, the energy balance, equation 2.56, which involves a term representing the change in the enthalpy, may be employed ... 

Consider a section of uniform cylindrical pipe of length L and radius R, inclined upward at an angle 0 to the horizontal, as shown in Fig. 6-2. The steady-state energy balance (or Bernoulli equation) applied to an incompressible fluid flowing in a uniform pipe can be written... 

A storage tank is shown in Figure 4-5. A hole develops at a height hL below the fluid level. The flow of liquid through this hole is represented by the mechanical energy balance (Equation 4-1) and the incompressible assumption, as shown in Equation 4-2. 

The equation of motion and the equation of energy balance can also be time averaged according to the procedure indicated above (SI, pp. 336 et seq. G7, pp. 191 et seq. pp. 646 et seq.). In this averaging process there arises in the equation of motion an additional component to the stress tensor t(,) which may be written formally in terms of a turbulent (eddy) coefficient of viscosity m(I) and in the equation of energy balance there appears an additional contribution to the energy flux q(1), which may be written formally in terms of the turbulent (eddy) coefficient of thermal conductivity Hence for an incompressible fluid, the x components of the fluxes may be written... 

An energy balance for an incompressible fluid in turbulent flow is given by ... 

The second term in the mechanical-energy balance. Equation 5.1, is the change in potential energy and requires no comment. The third term is "pressme work" and its evaluation depends on whether the fluid is compressible or incompressible. Because the increase in pressure across the fan is small, we treat the flow as essentially incompressible. Thus, the fluid density may be removed from the integral sign and the mechanical energy balance becomes... 

Starting with the open system balance equation, derive the steady-state mechanical energy balance equation (Equation 7.7-2) for an incompressible fluid and simplify the equation further to derive the Bernoulli equation. List all the assumptions made in the derivation of the latter equation. 

Equation 7.7-2 is referred to as the mechanical energy balance. Once again, it is vaiid for steady-state flow of an incompressible fluid. 

The energy required to pump a liquid food through a pipe line can be calculated from the mechanical eneigy balance (MEB) equation. The MEB equation can be used to analyze pipe flow systems. For the steady-state flow of an incompressible fluid, the MEB can be written as follows (Brodkey, 1%7) ... 

We shall employ the steady-state mechanical energy balance for an incompressible fluid, discussed in Chap. 4, to relate o and A, See Fig. E6.2b. Recall from Eq. (4.30) with W — 0 and = 0 that... 

If the frequently used case of an incompressible Newtonian fluid, g = const, dg/dt = 0, of constant viscosity is presumed, the continuity equation, momentum and energy balances are transformed into... 

Problem 2-15. Derivation of Transport Equation for a Sedimenting Suspension. There are many parallels among momentum, mass, and energy transport because all three are derived from similar conservation laws. In this problem we derive a microscopic balance describing the concentration distribution (x, t) of a very dilute suspension of small particles suspended in an incompressible fluid undergoing unsteady flow. [Note cj>(. t) is the local volume fraction of particles in the fluid (i.e. volume of particles/volume of fluid) and hence is dimensionless. ... 

To use the energy balances, we will need to relate the energy to more easily measurable properties, such as temperature and pressure (and in later chapters, when we consider mixtures, to composition as well). The interrelationships between energy, temperature, pressure, and composition can be complicated, and we will develop this in stages. In this chapter and in Chapters 4,5, and 6 we will consider only pure fluids, so composition is not a variable. Then, in Chapters 8 to 15, mixtures will be considered. Also, here and in Chapters 4 and 5 we will consider only the simple ideal gas and incompressible liquids and solids for which the equations relating the energy, temperature, and pressure are simple, or fluids for which charts and tables interrelating these properties are available. Then, in Chapter 6, we will discuss how such tables and charts are prepared. 

If w e consider the constant-volume reactor with incompressible fluid (a = 0,Cv = Cp), Equation 6.16 reduces to Equation 6.15 as it should because Equation 6.15 is valid for any reactor operation with an incompressible fluid. We also notice that, in the constant-pressure case, the same energy balance applies for any fluid mixture (ideal gas, incompressible fluid, etc.), and that this balance is the same as the balance for an incompressible fluid in a constant-volume reactor. Although the same final balances are obtained for these two cases, the physical situations they describe are completely different. 

The energy balance for steady, incompressible flow, which is called Bernoulli s equation, is probably the most useful single equation in fluid mechanics. 

This equation now expresses the macroscopic energy balance for isothermal flow in a pipe. It may further be simplified in the special case of an incompressible fluid. In this case Fis constant and the energy balance equation becomes... 

The fact that frictional losses represent a small contribution to the overall energy balance also justifies the initial assumptions of the calculation. Since pressure drop is small, steam may indeed be treated as an incompressible fluid. The amount of work that is dissipated is correspondingly small (3.319 kJ/kg) therefore, the temperature rise that would be expected (recall that the pipe is insulated) must also be small. This temperature rise maybe estimated by assuming that the enthalpy of steam at the exit has increased by the amount of frictional losses. Interpolation in the steam tables shows that the expected temperature rise over the entire pipe length is about 4 °C. 

To derive the equation for the venturi meter, friction is neglected and the pipe is assumed horizontal. Assuming turbulent flow and writing the mechanical-energy-balanCe equation (2.7-28) between points 1 and 2 for an incompressible fluid. 

Applying the overall energy balance equation, (1.4.16), to this system, we will assume an incompressible fluid such that E = U — H. This means that U is related to the temperature >y U = Cp T — Tr), where Tr is the reference temperature, and that the kinetic energy and potential energy terms are negligible. The energy balance is therefore... 

Dil man et al [8] developed a model of fluid movement in a channel with permeable walls from the macroscopic mechanical energy balance. This was later extended to apply to a radial-flow reactor by Genkin et al [3] for the CF-flow case. The equations were valid for isothermal incompressible flow in a horizontal cylindrical channel they were extended by Kaye [5] to eliminate the restriction to a horizontal tube and to cover all four operating modes. Chang and Calo [2] have also presented an extension to include both axial and radial flow in the catalyst bed, although they concluded that axial flow was small. 

For steady flow of essentially incompressible fluids under conditions where friction is negligible, in the absence of external work effects it is found experimentally that there is little error in an energy balance involving only the terms for the mechanical forms of energy."... 

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