Pressure drop Incompressible fluid

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. 

Porous Media Packed beds of granular solids are one type of the general class referred to as porous media, which include geological formations such as petroleum reservoirs and aquifers, manufactured materials such as sintered metals and porous catalysts, burning coal or char particles, and textile fabrics, to name a few. Pressure drop for incompressible flow across a porous medium has the same quahtative behavior as that given by Leva s correlation in the preceding. At low Reynolds numbers, viscous forces dominate and pressure drop is proportional to fluid viscosity and superficial velocity, and at high Reynolds numbers, pressure drop is proportional to fluid density and to the square of superficial velocity. 

Permeability is normally determined using linear flow in the incompressible or compressible form, depending on whether a liquid or gas is used as the flowing fluid. The volumetric flowrate Q (or Q ,) is determined at several pressure drops. Q (or Q ,) is plotted versus the average pressure p . The slope of this line will yield the fluid conductivity K or, if the fluid density and viscosity are known, it provides the intrinsic permeability k. For gases, the fluid conductivity depends on pressure, so that... 

Because there are no moving parts in the Kenics mixer, only the processed materials are in motion. Therefore, the only energy required for the mixer is the energy required to overcome the pressure drop (AP). The general equation for calculating the pressure drop in an empty pipe for isothermal incompressible fluids is given by... 

Pressure Drop in Straight Pipe Incompressible Fluid... 

Scope, 52 Basis, 52 Compressible Flow Vapors and Gases, 54 Factors of Safety for Design Basis, 56 Pipe, Fittings, and Valves, 56 Pipe, 56 Usual Industry Pipe Sizes and Classes Practice, 59 Total Line Pressure Drop, 64 Background Information, 64 Reynolds Number, R,. (Sometimes used Nr ), 67 Friction Factor, f, 68 Pipe—Relative Roughness, 68 Pressure Drop in Fittings, Valves, Connections Incompressible Fluid, 71 Common Denominator for Use of K Factors in a System of Varying Sizes of Internal Dimensions, 72 Validity of K Values,... 

Flow through chokes and nozzles is a special case of fluid dynamics. For incompressible fluids the problem can be handled by mass conservation and Bernoulli s equation. Bernoulli s equation is solved for the pressure drop across the choke, assuming that the velocity of approach and the vertical displacement are negligible. The velocity term is replaced by the volumetric flow rate times the area at the choke throat to yield... 

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 ... 

If the pressure drop in the pipe is a small proportion of the inlet pressure, the first term is negligible and the fluid may be treated as an incompressible fluid at the mean pressure in the pipe. 

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 ... 

In a synthetic ammonia plant the hydrogen is fed through a 50 mm steel pipe to the converters. The pressure drop over the 30 m length of pipe is 500 kN/m2, the pressure at the downstream end being 7.5 MN/m2. What power is required in order to overcome friction losses in the pipe Assume isothermal expansion of the gas at 298 K. What error is introduced by assuming the gas to be an incompressible fluid of density equal to that at the mean pressure in the pipe /r — 0.02 mN s/m2. 

As will be outlined below, the computation of compressible flow is significantly more challenging than the corresponding problem for incompressible flow. In order to reduce the computational effort, within a CED model a fluid medium should be treated as incompressible whenever possible. A rule of thumb often found in the literature and used as a criterion for the incompressibility assumption to be valid is based on the Mach number of the flow. The Mach number is defined as the ratio of the local flow velocity and the speed of sound. The rule states that if the Mach number is below 0.3 in the whole flow domain, the flow may be treated as incompressible [84], In practice, this rule has to be supplemented by a few additional criteria [3], Especially for micro flows it is important to consider also the total pressure drop as a criterion for incompressibility. In a long micro channel the Mach number may be well below 0.3, but owing to the small hydraulic diameter of the channel a large pressure drop may be obtained. A pressure drop of a few atmospheres for a gas flow clearly indicates that compressibility effects should be taken into account. 

The viscosities of supercritical fluids are intermediate between those of a gas and a liquid, but this tine, are much closer to those of a gas than a liquid. For a fixed column pressure drop much longer columns or higher flow rates are possible in SFC compared to liquid chromatography. Liquids, however, are virtually incompressible while gases and... 

Pressure drop in the transmission pipes is a combination of pressure losses in the pipes and pipe fittings7. Pipe fittings include bends, isolation valves, control valves, orifice plates, expansions, reductions, and so on. If the fluid is assumed to be incompressible and the change in kinetic energy from inlet to outlet is neglected, then ... 

For a fan the changes in pressure and temperature are small enough that the incompressible fluid flow equations given for pumps may be used. In fact, for most gas systems, if the pressure drop is less than 40% of absolute upstream pressure the fluids can be treated as incompressible.30... 

The principles of conservation of mass and momentum must be applied to each phase to determine the pressure drop and holdup in two phase systems. The differential equations used to model these principles have been solved only for laminar flows of incompressible, Newtonian fluids, with constant holdups. For this case, the momentum equations become... 

Martinelli and Nelson (M7) developed a procedure for calculating the pressure drop in tubular systems with forced-circulation boiling. The procedure, which includes the accelerative effects due to phase change while assuming each phase is an incompressible fluid, is an extrapolation of the Lockhart and Martinelli x parameter correlation. Other pressure drop calculation procedures have been proposed for forced-circulation phase-change systems however, these suffer severe shortcomings, and have not proved more accurate than the Martinelli and Nelson method. 

Now the pressure drops along the pipe because of energy dissipation (e.g., friction), just as for an incompressible fluid. However, because the density decreases with decreasing pressure and the product of the density and velocity must be constant, the velocity must increase as the gas moves through the pipe. This increase in velocity corresponds to an increase in kinetic energy per unit mass of gas, which also results in a drop in temperature. There is a limit as to how high the velocity can get in a straight pipe, however, which we will discuss shortly. 

Normally, the pressure drop in a tubular reactor is small compared with the operating pressure and in this case, the fluid can be treated as incompressible. Then the pressure drop in a tube of length L and diameter d is given by... 

Eqs. (3.448) and (3.449) assume constant fluid density, in other words, diey hold for incompressible flow. However, in compressible flow, die density of the fluid is a function of the pressure drop (see Section 5.3.4). Then, we have to use the differential forms of... 

The pressure drop AP (Pa) in turbulent flow of an incompressible fluid with density p (kg m ) through a tube of length L (m) and diameter d (m) at an average velocity of v (m s ) is given by Equation 2.10, namely, the Fanning equation ... 

The most important assumptions when solving this problem are a steady, incompressible and isothermal flow. Let us now consider a power-law fluid, but neglect the elastic effects. Furthermore, for the solution of this specific problem, let us assume that the flow is primarily driven by drag and that there are no significant pressure drops across the die. 

Frictional dissipation of mechanical energy can result in significant heating of fluids, particularly for very viscous liquids in small channels. Under adiabatic conditions, the bulk liquid temperature rise is given by AT=AP/CV p for incompressible flow through a channel of constant cross-sectional area. For flow of polymers, this amounts to about 4°C per 10 MPa pressure drop, while for hydrocarbon liquids it is about... 

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