Exitance equations

For pipe entrances and exits Equation 4-38 is modified to account for the change in kinetic energy ... 

Papanastasiou et al. (1992) suggested that in order to generate realistic solutions for Navier-Stokes equations the exit conditions should be kept free (i.e. no outflow conditions should be imposed). In this approach application of Green s theorem to the equations corresponding to the exit boundary nodes is avoided. This is eqvrivalent to imposing no exit conditions if elements with... 

Attenuation of radiation as it passes through the sample leads to a transmittance of less than 1. As described, equation 10.1 does not distinguish between the different ways in which the attenuation of radiation occurs. Besides absorption by the analyte, several additional phenomena contribute to the net attenuation of radiation, including reflection and absorption by the sample container, absorption by components of the sample matrix other than the analyte, and the scattering of radiation. To compensate for this loss of the electromagnetic radiation s power, we use a method blank (Figure 10.20b). The radiation s power exiting from the method blank is taken to be Pq. 

There are certain limits to how far b can be increased and d can be reduced however, if the contact time in a weU-mixed extractor can be maintained at several minutes, it can usually be assumed that equiUbrium between the exit phases is attained, justifying the use of the equiUbrium stage concept represented by Figure 1 and equation 1. 

Entrance andExit SpanXireas. The thermal design methods presented assume that the temperature of the sheUside fluid at the entrance end of aU tubes is uniform and the same as the inlet temperature, except for cross-flow heat exchangers. This phenomenon results from the one-dimensional analysis method used in the development of the design equations. In reaUty, the temperature of the sheUside fluid away from the bundle entrance is different from the inlet temperature because heat transfer takes place between the sheUside and tubeside fluids, as the sheUside fluid flows over the tubes to reach the region away from the bundle entrance in the entrance span of the tube bundle. A similar effect takes place in the exit span of the tube bundle (12). 

AM(from equation 46) + AM(bypass flow) + AM(entrance/exit effects) + AM (baffle plates)... 

Droplet trajectories for limiting cases can be calculated by combining the equations of motion with the droplet evaporation rate equation to assess the likelihood that drops exit or hit the wall before evaporating. It is best to consider upper bound droplet sizes in addition to the mean size in these calculations. If desired, an instantaneous value for the evaporation rate constant may also be used based on an instantaneous Reynolds number calculated not from the terminal velocity but at a resultant velocity. In this case, equation 37 is substituted for equation 32 ... 

For most cooling towers in the United Kingdom, the exit air is saturated at a temperature close to the mean water temperature in the tower. Hence, if the water temperatures and the air inlet conditions are known, AH, AT, and AT can all be calculated, and Tcan be deterrnined. It was found that the quantity C was approximately constant for these towers, ca 0.4—0.5 (34). If the value of C is known for a given tower, then the left side of equation 49 can be computed and, setting this equal to Z9, the allowable Hquid flow rate can be found. Alternatively, when and air-inlet conditions are given, the... 

An equation representing an energy balance on a combustion chamber of two surface zones, a heat sink Ai at temperature T, and a refractory surface A assumed radiatively adiabatic at Tr, inmost simply solved if the total enthalpy input H is expressed as rhCJYTv rh is the mass rate of fuel plus air and Tp is a pseudoadiabatic flame temperature based on a mean specific heat from base temperature up to the gas exit temperature Te rather than up to Tp/The heat transfer rate out of the gas is then H— — T ) or rhCp(T f — Te). The... 

Though this is a quartic equation, it is capable of explicit solution because of the absence of second and third degree terms. Trial-and-error enters, however, because (GSi)r and are mild functions of Tg and related Te, respectively, and aprehminary guess of Tg is necessaiy. An ambiguity can exist in interpretation of terms. If part of the enclosure surface consists of screen tubes over the chamber-gas exit to a convection section, radiative transfer to those tubes is included in the chamber energy balance, but convection is not, because it has no effect on chamber gas temperature. 

Here g is the gravity vector and tu is the force per unit area exerted by the surroundings on the fluid in the control volume. The integrand of the area integr on the left-hand side of Eq. (6-10) is nonzero only on the entrance and exit portions of the control volume boundary. For the special case of steady flow at a mass flow rate m through a control volume fixed in space with one inlet and one outlet, (Fig. 6-4) with the inlet and outlet velocity vectors perpendicular to planar inlet and outlet surfaces, giving average velocity vectors Vi and V9, the momentum equation becomes... 

A useful simphfication of the total energy equation applies to a particular set of assumptions. These are a control volume with fixed solid boundaries, except for those producing shaft work, steady state conditions, and mass flow at a rate m through a single planar entrance and a single planar exit (Fig. 6-4), to whi(m the velocity vectors are perpendicular. As with Eq. (6-11), it is assumed that the stress vector tu is normal to the entrance and exit surfaces and may be approximated by the pressure p. The equivalent pressure, p + pgz, is assumed to be uniform across the entrance and exit. The average velocity at the entrance and exit surfaces is denoted by V. Subscripts 1 and 2 denote the entrance and exit, respectively. 

Selecting the inlet and outlet surfaces 1 and 2 as shown, the continuity equation Eq. (6-9) can he used to find the exit velocity V2 = ViAi/A2. The mass flow rate is obtained by m = pViAi. 

Expansion and Exit Losses For ducts of any cross section, the frictional loss for a sudden enlargement (Fig. 6-13c) with turbulent flow is given by the Borda-Carnot equation ... 

Equation (6-95) is valid for incompressible flow. For compressible flows, see Benedict, Wyler, Dudek, and Gleed (J. E/ig. Power, 98, 327-334 [1976]). For an infinite expansion, A1/A2 = 0, Eq. (6-95) shows that the exit loss from a pipe is 1 velocity head. This result is easily deduced from the mechanic energy balance Eq. (6-90), noting that Pi =pg. This exit loss is due to the dissipation of the discharged jet there is no pressure drop at the exit. 

These equations are consistent with the isentropic relations for a perfect gas p/po = (p/po), T/To = p/poY. Equation (6-116) is valid for adiabatic flows with or without friction it does not require isentropic flow However, Eqs. (6-115) and (6-117) do require isentropic flow The exit Mach number Mi may not exceed unity. At Mi = 1, the flow is said to be choked, sonic, or critical. When the flow is choked, the pressure at the exit is greater than the pressure of the surroundings into which the gas flow discharges. The pressure drops from the exit pressure to the pressure of the surroundings in a series of shocks which are highly nonisentropic. Sonic flow conditions are denoted by sonic exit conditions are found by substituting Mi = Mf = 1 into Eqs. (6-115) to (6-118). 

Equation (6-128) does not require fric tionless (isentropic) flow. The sonic mass flux through the throat is given by Eq. (6-122). With A set equal to the nozzle exit area, the exit Mach number, pressure, and temperature may be calculated. Only if the exit pressure equals the ambient discharge pressure is the ultimate expansion velocity reached in the nozzle. Expansion will be incomplete if the exit pressure exceeds the ambient discharge pressure shocks will occur outside the nozzle. If the calculated exit pressure is less than the ambient discharge pressure, the nozzle is overexpanded and compression shocks within the expanding portion will result. 

Proportional Element First, consider the outflow through the exit valve on the tank. If the flow through the line is turbulent, then Bernoulh s equation can be used to relate the flow rate through the valve to the pressure drop across the valve as ... 

For proper use of the equations, the chamber shape must conform to the spray pattern. With cocurrent gas-spray flow, the angle of spread of single-fluid pressure nozzles and two-fluid pneumatic nozzles is such that wall impingement wiU occur at a distance approximately four chamber diameters below the nozzle therefore, chambers employing these atomizers should have vertical height-to-diameter ratios of at least 4 and, more usually, 5. The discharge cone below the vertical portion should have a slope of at least 60°, to minimize settling accumulations, and is used entirely to accelerate gas and solids for entty into the exit duct. 

Although Eq. (14-31) is convenient for computing the composition of the exit gas as a function of the number of theoretical stages, an alternative equation derived by Colburn [Tran.s. Am. Jn.st. Chem. Eng., 35, 211 (1939)] is more useful when the number of theoretical plates is the unknown ... 

The conversion achieved in the vessel is obtained by the solution of the differential equation at the exit of the vessel where the hfe expectation is t = 0. The starting point for the integration is tj). When integrating numerically, however, the RTD becomes essentially 0 by the time becomes 3 or 4, and the value of the integral beyond that point becomes nil. Accordingly, the integration interval is from (f, t,<2> or 4) to (/effluent, = 0)-jl IS found from Eq. (23-46). 

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