Ejectors Calculations

In dry compressors, shaft end seals are generally one of five type.s. These are labyrinth, restrictive ring, mechanical contact, liquid film, and dry gas seal. The labyrinth type is the most simple but has the highest leakage. The labyrinth seal is generally ported at an axial point between the seals in order to use an eductor or ejector to control leakage and direct it to the suction or a suitable disposal area. Alternatively, a buffer gas is used to prevent the loss of process gas. Appendix D presents a calculation method for use with labyrinth seals. 

Calculating the Pressure Loss of an Ejector in a Pneumatic Conveying System 1353... 

Reprinted from M. Lampinen, Calculation Methods for Determining the Pressure Loss of Two-Phase Pipe Flow and Ejectors in Pneumatic Conveying Systems, Acta Polytechnica Scandinavica, Mechanical Engineering Series No. 99, published by the Finnish Academy of Technology, Helsinki, 1991. 

As another example of calculation and dimensioning of pneumatic conveying systems we consider an ejector shown in Fig. 14.20. In fluidized bed combus tion systems a part of the ash is circulated with the hot flue gas. The task of the ejector, is to increase the pressure of the circulating gas to compensate the pressure losses of the circulation flow. The motivation for using an ejector, rather than a compressor, is the high temperature of the flue gas. The energy... 

TABLE 14.3. Calculations for an Ejector of a Pneumatic Conveying System, According to Fig. 14.20... 

For a mixture of steam and air handled by an ejector, the temperature of the mixture in the ejector mixing chamber is calculated by [11] ... 

Reference [11] provides a complete procedure for testing ejector units in vacuum service, and the charts and calculation procedures for the tests. 

Calculate pressure drop from this point to the process location of the suction flange of the first stage ejector. 

Dusts, particle sizes, 225 Dusts, hazard class, 521-523 Explosion characteristics, 524 Efficiency, centrifugal pumps, 200 Ejector control, 380 Ejector systems, 343, 344, 351 Air inleakage, table, 366, 367 Applications, 345 Calculations, 359-366 Chilled water refrigeration, 350 Comparison guide, 357, 375 Evacuation lime, 380, 381 Charts, 382 Example, 381 Features, 345... 

Installation arrangements, 351 Pump-down time, 380 Selection procedure, 374 Specification form, 377 Specifications, 373 Steam jet comparison, 356 Types of loads, 359 Ejectors, 346 Applications, 353 Barometric condenser, 249, 376 Booster, 370 Calculations Actual air capacity, 362 Air equivalent, 360... 

Ejectors, steam/water requirements, 371 Electrical charge on tanks, 537 Electrical precipaiaiors, 280 Applications, 280, 282 Concept of operation, 281 Emergency relief, 450 Engineering, plant development, 46 Equipment symbols, 19—2 L Abbreviations, 25 Instruments, 21, 26. 29 Piping, 22 Valve codes, 26 Equivalent feel (flow), 86 Estimated design calculation time,... 

Other pieces may have to be elevated to enable the system to operate. A steam jet ejector with an intercondenser that is used to produce a vacuum must be located above a 34 ft (10 m) barometric leg. Condensate receivers and holding tanks frequently must be located high enough to provide an adequate net positive suction head (NPSH) for the pump below. For many pumps an NPSH of at least 14 ft (4.2 m) H2O is desirable. Others can operate when the NPSH is only 6 ft (2 m) H2O. See Chapter 8 for a method of calculating NPSH. 

Ejector. Gas chamber pressure needs to be known in order to calculate ejector power input. A semiempirical equation was developed by Henzler (10) that related the entrainment ratio to other system variables ... 

Gas chamber pressure, ps, can be calculated from this equation through iteration. Factor B depends on the mixing tube/nozzle diameter ratio for a given ejector type and needs to be determined experimentally. 

The power input, required for calculation of the mass transfer coefficient, is calculated from Eqs. (36) and (37). For a motionless mixer, the power comes from the gas and liquid phases for the ejector, power comes from the liquid only. 

Detailed design and technoeconomic calculations in close association with manufacturers of compressors, steam jet ejectors, and steam dryers and with utility companies are required to make confident conclusions regarding the optimal steam reuse and recycle systems for steam dryers. 

The above optimization problem is applicable to systems involving air and water only. It ignores vapour superheat at the ejector inlet, vapour sub-cooling and liquid sub-cooling in condenser area calculations, and also assumes a simple log-mean temperature driving force (LMTD) without any correction factor. Hence, condenser area calculations are approximate but sufficient for optimization. Actual condenser design is quite complex for systems... 

Calculate the entrainment velocity U, the velocity U at the outlet of the hydro-ejector, and the numerical values of the pressure. Calculations will be facilitated by bringing in the ratio of the area of the primary jet to that of the hydroejector tube, 5 = Dq/Di = 1/4, as well as the velocity ratio, U/Ue.. 

Calculate the flow rate of the hydro-ejector, as well as the rate of momentum at the outlet of the hydro-ejector (plane P ). Explain one benefit of the hydro-ejector by comparing it to the configuration without a secondary supply tube as studied in parti. 

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