{"id":249,"date":"2015-11-02T08:29:43","date_gmt":"2015-11-02T08:29:43","guid":{"rendered":"http:\/\/ejektor.co.rs\/?p=249"},"modified":"2017-01-04T18:19:01","modified_gmt":"2017-01-04T18:19:01","slug":"ejector-compressors","status":"publish","type":"post","link":"https:\/\/ejektor.co.rs\/?p=249","title":{"rendered":"EJECTOR COMPRESSORS"},"content":{"rendered":"

Depending on the type of driving fluid ejector compressors are divided into:<\/p>\n

    \n
  1. Ejector hydro compressors<\/a><\/li>\n
  2. Ejector gas compressors<\/a><\/li>\n
  3. Ejector vapor compressors<\/a><\/li>\n<\/ol>\n

    Operating principle:<\/dfn><\/strong><\/p>\n

    All ejector compressors operate on the principle of using high pressure energy of the driving fluid for pumping in and compressing pumped gases. The pressure of the fluid mixture at the ejector output depends on the input pressures of the driving and pumped fluid and their mass flow ratio.<\/p>\n

    <\/a>6.1 Ejector hydro compressors (liquid-gas)<\/h2>\n

    Liquid, most often water is used as the driving fluid for these compressors for pumping in and compressing gases (most often air). Depending on the pressure of the driving liquid and mass ratio between the gas-liquid flow it is possible to compress gases to pressures higher than 10 bars. The volume ratio between the gas-liquid flow can be realized within the limits of 0 \u2013 3 (Vg<\/sub>\/Vt<\/sub>, [m3<\/sup>\/m3<\/sup>]).<\/p>\n

    Compressed gas exiting the ejector is practically washed and partly cleansed from various pollutants. This compression procedure practically removes dust and other mechanical pollutants from gas and also droplets and evaporates of other liquids (water, oil, etc.).<\/p>\n

    Application:<\/dfn><\/strong><\/p>\n

    They are used for compression of gas requiring relatively low flows and large pressures (for example ozoning drinking water, figure 6.1), pumping in compressed air into a hydrofor (figure 6.2) and many other processes.<\/p>\n

    They are especially applied in the food industry that requires clean compressed air with no dirt and compressed oil content. The relative moisture of such compressed air is close to 100% that is also desirable in the food industry.<\/p>\n

    Diagram 6.1 shows the dependence between the output pressure p3<\/sub> on the pump input pressure p1<\/sub> = pp<\/sub> and the volume ratio between the pumped in air and water V2<\/sub>\/V1<\/sub>.<\/p>\n

    \n
    \n
    \"6-1\"<\/a>
    \nFigure 6.1 Ozoning of drinking water<\/strong><\/div>\n
    \"6-2\"<\/a>
    \nFigure 6.2 Pumping air into a hydrofor<\/strong><\/div>\n<\/div>\n
    \n
    \"6-3\"<\/a>
    \nFigure 6.3 Open system for air compression<\/strong><\/div>\n
    \"6-4\"<\/a>
    \nFigure 6.4 Circulation (closed) system for air compression<\/strong><\/div>\n<\/div>\n<\/div>\n

    Example 6.1<\/dfn><\/strong><\/p>\n

    Data: <\/dfn>Air using water needs to be pumped in a vessel (figure 6.3) under pressure p3<\/sub> = 1.6 bar. The pressure of driving water in front of the ejector is p1<\/sub> = 10 bar.<\/p>\n

    How many cubic meters of air can be pumped in with one cubic meter of water?<\/p>\n

    Solution: <\/dfn>For the reservoir pressure of p3<\/sub> = 1.6 bar and water pressure in front of the ejector of p1<\/sub> = 10 bar, from diagram 6.1 one reads that V2<\/sub>\/V1<\/sub> = 1.1 that means that 1.1 m3 <\/sup>of air can be pumped in with 1 m3<\/sup> of water.<\/p>\n

    Diagram 6.2 shows the dependence of the ratio between manometer pressures at the ejector output and pump pressure p3<\/sub>\/pp<\/sub> on the ratio between the volume flow of air and water, for operating conditions presented in figures 6.3 and 6.4.<\/p>\n

    The maximal pressure that can be reached in the vessel according to the connection scheme presented in figure 6.3 is 0.7 of the driving pressure of pump pp<\/sub>, i.e. pipeline pressure p1<\/sub> (p3<\/sub> =0.7\u00d7p1 <\/sub>= 0.7\u00d7pp<\/sub>) and according to the connection scheme given in figure 6.4 is 2.44 times the pump pressure pp<\/sub> (2.44\u00d7pp<\/sub>)<\/p>\n

    Example 6.2<\/dfn><\/strong><\/p>\n

    Data:<\/dfn> The pressure ratio between the reservoir and pump is p3<\/sub>\/pp<\/sub> = 0.25. The flow for the connection schemes given in figures 6.3 and 6.4 is required?<\/p>\n

    Solution:<\/dfn> For the pressure ratio of p3<\/sub>\/pp<\/sub> = 0.25 the volume flow ratio is read from figure 6.3 as V2<\/sub>\/V1<\/sub> = V2<\/sub>\/Vp<\/sub> = 0.75 and according to figure 6.4 V2<\/sub>\/V1<\/sub> = 0.9.<\/p>\n

    \n
    \n
    \"kompresori02\"<\/a><\/div>\n
    \"kompresori03\"<\/a><\/div>\n<\/div>\n<\/div>\n

    6.2 Ejector gas compressors<\/h2>\n

    These compressors are used for mixing and compressing gases. Compressed gas (most often air) is used as the driving fluid for these compressors. It is used for pumping, mixing and compressing the pumped gas. Using the Laval nozzle the output rates of the driving gas rise to speeds above the speed of sound, so they can be efficiently used for attaining relatively high output pressures.<\/p>\n\n\n\n\n
    \"6-5\"<\/a><\/td>\n<\/tr>\n
    Figure 6.5 Mixing combustible gas and air<\/strong><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

    Application:<\/dfn><\/strong><\/p>\n

    They are used for extraction of natural undergoing gases with low pressure using other or similar gases present at higher pressures, for mixing compressed gases with different pressure in order to obtain a mixture with a corresponding pressure, for obtaining gas sinter under very high pressures (several hundred bars), for example for obtaining ammoniac.<\/p>\n

    Figure 6.5 shows the scheme of pumping, mixing and compressing combustible gas and air.<\/p>\n

    Diagram 6.3 shows the dependence of the ratio between the absolute pressure of the driving and pumped fluid (p1<\/sub>\/p2<\/sub>) on the ratio between mass flows (m2<\/sub>\/m1<\/sub>) for t1<\/sub> = t2<\/sub>. The diagram can be used for different temperatures t1<\/sub> and t2<\/sub> though the read values of m must be multiplied with (T1<\/sub>\/T2<\/sub>)0.5<\/sup> (T1<\/sub> and T2<\/sub>are absolute temperatures of the driving and pumped gas).<\/p>\n

    Example 6.3<\/dfn><\/strong><\/p>\n

    Data:<\/dfn> The driving pressure at the ejector input p1<\/sub> = 4 barabs<\/sub>, pumped pressure p2<\/sub> = 0.9 barabs<\/sub>, pressure at the ejector output is atmospheric p3<\/sub> = 1 barabs<\/sub>. What is the mass flow ratio?<\/p>\n

    Solution: <\/dfn>For t1<\/sub> = t2<\/sub> and pressure ratios p1<\/sub>\/p2<\/sub> = 4\/0.9 = 4.4 and p3<\/sub>\/p2<\/sub> = 1\/0.9 = 1.11. From diagram 6.3 m = m2<\/sub>\/m1<\/sub> = 2.4.<\/p>\n\n\n\n
    \"kompresori05\"<\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

    6.3 Ejector vapor compressors<\/h2>\n

    Saturated or pre-heated water vapor is used as the driving fluid for these compressors. Driving vapor, passing through the nozzle expands to very low pressures and enters the ejector mixing chamber at the speed of sound or higher. In the mixing chamber particles of the driving liquid are taken up, compressed and take along vapor from the inlet port. The ratio between the output and input pressure is 1-10. Higher values are related to low pumping pressures of 0.01-0.1 barabs<\/sub>.<\/p>\n

    Application:<\/dfn><\/strong><\/p>\n

    Using fresh vapor as the driving fluid it is possible to pump in and compress used vapor and return it back into the process. Ejector vapor compressors are used in the following processes:<\/p>\n