{"id":1538,"date":"2006-01-01T16:00:47","date_gmt":"2006-01-01T22:00:47","guid":{"rendered":"http:\/\/www.jmcampbell.com\/tip-of-the-month\/?p=1538"},"modified":"2012-12-07T10:52:15","modified_gmt":"2012-12-07T16:52:15","slug":"finding-the-optimum-compressor-interstage-pressure","status":"publish","type":"post","link":"http:\/\/www.jmcampbell.com\/tip-of-the-month\/2006\/01\/finding-the-optimum-compressor-interstage-pressure\/","title":{"rendered":"Finding the Optimum Compressor Interstage Pressure"},"content":{"rendered":"<p>In this Tip of the Month, we will show how to determine the optimum interstage pressure for a two-stage compression process. We will also study other operating condition such as feed temperature, heavy end in the feed, and water moisture. For this purpose, we used a commercial simulation package and the SRK EoS for the prediction of phase behavior and thermodynamic properties.<\/p>\n<p style=\"text-align: center;\"><img data-recalc-dims=\"1\" decoding=\"async\" loading=\"lazy\" class=\"size-full wp-image-1549 alignnone\" title=\"Table 1\" src=\"https:\/\/i0.wp.com\/www.jmcampbell.com\/tip-of-the-month\/wp-content\/uploads\/2012\/12\/Table-12.png?resize=234%2C261\" alt=\"\" width=\"234\" height=\"261\" \/><\/p>\n<p style=\"text-align: left;\">\nThe gas mixture with the composition shown in Table1 at\u00a037 \u00b0C (98.6 \u00b0F) and 31 bar(g) (450 psig) on a dry basis is compressed in two stages for injection into an oil field as a means of enhancing production. The process flow diagram is shown in Figure 1. The injection pressure is 131 bar (g) (1900 psig) and temperature is 65 \u00b0C (149 \u00b0F). The gas rate for stream 2 is 6.792&#215;10<sup>6<\/sup> std m<sup>3<\/sup>\/d (240 MMSCFD). The suggested isentropic efficiency is 72 percent and a mechanical efficiency for each stage of compressor is 80 percent. The inlet temperature of each compressor stage should not exceed 56 \u00b0C (132.8 \u00b0F). The feed gas is saturated with water and 5 psi (34 kPa) pressure drop is allowed between each compressor discharge and exit of the flash separator.<\/p>\n<p align=\"center\"><img data-recalc-dims=\"1\" decoding=\"async\" loading=\"lazy\" class=\"aligncenter size-full wp-image-1548\" title=\"Figure 1\" src=\"https:\/\/i0.wp.com\/www.jmcampbell.com\/tip-of-the-month\/wp-content\/uploads\/2012\/12\/Figure-11.png?resize=610%2C191\" alt=\"\" width=\"610\" height=\"191\" srcset=\"https:\/\/i0.wp.com\/www.jmcampbell.com\/tip-of-the-month\/wp-content\/uploads\/2012\/12\/Figure-11.png?w=610 610w, https:\/\/i0.wp.com\/www.jmcampbell.com\/tip-of-the-month\/wp-content\/uploads\/2012\/12\/Figure-11.png?resize=300%2C93 300w\" sizes=\"auto, (max-width: 610px) 100vw, 610px\" \/><\/p>\n<p><strong>Phase Envelope \u2013 <\/strong>The first step is to determine the state of the inlet to the 1<sup>st<\/sup> stage suction scrubber. The phase envelope for the feed gas after being saturated with water (stream 2) is shown in Figure 2. This figure also presents the phase envelope for stream 6 which is the vapor stream at the suction to the first stage of compression. The red circle displays the condition at the first stage suction.<strong><\/strong><\/p>\n<p><strong>Optimization Scenarios<\/strong>-Figure 3 presents the variation of compression ratios as a function of 1<sup>st<\/sup> stage discharge pressure. From this figure, it can be seen that equal compression ratios of 2.04 is obtained at a pressure of 65.2 bar. The ideal optimum interstage pressure for equal compression ratio is also found to be <a href=\"https:\/\/i0.wp.com\/www.jmcampbell.com\/tip-of-the-month\/wp-content\/uploads\/2012\/12\/Equation.png\"><img data-recalc-dims=\"1\" decoding=\"async\" loading=\"lazy\" class=\"alignnone  wp-image-1547\" title=\"Equation\" src=\"https:\/\/i0.wp.com\/www.jmcampbell.com\/tip-of-the-month\/wp-content\/uploads\/2012\/12\/Equation.png?resize=150%2C20\" alt=\"\" width=\"150\" height=\"20\" \/><\/a>\u00a0bar. Figure 4 presents the variation of each heat exchanger cooling load and each stage compression power requirement as a function of 1<sup>st<\/sup> stage discharge pressure. These variations are almost linear.<\/p>\n<p style=\"text-align: center;\"><img data-recalc-dims=\"1\" decoding=\"async\" loading=\"lazy\" class=\"aligncenter size-full wp-image-1546 alignnone\" title=\"Figures 2&amp;3\" src=\"https:\/\/i0.wp.com\/www.jmcampbell.com\/tip-of-the-month\/wp-content\/uploads\/2012\/12\/Figures-23-e1354746093200.png?resize=600%2C196\" alt=\"\" width=\"600\" height=\"196\" \/><br \/>\n<img data-recalc-dims=\"1\" decoding=\"async\" loading=\"lazy\" class=\"size-full wp-image-1545 aligncenter\" title=\"Figure 4\" src=\"https:\/\/i0.wp.com\/www.jmcampbell.com\/tip-of-the-month\/wp-content\/uploads\/2012\/12\/Figure-4.png?resize=336%2C219\" alt=\"\" width=\"336\" height=\"219\" srcset=\"https:\/\/i0.wp.com\/www.jmcampbell.com\/tip-of-the-month\/wp-content\/uploads\/2012\/12\/Figure-4.png?w=336 336w, https:\/\/i0.wp.com\/www.jmcampbell.com\/tip-of-the-month\/wp-content\/uploads\/2012\/12\/Figure-4.png?resize=300%2C195 300w\" sizes=\"auto, (max-width: 336px) 100vw, 336px\" \/><\/p>\n<p style=\"text-align: left;\">\nTable 2 presents the simulation results for two cases of optimizations. In case A , the total power requirements was minimized by finding the 1<sup>st<\/sup> stage discharge pressure with the constraint of equal stage compression ratio. This results in approximately equal compression power requirements for the two stages. It should be noted that the slight difference in compression ratio and stage compression power is due to the 34 kPa (5 psi) pressure-drop between discharge of 1<sup>st<\/sup> stage and suction of 2<sup>nd<\/sup> stage. However, in Case B, the total energy requirement was minimized by finding the 1<sup>st<\/sup> stage discharge pressure without the constraint of equal stage compression ratios. The results summarized in Table 2 indicate that there is a big difference between the case A and case B 1<sup>st<\/sup> stage discharge pressures. It can also be seen that the case A total power requirement (W<sub>1<\/sub>+W<sub>2<\/sub>) is clearly larger than case A (about 40 % higher).<\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: center;\"><img data-recalc-dims=\"1\" decoding=\"async\" loading=\"lazy\" class=\"size-full wp-image-1543\" title=\"Table 2\" src=\"https:\/\/i0.wp.com\/www.jmcampbell.com\/tip-of-the-month\/wp-content\/uploads\/2012\/12\/Table-2-e1354746185569.png?resize=600%2C126\" alt=\"\" width=\"600\" height=\"126\" \/><\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: center;\"><img data-recalc-dims=\"1\" decoding=\"async\" loading=\"lazy\" class=\"size-full wp-image-1544 aligncenter\" title=\"Figure 5\" src=\"https:\/\/i0.wp.com\/www.jmcampbell.com\/tip-of-the-month\/wp-content\/uploads\/2012\/12\/Figure-5.png?resize=360%2C255\" alt=\"\" width=\"360\" height=\"255\" srcset=\"https:\/\/i0.wp.com\/www.jmcampbell.com\/tip-of-the-month\/wp-content\/uploads\/2012\/12\/Figure-5.png?w=360 360w, https:\/\/i0.wp.com\/www.jmcampbell.com\/tip-of-the-month\/wp-content\/uploads\/2012\/12\/Figure-5.png?resize=300%2C212 300w\" sizes=\"auto, (max-width: 360px) 100vw, 360px\" \/><\/p>\n<p>The variation of total compression power and total cooling load requirement as a function of 1<sup>st<\/sup> stage discharge pressure are shown on the left hand side y-axis of Figure 5. This figure indicates clearly that the minimum power requirement occurs when the 1<sup>st<\/sup> stage discharge pressure is 77.8 bars. Figure 5 also provides an indication that the total power requirement changes very little for 1<sup>st<\/sup> Stage discharge pressures between about 76 bars and 80 bars.<\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: center;\"><img data-recalc-dims=\"1\" decoding=\"async\" loading=\"lazy\" class=\"size-full wp-image-1542 aligncenter\" title=\"Table 3\" src=\"https:\/\/i0.wp.com\/www.jmcampbell.com\/tip-of-the-month\/wp-content\/uploads\/2012\/12\/Table-3.png?resize=509%2C359\" alt=\"\" width=\"509\" height=\"359\" srcset=\"https:\/\/i0.wp.com\/www.jmcampbell.com\/tip-of-the-month\/wp-content\/uploads\/2012\/12\/Table-3.png?w=509 509w, https:\/\/i0.wp.com\/www.jmcampbell.com\/tip-of-the-month\/wp-content\/uploads\/2012\/12\/Table-3.png?resize=300%2C211 300w\" sizes=\"auto, (max-width: 509px) 100vw, 509px\" \/><\/p>\n<p><strong>Effect of Water Vapor in the Feed-<\/strong>The detail of simulation results based on unequal compression ratio for the two options of wet feed (saturated with water vapor) and dry feed is shown in Table 3. As can be seen from this table, water vapor has little effect on the performance of the process. The 0.23 % increased compression power requirement for the wet feed is due to 0.28 % increase in feed flow rate for the presence of water. It should be noted that the dry feed flow rate is 11953 kmol\/hr and the wet feed flow rate is 11987 kmol\/hr (34 kmol water\/hr + 11953 kmol of dry gas\/hr).<\/p>\n<p style=\"text-align: center;\"><strong><img data-recalc-dims=\"1\" decoding=\"async\" loading=\"lazy\" class=\"size-full wp-image-1541 aligncenter alignnone\" title=\"Table 4\" src=\"https:\/\/i0.wp.com\/www.jmcampbell.com\/tip-of-the-month\/wp-content\/uploads\/2012\/12\/Table-4.png?resize=508%2C363\" alt=\"\" width=\"508\" height=\"363\" srcset=\"https:\/\/i0.wp.com\/www.jmcampbell.com\/tip-of-the-month\/wp-content\/uploads\/2012\/12\/Table-4.png?w=508 508w, https:\/\/i0.wp.com\/www.jmcampbell.com\/tip-of-the-month\/wp-content\/uploads\/2012\/12\/Table-4.png?resize=300%2C214 300w\" sizes=\"auto, (max-width: 508px) 100vw, 508px\" \/><\/strong><\/p>\n<p style=\"text-align: left;\"><strong><br \/>\nEffect of Heavy Ends in the Feed-<\/strong>In order to study the effect of heavy ends on the performance of the process, normal octane (nC<sub>8<\/sub>H<sub>18<\/sub>) was replaced with normal decane (nC<sub>10<\/sub>H<sub>22<\/sub>) and the simulation was repeated.\u00a0The detail of simulation results based on unequal compression ratio for these two options of heavy ends is shown in Table 4. As can be seen from this table, the total compression power requirements decreases slightly for the case using nC<sub>10<\/sub>H<sub>22<\/sub> due to the fact that more of the heavy component is removed in the first separator. The compressor power for the stages and heat exchanger duties are not affected by the presence of heavier components in the feed stream. In other words, the feed flow rate to the compressor decreases when nC<sub>8<\/sub>H<sub>18<\/sub> is replaced by nC<sub>10<\/sub>H<sub>22<\/sub>.<\/p>\n<p><strong>Effect of Feed Temperature &#8211;\u00a0<\/strong>In order to study the effect of feed temperature on the performance of the process, the feed temperature was increased from 37 \u00b0C to 56 \u00b0C and the simulation results are shown in Table 5.<\/p>\n<p style=\"text-align: center;\"><img data-recalc-dims=\"1\" decoding=\"async\" loading=\"lazy\" class=\"size-full wp-image-1540 aligncenter\" title=\"Table 5\" src=\"https:\/\/i0.wp.com\/www.jmcampbell.com\/tip-of-the-month\/wp-content\/uploads\/2012\/12\/Table-5.png?resize=519%2C362\" alt=\"\" width=\"519\" height=\"362\" srcset=\"https:\/\/i0.wp.com\/www.jmcampbell.com\/tip-of-the-month\/wp-content\/uploads\/2012\/12\/Table-5.png?w=519 519w, https:\/\/i0.wp.com\/www.jmcampbell.com\/tip-of-the-month\/wp-content\/uploads\/2012\/12\/Table-5.png?resize=300%2C209 300w\" sizes=\"auto, (max-width: 519px) 100vw, 519px\" \/><\/p>\n<p><strong>Table 5. The effect of feed temperature on the performance of the process<\/strong><\/p>\n<p>The feed at 56 \u00b0C represents the actual condition during the summer season. As can be seen from this table, the warmer feed requires an increase of 5.34 % in total compression power consumption. So the feed temperature is an important parameter and its variation, especially due to seasonal change, should be taken into considerations.<strong> <\/strong><\/p>\n<p>For the unequal compression ratio case having compression ratios of 2.397 and 1.733 for stages 1 and 2, respectively, the variation of energy requirement with feed temperature is shown in Figure 6. Stage 2 was not affected with the variation of feed temperature; therefore, the compression power and cooling load for stage 2 remained constant at 7.254 MW and 8.407 MW, respectively. Since, the compression ratio was constant, the compression power requirement for stage 1 varied from 12.2 to 13.28 MW; however, the cooling load varied drastically from 10.85 to 14.83 MW.<\/p>\n<p style=\"text-align: center;\"><img data-recalc-dims=\"1\" decoding=\"async\" loading=\"lazy\" class=\"size-full wp-image-1539 aligncenter alignnone\" title=\"Figure 6\" src=\"https:\/\/i0.wp.com\/www.jmcampbell.com\/tip-of-the-month\/wp-content\/uploads\/2012\/12\/Figure-6.png?resize=349%2C244\" alt=\"\" width=\"349\" height=\"244\" srcset=\"https:\/\/i0.wp.com\/www.jmcampbell.com\/tip-of-the-month\/wp-content\/uploads\/2012\/12\/Figure-6.png?w=349 349w, https:\/\/i0.wp.com\/www.jmcampbell.com\/tip-of-the-month\/wp-content\/uploads\/2012\/12\/Figure-6.png?resize=300%2C209 300w\" sizes=\"auto, (max-width: 349px) 100vw, 349px\" \/><\/p>\n<p style=\"text-align: left;\">\nIn the light of preceding discussion, the following tips are suggested:<\/p>\n<ul>\n<li>Be sure to check the phase of the compressor suction stream. This also includes the interstage condition to ensure that liquid does not enter the compressor.<\/li>\n<li>If economically possible, lower the interstage suction temperature since this will reduce the overall compression power requirement.<\/li>\n<li>Be sure to check the water content at the interstage conditions since there may be water drop out which would impact equipment performance.<\/li>\n<li>The choice of equal pressure ratios for minimizing the compression power requirement is close to an optimum choice when the suction temperatures are equal.<\/li>\n<li>Characterization of the heavy ends (C<sub>7+<\/sub>) does not greatly impact the compressor power requirement since heavy components are mostly removed in the inlet scrubber.<\/li>\n<li>Characterization of the C<sub>7+<\/sub> will impact the condensation that takes place in the inlet suction scrubber and thus the molecular weight of the compressed gas will be affected.<\/li>\n<\/ul>\n<p style=\"text-align: left;\" align=\"right\"><em>Dr. Mahmood Moshfeghian<\/em><\/p>\n<p>&nbsp;<\/p>\n","protected":false},"excerpt":{"rendered":"<p>In this Tip of the Month, we will show how to determine the optimum interstage pressure for a two-stage compression process. We will also study other operating condition such as feed temperature, heavy end in the feed, and water moisture. For this purpose, we used a commercial simulation package and the SRK EoS for the [&hellip;]<\/p>\n","protected":false},"author":23,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"nf_dc_page":"","_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"_jetpack_newsletter_access":"","_jetpack_dont_email_post_to_subs":false,"_jetpack_newsletter_tier_id":0,"_jetpack_memberships_contains_paywalled_content":false,"_jetpack_feature_clip_id":0,"_jetpack_memberships_contains_paid_content":false,"footnotes":"","jetpack_publicize_message":"","jetpack_publicize_feature_enabled":true,"jetpack_social_post_already_shared":false,"jetpack_social_options":{"image_generator_settings":{"template":"highway","default_image_id":0,"font":"","enabled":false},"version":2},"jetpack_post_was_ever_published":false},"categories":[3,10],"tags":[],"coauthors":[15],"class_list":["post-1538","post","type-post","status-publish","format-standard","hentry","category-gas-processing","category-process-facilities"],"jetpack_publicize_connections":[],"jetpack_featured_media_url":"","jetpack_shortlink":"https:\/\/wp.me\/p1pQc4-oO","jetpack_sharing_enabled":true,"_links":{"self":[{"href":"http:\/\/www.jmcampbell.com\/tip-of-the-month\/wp-json\/wp\/v2\/posts\/1538","targetHints":{"allow":["GET"]}}],"collection":[{"href":"http:\/\/www.jmcampbell.com\/tip-of-the-month\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"http:\/\/www.jmcampbell.com\/tip-of-the-month\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"http:\/\/www.jmcampbell.com\/tip-of-the-month\/wp-json\/wp\/v2\/users\/23"}],"replies":[{"embeddable":true,"href":"http:\/\/www.jmcampbell.com\/tip-of-the-month\/wp-json\/wp\/v2\/comments?post=1538"}],"version-history":[{"count":11,"href":"http:\/\/www.jmcampbell.com\/tip-of-the-month\/wp-json\/wp\/v2\/posts\/1538\/revisions"}],"predecessor-version":[{"id":1551,"href":"http:\/\/www.jmcampbell.com\/tip-of-the-month\/wp-json\/wp\/v2\/posts\/1538\/revisions\/1551"}],"wp:attachment":[{"href":"http:\/\/www.jmcampbell.com\/tip-of-the-month\/wp-json\/wp\/v2\/media?parent=1538"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"http:\/\/www.jmcampbell.com\/tip-of-the-month\/wp-json\/wp\/v2\/categories?post=1538"},{"taxonomy":"post_tag","embeddable":true,"href":"http:\/\/www.jmcampbell.com\/tip-of-the-month\/wp-json\/wp\/v2\/tags?post=1538"},{"taxonomy":"author","embeddable":true,"href":"http:\/\/www.jmcampbell.com\/tip-of-the-month\/wp-json\/wp\/v2\/coauthors?post=1538"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}