{"id":2396,"date":"2023-06-13T09:16:28","date_gmt":"2023-06-13T09:16:28","guid":{"rendered":"https:\/\/fairconditioning.org\/fc-activity\/?post_type=activity&p=2396"},"modified":"2023-06-13T09:16:28","modified_gmt":"2023-06-13T09:16:28","slug":"structure-cooling-system-design","status":"publish","type":"activity","link":"https:\/\/fairconditioning.org\/fc-activity\/activity\/structure-cooling-system-design\/","title":{"rendered":"Structure Cooling: System Design"},"content":{"rendered":"\n
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Objective<\/h5>\n\n\n\n

Estimating Structure Cooling Water Flow Rate Requirements for Concrete Slab<\/p>\n\n\n\n

Outcome<\/h5>\n\n\n\n

Students will have hands-on experience of determining the water required for structure cooling and enhance their learning about sustainable cooling systems and technologies which can help them integrate it into their designs and reduce the use of Air Conditioners.<\/p>\n\n\n\n

Prerequisites<\/h5>\n\n\n\n

Knowledge required about Structure Cooling system and intermediate level knowledge about calculations.<\/p>\n\n\n\n

Given Data<\/h5>\n\n\n\n

Design Conditions:<\/p>\n\n\n\n

Rooftop Temperature = 45\u00b0C<\/p>\n\n\n\n

Solar Radiation (peak) = 500 W\/m2<\/p>\n\n\n\n

10,000 sq. ft. slab<\/p>\n\n\n\n

Wet Bulb Temperature  = 20\u00b0C<\/p>\n\n\n\n

Pipe Material = Polypropylene<\/p>\n\n\n\n

Pipe Wall Thickness = 3 mm<\/p>\n\n\n\n

Pipe Inside Diameter = 25 mm<\/p>\n\n\n\n

Total Pipe Length = 6200 m<\/p>\n\n\n\n

Conversion Factors & Constants:<\/p>\n\n\n\n

1 m2<\/sup> = 10.76 ft2<\/sup><\/p>\n\n\n\n

Specific Heat Capacity (Water) =4.186 kJ\/kg \u1d52C<\/p>\n\n\n\n

Thermal Conductivity (Polypropylene) = 0.44 W\/m \u1d52C<\/p>\n\n\n\n

Calculations<\/h5>\n\n\n\n

Step 1: <\/strong>Calculate Rooftop Heat Load (Q, Solar) = 500 wm\u00b2 x 10,000 ft\u00b2 x 1 m\u00b210.76 ft2 = 4,64,684 W<\/p>\n\n\n\n

Step 2: <\/strong>Calculate Pipe Area: A = 2\u03c0rl= 2*3.14*25m*6200m1000*2 = 486.9 m\u00b2<\/p>\n\n\n\n

Step 3: <\/strong>Calculate R = d (thickness)\ud835\udf7a(thermal conductivity) = 0.003m0.44 Wm\u00b0C = 0.0068 m\u00b2CW<\/p>\n\n\n\n

Step 4: <\/strong>Calculate U = 1R = 10.0068= 146.67 Wm\u00b2\u00b0C<\/p>\n\n\n\n

Step 5: <\/strong>Calculate LMTD based on known Heat Transfer to Water (Q, Solar) = U x A x LMTD<\/p>\n\n\n\n

4,64,684 W= 146.67 Wm\u00b2\u00b0C  x 486.9 m\u00b2 x LMTD <\/p>\n\n\n\n

LMTD =6.5 \u00b0C<\/p>\n\n\n\n

Step 6: <\/strong>Calculate \u2206T1 = 45-20 = 25\u00b0C<\/p>\n\n\n\n

Step 7: <\/strong>Calculate \u2206T2 from LTMD =\u2206T1 – \u2206T2ln \u2206T1\u2206T2<\/p>\n\n\n\n

6.5= (25-\u2206T2)ln 25\u2206T2<\/p>\n\n\n\n

Using Spreadsheet: \u2206T2 = 0.6<\/p>\n\n\n\n

Step 8: <\/strong>Calculate water exit temperature TCOLD-OUT<\/sub>, assuming constant slab temperature:<\/p>\n\n\n\n

\u2206T2 = 45-TCOLD OUT<\/sub> <\/p>\n\n\n\n

TCOLD OUT<\/sub> = 44.4<\/p>\n\n\n\n

References<\/h5>\n\n\n\n

Fairconditioning Structure cooling ppt and animation video.<\/p>\n<\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

Objective Estimating Structure Cooling Water Flow Rate Requirements for Concrete Slab Outcome Students will have hands-on experience of determining the water required for structure cooling and enhance their learning about sustainable cooling systems and technologies which can help them integrate it into their designs and reduce the use of Air Conditioners. Prerequisites Knowledge required about […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","template":"","meta":{"inline_featured_image":false},"activity_subject":[124,125],"academic_year":[69],"complexity":[60],"content_tags":[434,494,500],"activity_type":[371],"activity_duration":[174],"acf":[],"_links":{"self":[{"href":"https:\/\/fairconditioning.org\/fc-activity\/wp-json\/wp\/v2\/activity\/2396"}],"collection":[{"href":"https:\/\/fairconditioning.org\/fc-activity\/wp-json\/wp\/v2\/activity"}],"about":[{"href":"https:\/\/fairconditioning.org\/fc-activity\/wp-json\/wp\/v2\/types\/activity"}],"author":[{"embeddable":true,"href":"https:\/\/fairconditioning.org\/fc-activity\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/fairconditioning.org\/fc-activity\/wp-json\/wp\/v2\/comments?post=2396"}],"version-history":[{"count":1,"href":"https:\/\/fairconditioning.org\/fc-activity\/wp-json\/wp\/v2\/activity\/2396\/revisions"}],"predecessor-version":[{"id":2397,"href":"https:\/\/fairconditioning.org\/fc-activity\/wp-json\/wp\/v2\/activity\/2396\/revisions\/2397"}],"wp:attachment":[{"href":"https:\/\/fairconditioning.org\/fc-activity\/wp-json\/wp\/v2\/media?parent=2396"}],"wp:term":[{"taxonomy":"activity_subject","embeddable":true,"href":"https:\/\/fairconditioning.org\/fc-activity\/wp-json\/wp\/v2\/activity_subject?post=2396"},{"taxonomy":"academic_year","embeddable":true,"href":"https:\/\/fairconditioning.org\/fc-activity\/wp-json\/wp\/v2\/academic_year?post=2396"},{"taxonomy":"complexity","embeddable":true,"href":"https:\/\/fairconditioning.org\/fc-activity\/wp-json\/wp\/v2\/complexity?post=2396"},{"taxonomy":"content_tags","embeddable":true,"href":"https:\/\/fairconditioning.org\/fc-activity\/wp-json\/wp\/v2\/content_tags?post=2396"},{"taxonomy":"activity_type","embeddable":true,"href":"https:\/\/fairconditioning.org\/fc-activity\/wp-json\/wp\/v2\/activity_type?post=2396"},{"taxonomy":"activity_duration","embeddable":true,"href":"https:\/\/fairconditioning.org\/fc-activity\/wp-json\/wp\/v2\/activity_duration?post=2396"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}