{"id":2400,"date":"2023-06-13T11:04:22","date_gmt":"2023-06-13T11:04:22","guid":{"rendered":"https:\/\/fairconditioning.org\/fc-activity\/?post_type=activity&p=2400"},"modified":"2023-06-13T11:05:42","modified_gmt":"2023-06-13T11:05:42","slug":"water-or-air-better-cooling-medium","status":"publish","type":"activity","link":"https:\/\/fairconditioning.org\/fc-activity\/activity\/water-or-air-better-cooling-medium\/","title":{"rendered":"Water or Air: Better Cooling Medium?"},"content":{"rendered":"\n
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Objective<\/h5>\n\n\n\n

To compare the effectiveness of air vs water as a cooling medium.<\/p>\n\n\n\n

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

Students will realize that water is a much better cooling medium than air. Students will realize that air conditioning is an inefficient method to cool buildings.<\/p>\n\n\n\n

Requirements<\/h5>\n\n\n\n

None<\/p>\n\n\n\n

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

 <\/em><\/strong>A conceptual knowledge of the following:<\/p>\n\n\n\n

    \n
  1. Density<\/li>\n\n\n\n
  2. Specific Heat Capacity<\/li>\n\n\n\n
  3. Volumetric Heat Capacity<\/li>\n\n\n\n
  4. Thermal Conductivity<\/li>\n\n\n\n
  5. Functioning of Airconditioning and Water coolers<\/li>\n<\/ol>\n\n\n\n
    Procedure<\/h5>\n\n\n\n

    Step 1: <\/em><\/strong>Find out the thermal conductivity (in W\/m-K) of<\/p>\n\n\n\n

      \n
    1. air<\/li>\n\n\n\n
    2. water<\/li>\n<\/ol>\n\n\n\n

      Step 2:<\/em><\/strong> Estimate the thermal conductivity multiplier for water and air by dividing the higher thermal conductivity value with the lower one.<\/p>\n\n\n\n

      Step 3: <\/em><\/strong>Find out the density (in kg\/m3<\/sup>). of <\/p>\n\n\n\n

        \n
      1. air<\/li>\n\n\n\n
      2. water <\/li>\n<\/ol>\n\n\n\n

        Step 4: <\/em><\/strong>Find out the specific heat capacity (in kW\/kg-K) of<\/p>\n\n\n\n

          \n
        1. air<\/li>\n\n\n\n
        2. water<\/li>\n<\/ol>\n\n\n\n

          Step 5:<\/em><\/strong> Estimate the volumetric heat capacity (i.e multiplier of density and specific heat capacity) (in kW\/m3<\/sup>-K) of<\/p>\n\n\n\n

            \n
          1. air<\/li>\n\n\n\n
          2. water<\/li>\n<\/ol>\n\n\n\n

            Step 6:<\/em><\/strong> Estimate the volumetric heat capacity multiplier for water and air by dividing the higher volumetric heat capacity value with the lower one.<\/p>\n\n\n\n

            Step 7:<\/em><\/strong> Ask the students: \u2018Which one of water and air can be a better medium for cooling spaces?\u2019 <\/p>\n\n\n\n

            Calculations<\/h5>\n\n\n\n
            Table<\/h5>\n\n\n\n
            <\/td>Thermal Conductivity (K) (W\/m-K)<\/em><\/strong><\/td>Specific Heat Capacity (Cp<\/sub>) (kW\/kg-K)<\/em><\/strong><\/td>Density (\u2374) (kg\/m3<\/sup>)<\/em><\/strong><\/td>Volumetric Heat Capacity (Cp<\/sub>x\u2374) (VP<\/sub>) (kW\/m3<\/sup>-K)<\/em><\/strong><\/td><\/tr>
            Air<\/em><\/td>0.03<\/td>1.004<\/td>1.225<\/td>1.23<\/td><\/tr>
            Water<\/em><\/td>0.6<\/td>4.186<\/td>1000<\/td>4186<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

            Step 1: <\/em><\/strong>Thermal Conductivity Multiplier = KWater<\/sub>(W<\/sup>\u2044m-K<\/sub><\/box>)<\/sup>\u2044KAir<\/sub>(W<\/sup>\u2044m-K<\/sub><\/box>)<\/sub><\/box><\/p>\n\n\n\n

            = 0.60.03<\/p>\n\n\n\n

            = 20 times<\/p>\n\n\n\n

            Hence, Thermal Conductivity of water is 20 times that of air.<\/p>\n\n\n\n

            Step 2: <\/em><\/strong>Volumetric Heat Capacity Multiplier = VP-Water<\/sub>(kW<\/sup>\u2044m3<\/sup>-K<\/sub><\/box>)<\/sup>\u2044VP-Air<\/sub>(kW<\/sup>\u2044m3<\/sup>-K<\/sub><\/box>)<\/sub><\/box><\/p>\n\n\n\n

            = 41861.23<\/p>\n\n\n\n

            = 3403 times <\/p>\n\n\n\n

            Hence, Volumetric Heat Capacity of water is approximately 3400 times that of air.<\/p>\n\n\n\n

            Step 8: <\/strong>Discuss the following questions to students:<\/p>\n\n\n\n