Page 94 - Low Temperature Energy Systems with Applications of Renewable Energy
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Characteristics of low-temperature energy sources for heat pumps   83

           Example: sizing a European residential ground-source
           heat pump


           Case A. Determine the length of the ground collector of a heat pump for heating a new
                                                                   2
           residential building in Europe with dimensions: 20 m   20 m ¼ 400 m . You may as-
           sume that a radiant floor heating system is employed. The soil at a depth of 5 m is wet
                           3
           clay (r ¼ 600 kg/m , moisture content is 30%); coefficient of performance, 4 ¼ 3.5;

           W c /W w ¼ 0.4; W ec ¼ 4.9 kW; soil temperature is 8 S; and brine temperature is

           0 S, soil coefficient of thermal conductivity l ¼ 3.56 W/mK (Table 2.3 or standards
           VDI 4640).
              Solution. Let us determine the required heat output of the heat pump. According
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           to EU-Norm EN 12831, the heat consumption for new buildings is 30e50 W/m .
           So the required capacity of the heat pump is 400   50 ¼ 20,000 W ¼ 20 kW.
              Horizontal collector
              The specific heat flux from the ground to the horizontal pipe, from Eq. (2.4),is

                q 0 ¼½ð1:4   3:56Þ  0:5ފð0:4 þ 1Þð8   0Þ¼ 50:2W=m

              The thermal capacity of the reservoir is Q 0 ¼ Q hp   W ec , where Q hp is heat pump
           power, and W ec is electric power of the compressor.
              The required collector power is Q 0 ¼ 20e4.9 ¼ 15.1 kW.
              The length of the collector pipes, from Eq. (2.3),is


                  15100 3:5   1
               L ¼               ¼ 214:9m.
                   50:2    3:5
              Thus, for the construction of a ground collector, two loops of 108 m length are
           required.
              The necessary area of the site to accommodate the collector is

                       n   1
               A ¼ L   a
                         n
           where a is the distance between the pipes (assuming a ¼ 0.75e1 m), and n is the
           number of loops.

                            2   1      2
               A ¼ 215   0:75    ¼ 81 m
                              2
              If three contours of length of 40 m are used, the area of the plot will be

                            3   1       2
               A ¼ 215   0:75    ¼ 108 m
                              3
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