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Thus given an example from a well in Texas, with an input temperature of 121°C (250°F) and an outlet temperature of 107°C (225°F) we can use the equations to determine the new kW value. Using Figure 7.3, T1 = 121°C and the correlated kW per kg/s for 40°C outlet temperature (halfway between the 35 and 45°C curves) is 31 kW. Now using the equations the new value is: kW = 31 + (0.098701 - 0.0039645*121°C) * (107°C-40°C) kW = 5.5 per kg/s flow rate In a second related example, if the outlet temperature (T2) is decreased to 82°C (180°F) then the kW value increases to 14.9 kW per kg/s flow rate. More energy is being extracted for the same amount of flow. From this equation it shows that for the same drop in temperature, the higher inlet temperature (T1) there is an increased amount of power (W) which can be produced. Since the well temperature of 1000s of wells can not be easily used in determining the potential power, the tables give multiple temperature values to use as a range. Depending on if the average temperature of a group of wells is known, or estimated from the geology and heat flow, it is helpful to have more than one temperature estimate to give the possible range for a project. Table 7.3 gives initial input temperatures (T1) to start from in estimating the total flow. From there the total flow rate can be manipulated, i.e., if you want to consider 50% of the wells at 100°C and 50% of the wells at 180°C. One way to convert from bbl to kg/s, divide the bbl by 365.24 days/year to get bbls/day. Next bbls/day *.0004861 = gal/sec. From gal/sec * 3785.411784 to get cm3/s. 1 gram of water is equal to 1 cm3. Lastly multiply by .001 to get kg/s. Example conversion: Using Texas 12,097,990,120 bbl/year = 3312339.64 bbls/day 3312339.64 bbls/day *0.0004861 = 16,101.28 gal/sec 16,101.28 gal/sec * 3785.411784 = 60,949,990 cm3/sec 60,949,990.49 cm3/sec = 60,949,990.49 g/sec 60,949,990.49 g/sec *.001 = 60,949.99 kg/sec 77PDF Image | Texas Geothermal Assessment for the I35 Corridor East
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