{"id":62,"date":"2022-12-25T19:16:06","date_gmt":"2022-12-25T19:16:06","guid":{"rendered":"https:\/\/books.gw-project.org\/distributed-fiber-optic-hydrogeophysics\/chapter\/solution-exercise-1\/"},"modified":"2022-12-27T04:48:22","modified_gmt":"2022-12-27T04:48:22","slug":"solution-exercise-1","status":"publish","type":"chapter","link":"https:\/\/books.gw-project.org\/distributed-fiber-optic-hydrogeophysics\/chapter\/solution-exercise-1\/","title":{"raw":"Solution Exercise 1","rendered":"Solution Exercise 1"},"content":{"raw":"
\r\n
    \r\n \t
  1. At ~230\u00a0m down the fiber, the stream temperature abruptly cools during the warmest part of the day (15:00\u00a0hours) and abruptly warms during the coolest part of the day (3:00\u00a0hours). This cooling and warming carries on downstream. To cool the stream temperature in the middle of the day, cooler groundwater must be entering at ~230\u00a0m. At ~350\u00a0m down the stream, the fiber records a significant rise in \u201cstream\u201d temperature during the day and a significant cooling at night, but these changes are not carried downstream, and the stream returns to the temperature upstream of the excursions. In this case, the fiber is probably either out of the stream and in the air or very, very close to the stream surface, where it can warm by solar radiation. This may be an intentional placement of the cable above the water or accidental after initial installation. In either case, the excursions in temperature match the timing of air temperature and solar radiation, in contrast to inflowing groundwater which is opposite in phase from the daily warming and cooling.<\/li>\r\n \t
  2. At 3:00 and 15:00 hours, the stream temperature changes abruptly at ~230 meters indicating a groundwater inflow. At 10:00 hours, there is no change in stream temperature at 230 meters, indicating that the groundwater and stream water are at the same temperature. In this case, from the figure, the stream temperature is ~13.5 \u00b0C, and so the groundwater temperature must also be ~13.5 \u00b0C (T<\/em>g<\/em><\/sub>\u00a0=\u00a013.5\u00a0\u00b0C).<\/li>\r\n \t
  3. For the 15:00-hour profile, the upstream temperature, T<\/em>i<\/em><\/sub>, is 16\u00a0\u00b0C while the downstream temperature T<\/em>o<\/em><\/sub> is 15\u00a0\u00b0C and the upstream inflow is given as Q<\/em>i<\/em><\/sub>\u00a0=\u00a00.4\u00a0m3<\/sup>\/s, Assuming the groundwater temperature is 13.5 \u00b0C, from Selker and others (2006), the energy balance reduces to the following.<\/li>\r\n<\/ol>\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
    [latex]\\displaystyle Q_{i}T_{i}+Q_{g}T_{g}=Q_{o}T_{o}=(Q_{i}+Q_{g})T_{o}[\/latex]<\/td>\r\n<\/tr>\r\n
    or<\/td>\r\n<\/tr>\r\n
    [latex]\\displaystyle \\frac{Q_{i}}{Q_{g}}=\\frac{(T_{o}-T_{g})}{(T_{i}-T_{o})}[\/latex]<\/td>\r\n<\/tr>\r\n
    or<\/td>\r\n<\/tr>\r\n
    [latex]\\displaystyle Q_{g}=Q_{i}\\left ( \\frac{T_{i}-T_{o}}{T_{o}-T_{g}}\\right )[\/latex]<\/td>\r\n<\/tr>\r\n
    Substituting<\/td>\r\n<\/tr>\r\n
    [latex]\\displaystyle Q_{g}=0.4\\frac{\\textrm{m}^{3}}{\\textrm{s}}\\frac{16\\ ^{\\textrm{o}}\\textrm{C}-15\\ ^{\\textrm{o}}\\textrm{C}}{15\\ ^{\\textrm{o}}\\textrm{C}-13.5\\ ^{\\textrm{o}}\\textrm{C}}[\/latex] = [latex]\\displaystyle 0.4\\frac{\\textrm{m}^{3}}{\\textrm{s}}0.67=0.267\\frac{\\textrm{m}^{3}}{\\textrm{s}}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n

    Note that this is quite a large inflow of groundwater and would likely be visible as an increase in streamflow.<\/p>\r\n\r\n

      \r\n \t
    1. If the DTS can resolve, at best, differences in temperature from one location to the next of 0.05\u00a0\u00b0C, then we can begin to estimate the smallest flux measurable under \u201ctypical\u201d groundwater temperatures. For example, if groundwater is 5\u00a0\u00b0C different than the stream temperatures, then what is the groundwater inflow needed to lower (or raise) the stream temperature by 0.05\u00a0\u00b0C. We can express the downstream temperature T<\/em>o<\/em><\/sub> as the upstream temperature T<\/em>i<\/em><\/sub> plus the measurable difference of \u22120.1 \u00b0C (we use a negative difference here assuming the groundwater is colder than the stream, which is common during summer daytime hours, and we use 0.1 instead of 0.05 because we can only measure the temperature +\/\u22120.05 so if we are 0.05 too low on the upstream and 0.05 too high on the downstream then there appears to be no inflow), then simplify the heat balance equation presented at the end of part (c) as follows.<\/li>\r\n<\/ol>\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
      [latex]\\displaystyle Q_{g}=Q_{i}\\left ( \\frac{T_{i}-T_{o}}{T_{o}-T_{g}} \\right )[\/latex]<\/td>\r\n<\/tr>\r\n
      [latex]\\displaystyle Q_{g}=Q_{i}\\left ( \\frac{T_{i}-(T_{i}+\\;-0.10\\;^{\\textrm{o}}\\textrm{C})}{T_{o}-T_{g}} \\right )[\/latex]<\/td>\r\n<\/tr>\r\n
      [latex]\\displaystyle Q_{g}=Q_{i}\\left ( \\frac{0.10\\;^{\\textrm{o}}\\textrm{C}}{T_{o}-T_{g}} \\right )[\/latex]<\/td>\r\n<\/tr>\r\n
      or<\/td>\r\n<\/tr>\r\n
      [latex]\\displaystyle \\frac{Q_{g}}{Q_{i}}=\\frac{0.1\\;^{\\textrm{o}}\\textrm{C}}{15\\;^{\\textrm{o}}\\textrm{C}-5\\;^{\\textrm{o}}\\textrm{C}}=0.01\\;\\textrm{or}\\;1.0%[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n

      Thus, for streamflow Q<\/em>i<\/em><\/sub>\u00a0of 1\u00a0m3<\/sup>\/s, we can resolve 0.01 m3<\/sup>\/s of groundwater inflow Q<\/em>g<\/sub> or ~10 liters per second. For a larger river of 10 m3<\/sup>\/s, the best we can resolve is 0.1 m3<\/sup>\/s, which is a large groundwater inflow. As stream volume grows, the ability of DTS to detect small groundwater inflow decreases due to the resolution of the DTS, unlike chemical tracers that can be detected in the parts per million. DTS typically detects changes in the \u201cparts per 100\u201d.<\/p>\r\n

      Return to Exercise 1<\/a><\/p>\r\n

      <\/p>\r\n\r\n<\/div>","rendered":"

      \n
        \n
      1. At ~230\u00a0m down the fiber, the stream temperature abruptly cools during the warmest part of the day (15:00\u00a0hours) and abruptly warms during the coolest part of the day (3:00\u00a0hours). This cooling and warming carries on downstream. To cool the stream temperature in the middle of the day, cooler groundwater must be entering at ~230\u00a0m. At ~350\u00a0m down the stream, the fiber records a significant rise in \u201cstream\u201d temperature during the day and a significant cooling at night, but these changes are not carried downstream, and the stream returns to the temperature upstream of the excursions. In this case, the fiber is probably either out of the stream and in the air or very, very close to the stream surface, where it can warm by solar radiation. This may be an intentional placement of the cable above the water or accidental after initial installation. In either case, the excursions in temperature match the timing of air temperature and solar radiation, in contrast to inflowing groundwater which is opposite in phase from the daily warming and cooling.<\/li>\n
      2. At 3:00 and 15:00 hours, the stream temperature changes abruptly at ~230 meters indicating a groundwater inflow. At 10:00 hours, there is no change in stream temperature at 230 meters, indicating that the groundwater and stream water are at the same temperature. In this case, from the figure, the stream temperature is ~13.5 \u00b0C, and so the groundwater temperature must also be ~13.5 \u00b0C (T<\/em>g<\/em><\/sub>\u00a0=\u00a013.5\u00a0\u00b0C).<\/li>\n
      3. For the 15:00-hour profile, the upstream temperature, T<\/em>i<\/em><\/sub>, is 16\u00a0\u00b0C while the downstream temperature T<\/em>o<\/em><\/sub> is 15\u00a0\u00b0C and the upstream inflow is given as Q<\/em>i<\/em><\/sub>\u00a0=\u00a00.4\u00a0m3<\/sup>\/s, Assuming the groundwater temperature is 13.5 \u00b0C, from Selker and others (2006), the energy balance reduces to the following.<\/li>\n<\/ol>\n\n\n\n\n\n\n\n\n\n
        \"\displaystyle Q_{i}T_{i}+Q_{g}T_{g}=Q_{o}T_{o}=(Q_{i}+Q_{g})T_{o}\"<\/td>\n<\/tr>\n
        or<\/td>\n<\/tr>\n
        \"\displaystyle \frac{Q_{i}}{Q_{g}}=\frac{(T_{o}-T_{g})}{(T_{i}-T_{o})}\"<\/td>\n<\/tr>\n
        or<\/td>\n<\/tr>\n
        \"\displaystyle Q_{g}=Q_{i}\left ( \frac{T_{i}-T_{o}}{T_{o}-T_{g}}\right )\"<\/td>\n<\/tr>\n
        Substituting<\/td>\n<\/tr>\n
        \"\displaystyle Q_{g}=0.4\frac{\textrm{m}^{3}}{\textrm{s}}\frac{16\ ^{\textrm{o}}\textrm{C}-15\ ^{\textrm{o}}\textrm{C}}{15\ ^{\textrm{o}}\textrm{C}-13.5\ ^{\textrm{o}}\textrm{C}}\" = \"\displaystyle 0.4\frac{\textrm{m}^{3}}{\textrm{s}}0.67=0.267\frac{\textrm{m}^{3}}{\textrm{s}}\"<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

        Note that this is quite a large inflow of groundwater and would likely be visible as an increase in streamflow.<\/p>\n

          \n
        1. If the DTS can resolve, at best, differences in temperature from one location to the next of 0.05\u00a0\u00b0C, then we can begin to estimate the smallest flux measurable under \u201ctypical\u201d groundwater temperatures. For example, if groundwater is 5\u00a0\u00b0C different than the stream temperatures, then what is the groundwater inflow needed to lower (or raise) the stream temperature by 0.05\u00a0\u00b0C. We can express the downstream temperature T<\/em>o<\/em><\/sub> as the upstream temperature T<\/em>i<\/em><\/sub> plus the measurable difference of \u22120.1 \u00b0C (we use a negative difference here assuming the groundwater is colder than the stream, which is common during summer daytime hours, and we use 0.1 instead of 0.05 because we can only measure the temperature +\/\u22120.05 so if we are 0.05 too low on the upstream and 0.05 too high on the downstream then there appears to be no inflow), then simplify the heat balance equation presented at the end of part (c) as follows.<\/li>\n<\/ol>\n\n\n\n\n\n\n\n
          \"\displaystyle Q_{g}=Q_{i}\left ( \frac{T_{i}-T_{o}}{T_{o}-T_{g}} \right )\"<\/td>\n<\/tr>\n
          \"\displaystyle Q_{g}=Q_{i}\left ( \frac{T_{i}-(T_{i}+\;-0.10\;^{\textrm{o}}\textrm{C})}{T_{o}-T_{g}} \right )\"<\/td>\n<\/tr>\n
          \"\displaystyle Q_{g}=Q_{i}\left ( \frac{0.10\;^{\textrm{o}}\textrm{C}}{T_{o}-T_{g}} \right )\"<\/td>\n<\/tr>\n
          or<\/td>\n<\/tr>\n
          \"\displaystyle \frac{Q_{g}}{Q_{i}}=\frac{0.1\;^{\textrm{o}}\textrm{C}}{15\;^{\textrm{o}}\textrm{C}-5\;^{\textrm{o}}\textrm{C}}=0.01\;\textrm{or}\;1.0%\"<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

          Thus, for streamflow Q<\/em>i<\/em><\/sub>\u00a0of 1\u00a0m3<\/sup>\/s, we can resolve 0.01 m3<\/sup>\/s of groundwater inflow Q<\/em>g<\/sub> or ~10 liters per second. For a larger river of 10 m3<\/sup>\/s, the best we can resolve is 0.1 m3<\/sup>\/s, which is a large groundwater inflow. As stream volume grows, the ability of DTS to detect small groundwater inflow decreases due to the resolution of the DTS, unlike chemical tracers that can be detected in the parts per million. DTS typically detects changes in the \u201cparts per 100\u201d.<\/p>\n

          Return to Exercise 1<\/a><\/p>\n

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