Whitney Kihlstrom
Q1: The parameters, a and b, in the linear regression can be estimated from the graph below. The y-intercept, a, is approximately zero while the slope, b, is roughly around 0.85. The linear regression equation fits the monthly stream flow data very well, as evidenced by the R-squared value of 0.9971.Overall, the linear regression line fits the observed data well, but the coefficients of the regression line are determined dominantly by high flow data. As a result, this graph shows a systematic bias for low flow less than 50 mm/month. The bias can introduce serious errors in calculating monthly stream discharge of undisturbed WS17 during dry season. 2.5/3
Q2: I think this assumption is acceptable as evidenced by how well the linear regression equation fits the monthly stream flow data. The undisturbed WS17 data is found from the linear regression line of WS17 versus WS18 before the disturbance. However, there are many potential sources of error when extrapolating the relationship derived in the short period to the entire data collection period. Firstly, the three years from which the relationship is extrapolated could be particularly dry, thus not applicable to average or wet years. This can be attributed to differing precipitation patterns that result in varying amounts of precipitation. Additionally, the three years may have experienced high flows, which would prove problematic in extrapolating average or low flows. Refer to a comment in Q1. Lastly, the linear regression is not perfect (i.e. R-squared does not equal 1), producing another source of error in the extrapolation.
If natural disturbance such as hurricanes or insect infection influences the hydrologic response of one of the two watersheds after the clear-cutting, the derived regression relationship is no longer valid. In fact, the hemlock trees of WS18 were seriously infected by the introduction of hemlock woolly adelgid (Adelges tsugae), an exotic invasive insect, in 2002 (
1.5/2
Q3:
Correct. 3/3
Q4: After the clear-cutting in 1940, the annual water yields of WS17 increased dramatically as evidenced by the large negative difference between undisturbed WS17 and disturbed WS17. After the pine plantation in 1955, it took approximately twenty years for the annual water yields to stabilize. During the transient period, the annual water yields of disturbed WS17 approached those of undisturbed WS17. This resulted in a less varying difference and thus a more stable trend in annual water yields. The decrease in annual water yields of disturbed WS17 can be attributed to the maturation of the pine plantation. Specifically, roots do not intercept but absorb and leaves of vegetation intercepted more water andAha! you want to say this with roots the macropores allowed more water to infiltrate rather than run off into the stream. Once the crown closed, the rate of decreased stream flow lessened and the graph leveled out. 4/4
Q5: The evaporation rate of mature pine forest ishigher than that of mature deciduousoak forest as evidenced by the graph above. In the winter, pine trees tend to intercept more water than deciduous oaks, which lose their leaves. Thisresults in a higher rate of evaporation and thus less water reaching streams. This trend also holds true for the growing season,becausethe leaf area of conifers is greater than that of deciduous trees, explaining why stream flow is less in disturbed WS17 than undisturbed WS17. And, More effective leaf structure of coniferous forest for interception than deciduous tree. 3/3
Q6: The seasonal change of monthly discharge difference is greatest in the winter, when coniferous trees have their leaves anddeciduous trees do not. The opposite holds true for the spring and summer, when both coniferous and deciduous trees have leaves. It is important to note that the leaf surface area of conifers is larger than that of deciduous trees, serving as one reason there is a positive difference in monthly stream discharge.
In April and May, the deciduous are just leafing out. At this time, stronger solar radiation increases ET, which makes the difference larger.
Correct.
Total: Excellent.
17.5/19= / 9.2