Data Analysis for Hydrology

Sophisticated data analysis will help you spot patterns, trends and relationships in your results. Data analysis can be qualitative and/or quantitative, and may include statistical tests. An example of a statistical test is outlined below.

Water balance

The water balance of a drainage basin is calculated as

\(P–Q–G–ΔS–E = 0\)

• \(P\) is precipitation
• \(Q\) is stream discharge
• \(G\) is groundwater discharge
• \(ΔS\) is change in storage
• \(E\) is evapotranspiration

Use secondary data and extrapolated primary data to calculate the annual water balance.

Analysing drainage basins: circularity and elongation

Both the size and the shape of the drainage basin are important in determining the runoff response to rainfall. The circularity and elongation of the drainage basin can be assesed, using secondary data collected from maps. Useful maps of small drainage basins are available at the Environment Agency’s Catchment Data Explorer.

(a) Circularity ratio

The circularity ratio the ratio of the area of a circle of the same area as the basin to the basin perimeter.

\(\mathsf{Circularity\;ratio = \frac{Basin\;area}{Area \;of\;circle\;with\;same\;basin\;perimeter}}\)

Using the maths of a circle (radius, area and circumference), this can be expressed as

\(\mathsf{Circularity\;ratio = \frac{(4\times\pi\times A)}{(P)^2}}\)

• \(A\) = basin area
• \(P\) = basin perimeter

To calculate the circularity ratio, use map data to find the

• area of the drainage basin
• length of the drainage basin perimeter

make sure that the same units (metres and square metres or kilometres and square kilometres) are used for length and area

Worked example

The River Redlake drainage basin in Shropshire is shown below.

The Redlake drainage basin has an area of \(27.215\; km^2\) and a perimeter length of \(42.55\; km\). What is its circularity ratio?

\(\mathsf{Circularity\;ratio = \frac{(4\times\pi\times A)}{(P)^2}}\) \(\mathsf{Circularity\;ratio = \frac{(4\times\pi\times 27.215)}{(42.55)^2}}\) \(\mathsf{Circularity\;ratio} = 0.19\)

Therefore the drainage basin is not at all circular. Of course this is fairly obvious from the map, but it would be a useful calculation if you were comparing two or more basins.

(b) Elongation ratio

The elongation ratio is the ratio of diameter of a circle of the same area as the basin to the maximum basin length. The maximum basin length is a straight line from the mouth of a stream to the furthest point on the watershed of its basin.

The ratio runs from 1.0 (for a perfect circle) towards zero (the more elongated the basin is).

\(\mathsf{Elongation\;ratio = \frac{1}{L} \times \sqrt{\frac{4}{π} \times A}}\)

• \(A\) = basin area
• \(L\) = maximum basin length

Worked example

The River Redlake drainage basin in Shropshire is shown below.
The Redlake drainage basin has an area of \(27.215 \;km^2\) and a maximum basin length of \(15.95 \;km\)(shown by the red line). What is its elongation ratio?

\(\mathsf{Elongation\;ratio = \frac{1}{L} \times \sqrt{\frac{4}{π} \times A}}\) \(\mathsf{Elongation\;ratio} = \frac{1}{15.95} \times \sqrt{\frac{4}{π} \times 27.215}\) \(\mathsf{Elongation\;ratio} = 0.37\)

Mann Whitney U test

Mann Whitney U is a statistical test that is used either to test whether there is a significant difference between the medians of two sets of data.

The Mann Whitney U test can only be used if there are at least 6 pairs of data. It does not require a normal distribution.

There are 3 steps to take when using the Mann Whitney U test

Step 1. State the null hypothesis

There is no significant difference between _______ and _______

Step 2. Calculate the Mann Whitney U statistic

\(U_1= n_1 \times n_2 + 0.5 n_2 (n_2 + 1)\;- ∑ R_2\) \(U_2 = n_1 \times n_2 + 0.5 n_1 (n_1 + 1)\;- ∑ R_1\)

• \(n_1\) is the number of values of \(x_1\)
• \(n_2\) is the number of values of \(x_2\)
• \(R_1\) is the ranks given to \(x_1\)
• \(R_2\) is the ranks given to \(x_2\)

Step 3. Test the significance of the result

Compare the value of U against the critical value for U at a confidence level of 95% / significance value of P = 0.05.

If U is equal to or smaller than the critical value (p=0.05) the REJECT the null hypothesis. There is a SIGNIFICANT difference between the 2 data sets.

If U is greater than the critical value, then ACCEPT the null hypothesis. There is NOT a significant difference between the 2 data sets.

Worked example

A geographer was interested in whether there was a difference in soil infiltration rates for two land-uses within a small drainage basin: an area of deciduous woodland and an area of evergreen woodland.

Eight readings of infiltration rate were taken at random locations in each site. Here are the results.

Infiltration rate (cm / minute) in deciduous woodland	Infiltration rate (cm / minute) in evergreen woodland
0.5	0.2
0.7	0.1
0.4	0.4
0.4	0.4
0.6	0.3
0.5	0.5
0.6	0.2
0.3	0.2

Step 1. State the null hypothesis

There is no significant difference in infiltration rate between deciduous woodland and evergreen woodland within the drainage basin

Step 2. Calculate Mann Whitney U statistic.

(a) Give each result a rank. Calculate the sum of the ranks for the two columns.

Deciduous woodland	Evergreen woodland
Infiltration rate (cm / minute)	rank	Infiltration rate (cm / minute)	rank
0.5	12	0.2	3
0.7	16	0.1	1
0.4	8.5	0.4	8.5
0.4	8.5	0.4	8.5
0.6	14.5	0.3	5.5
0.5	12	0.5	12
0.6	14.5	0.2	3
0.3	5.5	0.2	3
SUM	91.5	SUM	44.5

(b) Calculate \(∑R_1\)and \(∑R_2\)

\(∑R_1\) is the sum of the ranks in the first column (deciduous woodland) = \(91.5\)

\(∑R_2\) is the sum of the ranks in the first column (evergreen woodland) = \(44.5\)

\(n_1 = 8\) and \(n_2 = 8\)

(c) Calculate \(U_1\) and \(U_2\)

\(U_1 = 8 \times 8 + 0.5 \times 44.5 \times (8 + 1) \;– 91.5 = 55.5\) \(U_2 = 8 \times 8 + 0.5 \times 91.5 \times (8 + 1)\;– 44.5 = 8.5\)

Step 3. Test the significance of the result

In this example, \(U_1 = 55.5\) and \(U_2 = 8.5\)

Select the smaller of the values. \(U_2\) is smaller, so \(U=8.5\)

Use a set of Mann-Whitney U critical value tables (example here) to find the critical value at the 95% confidence level.

The critical value at \(p=0.05\) significance level for \(n_1=8\) and \(n_2=8\) is \(13\). Since our calculated value of \(8.5 < 13.5\), the null hypothesis can be rejected.

In conclusion, there is a significant difference in infiltration rate between deciduous and evergreen woodland within the drainage basin.

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