Predicting Stream and River Flooding This exercise introduces techniques that ge
ID: 107866 • Letter: P
Question
Predicting Stream and River Flooding
This exercise introduces techniques that geologists use to predict stream and river floods. You will construct a flood-frequency curve from stream discharge data in the table on the following page. You will rank the maximum stream discharges per year, calculate the recurrence interval (the average interval of time (in years) within which a flood of a given magnitude will occur) for each maximum stream discharge (the amount of water, measured in cubic feet per second (cfs) , that passes a point on the stream during a flood), and then plot the recurrence intervals vs. the discharges. You will then be able to predict discharges for large floods and their expected frequency of occurrence.
It is normal for streams to flood. Damage caused by stream flooding is the result of constructing buildings and roads on natural flood plains. Damage and the resulting monetary loss can be prevented if structures are built outside the flood plain, or of dikes, artificial levees, retention ponds, and other drainage-control modifications are built to contrail flooding. Either solution requires that planners the expected frequency and amount of stream flooding. If this information is known, then restrictions can be placed on construction within flood plains.
A flood having a recurrence interval of 10 years, called a 10-year flood, is one that has a 10% chance of occurring in any year. In other words, it is expected to occur once every ten years. A flood having a recurrence interval of 100 years, called a 100-year flood, has a 1% chance of occurring in any year and is expected to occur once every 100 years. It is important to calculate the predicted discharge of a 100-year flood for any stream on whose floodplain structures will be built, since this number is usually used by planners as a reasonable limit for flood plain management. The reasoning for this is that it is prudent to allow construction in areas that will only flood once every 100 years. Another way of stating this is that there is only a 1% chance that structures will be damaged or destroyed by flooding, a risk level acceptable to governments and insurance companies.
Calculations of recurrence interval for any stream can be made from the following equation:
R = (N+1)/m
R = Recurrence Interval N = number of years of record m = peak discharge rank
Instructions:
On the following page, you will find a table of annual stream peak discharges from 2000 to 2010.
1. Determine the rank of each (1 is greatest discharge, 11 is smallest discharge)
2. Using the above equation, calculate the recurrence interval and enter that data into the table. (Note: For recurrence interval calculation, N = 11.)
3. Manually plot each year as a point on the graph of Recurrence Interval vs. Discharge. You will need to print the graph in order to do this. (Note that recurrence interval is a logarithmic scale. This helps compress the data into more of a straight line.)
4. Draw a best-fit straight line through your data points which best estimates the trend of the data. (Do not ‘connect the dots’.) Extend this straight line to 100 years. This is the flood-frequency curve.
5. Use the flood-frequency curve to answer the questions which follow the table.
6. Type your answers in the Table on the following page and after manually plotting your data on the graph (you will need to print the graph in order to plot the data), save this document as a Word file and upload it to the designated D2L dropbox.
Year
Discharge (cfs)
Rank
Recurrence Interval (years)
2000
800
2001
1080
2002 (example)
692
11
N = (11 + 1)/11 = 1.1
2003
1380
2004
1280
2005
889
2006 (example)
2030
1
N = (11 + 1)/1 = 12
2007
890
2008
1480
2009
1600
2010
1330
What would be the discharge during a 20-year flood?
What would be the discharge during a 100-year flood?
What is the recurrence interval for a flood with a discharge of 1800 cfs?
What is the recurrence interval of a flood with discharge of 2500 cfs?
Year
Discharge (cfs)
Rank
Recurrence Interval (years)
2000
800
2001
1080
2002 (example)
692
11
N = (11 + 1)/11 = 1.1
2003
1380
2004
1280
2005
889
2006 (example)
2030
1
N = (11 + 1)/1 = 12
2007
890
2008
1480
2009
1600
2010
1330
Explanation / Answer
Flooding occurs in known floodplains when prolonged rainfall over several days, intense rainfall over a short period of time, or a debris jam causes a river or stream to overflow and flood the surrounding area. Severe thunderstorms can bring heavy rain in the spring and summer; or tropical cyclones can bring intense rainfall to the coastal and inland states in the summer and fall. Even though you may never have heard of "recurrence interval", it may be familiar to you. When a major flood occurs, you might have heard that the stream stage reached the "100-year flood level". This means that a flood of that magnitude has a 1 in 100 chance of occurring in any year. We would discuss the following steps to calculate various descriptions asked in the questions.
-Obtain streamflow data to get a rank
-Organize the information in a table.
-Rank the data from largest discharge to smallest discharge. Add a column for Rank and number each streamflow value from 1 to n (the total number of values in your dataset).
-Create a column with the log of each max or peak streamflow using the Excel formula {log (Q)} and copy command.
-Calculate the Average Max Q or Peak Q and the Average of the log (Q)
-Create a column with the excel formula {(log Q – avg(logQ))^2}
-Create a column with the excel formula {(log Q – avg(logQ))^3
-Create a column with the return period (Tr) for each discharge using Excel formula {(n+1)/m}. Where n = the number of values in the dataset and m = the rank.
-Complete the table with a final column showing the exceedence probability of each discharge using the excel formula {=1/Return Period or 1/Tr} and the copy command.
-Calculate the Sum for the {(logQ – avg(logQ))^2} and the {(logQ – avg(logQ))^3} columns.
-Calculate the variance, standard deviation, and skew coefficient as follows:
Variance = (logQ – avg (Log Q) )^2 / n-1
Standard deviation = sigma log Q = square root (variance)
Skew coefficient = n * (logQ – avg (logQ) )^3 / (n-1)(n-2)(sigma log Q ^ 3
-To calculate k values follow the following steps:
(a)Use the frequency factor table and the skew coefficient to find the k values for the 2,5,10,25,50,100, and 200 recurrence intervals.
(b)If the skew coefficient is between two given skew coefficients in the table than you can linearly extrapolate between the two numbers to get the appropriate k value.
-Using the general equation, list the discharges associated with each recurrence interval the general equation would be = log QTr -= avg(log Q) + [K(Tr, Cs)] * sigma logQ
-Create table of Discharge values found using the log – Pearson analysis
-Next step is to create a plot. A comparison of flood frequency analysis completed using mean daily data versus instantaneous discharge data. As per the values found we see the result is more conservative estimation of the discharges associated with each return period.
Year
Discharge (cfs)
Rank
Recurrence Interval (years)
2000
800
10
N = (11+1)/10 =1.2
2001
1080
7
N = (11+1)/7 = 1.71
2002 (example)
692
11
N = (11 + 1)/11 = 1.1
2003
1380
4
N = (11+1)/4 = 3
2004
1280
6
N = (11+1)/6 = 2
2005
889
9
N = (11+1)/9 = 1.33
2006 (example)
2030
1
N = (11 + 1)/1 = 12
2007
890
8
N = (11+1)/8 = 1.5
2008
1480
3
N = (11+1) / 3 = 4
2009
1600
2
N = (11+1) / 2 = 6
2010
1330
5
N = (11+1) / 5 = 2.4
Q-What would be the discharge during a 20-year flood?
ANS – 2660
Q-What would be the discharge during a 100-year flood?
ANS – 13300
Q-What is the recurrence interval for a flood with a discharge of 1800 cfs?
ANS – For 1800cfs discharge, the rank would be 2 leading to recurrence interval of N = (11+1) / 2 = 6.
Q- What is the recurrence interval of a flood with discharge of 2500 cfs?
ANS - For 2500cfs discharge, the rank would be 1 leading to recurrence interval of N = (11 + 1)/1 = 12.
Year
Discharge (cfs)
Rank
Recurrence Interval (years)
2000
800
10
N = (11+1)/10 =1.2
2001
1080
7
N = (11+1)/7 = 1.71
2002 (example)
692
11
N = (11 + 1)/11 = 1.1
2003
1380
4
N = (11+1)/4 = 3
2004
1280
6
N = (11+1)/6 = 2
2005
889
9
N = (11+1)/9 = 1.33
2006 (example)
2030
1
N = (11 + 1)/1 = 12
2007
890
8
N = (11+1)/8 = 1.5
2008
1480
3
N = (11+1) / 3 = 4
2009
1600
2
N = (11+1) / 2 = 6
2010
1330
5
N = (11+1) / 5 = 2.4