John Walter/Homework Set #1 FABE 2720/January 24, 2014

Problem Statement 1: Compare the annual precipitation and temperature of four cities within different countries that share a close latitudinal plane. I chose to analyze the following four cities: Akita, Japan; Hungnam, North Korea; Columbus, Ohio (USA) and Rome, Italy. After several minutes of searching the web I found a site that provides world weather information: http://www.worldweatheronline.com. I compared the data for Columbus, Ohio provided by this site with data available on: www.weather.com. Both were comparable to each other, so I concluded that the site would provide data I could use to answer this statement. The Weather Channel site was not an option to use because it did not have average temperature and precipitation data for regions outside the USA. In addition, World Weather Online allowed the user to switch between both metric and IPS, which was definitely helpful in generating the data. The following graphs were obtained from http://www.worldweatheronline.com for each region:

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John Walter/Homework Set #1 FABE 2720/January 24, 2014

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John Walter/Homework Set #1 FABE 2720/January 24, 2014

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John Walter/Homework Set #1 FABE 2720/January 24, 2014

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John Walter/Homework Set #1 FABE 2720/January 24, 2014

Data from these graphs was used to generate the following tables in excel: CITY AVERAGE TEMPERATURE Akita, Hungnam, Columbus, MONTH Rome, Italy Japan N. Korea Ohio 0 -2 0 8 January 0.5 0.5 0.5 8.5 February 4 5 5.5 11.5 March 9.5 11.5 12 13.5 April 14.5 16 17 18.5 May 19 20 22.5 23 June 23.5 23 24 25.5 July 25 23.5 23.5 25.5 August 20.5 19 19.5 21.5 September 14.5 13.5 13.5 17.5 October 8 7 7 12.5 November 3 1 0.5 8.5 December 11.83333333 11.5 12.125 16.16666667 MAT (C) *MAT = Mean Annual Temperature Table 1.1: Showing average temperature data: obtained from www.worldweatheronline.com 5

John Walter/Homework Set #1 FABE 2720/January 24, 2014

The charts obtained from World Weather Online did not provide the mean temperature for each month, so it was necessary to use both high and low averages. Therefore, I used the following equation to calculate the Mean Annual Temperature: AHT = Average high temperature per month ALT = Average low temperature per month n = number of months (𝐴𝐴𝐴𝐴𝐴𝐴 + 𝐴𝐴𝐴𝐴𝐴𝐴) (𝐴𝐴𝐴𝐴𝐴𝐴 + 𝐴𝐴𝐴𝐴𝐴𝐴) 𝑀𝑀𝑀𝑀𝑀𝑀 = � + ⋯+ � /𝑛𝑛 2 2

Equation: 1.1 (from Environmental Hydrology 2nd, Edition)

CITY AVERAGE PRECIPITATION Akita, Hungnam, Columbus, MONTH Rome, Italy Japan N. Korea Ohio 126 30.5 83.7 24.8 January 93.9 27.1 58.5 20.5 February 104.6 31.5 85.5 19.4 March 120.9 45 98.7 27.5 April 130.2 109.1 110 18.2 May 134.3 144.4 111.2 5.5 June 202 255.5 125.5 6.6 July 181.3 265.7 103.1 16.1 August 188.8 167.9 82.3 25.8 September 173.2 76 72.5 38.5 October 197.5 62.4 77.2 49.3 November 165.7 30.9 80.7 37.7 December 1818.4 1246 1088.9 289.9 MAP (mm) *MAP = Mean Annual Precipitation Table 1.2: Showing average precipitation data: obtained from www.worldweatheronline.com Calculating the Mean Annual Precipitation was slightly easier, because World Weather Online provided exact values for each bar on their precipitation graph. All I needed was to rest the pointer over each bar and wait for the information to show up. I used the following equation to calculate the MAP: PPM = Precipitation per month n = number of months

𝑀𝑀𝑀𝑀𝑀𝑀 = (𝑃𝑃𝑃𝑃𝑃𝑃 + ⋯ + 𝑃𝑃𝑃𝑃𝑃𝑃) ∗ 𝑛𝑛

Equation: 1.1 (from Environmental Hydrology 2nd, Edition)

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John Walter/Homework Set #1 FABE 2720/January 24, 2014

Answer: Below are the results obtained from both research and calculations performed. I obtained the climate zone and biome information from Environmental Hydrology Table 2.2 on Page 31. I placed them into a new table to summarize and for clarification.

LOCATION Akita, Japan Hungnam, N. Korea Columbus, Ohio Rome, Italy

CITY PRECIPITATION COMPARISON MAP CLIMATE LATITUDE MAT (C) (mm) ZONE Warm 11.83333333 1818.4 Temperate 39° 43' 6.96" Warm 11.5 1246 Temperate 39°49′54″ N Warm 39° 59' 0" 12.125 1088.9 Temperate Warm 41° 54' 36" 16.16666667 289.9 Temperate

BIOME TYPE Temp. Deciduous and Evergreen Temp. Deciduous and Evergreen Temp. Woodlands and Grasslands Cool semi-deserts

Sources: Latitude: www.google.com Mean Annual Temperature & Precipitation: www.worldweatheronline.com Table 1.3: Summarizing all data requested per the problem statement Problem Statement 2: What agricultural crops are grown in rural areas near the four cities you considered in Problem 1? Are the crops irrigated, and if so, why?

CROPS

AGRICULTURAL CROPS FOUND Hungnam, N. Columbus, Akita, Japan Korea Ohio asparagus rice rice *gakko corn sugar beets *sake millet cabbages broccoli potatoes wheat brussel sweet potatoes sorghum sprouts cauliflower cabbage potatoes chinese taro carrots cabbage carrots turnips sweet corn cucumbers turnips onions radishes eggplant lecttuce sugar beets melons fruit 7

Rome, Italy wheat maize rice olives grapes peaches peaches citrus sugar beets tomatoes mushrooms truffles

John Walter/Homework Set #1 FABE 2720/January 24, 2014

onions strawberries garlic raspberries peas chestnuts beans hazelnuts peppers potatoes pumpkins squash tomatoes wheat leafy greens Information for Japan obtained from: http://wiki.answers.com; bold=comfirmed for specific region: http://www.mesay.biz/info_akita.htm Information for N. Korea obtained from: http://wiki.answers.com/Q/Crops_in_North_Korea Information for Columbus obtained from:http://ohioline.osu.edu/lines/vcrop.html Information for Rome obtained from: ask.com; http://www.nationsencyclopedia.com/Europe/Italy-AGRICULTURE.html Table 2.1: List of various crops grown in the rural areas Answer: I didn’t have trouble finding crops grown in Columbus, but for the other three locations I could not find specific crop listings, except for Akita, Japan. I listed those crops in italics and with an asterisk. I decided to search for crops grown by the respective country and found that a lot of the same crops showed up, so I placed these crops in bold to highlight these similarities. Regarding the use of irrigation, the answer is yes for all regions. The data from problem statement 1 indicates all locations fall within warm temperate zones. Therefore, each area would need some source of irrigation used to maximize healthy plant growth. The area that would need more irrigation compared to the other three others is clear when looking at Table 1.2 with Rome requiring the most, which is due to its biomes being a cool semi-desert.

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John Walter/Homework Set #1 FABE 2720/January 24, 2014

Problem Statement 3: The average rainfall for a 3-h storm was estimated as 1.0 in. What would be the average rainfall amount and average intensity if the watershed area is 200 mi2? ∑𝑛𝑛𝑖𝑖=1 𝐴𝐴𝑖𝑖 𝑃𝑃𝑖𝑖 200𝑚𝑚𝑚𝑚 2 ∗ 1.0𝑖𝑖𝑖𝑖 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 = 𝑃𝑃 = =� � = 1.0 𝑖𝑖𝑖𝑖 ∑𝑛𝑛𝑖𝑖=1 𝐴𝐴𝑖𝑖 200𝑚𝑚𝑚𝑚 2

Equation: 2.1 (from Environmental Hydrology 2nd, Edition) 𝐴𝐴𝑖𝑖 = area of the ith polygon 𝑃𝑃𝑖𝑖 = precipitation within that polygon P = average depth of rainfall

𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 =

𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 1.0𝑖𝑖𝑖𝑖 = = 0.33𝑖𝑖𝑖𝑖𝑖𝑖ℎ𝑒𝑒𝑒𝑒/ℎ𝑜𝑜𝑜𝑜𝑜𝑜 3ℎ𝑟𝑟 𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷

Equation obtained from Environmental Hydrology 2nd, Edition, Example 2.3 on page 42 Answer: The average rainfall amount within the 200 𝑚𝑚𝑚𝑚 2 watershed area is 1.0 in. The average intensity for a 3 hour storm with a maximum depth of 1.0 inch is 0.33in/hr.

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John Walter/Homework Set #1 FABE 2720/January 24, 2014

Problem Statement 4: Compute the average rainfall for a 1000-acre watershed by the arithmetic average method and by the Thiessen method with data from Table 1. Rain Gauge

Acres per Gauge

Rainfall (in)

A B C D

400 250 200 150

2.1 2.9 2.3 2.7

Table 4.1: Rainfall data over an area Thiessen Method: ∑𝑛𝑛𝑖𝑖=1 𝐴𝐴𝑖𝑖 𝑃𝑃𝑖𝑖 400 ∗ 2.1 + 250 ∗ 2.9 + 200 ∗ 2.3 + 150 ∗ 2.7 𝑃𝑃 = =� � = 2.43 𝑖𝑖𝑖𝑖𝑖𝑖ℎ𝑒𝑒𝑒𝑒 𝑛𝑛 ∑𝑖𝑖=1 𝐴𝐴𝑖𝑖 400 + 250 + 200 + 150

Equation: 2.1 (from Environmental Hydrology 2nd, Edition) 𝐴𝐴𝑖𝑖 = area of the ith polygon 𝑃𝑃𝑖𝑖 = precipitation within that polygon P = average depth of rainfall Arithmetic Method: 𝑃𝑃 =

∑𝑛𝑛𝑖𝑖=1 𝑃𝑃𝑖𝑖 2.1 + 2.9 + 2.3 + 2.7 =� � = 2.5 𝑖𝑖𝑖𝑖𝑖𝑖ℎ𝑒𝑒𝑒𝑒 𝑛𝑛 4

Equation obtained from Environmental Hydrology 2nd, Edition, Example 2.4 on page 44 𝑛𝑛 = number of observations 𝑃𝑃𝑖𝑖 = precipitation within that polygon P = average depth of rainfall Note: No diagram was included with this problem, because the acreage was already provided. Answer: The average rainfall for a 1000 acre watershed using the Thiessen Method is 2.43 inches. The average rainfall for a 1000 acre watershed using the Arithmetic Method is 2.5 inches.

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John Walter/Homework Set #1 FABE 2720/January 24, 2014

Problem Statement 5 & 6: Calculate the expected annual rainfall for return periods of 2, 20 and 100 years using the Hazen and Weibull Methods given yearly rainfall information. Below is a table containing the given year and annual rainfall information and also the calculated information to complete these tasks. Rank

Probability of Occurrence

Return Period Hazen

Return Period Weibull

42.5

1

2.5

40.00

21.00

2010

41.2

2

7.5

13.33

10.50

2006 1997 2003 1996 2013 2007 1998 2004 1995 1994 2012 2001 2011 2002 2009 2000 1999 2008

39.5 39 35.1 34.4 33.4 32.5 32.3 31.5 31.2 28.2 28.1 28 28 27.3 27.2 25.4 23.6 20.5

3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

12.5 17.5 22.5 27.5 32.5 37.5 42.5 47.5 52.5 57.5 62.5 67.5 72.5 77.5 82.5 87.5 92.5 97.5

8.00 5.71 4.44 3.64 3.08 2.67 2.35 2.11 1.90 1.74 1.60 1.48 1.38 1.29 1.21 1.14 1.08 1.03

7.00 5.25 4.20 3.50 3.00 2.63 2.33 2.10 1.91 1.75 1.62 1.50 1.40 1.31 1.24 1.17 1.11 1.05

Year

Rainfall (mm)

2005

Table 5.1: Given data and calculated data Year

Annual Hazen

Year

Annual Weibull

2 20 100

29.76 43.18 52.57

2 20 100

29.06 43.88 54.24

Table 5.2 & 5.3: Calculated annual rainfall predictions for 2, 20 and 100 year

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John Walter/Homework Set #1 FABE 2720/January 24, 2014

ANNUAL RAINFALL USING HAZEN METHOD 100

RAINFALL (mm)

y = 5.8319ln(x) + 25.714 R² = 0.8733

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Return Period Hazen Log. (Return Period Hazen)

1 1.00

10.00

100.00

YEAR

Graph 5.1: Represents the predicted level of rainfall using Hazen Method

ANNUAL RAINFALL USING WEIBULL METHOD 100

RAINFALL (mm)

y = 6.438ln(x) + 24.593 R² = 0.9058

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Return Period Weibull Log. (Return Period Weibull)

1 1.00

10.00

100.00

YEAR

Graph 5.2: Represents the predicted level of rainfall using Weibull Method 12

John Walter/Homework Set #1 FABE 2720/January 24, 2014

The following equations were used in calculating the data for comparison. Hazen Method: 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 𝑜𝑜𝑜𝑜 𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂: 𝐹𝐹𝑎𝑎 =

100(2𝑛𝑛 − 1) 2𝑦𝑦

Equation: 2.2 (from Environmental Hydrology 2nd, Edition) n = rank y = number of occurence 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃: 𝑇𝑇 =

100 𝐹𝐹𝑎𝑎

Equation: from Environmental Hydrology 2nd, Edition page 46 n = rank y = number of occurence Weibull Method: 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 𝑜𝑜𝑜𝑜 𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂: 𝐹𝐹𝑎𝑎 =

100(2𝑛𝑛 − 1) 2𝑦𝑦

Equation: 2.2 (from Environmental Hydrology 2nd, Edition) n = rank y = number of occurence 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃: 𝑇𝑇 =

𝑦𝑦 + 1 𝑛𝑛

Equation: 2.3 (from Environmental Hydrology 2nd, Edition) n = rank y = number of occurrence Answer: After entering in all information in excel and comparing the results shown in Table’s 5.2 and 5.3, its clear that both methods show close predictions for 2, 20 and 100 years into the future.

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John Walter/Homework Set #1 FABE 2720/January 24, 2014

Problem Statement 7: An irrigation reservoir was designed to withstand a 50-year storm event. During the first 5 years, the reservoir withstood the design storm runoff once. What was the probability that this reservoir would experience this 50-year storm at least once in those 5 years? 𝑃𝑃 (𝑇𝑇, 𝑛𝑛) = 1 − [1 −

1 𝑛𝑛 1 ] = 1 − [1 − ]5 = .09607 ∗ 100% = 9.6% 𝑇𝑇 50

Equation: 2.4 (from Environmental Hydrology 2nd, Edition) n = years that event will occur T = projected return period storm event Answer: There is a 9.6% probability that a 50 year storm event will occur in 5 years. Problem Statement 8:

Rainfall rate–duration frequency data for Columbus, OH, are presented in Table 2.12. How well does the 100-year, 6-h duration value compare to estimates obtained using Figure 2.17 and Figure 2.18? What percentage of the PMP for Columbus, from Figure 2.21, is your value? If you live below a dam designed based on 100-year values, should you be concerned? Table 2.12 does not list data for 6-h, but it does show 4 and 8 hours at 100 years. By taking the average 8h = 3.64 and 4h = 3.30: 6ℎ =

8ℎ + 4ℎ 3.64 + 3.30 = = 3.47𝑖𝑖𝑖𝑖𝑖𝑖ℎ𝑒𝑒𝑒𝑒 2 2

Furthermore, Table 2.12 does not list units. I assumed units are in inches because of data shown on Figure 2.21 for PMP at 6-h duration is in inches. Figure 2.17 shows an intensity value of approximately 0.87 inches/h at 6h duration for 100-year.

Using Figure 2.18:

6ℎ = 0.87

𝑖𝑖𝑖𝑖𝑖𝑖ℎ𝑒𝑒𝑒𝑒 ∗ 6ℎ = 5.22 𝑖𝑖𝑖𝑖𝑖𝑖ℎ𝑒𝑒𝑒𝑒 (𝑖𝑖𝑖𝑖 𝑆𝑆𝑆𝑆. 𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿) ℎ

5.22𝑖𝑖𝑖𝑖𝑖𝑖ℎ𝑒𝑒𝑒𝑒 ∗ .75 = 3.92 𝑖𝑖𝑖𝑖𝑖𝑖ℎ𝑒𝑒𝑒𝑒

Figure 2.21 shows Probable Maximum Precipitation for 6-h duration to be around 25 inches. 3.47𝑖𝑖𝑖𝑖𝑖𝑖ℎ𝑒𝑒𝑒𝑒 = .1388 ∗ 100% = 13.9% 25 𝑖𝑖𝑖𝑖𝑖𝑖ℎ𝑒𝑒𝑒𝑒

Answer: The 100-year, 6-h duration values are approximately off by .6 in, which results around an 11.5% discrepancy between the Figure 2.17 and Table 2.12. The percentage of the PMP for Columbus is 13.9%. You should be concerned because there is a very good chance of flooding. 14

John Walter/Homework Set #1 FABE 2720/January 24, 2014

Problem Statement 9: The combined weight of a soil sample and a tin is 50 g. The tin weighs 20 g. After oven-drying, the soil sample and the tin have a weight of 46 g. Determine the soil water content (by weight) of the soil. In order to use Equation 3.7 from Environmental Hydrology 2nd Edition, we must first find the mass of the sample with water by the following steps: 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝑜𝑜𝑜𝑜 𝑤𝑤𝑤𝑤𝑤𝑤 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝑜𝑜𝑜𝑜 𝑤𝑤𝑤𝑤𝑤𝑤 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑤𝑤𝑤𝑤𝑤𝑤ℎ 𝑡𝑡𝑡𝑡𝑡𝑡 − 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝑜𝑜𝑜𝑜 𝑡𝑡𝑡𝑡𝑡𝑡 = 50𝑔𝑔 − 20𝑔𝑔 = 30𝑔𝑔

𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝑜𝑜𝑜𝑜 𝑑𝑑𝑑𝑑𝑑𝑑 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝑜𝑜𝑜𝑜 𝑑𝑑𝑑𝑑𝑑𝑑 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑤𝑤𝑤𝑤𝑤𝑤ℎ 𝑡𝑡𝑡𝑡𝑡𝑡 − 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝑜𝑜𝑜𝑜 𝑡𝑡𝑡𝑡𝑡𝑡 = 46𝑔𝑔 − 20𝑔𝑔 = 26𝑔𝑔

𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝑜𝑜𝑜𝑜 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 = 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝑜𝑜𝑜𝑜 𝑤𝑤𝑤𝑤𝑤𝑤 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 − 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝑜𝑜𝑜𝑜 𝑑𝑑𝑑𝑑𝑑𝑑 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = 30𝑔𝑔 − 46𝑔𝑔 = 4𝑔𝑔

𝑚𝑚𝑊𝑊 4𝑔𝑔 ∗ 100% = 100% = 15% 𝑚𝑚𝑠𝑠 26𝑔𝑔 Equation: 3.7 (from Environmental Hydrology 2nd, Edition) 𝑚𝑚𝑊𝑊 = mass of the water 𝑚𝑚𝑠𝑠 = mass of the dry soil 𝜃𝜃𝑔𝑔 = gravimetric soil water content 𝜃𝜃𝑔𝑔 =

Answer: The water soil content by mass (or weight) is 15%.

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John Walter/Homework Set #1 FABE 2720/January 24, 2014

Problem Statement 10: Determine the porosity of a soil profile if it has a dry bulk density of 1.8 g/cc, and the density of the soil particles is 2.7 g/cc. What would the bulk density of the soil be if it was saturated? To solve for porosity need to rewrite equation 3.10

𝜌𝜌𝑑𝑑𝑑𝑑𝑑𝑑 = 𝜌𝜌𝑝𝑝 ∗ (1 − 𝑛𝑛)

2.7𝑔𝑔 1.8𝑔𝑔 = ∗ (1 − 𝑛𝑛) 𝑐𝑐𝑐𝑐 𝑐𝑐𝑐𝑐

𝑛𝑛 = 0.33 Equation: 3.10 (from Environmental Hydrology 2nd, Edition) 𝜌𝜌𝑑𝑑𝑑𝑑𝑑𝑑 = dry bulk density 𝜌𝜌𝑝𝑝 = density of the soil particles n = porosity Since the sample is considered saturated, 𝑉𝑉𝑊𝑊 = 𝑉𝑉𝑙𝑙 . Using equation 3.3 and rewriting: 𝑉𝑉𝑊𝑊 = 𝑛𝑛 ∗ 𝑉𝑉𝑙𝑙

The mass of the water is equal to the density of the water times the volume and the mass of the soil is density of the soil times the dry bulk density. Combining equations 3.6 and 3.8: 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 𝑜𝑜𝑜𝑜 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 =

1000𝑘𝑘𝑘𝑘 1000𝑔𝑔 1𝑔𝑔 ∗ = 3 3 𝑚𝑚 (100𝑐𝑐𝑐𝑐) 𝑐𝑐𝑐𝑐

1𝑔𝑔 1.8𝑔𝑔 0.33 ∗ 𝑐𝑐𝑐𝑐 ∗ 𝑉𝑉𝑡𝑡 + 𝑐𝑐𝑐𝑐 ∗ 𝑉𝑉𝑡𝑡 + 0 2.13𝑔𝑔 𝑚𝑚𝑡𝑡 𝑚𝑚𝑎𝑎 + 𝑚𝑚𝑊𝑊 + 𝑚𝑚𝑠𝑠 𝜌𝜌𝑏𝑏 = = = = 𝑉𝑉𝑡𝑡 𝑉𝑉𝑡𝑡 𝑉𝑉𝑡𝑡 𝑐𝑐𝑐𝑐 𝜌𝜌𝑏𝑏 = bulk density 𝑚𝑚𝑡𝑡 = total mass 𝑉𝑉𝑡𝑡 = total volume 𝑚𝑚𝑎𝑎 = mass of air 𝑚𝑚𝑊𝑊 = mass of water 𝑚𝑚𝑠𝑠 = mass of solids n = porosity Answer: 2.13𝑔𝑔 The porosity ‘n’ is equal to 0.33. 𝑉𝑉𝑡𝑡 in the numerator and denominator cancel, so 𝜌𝜌𝑏𝑏 = 𝑐𝑐𝑐𝑐 .

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John Walter/Homework Set #1 FABE 2720/January 24, 2014

Problem Statement 11: The field capacity of a soil is 35%, and the wilting point is 15% by weight. The dry bulk density of the soil is 78 lb/ft3. Determine the available water content in percentage by weight and by volume. Also determine the plant available soil water content in the top 2 ft of soil. Express the answer as a depth of water in inches. Using equation 3.13 to determine the available water content in percentage by weight:

0.20𝑔𝑔 𝑜𝑜𝑜𝑜 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 ∗ 100% = 20% 𝑏𝑏𝑏𝑏 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤ℎ𝑡𝑡 𝑔𝑔 𝑜𝑜𝑜𝑜 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 Equation: 3.13 (from Environmental Hydrology 2nd, Edition) 𝜃𝜃𝑎𝑎 = available water content 𝜃𝜃𝑓𝑓𝑓𝑓 = the soil water contents at field capacity 𝜃𝜃𝑤𝑤𝑤𝑤 = the soil water contents at wilting point 𝜃𝜃𝑎𝑎 = 𝜃𝜃𝑓𝑓𝑓𝑓 − 𝜃𝜃𝑤𝑤𝑤𝑤 = 0.35 − (0.15) =

Using equation 3.13 to determine the available water content in percentage by volume: Conversion of 35% by weight to by volume:

1 𝜃𝜃𝑣𝑣 = 𝜃𝜃𝑔𝑔 ∗ 𝜌𝜌𝑑𝑑𝑑𝑑𝑑𝑑 ∗ � � = 𝜌𝜌𝑊𝑊

(𝑚𝑚)3 . 35 𝑔𝑔 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 78 𝑙𝑙𝑙𝑙 (3.28𝑓𝑓𝑓𝑓)3 453.59𝑔𝑔 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 ∗[ 3 ∗� � ∗ � �∗� � 3 3 𝑔𝑔 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 1 𝑙𝑙𝑙𝑙 (𝑚𝑚) (100𝑐𝑐𝑐𝑐) 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑓𝑓𝑓𝑓 = 𝑔𝑔 𝑐𝑐𝑐𝑐3 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝜃𝜃𝑣𝑣 = water by volume 𝜃𝜃𝑔𝑔 = water by weight 𝜌𝜌𝑑𝑑𝑑𝑑𝑑𝑑 = bulk density of soil 𝜌𝜌𝑊𝑊 = density of water

0.437𝑐𝑐𝑐𝑐3 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 = 𝑐𝑐𝑐𝑐3 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠

Conversion of 15% by weight to by volume:

1 𝜃𝜃𝑣𝑣 = 𝜃𝜃𝑔𝑔 ∗ 𝜌𝜌𝑑𝑑𝑑𝑑𝑑𝑑 ∗ � � = 𝜌𝜌𝑊𝑊

(𝑚𝑚)3 . 15 𝑔𝑔 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 78 𝑙𝑙𝑙𝑙 (3.28𝑓𝑓𝑓𝑓)3 453.59𝑔𝑔 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 ∗[ 3 ∗� �∗� �∗� � 3 𝑔𝑔 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 1 𝑙𝑙𝑙𝑙 𝑓𝑓𝑓𝑓 (𝑚𝑚) (100𝑐𝑐𝑐𝑐)3 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = 𝑔𝑔 3 𝑐𝑐𝑐𝑐 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 0.187𝑐𝑐𝑐𝑐3 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 = 𝑐𝑐𝑐𝑐3 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 17

John Walter/Homework Set #1 FABE 2720/January 24, 2014

𝜃𝜃𝑣𝑣 = water by volume 𝜃𝜃𝑔𝑔 = water by weight 𝜌𝜌𝑑𝑑𝑑𝑑𝑑𝑑 = bulk density of soil 𝜌𝜌𝑊𝑊 = density of water

0.25 𝑐𝑐𝑐𝑐3 𝑜𝑜𝑜𝑜 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 ∗ 100% = 25% 𝑏𝑏𝑏𝑏 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 𝑐𝑐𝑐𝑐3 𝑜𝑜𝑜𝑜 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 Equation: 3.13 (from Environmental Hydrology 2nd, Edition) 𝜃𝜃𝑎𝑎 = available water content 𝜃𝜃𝑓𝑓𝑓𝑓 = the soil water contents at field capacity 𝜃𝜃𝑤𝑤𝑤𝑤 = the soil water contents at wilting point 𝜃𝜃𝑎𝑎 = 𝜃𝜃𝑓𝑓𝑓𝑓 − 𝜃𝜃𝑤𝑤𝑤𝑤 = 0.437 − (0.187) =

Using equation 3.13 to determine the plant available soil water content within the top 2 ft of soil:

12𝑖𝑖𝑖𝑖 � = 6𝑖𝑖𝑖𝑖 1𝑓𝑓𝑓𝑓 Equation: 3.13 (from Environmental Hydrology 2nd, Edition) 𝜃𝜃𝑝𝑝𝑝𝑝𝑝𝑝 = volumetric or gravitational plant available soil water content 𝜃𝜃𝑓𝑓𝑓𝑓 = the soil water contents at field capacity 𝜃𝜃𝑤𝑤𝑤𝑤 = the soil water contents at wilting point 𝜃𝜃𝑝𝑝𝑝𝑝𝑝𝑝 = 𝜃𝜃𝑓𝑓𝑓𝑓 − 𝜃𝜃𝑤𝑤𝑤𝑤 = 2𝑓𝑓𝑡𝑡(0.437) − 2𝑓𝑓𝑓𝑓(0.187) = 0.5𝑓𝑓𝑓𝑓 ∗ �

Answer: The available water content in percentage of weight is 20%. The available water content in percentage by volume is 25%. The available soil water content in the top 2ft of soil is 6in.

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John Walter/Homework Set #1 FABE 2720/January 24, 2014

Problem Statement 12: The root zone (l m deep) of a silt clay loam soil has an average dry bulk density of 1.35 g/cc, and the specific gravity of the soil particles is 2.7. Based on laboratory measurements, it has been established that the degree of saturation at field capacity and wilting point are 65% and 35%, respectively. Irrigation is initiated by an automatic irrometer system when 40% of the plant available water has been depleted. The profile is irrigated at a constant rate of 10 mm/h for 6 h. Using the Green–Ampt equation, determine (a) the infiltration rate at the end of the irrigation application period; (b) the volume of runoff; and (c) the soil water content of the root zone when gravity flow has ceased. Based on a double-ring infiltration test, the Green–Ampt parameters have been determined as A = 30 mm/h and B = 3 mm/h. Hint: The problem is best solved on a spreadsheet and will require several iterations. Divide the irrigation application time into 30-min time blocks. Assume the initial infiltration rate is the irrigation application rate. Obtain a first estimate of the accumulative infiltration in the first 30 min by assuming a constant infiltration rate. You will then need to estimate infiltration rates at the end of each time block and compare them with values determined from Equation 3.16. Using the constant irrigation rate of 10mm/h over the entire 6h time frame and using the equation 3.17 we can generate the following table. 𝐴𝐴 + 𝐵𝐵 𝐹𝐹 Equation: 3.17 (from Environmental Hydrology 2nd, Edition) 𝑓𝑓 = infiltration rate F = accumulative filtration 𝐴𝐴 = Green-Ampt parameter 𝐵𝐵= Green-Ampt parameter 𝑓𝑓 =

f-rate F (Cumulative f-rate (based (Cumulative Irrigation filtration F(Cumulative on filtration Time rate rate based ) F(Cumulative based from (mm/h) from )) constant constant Irrigation) Irrigation) 0.50 1.00 1.50 2.00 2.50 3.00 3.50 4.00 4.50 5.00

10.00 10.00 10.00 10.00 10.00 10.00 10.00 10.00 10.00 10.00

5.00 10.00 15.00 20.00 25.00 30.00 35.00 40.00 45.00 50.00

9.00 6.00 5.00 4.50 4.20 4.00 3.86 3.75 3.67 3.60

4.50 6.00 7.50 9.00 10.50 12.00 13.50 15.00 16.50 18.00 19

9.67 8.00 7.00 6.33 5.86 5.50 5.22 5.00 4.82 4.67

Vol. Runoff

-0.50 -4.00 -7.50 -11.00 -14.50 -18.00 -21.50 -25.00 -28.50 -32.00

John Walter/Homework Set #1 FABE 2720/January 24, 2014 5.50 6.00

10.00 10.00

55.00 60.00

3.55 3.50

19.50 21.00

4.54 4.43

-35.50 -39.00

Table 12.1: Calculated infiltration data based off constant irrigation over a 6h period.

FILTRATION RATE VS TIME 12.00 10.00

Time (h)

8.00

f-rate (Cumulative filtration rate based from constant Irrigation)

y = -2.106ln(x) + 7.9673 R² = 0.9907

6.00

f-rate (based on F(Cumulative ))

Log. (f-rate (Cumulative filtration rate based from constant Irrigation))

4.00 2.00

y = -1.959ln(x) + 6.4563 R² = 0.8783

Log. (f-rate (based on F(Cumulative )))

0.00 0.00

2.00

4.00

6.00

8.00

Filtration Rate (mm/h)

Graph 12.1: Graphical representation of the data from Table 12.1. Soil Water Content at Root Zone: Assuming that the field capacity and wilting point are by weight… Using equation 3.13 to determine the available water content in percentage by volume: Conversion of 15% by weight to by volume:

1 𝜃𝜃𝑣𝑣 = 𝜃𝜃𝑔𝑔 ∗ 𝜌𝜌𝑑𝑑𝑑𝑑𝑑𝑑 ∗ � � = 𝜌𝜌𝑊𝑊

𝜃𝜃𝑣𝑣 = water by volume 𝜃𝜃𝑔𝑔 = water by weight 𝜌𝜌𝑑𝑑𝑑𝑑𝑑𝑑 = bulk density of soil 𝜌𝜌𝑊𝑊 = density of water

. 65 𝑔𝑔 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 1.35 𝑔𝑔 ∗ 𝑐𝑐𝑐𝑐3 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑔𝑔 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑐𝑐𝑐𝑐3 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = .8775 𝑔𝑔 𝑐𝑐𝑐𝑐3 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑐𝑐𝑐𝑐3 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤

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John Walter/Homework Set #1 FABE 2720/January 24, 2014 Conversion of 35% by weight to by volume:

1 𝜃𝜃𝑣𝑣 = 𝜃𝜃𝑔𝑔 ∗ 𝜌𝜌𝑑𝑑𝑑𝑑𝑑𝑑 ∗ � � = 𝜌𝜌𝑊𝑊

𝜃𝜃𝑣𝑣 = water by volume 𝜃𝜃𝑔𝑔 = water by weight 𝜌𝜌𝑑𝑑𝑑𝑑𝑑𝑑 = bulk density of soil 𝜌𝜌𝑊𝑊 = density of water

. 35 𝑔𝑔 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 1.35 𝑔𝑔 ∗ 𝑐𝑐𝑐𝑐3 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑔𝑔 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑐𝑐𝑐𝑐3 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = .4725 𝑔𝑔 𝑐𝑐𝑐𝑐3 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 3 𝑐𝑐𝑐𝑐 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤

Using equation 3.13 and specific gravity of 2.7 to determine the soil water content of the root zone at 1m depth:

𝜃𝜃𝑝𝑝𝑝𝑝𝑝𝑝 = 𝜃𝜃𝑓𝑓𝑓𝑓 − 𝜃𝜃𝑤𝑤𝑤𝑤 = �1𝑚𝑚(0.8775) − 1𝑚𝑚(0.4725)�2.7 = 1.094 𝑚𝑚 Equation: 3.13 (from Environmental Hydrology 2nd, Edition) 𝜃𝜃𝑝𝑝𝑝𝑝𝑝𝑝 = volumetric or gravitational plant available soil water content 𝜃𝜃𝑓𝑓𝑓𝑓 = the soil water contents at field capacity 𝜃𝜃𝑤𝑤𝑤𝑤 = the soil water contents at wilting point

Answer: The infiltration rate at the end of the irrigation application period is 60 mm. The volume run-off based from F(cumulative based from irrigation application) – F(Cumulative) increases from 0.5 mm to 39 mm over the 6h period. The soil water content of the root zone at 1m depth is 1.094m.

21