CHAPTER 2 SOLUTIONS. Given: 333 houses; AWWA household average demand. a. The AWWA average household water use is 1,320 L/d

Transcription

1 CHAPTER 2 SOLUTIONS 2-1 Subdivision Demand Given: 333 houses; AWWA household average demand a. The AWWA average household water use is 1,320 L/d b. For the average day (333 houses)(1,320 L/d) = 439, 560 L/d or 440,000 L/d or 440 m 3 /d c. With the assumption of 3 residents per house the population of the subdivision (333 houses)(3 people/house) = 999 people d. From Figure 2-1 and a population of 1,000, read a ratio of 5.5 for peak to average. The maximum day demand is then on the order of (5.5)(440,000 L/d) = 2,420,000 L/d or 2,400,000 L/d or 2,420 m 3 /d 2-2 Gross estimate of average day Given: Population of 7,000; location is Arizona; Hutson s data applies a. Using the website search to find b. At Arizona find under public water supply population = 4,870,000 total water = 1.08 x 10 9 gal/d c. Calculate per capita demand as 1.08 x x gal / d = gal/d x person people

2 d. For 7000 people 2-3 Ski Lift Development ( gal/d x person)(7,000 people)(3.79 x 10 3 m 3 /gal) = 5,884 m 3 Given: 250 room hotel; restaurant to seat 250 people; dormitory-style quarters for 25 people; hotel occupancy of 2 people/room a. From Table 2-2, assume 190 L/guest x d in hotel. The average water demand is estimated to be (250 rooms)(2 people/room)(190 L/guest x d) = 95,000 L/d b. From Table 2-2, assume 35 L/customer x d and that the restaurant serves 3 meals per day per seat (250 seats)(3 meals)(35 L/customer x d) = 26, 250 L/d c. From Table 2-4 with 25 staff in dormitory (150 L/person x d)(25 people) = 3,750 L/d d. Total average day 95,000 L/d + 26, 250 L/d + 3,750 L/d = 125,000 L/d 2-4 Complete series analysis for Squannacook River Given: Mean monthly discharge for the period Jan 1951 through Dec 1969 a. Using a spreadsheet type in monthly discharge b. Set up 3 columns: Rank Monthly discharge % of time equaled or exceeded c. Sort the monthly discharge data in descending order d. Calculate the % of time equaled or exceeded as

3 Rank Total # of data point s x 100% For example, for the data point with rank no.1 (14.73 m 3 /s) 1 1 = (19 )(12 ) years months % of time equaled or exceeded = 228 (100%) = 0.44% e. Plot mean monthly discharge (y-axis) against % of time (x-axis) f. From the data columns at 50.0% find the discharge = 2.18 m 3 /s g. The safe yield with a 6.0% restriction is (0.06)(2.18 m 3 /s) =.131 m 3 /s h. The plot of the yield curve is shown below

4 Problem 2-4 Complete Series Analysis for Squannacook River Monthly discharge, m y = e x R 2 = Se ries 1 Expon. (Series 1) Percent of time equaled or exceeded 2-5 Complete series analysis for Clear Fork Trinity River Given: Mean monthly discharge for the period Oct 1940 through Sep 1970 a. Using a spreadsheet type in monthly discharge b. Set up 3 columns: Rank Monthly discharge % of time equaled or exceeded c. Sort the monthly discharge data in descending order 1 1 d. = (30 )(12 ) years months % of time equaled or exceeded = (100%) = 0.246% 360

5 e. Plot mean monthly discharge (y-axis) against % of time (x-axis) f. From the data column at 50.0% find the discharge = m 3 /s g. The safe yield with a 3.0% restriction is (0.03)(0.714 m 3 /s) = m 3 /s h. The plot of the yield curve is shown below Problem 2-5 Complete Series for Clear Fork Trinity River Mean monthly discharge, m3/s Pe rce nt of tim e e quale d or e xce e de d 2-6 Annual Minima for Clear Fork Trinity River Given: Mean monthly discharge for the period Oct 1940 through Sep 1970 a. Using a spreadsheet type in monthly discharge b. The data are already arranged by hydrologic year. By convention the first hydrologic year is called the 1941 year. c. For each hydrologic year find the minimum value.

6 d. Set up the following columns Sort Rank Return period e. Perform the sort. Assign the lowest discharge a row of 1. f. T = = 31 years 1 g. The restriction is 3%; set up the columns and calculate Annual minima 3% of minima h. By observation, the mean monthly discharge never exceeds m 3 /s. Thus, storage will be required. i. The mean annual drought is about 0.06 m 3 /s. Annual Return Annual 6% of Minima Sort Rank Period, Minima Minima for year years by rank

7 2-7 Annual Minima for Squannacook River Given: Mean monthly discharge for the period Jan 1951 to Dec 1969 a. Using a spreadsheet type in monthly discharge b. Rearrange data into hydrologic years (Oct through Sep). The first 9 months of 1951 and the last 3 months of 1969 cannot be used. The first hydrologic year begins with Oct By convention it is called the 1952 hydrologic year. c. For each hydrologic year find the minimum value. d. Set up the following columns Sort Rank Return period e. Perform the sort. Assign the lowest discharge a row of 1. f. Calculate the return period as T = n + 1 m For the first value T = = 19 years g. The restriction is 6%; set up two columns and calculate Annual minima 6% of minima h. By observation, the mean monthly discharge never exceeds m 3 /s. Thus, storage will be required. i. From the Gumbel plot below, the mean annual drought is between 0.36 and 0.42 m 3 /s. The scatter of the data precludes a more definitive value. Annual Return Annual 3% of Minima Sort Rank Period Minima Minima

8 for year by rank Continuation of Example 2-3 Given: Example 2-3 Month Q in, (0.05)(Q in) (0.05)(Q in)(δt) Qout, (Qout)(δt) S Σ( S) m 3 /s 10 6 m 3 m 3 /s 10 6 m m m Dec

9 1997 Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Jan Feb Mar Apr

10 May Jun Jul Aug Sep Oct Nov Dec Jan Feb Mar Apr The reservoir is full The maximum storage required remains at x 10 6 m 3 from FEB 1996 The maximum storage required for this problem is x 10 6 m 3 in FEB Storage for the Hoko River Given: Mean monthly discharge data for the period Jan 1963 through Dec 1973 Month Q in, (0.05)(Q in) (0.05)(Q in)(δt) Qout, (Qout)(δt) S Σ( S) m 3 /s 10 6 m 3 m 3 /s 10 6 m m m Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec

11 1970 Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Jan Feb Mar Apr May

12 Jun Jul Aug Sep Oct Nov Dec The maximum storage required is x 10 6 m 3 in SEP 1970 Note September 1970 Σ( S) 10 6 m 3 is the maximum deficit 2.10 Eudora s Maximum Sustained Pumping Rate Given: D = 10.0 m; piezometric surface = 40.0 m above bottom confining layer; medium san aquifier; non-pumping wellis m from the pumping well; drawdown at non-pumping well is 1.00 m, the pumping well is 1.0 m in diameter a. From Table 2-9, select hydraulic conductivity (K) of 1.5 x 10 4 m/s b. Estimate T as T = KD = (1.5 x 10 4 m/s)(10 m) = 1.5 x 10 3 m 2 /s c. Calculate h 1 and h 2 from the following sketch SKETCH GOES HERE! h 2 = 40.0 m 1.0 m = 39.0 m h 1 = 10.0 m d. Estimate maximum allowable sustained pumping rate using Equation 2-6 ( x 3 m 2 s ) ( m m ) 2π / Q = 200m ln 0.5 m m / s = = m 3 /s or m 3 /s 5.99

13 This is greater than the desired m 3 /s therefore the larger pumping rate is acceptable Maximum sustained pumping rate in confined aquifier Given: Well boring log; well diameter = 1.0 m; 2 m safety factor; drawdown 100m for pumping wellis 0.0 m a. From Table 2-9, select hydraulic conductivity of fractured rock which is the aquifier based on the well log K = 5.8 x 10 5 m/s b. Estimate T as T = KD = (5.8 x 10 5 m/s)(55 m ) = m 2 /s c. Calculate h 1 and h 2 from the following sketch SKETCH GOES HERE! h 2 = 10 m + 2 m + 55 m = 67 m h 1 = 55 m + 2 m safety = 57 m Q = ( m 2 s ) ( m m ) 2π / m ln 0.5 m = = m 3 /s Estimated maximum sustainable pumping rate is m 3 /s 2.12 Cumulative frequency distribution for Lake Michigan turbidity Given: Mean monthly turbidity for 2005 through 2007 a. Arrange data and sort in ascending order b. Compute percent less than. For example, for start value:

14 1 * ( ) % less than = 100 % = 2.78% 36 c. Spreadsheet comparability and plot are on following page d. The report should note the following: About half the time (50%) the variability is between 0.5 and 1.5 NTV. The overall range of variability is more than 10 fold. The plant operator will need substantial flexibility in operation to care for this range of variation. Because only monthly summaries were provided, a more thorough analysis of daily values may reveal an even greater variation in turbidity. Analysis of daily values is recommended.

15 Turbidity, Sort Percent NTU Less Than Cumulative frequency distribution for Alma river turbidity Given: Daily turbidity for Alma

16 a. Arrange data and sort in ascending order b. Compute percent less than. For example, for start value: 1 * ( ) % less than = 100 % = 3.23% 31 c. Spreadsheet comparability and plot are on following page d. The report should note the following: About 65% of the data fall between 2 and 4 NTV. The overall range of turbidity is 30 fold. The plant operator will need substantial flexibility in operation to care for this range of variation. Because only one month of data was provided, a more thorough analysis is recommended. Analysis of at least one year of daily turbidity is recommended.

17 Day Turbidity, Sort Percent NTU Less Than