Examples VVR145 water

Joakim Malm, Magnus Persson and Rolf Larsson December 2010 Dept. of Water Resources Engineering LTH/Lund University

Examples VVR145 water

CONTENTS A WATER BALANCE B FLUID PROPERTIES C PRECIPITATION AND EVAPORATION D INFILTRATION, SOIL WATER E GROUNDWATER F HYDROGRAPH ANALYSIS G FREQUENCY ANALYSIS H HYDROSTATICS I FUNDAMENTAL EQUATIONS J PIPE FLOW K URBAN HYDROLOGY FIGURES/TABLES ANSWERS A-K SOLUTIONS A-K - - - - - - APPENDIX Physical properties of water Viscosities of some common fluids Approximate properties of some common liquids at standard atmospheric pressure Some figures and tables

Vatten (VVR145) - Examples

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BRUKSANVISNING Vatten som det lrs ut vid LTH/Lunds Universitet genom Avd Teknisk Vattenresurslra r i huvudsak ett ingenjrsmne. Det innebr att studenterna ska lra sig att gra berkningar fr att kunna avge kvantitativa svar p olika vattenrelaterade frgor. Problemlsning r drfr ett centralt omrde inom kursen. Vi vill hr betona ngra aspekter p problemlsning som inte alltid uppmrksammas: I denna disciplin, liksom fr den del i de flesta andra, r ofta de huvudsakliga

svrigheterna inte relaterade till att besvara vldefinierade frgor med fullstndiga och entydiga frutsttningar. Snarare r det alltid svrast att stlla rtt frga, och att ta fram de data som erfordras fr att besvara denna frga. Det r drfr vr ambition att lta flera av problemen i denna problemsamling vara ppna i den bemrkelsen att det erfordras vissa preciseringar och antaganden av studenten innan den egentliga problemlsningen tar vid. Detta gller speciellt kapitel A, C-G, och K. Fljaktligen r inte alla svar entydiga.

Inom all ingenjrsvetenskap, r alla svar behftade med oskerhet. Studenterna

mste vnja sig vid att begrunda vilka faktorer som pverkar noggranheten i avgivna svar och ppeka vilka dessa r. Svaren fr heller aldrig avges med verdrivet hg (eller lg) numerisk noggranhet.


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A WATER BALANCE (* denotes more difficult problems) A1 For a lake with an area of 14.3 km2, the following data has been recorded during a specific year; Precipitation 695 mm Average inflow 1.39 m3/s Average outflow 1.56 m3/s Assume that there is no net water exchange between the lake and the groundwater. Determine the evaporation during this year. A2 A pipe leading storm water from a parking lot (area = 2500 m2) in Lund is to be constructed. The design storm is supposed to have an intensity of 60 mm/h. Determine the design flow (in m3/s) and the dimension of the pipe if the water velocity in the pipe is assumed to be 1 m/s. A3* The evaporation from a lake is to be calculated by the water budget method. Inflow to the lake is mainly through three small rivers A, B and C. The outflow is through river D. Calculate the evaporation from the lake surface during summer (May - August) if the water level was at +571.04 (m) on May 1 and +571.10 on August 31. The surface area of the lake is 100 km2. The precipitation during the period was 100 mm. The average flows in the rivers are given below. Be careful when choosing the units in the water balance! River Catchment ( km2)* Q, average (m3/s) A 140 17.0 B 110 13.0 C 120 15.0 D 43.0 *the area of the sub catchments A-D. A4 Lake X has a surface area of 7.08105 m2. During a given month, the inflow to the lake is 1.5 m3/s while the outflow is 1.25 m3/s. The increase in stored volume is measured / calculated at 5.35105 m3. Precipitation during the same month is measured at 225 mm. Calculate the evaporation (mm) for the lake.


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A5 A retention reservoir is located such that all runoff from the catchment passes the reservoir (c.f. fig.). The catchment area (reservoir not included) is 200 ha. The area of the reservoir is 2.0 ha. Infiltration though the lake sediments, which consists of clay, is negligible. Outflow is regulated and constant at Q = 0.045 m3/s. Average precipitation in June is 176 mm and average inflow to the reservoir during the same month is 95 000 m3. Evaporation from the lake surface is 2.8 mm/day as an average for June. a) What is the average June runoff coefficient? (a=Q/P) b) In June 1991 precipitation was slightly higher than normal, P = 220 mm. What total inflow to the reservoir may be expected for this specific month? c) What is the expected change in water level in the reservoir during this month?

Damm

Avr. omrde

A6 Use the enclosed flow data for Sege to answer the following questions: a) What is the annual runoff/discharge at Svedala in m3/s, mm/year, and l/(s km2)? b) Use the enclosed precipitation data for Sweden to estimate the runoff coefficient for this area.


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B FLUID PROPERTIES (* denotes more difficult problems)

Compressibility B1 What pressure must be applied to water to reduce its volume 1 % ?

Viscosity B2 A very large thin plate is centered in a gap of width 0.06 m with different oils of unknown viscosities above and below; one viscosity is twice the other. When the plate is pulled at a velocity of 0.3 m/s, the resulting force on one square meter of plate due to the viscous shear on both sides is 29 N. Assuming viscous flow, and neglecting all end effects, calculate the viscosities of the oils. B3 Calculate the approximate viscosity of the oil in the Fig.

B4 A vertical gap 25 mm wide of infinite extent contains oil of density 950 kg/m3 and viscosity 2.4 Pas. A metal plate 1.5 m 1.5 m 1.6 mm weighing 45 N is to be lifted through the gap at a constant speed of 0.06 m/s. Estimate the force required. B5* A circular disc of diameter d is slowly rotated in a liquid of large viscosity at a small distance h from a fixed surface. Derive an expression for the torque T necessary to maintain an angular velocity . Neglect centrifugal effects. B6* Calculate the approximate power lost in friction in this bearing.

Fig. B6

Fig. B3


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Surface tension B7 Calculate the maximum capillary rise of water (T = 20C) to be expected in a vertical glass tube, diameter 1 mm. B8 Calculate the maximum capillary rise of water (20C) to be expected between two vertical, clean glass plates spaced 1 mm apart. The contact angle between water and glass is 0o. B9 A soap bubble 50 mm in diameter contains a pressure (in excess of atmospheric) of 20 Pa. Calculate the tension in the soap film.


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C PRECIPITATION AND EVAPORATION (* denotes more difficult problems) C1 Calculate the areal precipitation for the rectangular catchment below, using a) Arithmetic mean b) Thiessen method The measured precipitation was A: 10 mm, B: 20 mm, C: 30 mm. C2 Use the idf-curve in the appendix and answer the following questions a) If the duration is 10 min, what is the maximum rainfall intensity for return periods

of T = 1, 2, 5 and 10 years [answer in mm/h]? b) Plot the intensities vs. T. c) If the return period is T = 5 years. What is the maximum intensity for durations of

60, 30 and 5 min [answer in mm/h]? C3 A catchment with area A = 88 km2, has a gage network according to fig. Use the precipitation data in the table to find the areal precipitation. a) Use the method of arithmetic average. b) Use the Thiessen method. Gage P (mm) A 13.0 B 17.1 C 22.9 D 15.3 E 14.1 F 12.6 G 11.6

2km 2 km 3 km 1 km

3 km

A C B


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C4 The water equivalent of a snow cover is determined by measuring the snow depth (d) and the snow density (). The density is calculated by taking snow samples using a snow tube that is pushed down from the snow surface to the ground, the snow tube is then lifted and weighted. Calculate the water equivalent using the following measured data (we=d*): (The weight is only measured in three points and the depth in all 11 points. Use average values of d and ) Inner diameter, snow tube: 46 mm location Depth (cm) Weight (g) 1 91 299 2 87 3 86 4 84 5 92 6 107 320 7 103 8 100 9 98 10 90 11 87 290 C5 Use the enclosed intensity-duration-frequency curve for rain in Malm to answer the following: a) What is the probability that during a given year a rain storm will occur giving more than 10 mm during 10 min? b) Assume that the rain in a) falls on a parking lot, the area of which is 2500 m2. If the runoff passes without losses and without time delay into a storm water system, what is the discharge in the pipe? C6 The evaporation from an evaporation pan beside a lake is measured daily. During one month the measured evaporation was Epan = 156 mm. The actual evaporation from the lake was Elake = 123 mm during the same month. a) What is the pan coefficient? b) If Epan = 6.7 mm during a specific day, what was the actual evaporation from the lake?


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C7 Calculate Epot according to Penman for a lake near Stockholm during September. Use the following data: Latitude 59. Temperature T = 10 C. Wind speed w = 2 m/s. n/N = 0.4, relative humidity RH = 0.65. a) Calculate the incoming radiation, Rin (mm/d). b) Calculate the reflected radiation, Rout (mm/d) c) Calculate H (mm/d) d) Calculate Ea (mm/d) e) Calculate the potential evaporation, Epot (mm/d) C8 If the air temperature is T = 28 C and the relative humidity RH = 70 %, what is a) Saturation vapour pressure (ea) b) The saturation deficit (ea-ed) c) The vapour pressure (ed) d) The dew point temperature Give pressures both in mm Hg, mbar and Pa. (Saturation vapour pressures are enclosed.) C9 Calculate potential evaporation (for a free water surface) according to Penman for: Place Month Temp(C) RH n/N Wind (m/s) Amsterdam(52o) July 18 0.5 0.5 1.2 RH = relative humidity n/N = number of sunshine hours / potential sunshine hours Some extrapolation is acceptable. C10 Use the enclosed meteorological data for Lund in this problem: a) Calculate potential evaporation for the western part of Skne (free water surface) for July 1980. Compare the result with the average value 1961-78 according to the enclosed table. b) What is the daily evaporation from Vomb lake, with an area of approximately 8 km2. c) Malm takes most of its drinking water from Vomb lake. What is the necessary extraction in l/s (Estimate the number of inhabitants in Malm and their consumption), compare with the answer in b).


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C11 In order to find the evaporation from a lake, an evaporation pan (class A pan) has been placed near the shore. The water level in the pan is observed daily and water is added if the level gets near 175 mm. Estimate the daily evaporation (i.e., the evaporation each day), if the pan coefficient is 0.70. Day

Precip. (mm) Water level(mm) Day Precip. (mm) Water level (mm)

1 0.0 203.2 6 0.0 201.7 2 5.8 205.9 7 0.5 196.7 3 14.2 217.5 8 0.3 192.4 4 1.3 214.3 9 0.0 187.6 5 0.3 208.3 10 0.0 182.8 C12 Krankesjn, a lake situated 20 km east of Lund, has an area of 3.4 km2 and an average depth of 0.7 m. Two small creaks flow in to the lake. The outflow is through labcken. During the sunny month of July 2006 the lake had 3.8 mm of precipitation, the inflow in the small creaks were 140 l/s each and the outflow in labcken was 240 l/s. In the table below measured values of the elevation of the lake surface are given in meters above sea level (masl). Date 2006-06-01 2006-06-22 2006-07-01 2006-07-31 2006-08-15 Water level (masl) 19.2 19.13 19.08 18.97 19.15 a) calculate the evaporation during July 2006. How would the evaporation be affected if b) the relative humidity was higher c) the wind speed was higher d) the cloudiness was higher


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D INFILTRATION, SOIL WATER (* denotes more difficult problems) D1 A heavy rainfall event in a catchment lasted 12 h and gave totally 55 mm of rain. The direct runoff corresponding to this rainfall was 30 mm. a) Calculate -index. b) During another rainfall event in the same catchment, 20 mm of rain came during

the first 6 h and 10 mm during the next 6 h. Calculate the losses according to the -index method. Use the calculated -index from a).

D2 The Horton equation f(t) describes how the infiltration capacity (mm/h) for the soil is changing with time as the soil becomes wetter. The accumulated infiltrated water (in mm) can be calculated using the integral of f(t), (F(t)).

ktcoc effftf

+= )()(

( )ktcoc ekfftftF += 1)(

where fo = infiltration capacity in dry soil (mm/h) fc = = infiltration capacity in wet soil (mm/h) k = time factor (h-1) Note that the Horton equation is only valid when water is available all the time at the soil surface (ponded conditions). In a specific soil the following parameters are known fo = 35 (mm/h) fc = 6 (mm/h) k = 2 (h-1) a) What is f after 15, 30, and 60 min? b) How much water has infiltrated after 15, 30, and 60 min?


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D3 In order to find the infiltration capacity of a certain soil type, experiments were made with an infiltrometer; a steel cylinder, D = 20 cm, set vertically in the soil. Water was supplied so that a constant water level was kept within the cylinder. a) Estimate the parameters of the Horton equation using the experimental data below. Time (min) Water supplied (ml) 0-10 105 10-20 69 20-30 50 30-40 39 40-50 33 50-60 30 60-70 28 70-80 27 80-90 27 b) Use the results from a) to calculate net precipitation from a storm that lasts 30

minutes with a total volume of 15 mm. (Net precipitation is P minus infiltrated volume).

D4 In a small catchment, infiltration tests have been performed and it has been found that the Horton equation is applicable with the following parameters: Io = 25 mm/h; Ic = 4 mm/h; k = 3 h

-1 Plot the potential infiltration and accumulated infiltration as a function of time. It may be assumed that the surface is covered with water the whole time. D5 A catchment has an area A = 2.26 km2. Find the -index for a given event with rain according to the table and with the direct runoff volume 5.6104 m3. Time (h) Rain intensity (mm/h) 0-2 7.1 2-5 11.7 5-7 5.6 7-10 3.6 10-12 1.5


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D6 A soil sample has a volume of 200 cm3 (= ml). Assume that the particle density is part = 2600 kg/m3. a) If the sample weighs 0.352 kg when it is oven dry, what is the dry bulk density? b) What is the particle volume? c) What is the void volume? d) What is the porosity? e) If the same sample weights 0.401 kg when it is wetted, what is the water content

in % by weight? f) What is the water content in % by volume? D7 A soil sample (cylinder) which measures D = 30 mm, L = 50 mm weighs 64.90 g. The sample is dried in an oven at T = 105 C after which it weighs 53.96 g. The sample is then saturated with water and allowed to drain for a couple of days. Measured drain water volume is 5.3 ml. The density of the soil particles is 2650 kg/ m3 a) Find the porosity and the initial moisture content (by volume %). b) Find the effective porosity (= specific yield) and the field capacity. c) Assuming the thickness of the root zone to be 1.0 m and the moisture content at the wilting point 10 %, how much water (as available water) can be stored in the soil? d) How much water must be supplied (m3/ha) to an agricultural field in Skne with this soil if the water content in the beginning of the growing season equals field capacity? Assume that the soil moisture content must not fall below (the wilting point + 25 % of available water). Precipitation during the growing season (May-July) is assumed to be 50 mm (conservative estimate) and the need of the crop equals the potential evapotranspiration according to Penman, see table/figure enclosed. D8 A soil sample (cylinder) measures D = 30 mm, L =50 mm and weighs 65.42 g. The soil particles have a density of 2650 kg/m3. After drying in oven the sample weighs 56.14 g. Find porosity, moisture content (in % weight of dry weight), and moisture content (% by volume).


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D9 An irrigation system in Skne is to be designed for an area of 10 ha and a soil sample is taken (cylinder) (D=100 mm, L=300 mm). It is saturated with water and thereafter left to drain for a couple of days. The amount of collected drained water is 365 ml. The porosity is 43 %. Assume that moisture content at the wilting point is 10 %. a) Find the field capacity and the effective porosity (or specific yield). b) Assume that the root zone is 0.5 m and find how much available water that can be

stored in the soil. c) Assume that the growing season is mainly May, June, and July and that the crop

needs water corresponding to the potential evapotranspiration according to Penman, see enclosed table (in appendix). Assume also that the soil has a moisture content equivalent to the field capacity on May 1st. The moisture content must not fall below an amount corresponding to (the wilting point + 25% of available water). Find the total volume (m3) of water that is needed if there is 50 mm of rain during the period. All losses may be neglected.

D10 In a catchment 125 mm rain fell during one month. The evaporation was 50 mm. How much of the rain reached the ground water? The following assumptions are made

The distance from the soil surface to the ground water table is 1.25 m The porosity is 50 %, field capacity 25 % and wilting point 5 % (by volume). The water content is assumed to be constant in the unsaturated zone No horizontal water movements in the unsaturated zone. a) The water content is equal to the field capacity at the beginning of the month. b) The water content is equal to the wilting point at the beginning of the month. c) If no water is leaving the saturated zone (not very realistic), how much would

the groundwater table increase in the above cases?


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E GROUNDWATER (* denotes more difficult problems) E1 In a large unconfined aquifer the water table has been lowered by approximately 3.0 m over an area of 200 km2, due to pumping for irrigation. Net extraction of water (irrigation minus percolation) is 1.2108 m3. Find the storage coefficient of the aquifer. E2* Henry Darcy and Charles Ritter performed a series of experiments to investigate the flow in saturated sand. Water was flowing through a vertical column filled with sand. The column had a diameter of 0.35 m. Four series were made using four different sand materials. The data is presented in Appendix. Mean pressure is the pressure at the top of the column, the bottom of the column had atmospheric pressure. Thickness of sand is the height of the column. Calculate K for the four different series. E3 In order to make a ground water investigation, three observation wells have been installed. The wells form an equilateral triangle with the side length 25 m. When the level of the ground water table was measured, the following results were obtained: +10.00, +9.46, +9.47 in a clockwise direction. a) What is the inclination of the ground water table? (Assuming that the water table

forms a plane). b) In order to find the conductivity, a tracer was injected in one of the wells, while the

concentration was measured in a new observation well 25 m away from the input well. If this was done in a rational way, which well was used and where was the new well located?

Outflow

Thickness of sand

Mean Pressure

Sand

col

umn


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c) The time between the input of the tracer and the registration of the tracer concentration peak in the downstream well was 100 hours. According to laboratory tests the effective porosity was 35 %. What was the hydraulic conductivity? Assume location according to b). E4 At an old landfill, an investigation of the risk for spreading of pollutants is to be made. Three observation wells are installed and the ground water table is measured (unconfined aquifer). The wells are located in the corners of an equilateral triangle with 100 m side, see fig. In one of the wells (A) the level is at +13.80 m. In the other wells the levels are +11.50 (B) and +12.50 (C) respectively. Estimate flow direction and the shortest possible time for a pollutant from the site to travel 100 m. Assume K = 1.0 10-6 m/s and effective porosity = 0.20.

A: +13.8

B: +11.5 C: +12.5 E5 An excavation 10 x 10 m and 3 m deep is to be made for a building site. The ground consists of a sand layer (10 m depth) with hydraulic conductivity K = 1.2*10-4 m/s. In order to be able to work on the site the ground water table has to be lowered by at least 1 m everywhere in the excavation from the original position 2 m under the surface. The plan is to use one well in the centre of the excavation. The pump to be used has a capacity of Q = 4 l/s. In consideration of the suction capacity of the pump the ground water table may not be lowered underneath a level of 3 m under the excavation floor. What is a suitable well diameter? It can be assumed that the well will fully penetrate the sand layer, and that the layer underneath is impermeable. E6 A well (diameter = 0.3 m) is to be constructed in the middle of an island (which can be considered circular with D = 1.6 km) in a lake. The aquifer is of thickness 15 m, consists of sandstone with K = 15 m/day and is overlain by clay. What is the highest constant flow rate that can be applied without lowering the piezometric surface more than 3 m at the well? E7 Water is pumped from a well fully penetrating a confined aquifer with thickness b = 20 m. Two observation wells are located at distance 100 m and 1000 m away from the pumped well respectively. Water is pumped at a constant discharge of Q = 0.2 m3/min. For steady conditions the drawdown is 8 m in one observation well and 2 m in the other one. Find the hydraulic conductivity and the transmissibility. E8 Two water bodies are separated with an earth dam.


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a) Find an expression for the ground water flow through the dam. b) calculate the flow though the dam if water level 1; h1 = 6 m, water level 2; h2 = 2 m, width of the earth dam 3 m, hydraulic conductivity = 10-5 m/s.

Earth dam Water level 1 Ground water table Water level 2 h x impermeable bedrock


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F HYDROGRAPH ANALYSIS (* denotes more difficult problems) F1 A catchment (A = 300 km2) gets 30 mm rain during one day. During the next five days the following flow is measured in the river that drains the catchment. Time (d) 0 1 2 3 4 5 Q (m3/s) 2 5 10 8 4 2 Assume that the baseflow is constant Qbase = 2 m3/s during the period. a) What was the total direct runoff during the period (m3)? b) What was the losses (mm)? F2 In a catchment the following one-hour unit hydrograph is known. Time (h) 0 1 2 3 4 QUH (m3/s) 0 3 2 1 0 a) Estimate the direct runoff if the effective rainfall during one h is 5 mm. b) Estimate the direct runoff if the effective rainfall is 5 mm during the first h and 10 mm during the second. c) What will the total flow be for the rainfall in b) if the baseflow is constant 1 m3/s? F3 During and after a rainfall event the following discharge was measured in a river. Plot the hydrograph both in lin/lin and lin/log scale. a) What was the baseflow if it can be assumed to be constant? b) What was the baseflow if it can be assumed to follow the recession curve? Tid (h) 0 1 2 3 4 5 6 Qtot (m3/s) 0.1 0.1 2.50 1.50 0.40 0.20 0.1 F4 A catchment in Egypt experiences a long period without rain. The discharge in the river which drains the catchment is 100 m3/s after 10 days without rain and 50 m3/s after another 30 days. What flow can be expected to occur on day 120 if there is no rainfall during this period?


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F5 The discharge from a catchment with area A = 1000 ha, was measured during a rain event. The discharge (m3/s) is given below for every 30 min starting at 10.00. a) Estimate the times for start and finish of the direct runoff. b) What is the base flow? c) What is the volume corresponding to the direct runoff? Q = 4.9, 4.7, 4.5, 7.3, 11.8, 17.2, 21.1, 25.5, 29.7, 32.4, 32.2, 30.9, 28.7, 26.2, 23.3, 20.5, 18.0, 15.7, 13.8, 12.3, 11.1, 10.0, 9.1, 8.4, 7.7, 7.2, 6.7, 6.3, 5.9, 5.5, 5.2, 4.9, 4.6, 4.4, 4.1, 3.9, 3.7, 3.5, 3.4, 3.2, 3.0 (It is a good idea to use Excel to solve this problem.) F6 The discharge values for a 30 min Unit Hydrograph is given below. a) What is the area of the catchment? b) Calculate and plot the hydrograph for a 60 minute rain with an effective intensity

of 10 mm/h. Time (min) Discharge (m3/s) 0 0 15 4.5 30 10 45 12.5 60 11 75 9 90 6.5 105 4 120 2.5 135 1 150 0


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F7 The one hour Unit Hydrograph for a catchment is given in the figure below. Find the maximum discharge and the time of peak flow for a storm that lasts 4 hours with an intensity of 2 mm/h. During the latter occasion, the base flow can be assumed to be 10 l/s.

F8 A catchment receives a heavy rain storm during 3 hours. The rain is distributed in time with 17 mm during the first hour, 22 mm during the second hour, and 14 mm during the last hour. The -index is 12 mm/h and the measured hydrograph is given below. a) Find the 1 hour Unit Hydrograph. (Assume the base flow to be constant.) Time (h) Discharge (m3/s) 0 0 1 1 2 5.5 3 12.4 4 15.4 5 13 6 9.6 7 6.2 8 3.05 9 0.9 10 0.1 11 0

0102030405060

0 10 20 30 40 50

Q [l

/s]

Time [h]


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F9 The runoff after a storm with an effective intensity of 2 mm/h, 4 mm/h, 1 mm/h during hours no. 1, 2 and 3, has been measured and is given in the table below. Find the Unit Hydrograph for a 1-hour storm with intensity 1 mm/h. Base flow is assumed to be constant 2 m3/s. Time (h)

0 1 2 3 4 5 6 7

Q (m3/s) 2 8 24 33 29 23 19 15.5 Time (h)

8 9 10 11 12

Q (m3/s) 12 8.5 5 2.5 2 F10 The Unit Hydrograph for a 1-hour storm with intensity 1 mm/h can be simplified as an isosceles (likbent) triangle with base 6 hours and maximum flow 12 m3/s after 3 hours. What is the likely discharge after a 3 hour storm with net intensity 5 mm/h? The base flow is 2 m3/s.


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G FREQUENCY ANALYSIS (* denotes more difficult problems) G1 A flow with a return period of 100 years for a river is Q100 = 123 m3/s a) What is the probability that a flow larger than Q100 will occur during a given year? b) What is the probability that the maximum flow during a given year will be smaller

than Q100? c) What is the probability that a flow smaller than Q100 will occur two years in a

row? d) What is the probability that a flow larger than Q100 will occur at least once during

a period of two years? e) What is the probability that a flow larger than Q100 will occur at least once during

a period of ten years? G2 Give the plotting positions according to Weibul [r/(N+1)] for the data given below. r Qmax (m3/s)

1990 10 1991 20 1992 24 1993 18 1994 12

G3 Make a frequency analysis for the annual maximum of daily discharge for the Trendlven at Junosuando (data enclosed). Use the period 1939-75. Find the 10, 50 and 100-year floods. Use the Gumbel extreme value distribution. Special plotting paper is enclosed. G4 Discharge data from Kttilsmla are enclosed. Use frequency analysis to find the value of the maximum daily discharge with recurrence interval of 10 and 100 years. a) Do this analysis, using 37 years of data, i.e. 1939-75. b) Do the same thing using 20 years of data (1939-58). "Gumbel-paper" is enclosed.


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H HYDROSTATICS (* denotes more difficult problems)

Pressure in a static fluid H1 The weight density ( = g) of water in the ocean may be calculated from the empirical relation = 0+ K(h)1/2, in which h = the depth (m) below the ocean surface. Derive an expression for the pressure at any point h and calculate weight density and pressure at a depth of 3220 m assuming 0 = 10 kN/m3, K = 7.08 N/m7/2. H2* If atmospheric pressure at the ground is 101.3 kPa where air density is 1.225 kg/m3, calculate the pressure 7.62 km above ground, assuming a) no density variation b) isothermal variation of density with pressure (p/ = const) and c) adiabatic variation of density with pressure (p/k = const. with k = 1.40 for air).

Manometry H3 With the manometer reading as shown, calculate px. (r.d = relative density i.e. = r.d. water). H4 Calculate px - py for this inverted U-tube manometer (r.d = relative density i.e. = r.d. water).

Fig. H3

Fig. H4


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H5 The manometer reading is 150 mm when the tank is empty (water surface at A). Calculate the manometer reading when the tank is filled with water. H6 Calculate the gauge reading. H7 In fig. l1 = 1.27 m, h = 0.51 m, l2 = 0.76 m, liquid 1 is water, 2 is benzene, and 3 mercury. Calculate px - py.

Fig. H7

Fig. H5

Fig. H6


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H8* The sketch shows a sectional view through a submarine. Calculate the depth of submergence, y. Assume that the weight density of sea water is 10.0 kN/m3. H9* An engineering department is evaluating a sophisticated $80,000 laser system to measure the small differences in water level between two large water storage tanks. You suggest that the job can be done with a $200 manometer arrangement, as shown in fig. Oil less dense than water can be used to give a 10:1 amplification of meniscus movement; a small difference in level between the tanks will cause 10 times as much deflection in the oil levels in the manometer. Determine the relative density of the oil required for 10:1 amplification.

Forces on plane surfaces

H10 A rectangular gate 1.8 m long and 1.2 m high lies in a vertical plane with its centre 2.1 m below a water surface. Calculate magnitude, direction and location of the total force on the gate. H11 A circular gate, 3 m in diameter, has its centre 2.5 m below the water surface and lies in a plane sloping at 60o from the horizontal. Calculate magnitude, direction, and location of total force on the gate.

Fig. H8

Fig. H9


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H12 This rectangular gate will open automatically when the depth of water, d, becomes large enough. What is the minimum depth that will cause the gate to open? H13 The water depth alongside the L-shaped dam is H = 4 m. Estimate the minimum possible width L of the horizontal limb if the dam does not overturn. Neglect seepage and the self weight of the dam.

H14 As water rises on the left side of the rectangular gate, the gate will open automatically. At what depth above the hinge will this occur? Neglect the mass of the gate.

Forces on curved surfaces

Fig. H12

Fig. H14

Fig. H13


27

H15 The quarter cylinder AB is 3 m long. Calculate magnitude, direction, and location of the resultant force of the water on AB. H16 If this weightless quarter-cylindrical gate is in static equilibrium, what is the ratio between 1 and 2? H17 If this solid concrete (24 kN/m3) overhang ABCD is added to the dam, what additional force (magnitude and direction) will be exerted on the dam? H18 The weightless sphere of diameter d is in equilibrium in the position shown. Calculate d as a function of 1, h1, 2, and h2.

Fig. H15

Fig. H17

Fig. H18

Fig. H16


28

I FUNDAMENTAL EQUATIONS (* denotes more difficult problems)

Laminar/turbulent flow Reynolds number

I1 When 0.0019 m3/s of water flow in a 76 mm pipeline at 20C, is the flow laminar or turbulent? I2 What is the maximum speed at which a spherical sand grain of diameter 0.254 mm may move through water (20C) and the flow regime be laminar? I3 What is the smallest diameter of pipeline that may be used to carry 6.3 l/s of jet fuel (JP-4) at 15C if the flow is to be laminar.

Continuity/Mass conservation equation I4 Water flows in a pipeline composed of 75 mm and 150 mm pipe. Calculate the mean velocity in the 75 mm pipe when that in the 150 mm pipe is 2.5 m/s. What is its ratio to the mean velocity in the 150 mm pipe? I5 Using the control volume in the fig. find the mixture flowrate and density if freshwater (1 = 1000 kg/m3) enters section 1 at 50 l/s, while saltwater (2 = 1030 kg/m3) enters section 2 at 25 l/s. I6 A 0.3 m by 0.5 m rectangular air duct carries a flow of 0.45 m3/s at a density of 2 kg/m3. Calculate the mean velocity in the duct. If the duct tapers to 0.15 m by 0.5 m size, what is the mean velocity in this section if the density is 1.5 kg/m3 there? I7 A pipe transports 200 kg/s of water. The pipe tees into a 5 cm diameter pipe and a 7 cm diameter pipe. If the average velocity in the smaller diameter pipe is 25 m/s, calculate the flowrate in the larger pipe.

Fig. I5

Fig. I7


29

I8 A tank has a hole in the bottom with a cross-sectional area of 0.0025 m2 and an inlet line on the side with a cross-sectional area of 0.0025 m2. The cross-sectional area of the tank is 0.1 m2. The velocity of the liquid flowing out of the bottom hole is

ghV 2= where h is the height of the water surface in the tank above the outlet. At a certain time the surface level in the tank is 1 m and rising at the rate of 0.1 cm/s. The liquid is incompressible. Determine the velocity of the liquid through the inlet.

Energy equation I9 If crude oil flows through this pipeline and its velocity at A is 2.4 m/s, where is the oil level in the open tube C?

Fig. I8

Fig. I9


30

I10* Water is flowing. The flow picture is axisymmetric. Calculate the flow rate and manometer reading. I11 Water is flowing. Calculate the required pipe diameter, d, for the two gages to read the same. V300 = 6.0 m/s

I12* For a flowrate of 2 m3/s of air ( = 12.0 N/m3) what is the largest A2 that will cause water to be drawn up to the piezometer opening? Neglect compressibility effects. I13 A siphon consisting of a 25 mm hose is used to drain water from a tank. The outlet end of the hose is 2.4 m below the water surface, and the bend in the hose is 0.9 m above the water surface. Calculate the pressure in the bend and flowrate.

Fig. I10

Fig. I11

Fig. I12


31

I14 Neglect losses and find the depth of water on the raised section of the rectangular channel. Assume uniform velocity profiles.

I15 If each gage shows the same reading for a flowrate of 28 l/s, what is the diameter of the constriction? (NOTE El. = elevation) I16* Channel and gate are 1 m wide (normal to the plane of the paper). Calculate q1, q2, and Q3. I17 Water at a flow rate of Q = 0.2 m3/s is supplied to the cylindrical water tank of diameter 1 m discharging as a jet through a round pipe of length 4 m and diameter 15 cm. How deep will the water in the tank be? Assume steady state (Fig on next page). I18* This two-dimensional gate contains a two-dimensional nozzle (slot). For a flow rate of 7.5 m3/s/m in the channel, predict the flow rates through the slot (free jet) and under the gate.

Fig. I15

Fig. I16

Fig. I14


32

I19* Water is flowing at 0.3 m3/s. Calculate the manometer reading a) using the sketch as shown, and b) when the pitot tube is at section 2 and the static pressure connection is at section 1. I20 Calculate the flowrate through this nozzle. I21 Calculate the flow rate through this pipeline.

Fig. I19

Fig. I17

Fig. I18

Fig. I21

Fig. I20


33

I22 Calculate the pump power.

I23 This turbine develops 75 kW when the flowrate is 0.6 m3/s. What flowrate may be expected if the turbine is removed? I24 Water is pumped from a large lake into an irrigation canal of rectangular cross section 3 m wide, producing the flow situation shown. Calculate the required pump power assuming ideal flow.

Fig. I22

Fig. I23

Fig. I24


34

I25 A water pump has an inlet and two outlets, all at the same elevation. What pressure does the gage to the right of the pump register? What pump power is required if the pump is 85% efficient. Neglect pipe losses.

I26 Compute the pump power required to maintain a flowrate of 0.1 m3/s if the barometric pressure is 98.6 kPa and the vapour pressure is 6.9 kPa. Calculate the maximum possible distance, x, for reliable operation. (NOTE El. = elevation)

I27 If cavitation is observed in the 50 mm section, what is the flowrate? Barometric pressure is 100 kPa. I28 A pump of what power is required to pump 0.56 m3/s of water from a reservoir of surface elevation 30 to one of surface elevation 75, if in the pump and pipeline 12 metres of head are lost? I29 In a 225 mm pipeline 0.14 m3/s of water are pumped from a reservoir of surface elevation 30.0 m over a hill of elevation 50.0. A pump of what power is required to maintain a pressure of 345 kPa on the hilltop if the head lost between reservoir and hilltop is 6 m?

Fig. I25

Fig. I26

Fig. I27


35

Momentum equation I30 A 100 mm nozzle is bolted (with 6 bolts) to the flange of a 300 mm horizontal pipeline and discharges water into the atmosphere. Calculate the tension load on each bolt when the pressure in the pipe is 600 kPa. Neglect vertical forces. I31 A 100 by 50 mm 180o pipe bend lies in a horizontal plane. Find the horizontal force of the water on the bend when the pressures in the 100 mm and 50 mm pipes are 105 kPa and 35 kPa respectively. I32 Assuming uniform velocity profiles, find the force F needed to hold the plug (cylindrical shape) in the pipe. Neglect viscous effects.

I33 A conical diverging tube is horizontal, 0.3 m long, has 75 mm throat diameter, 100 mm exit diameter, and discharges 28.3 l/s of water into the atmosphere. Calculate the magnitude and direction of the force components exerted by the water on the tube. I34 Calculate the force on the bolts. Water is flowing. Horizontal pipe.

Fig. I32

Fig. I34


36

I35* Two types of gasoline ( = 680 kg/m3) are blended by passing them through a horizontal wye. Calculate the magnitude and direction of the force exerted on the wye by the gasoline. The pressure p3 = 145 kPa. I36* Determine the total force and its direction on the horizontal T-section. Neglect viscous effects. Water is flowing. I37 This corrugated ramp is used as an energy dissipator in a two-dimensional open channel flow. For a flowrate of 5.4 m3/(sm) calculate the head lost, the power dissipated, and the horizontal component of force exerted by the water on the ramp.

Fig. I36

Fig. I37

Fig. I35


37

I38 Assuming hydrostatic pressure distributions and uniform velocity profiles at some short distance upstream and downstream the gate, find the force on the sluice gate. Neglect viscous effects. I39 The passage is 1.2 m wide normal to the paper. What will be the horizontal component of force exerted by the water on the structure? I40 Calculate the magnitude and direction of the vertical and horizontal components and the total force exerted on this stationary blade by a 50 mm jet of water moving at 15 m/s.

Fig. I38

Fig. I40

Fig. I39


38

Correction coefficients for momentum () and velocity head () I41 If the velocity profiles at the upstream and downstream ends of the mixing zone of a jet pump may be approximated as shown, and wall friction may be neglected, calculate the rise of pressure from section 1 to section 2, and the power lost in the mixing process. Water is flowing.

I42 Just downstream from the nozzle tip the velocity distribution is as shown. Calculate the flowrate past section 1, , , and the momentum flux. Assume water is flowing.

Fig. I41

Fig. I42


39

J PIPE FLOW (* denotes more difficult problems)

Laminar pipe flow J1 Glycerin at a temperature of 20C flows at a rate of 810-6 m3/s through a horizontal tube with a 30 mm diameter. What is the pressure drop in Pa per 10 m? The fluid properties for glycerin are given in the appendix below. J2 (exam 990116) A capillary tube of inside diameter 5 mm and length 4.3 m connects tank A to open container B. The liquid in A, B, and capillary CD is water having a density of 1000 kg/m3 and dynamic viscosity of 1.510-3 Pas. The gage pressure is pA = 35 kPa (relative). In which direction will the water flow? What is the flow rate? Assume steady flow and that all local losses may be neglected.

J3* (exam 990828) A hypodermic needle has an inside diameter of 0.3 mm and is 60 mm in length. If the piston moves to the right at a speed of 18 mm/s and there is no leakage, what force is needed on the piston? The medicine in the hypodermic has a viscosity of 0.9810-3 Pas and its density is 800 kgm-3. Consider the steady state flows in cylinder and needle. Neglect all local losses (like exit losses from the needle as well as losses at the juncture of the needle and cylinder) and the friction losses in the cylinder.

Fig. J2

Fig. J3


40

Turbulent pipe flow J4 If the turbulent velocity profile in a pipe 0.6 m in diameter may be approximated by v = 3.56 y1/7 (v in m/s, y in m) and the shear stress in the fluid 0.15 m from the pipe wall is 6.22 Pa, calculate the eddy viscosity, mixing length, and turbulence constant at this point. The density of the fluid is 900 kg/m3. J5 The thickness of the viscous sublayer is usually defined by the intersection of a laminar velocity profile (adjacent to the wall) with a turbulent velocity profile. If the former may be described by v = c1y and the latter by v = c2y1/7, derive an expression for the thickness of the film in terms of c1 and c2. J6 A horizontal rough pipe of 150 mm diameter carries water at 20C. It is observed that the fall of pressure along this pipe is 184 kPa per 100 m when the flowrate is 60 l/s. What size of smooth pipe would produce the same pressure drop for the same flowrate? J7 Water flows through a section of 300 mm pipeline 300 m long running from elevation 90 to elevation 75. A pressure gage at elevation 90 reads 275 kPa, and one at elevation 75 reads 345 kPa. Calculate head loss, direction of flow, and shear stress at the pipe wall and 75 mm from the pipe wall. If the flowrate is 0.14 m3/s, calculate the friction factor and friction velocity. J8 Water flows in a smooth pipeline at a Reynolds number of 106. After many years of use it is observed that half the original flowrate produces the same head loss as for the original flow. Estimate the size of the relative roughness of the deteriorated pipe. J9 A 75 mm smooth brass pipeline 30 m long carries 380 l/min of crude oil. Calculate the head loss when the oil is at a) 25C, b) 45C. J10 A 0.3 m pipeline 3.2 km long runs on an even grade between reservoirs of surface elevations 150 and 120, entering the reservoirs 10 m below their surfaces. The flowrate through the line is inadequate, and a pump is installed at elevation 125 to increase the capacity of the line. Assuming f to be 0.020, what pump power is required to pump 0.17 m3 /s downhill through the line? Sketch accurately the energy line before and after the pump is installed. What is the maximum dependable flowrate that may be obtained through the line? J11 Calculate the smallest reliable flowrate that can be pumped through this pipeline. D = 25 mm, f = 0.020, L = 2 x 45 m, Vertical distances are 7.5 m and 15 m respectively. Assume atmospheric pressure 101.3 kPa.

Fig. J11


41

Minor head losses J12 Water is flowing. Calculate the gage reading when V300 is 2.4 m/s. (NOTE El. = elevation)

J13 Experimental determination of minor losses and loss coefficients are made from measurements of the hydraulic grade lines in zones of established flow. Calculate the head loss and loss coefficient for this gradual enlargement from the data given in the figure.

Fig. J12

Fig. J13


42

J14 An irrigation siphon has the dimensions shown and is placed over a dike (Fig. b). Estimate the flowrate to be expected under a head of 0.3 m. Assume a re-entrant entrance (c.f. fig. J14a), a friction factor of 0.020, and bend loss coefficients of 0.20.

J15 A 300 mm pipeline (f=0.020) is horizontal and 60 m long and runs between two reservoirs. It leaves the high-level reservoir at a point 36 m below its surface and enters the low-level reservoir 6 m below its surface. (a) Assume standard square-edged entrance (see Fig. J14a) and exit and compute the flowrate to be expected. (b) A nozzle of 225 mm tip diameter is now attached to the downstream end of the pipe; neglecting head losses in the nozzle, what reduction of flowrate will be expected? (c) The nozzle is now replaced with a diffuser tube of 375 mm exit diameter. Assuming that the diffuser tube flows full and the losses therein may be neglected, what increase of flowrate is to be expected? J16 Water is circulated from a large tank, through a filter, and back to the tank. The power added to the water by the pump is 400 W. Determine the flowrate through the filter. The pipe has a length of 60 m, a diameter of 30 mm, and an equivalent sand roughness of 0.06 mm. The water temperature is 20C.

Fig. J14a

Fig. J14b

Fig. J16


43

Pipes in series J17 If the water surface elevation in reservoir B is 110 m, what must be the water surface elevation in reservoir A if a flow of 0.03 m3/s is to occur in the cast-iron pipe (ks = 0.25 mm)? Neglect local losses.

Pipes in parallel

J18 A 0.6 m pipeline branches into a 0.3 m and a 0.45 m pipe, each of which is 1.6 km long, and they rejoin to form a 0.45 m pipe. If 0.85 m3/s flow in the main pipe, how will the flow divide? Assume that f = 0.018 for both branches. J19 A straight 300 mm pipeline 5 km long is laid between two reservoirs of surface elevations 150 and 105 entering these reservoirs 9 m beneath their free surfaces. To increase the capacity of the line, a 300 mm line 2.5 km long is laid from the original lines midpoint to the lower reservoir. What increase in flowrate is gained by installing the new line? Assume that f = 0.020 for all pipes. J20 If the pump supplies 225 kW to the water, what flow rate will occur in the pipes? Assume a friction factor of 0.02 for both pipes. Neglect local losses. (NOTE El. = elevation)

Fig. J20

Fig. J17


44

J21 Determine the flow distribution of water in the parallel piping system. Pipe characteristics are given in the table below. Here K stands for the total local loss coefficient in each pipe. Pipe L(m) D(mm) f K 1 30 50 0.020 3 2 40 75 0.025 5 3 60 60 0.022 1

Branching pipes

J22 A 900 mm pipe divides into three 450 mm pipes at elevation 120. The 450 mm pipes (length, see table) runs to reservoirs with elevations according to table. When 1.4 m3 /s flows in the big pipe, how will the flow divide? Assume f = 0.017 in all pipes. Reservoir Elevation (m) Pipe length (m) A 90 3200 B 60 4800 C 30 6800 J23 Three reservoirs are connected with pipes via a connection point O at elevation 120. With the data according to the table calculate the flowrates in the lines. Assume f = 0.020 Reservoir Elevation (m) Pipe length (m) diameter (mm) A 150 1600 300 B 120 1600 200 C 90 2400 150 J24 A 0.3 m pipeline 600 m long leaves a reservoir of surface elevation 150 and runs to elevation 120, where it divides into two 150 mm lines each 300 m long, both of which discharge into the atmosphere, one at elevation 135, the other at elevation 105. Calculate the flow rates in the three pipes if all friction factors are 0.025. Neglect local losses.

Fig. J21


45

J25* For the system shown below, determine the water flow rates in each pipe and the piezometric head at the junction. The pump characteristic curve is 23020 QH p = and the water level in the reservoirs are z1 = 10 m, z2 = 20 m z3 = 18 m. Pipe characteristics are given in the table below where K stands for the total local loss coefficient in each pipe. Pipe L(m) D(mm) F K 1 30 240 0.020 2 2 60 200 0.015 0 3 90 160 0.025 0

Unsteady pipe flow J26* A pump is used to fill a tank from a reservoir. The head provided by the pump is

given by )1( 2max

2

0 QQhhp = , where h0 is 50 m, Q is the discharge through the pump

and Qmax is 2 m3/s. The friction factor f = 0.015 and the pipe diameter is 900 mm. Initially the water level in the tank is the same as the level in the reservoir. The cross-sectional area of the tank is 100 m2. How long time will it take to fill the tank to a height, h, of 40 m if the water level in the reservoir is assumed to be constant with time? Assume quasi-stationary flow through pipe and pump, i.e., that acceleration effects are negligible.

Fig. J25

Fig. J26


46

Pumps J27 Water is to be pumped by two centrifugal pumps through a pipeline connecting two reservoirs. The pumps can be operated in parallel or one at a time. The pipeline is 2000 m long, diameter 250 mm and equivalent sand roughness is 0.2 mm. The static lift is 25.0 m. Calculate the specific energy consumption (kWh/m3) both for single pump operation and when both pumps operate. The pump characteristics are Discharge Q (m3/s) 0.020 0.030 0.040 0.050 Head H (m) 55 47 35 20 Total efficiency (%) 78 80 72 60 J28 Water is pumped between two reservoirs with the same water surface elevation zo. Total pipe length is L = 2500 m, diameter D = 0.1 m and equivalent sand roughness k = 0.0001 m. The pump characteristics are given by the figure. What is the maximum permissible distance x from the upstream reservoir to the suction side of the pump, if the pressure must not be less than atmospheric. The local losses may be neglected. The temperature is 20C.

Fig. J28


47

J29 A pipeline is 3200 m long, with diameter 0.25 m and equivalent sand roughness k = 0.1 mm. The desired flow is 0.055 m3/s with two identical pumps running in parallel. a) There are four different pumps to choose between with characteristics according to the diagram. Which pump is to be chosen in order to meet the requirements? b) What pressure will there be at the maximum height after 1000 m when only one pump is running?

J30 Water (20C) is pumped between two reservoirs through two identical, parallel pipes each with a diameter of 0.2 m, length 1000 m, and equivalent sand roughness of 410-4 m. a) What flow is expected through the pump? b) How much energy (kWh) is needed to pump 1 m3 of water? The efficiency of the pump = 0.75

Fig. J29

Fig. J30


48

J31 Water is pumped between two reservoirs. The pump characteristics is shown in the figure. (a) What will the flowrate be if the valve is completely open? (b) To what degree must the valve be closed [expressed in terms of a head loss coefficient, kv (hv= kv(U2/2g))] for the flowrate to be reduced by 50%? J32* Cooling water to a thermal power plant is being pumped by three pumps operating in parallel (pump characteristics for one pump is given in the table below) through a 2000 m long, rectangular tunnel with equivalent sand roughness ks = 0.05 m. The tunnel starts and ends in the same sea. In the power plant, the water flows through a condenser at a level of +4 m. The condenser can be seen as a local loss, being 1.0(U2/2g) where U is the velocity in the tunnel. What will the pressure be at the end of the condenser (at a level of +4 m)? Account for inflow and outflow losses as well as friction in the tunnel. Assume rough turbulent flow. Discharge Q (m3/s) 0 5 10 15 20 Head H (m) 3 2.7 2.3 1.5 0.5

Fig. J31

Fig. J32


49

K URBAN HYDROLOGY AND WASTE WATER ENGINEERING (* denotes more difficult problems) K1 Rainstorms in urban areas lead to that large volumes of polluted storm water are lead to the treatment plant. One way to decrease this storm water runoff is to decrease the amount of impermeable surfaces within the city. One example of this is green roofs, i.e., roofs covered with a thin soil layer with low vegetation. Some of the rainwater falling on the green roofs will be stored in the soil layer and eventually evaporate and the runoff will be delayed. In Augustenborg, Malm, the runoff from a 1 x 4.7 m green roof was collected in a barrel with a diameter of 0.8 m. The rainfall and the water level in the barrel was recorded every 5 min. Calculate and plot the runoff from the green roof and from a corresponding ordinary roof where Q=P. Also, calculate how much the total runoff volume and the peak flow are reduced.

Time Rainfall Water level in the barrel [mm] [mm]

17:15 0.1 24.2 17:20 0.2 24.2 17:25 0 24.2 17:30 0.1 24.2 17:35 0 24.2 17:40 0.4 24.2 17:45 0.5 24.2 17:50 0.7 24.2 17:55 0.4 24.5 18:00 0.3 25.3 18:05 0.3 26.1 18:10 0.3 27 18:15 0.2 28 18:20 0.2 28.8 18:25 0.3 29.8 18:30 0.3 30.6 18:35 0.2 31.6 18:40 0.4 32.5 18:45 0.3 33.6 18:50 0.6 34.7 18:55 0.7 36.3 19:00 0.1 38 19:05 0.1 39.4 19:10 0 40.5 19:15 0 41.3 19:20 0 41.8 19:25 0 42.3 19:30 0 42.7 19:35 0 43 19:40 0 43


50

K2 a) A small rural catchment is situated along a stream. The outflow from the catchment consists of groundwater flow to the stream and surface runoff. Give the water balance equation for the catchment (no calculations). b) There are plans to urbanise the catchment. Give the annual water balance equation for the urbanised catchment. Which terms in the equation are changed and how are they changed (calculations needed)? Information about the catchment is given in the table below. c) How will the groundwater table change after urbanisation (calculations needed)? Information about the catchment, some further assumptions might be needed. area 1 km2

surface runoff before urbanisation 5% of the annual precipitation annual precipitation before urbanisation 600 mm annual evaporation before urbanisation 400 mm groundwater information unconfined aquifer, impermeable bedrock

at 15 m depth soil type sand fc 0.11 m

3 m-3 s 0.41 m

3 m-3 wp 0.04 m

3 m-3 K 1.1*10-4 m/s percentage impermeable surfaces after urbanisation

55%, the storm water runoff is led through a separated system to the stream

K3 A new industrial area is 25000 m2 large. It consists of 45 % impermeable surfaces. Conduits from the area is connected to a manhole A just downstream the area. If flooding only can be allowed once every five years in average, what is the design flow? Assume that the time of concentration for A is 10 minutes. Use the rational method. K4 a) Use the Manning formula to calculate the time for surface runoff from an asphalt surface with a slope of 1/30 when the rainfall intensity is 1 mm/min. The runoff length is 10 m. b) Calculate the surface runoff time for the above surface when the rainfall intensity is doubled.


51

K5 Calculate the design flow (in A) from the urban area in the figure below. The area is divided into three subareas. The concentration time for each area, surface runoff + pipe flow to the manholes B, C, and D respectively, is five minutes. The distance between A-B, B-C, and C-D is 300 m. Assume that the pipes A-D are sufficiently large to carry a 10-year flow. Subarea B is 9 ha, C is 12 ha and D is 7 ha. The runoff coefficient is 0.6, 0.7, and 0.4 for area B, C, and D, respectively. a) Calculate the design flow in A using the rational method for a 5 year rainfall. b) Calculate the design flow in A using the time-area method for a 5 year rainfall. c) If the area receives a 10 year rainfall with a duration of 15 min, calculate the volume of flooding at A. Assume that the pipe leading from A can handle a flow exactly equivalent to a 5-year rain.

A

C

B

D


52

ANSWERS A1 320 mm A2 0.042 m3/s, 0.23 m A3 253 mm A4 384 mm A5 a) 0.27 b) 119 * 103 m3 c) +254 mm A6 a) 0.287 m3/s, 174 mm/year, 5.5 l/(s km2) b) 0.25 B1 p = 22 Mpa B2 1 = 0.97 Pa s, 2 = 1.93 Pa s B3 0.60 Pas B4 66.8 N B5 T = D4/32h B6 P = 50.3 kW B7 h = 0.030 m B8 Max. rise is 15 mm B9 0.125 N/m C1 a) 20 mm b) 22 mm C2 a) 37, 46, 61, and 75 mm/h c) 82, 33, 21 mm/h C3 a) 16.2 mm b) 15.7 mm C4 180 mm C5 a) p=1/5, b) 0.042 m3/s C6 a) 0.79, b) 5.3 mm C7 a) 3.48, b) 2.02, c) 1.29, d) 1.77, e) 1.5 mm per day C8 a) 28.32 mm Hg, 37.67 mbar, 3.77 kPa b) 8.5 mm Hg, 11.31 mbar, 1.13 kPa c) 19.82 mm Hg, 26.37 mbar, 2.64 kPa d) 22.0 C C9 4.0 mm/d C10 a) 114 mm, b) 0.341 m3/s, c) The water consumption is of the same order of

magnitude as the evaporation. C11 See solution C12 a) 145 mm b) decrease, c) increase, d) decrease D1 a)2.08 mm/h, b) 22.5 mm D2 a) 23.6, 16.7, and 9.9 mm/h, b) 7.2, 12.2, and 18.5 mm D3 a) f0 = 0.41 mm/min, fc = 0.086 mm/min, k = 0.062 min-1, b) 8 mm D4 a) see solution. D5 -index = 5.1 mm/h D6 a) 1760 kg/m3, b) 135 cm3, c) 65 cm3, d) 0.32, e) 0.14 g/g, f) 0.25 m3/m3 D7 a) porosity = 0.423 m3 m-3 initial = 0.31 m3 m-3, b) effective porosity = 0.15 m3 m-3

field capacity = 0.273 m3 m-3, c) 173 mm, d) 1903 m3/ha D8 is 0.26 m3 m-3 or 0.16 g/g, porosity is 0.40 m3 m-3 D9 a) effective porosity 0.155 m3 m-3, field capacity 0.275 m3 m-3, b) 0.0875 m, c) 25500

m3 D10 a) 75 mm, b) 0 mm, c) in a) the groundwater table will rise 0.3 m, in b) no change E1 0.20 E2 1.66, 2.15 and 2.15 *10-4 m/s E3 K = 1 *10-3 m/s


53

E4 The flow direction is from A to B The shortest time for a substance to travel 100 m is 14 years.

E5 rw = 0.32 E6 5.7 l/s E7 K = 1.0 10-5 m/s, T = 2.0 10-4 (m2/s) E8 a) see solution b) 4.6 m3/day F1 a) 1.6 *106 m3, b) 24.5 mm F2 a) 0,15,10,5, and 0 m3/s, b) 0, 15, 40, 25, 10, and 0 m3/s, c) 1, 16, 41, 26, 11, and 1

m3/s F3 a) 0.1 m3/s, b) The base flow increases from (t = 2, Q = 0.1) to (t = 4, Q = 0.25), after

this point Q = Qbase F4 Flow after 120 days is 7.9 m3/s. F5 a) 11:00-22:30, b) see solution, c) 511560 m3, d) Peff = 51 mm. F6 a) 54.9 km2, b) see solution. F7 Maximum flow is 360 l/s after 7h. F8 see solution. F9 see solution. F10 see solution. G1 a) 0.01, b) 0.99, c) 0.9801, d) 0.0199, e) 0.0956 G2 0.17, 0.33, 0.50, 0.67, 0.83 G3 Q10 = 425 m3/s, Q50 = 550 m3/s, and Q100 = 630 m3/s G4 a) Q10 = 32 m3/s, Q100 = 49 m3/s, b) Q10 = 32 m3/s, Q100 = 47 m3/s H1 3.22 = 10.4 kN/m3, p3.22 = 33060 kPa H2 Pressure at 7.62 km is a) 9.7 kPa, b) 41.0 kPa, c) 35.6 kPa H3 px = 113.8 kPa H4 px-py = 5.15 kPa H5 380 mm H6 2.81 kPa H7 px-py = 62.0 kPa H8 y = 6.45 m H9 r.d. oil = 0.90 H10 F = 44.5 kN, lp = 2.16 m H11 F = 173.4 kN, lp = 3.08 m H12 d = 7.7 m H13 2.31 m H14 D = 2.60 m H15 Ftot = 184 kN, = 40.8 grad (through the axis of the quarter circle) H16 1/2 = 1/3 H17 Additional force is 3.27 MN H18 d = 3 (2h2-1h1)/(2+1) I1 Re = 31 800 -> turbulent flow I2 V = 3.9 mm/s I3 Dmin = 3.6 m I4 V75 = 10 m/s, V75/V150 = 4 I5 Q3 = 75 l/s, 3 = 1010 kg/m3 I6 3 m/s and 8 m/s I7 151 l/s I8 4.47 m/s I9 h = 1.5 m (above B)


54

I10 Q = 0.0082 m3/s, h = 0.40 m I11 D = 235 mm I12 A2,max = 0.022 m2 I13 Pressure in bend is 32.4 kPa, the flowrate is 3.4 l/s. I14 2.23 m I15 54.3 mm I16 q1 = 2.39 m3/s, q2 = 1.94 m3/s, Q3 = 0.45 m3/s, I17 2.53 m I18 qslot = 1.28 m3/s/m, qgate = 6.22 m3/s/m I19 a) manometer reading = 7.88 m, b) manometer reading = 1.56 m I20 240 l/s I21 60.6 l/s I22 P = 33.1 kW I23 Q = 1.26 m3/s I24 P = 383 kW I25 361.2 kPa, 28.5 kW I26 Pump power is 40.3 kW and max pump distance is 23.8 m I27 The flowrate is 34.1 l/s. I28 P = 313 kW I29 Pump power is 85 kW I30 F = 5.65 kN (in each bolt) I31 Fx = 1.26 kN I32 50 N I33 Force components: Horizontal: 17.3 N acting towards the right, Vertical: 17.8 N

downwards I34 9.0 Kn I35 1493 N directed downward right at 83 angle with the horizontal I36 766 N directed upwards-right at an angle 82.3 with the horizontal I37 hf = 1.4 m H2O, P = 73.8 kW/m, F = 14 kN/m I38 618 kN I39 509 N I40 Fx = 695 N, Fz = 95 N, Ftot = 701 N I41 p = 89.3 kPa, Ploss = 32.7 kW I42 0.166 m3/s, = 1.41, = 1.12, Momentum flux = 1740 N J1 6.1 kPa J2 The flow is directed from tank A to container B and the flowrate is 0.003 l/s. J3 The force needed on the piston is 2.25 N J4 = 2.40 Pa s, l = 0.032 m, = 0.214 J5 ysublayer = (c2/c1)7/6 J6 d = 0.115 m J7 hL = 7.9 m H2O, wall = 19.3 Pa, 75 = 9.64 Pa, f = 0.039, v* = 0.139 m/s J8 Relative roughness is 0.019 J9 Head loss is a) 1.19 m, b)1.08 m J10 Pump power is 54.9 kW, max. flowrate is 179 l/s J11 Q = 1.3 l/s J12 175.2 kPa J13 Head loss = 0.39 m H2O, Loss coefficient = 0.05 J14 Flowrate is 10.7 l/s J15 a) 0.73 m3/s, b) 0.62 m3/s, c) 0.77 m3/s J16 1.7 l/s J17 114 m J18 Q1 = 0.23 m3/s, Q2 = 0.62 m3/s


55

J19 Increase in flowrate = 27% or 31 l/s J20 100 l/s and 175 l/s J21 Q1 = 2.4 l/s, Q2 = 4.8 l/s, Q3 = 2.8 l/s J22 QA = 0.462 m3/s, QB = 0.473 m3/s, QC = 0.465 m3/s J23 Flowrates are in AO: 84 l/s, BO: 52 l/s, CO: 32 l/s J24 Q1 = 104 l/s, Q2 = 34 l/s, Q3 = 70 l/s J25 Q1 = 198 l/s, Q2 = 138 l/s, Q3 = 60 l/s, piezometric head at junction is 23.5 m J26 46.4 min. J27 Espec1 = 0.126 kWh/m3, Espec2 = 0.153 kWh/m3 J28 x < 210 m J29 a) choose pump b, b) Pressure is approx. 43 kPa. J30 a) Qpump = 0.065 m3/s, b) Energy needed to pump 1 m3 of water is 0.062 kWh J31 a) 8 l/s, b) kv = 3545 J32 -24.2 kPa K1 see solution K2 see solution K3 187 l/s K4 a)77 s, b) 58 s K5 a) 2.28 m3/s, b) 2.35 m3/s, c) 102 m3.


56

SOLUTIONS Section A A1 (P+Qin+GVin)-(E+Qout+GVout) = change in volume assume no change in volume and no net exchange between groundwater and lake: E = P+Qin-Qout Use mm over the lake as a common unit. Qin = 1.39*3600*24*365/14.3*106 = 3065.4 mm Qout = 1.56*3600*24*365/14.3*106 = 3440.3 mm E = 695 + 3065.4-3440.3 = 320 mm A2 Q = P*A = 2500*60*10-3/3600 = 0.042 m3/s If v = 1 m/s, then the cross sectional area will be 0.042 m2. The diameter of the pipe is then 2*(0.042/)0.5 = 0.23 m. A3 Water budget for the lake: (Qin + P) - (Qout + E) = S /t (Assuming exchange with groundwater is negligible) Use as common unit: m/(May - Aug); May - August = 123 days

Qin = (17.0 + 13.0 + 15.0) (123243600) / 100106 = 4.784 P = 0.100 Qout = 43 (123243600) / 100106= 4.571 S /t = 571.10 - 571.04 = 0.060

E = 4.784 + 0.100 - 4.571 - 0.060 = 0.253 Answer: Evaporation during this period is 253 mm A4 A=7.08 105 m2; Qin=1,5 m3/s; dM=+5.35 105 m3; P=225 mm; E=? Assuming exchange with groundwater is negligible Water balance: P+Qin-E-Qout=dM E=0.225+(1.5-1.25)*3600*24*30/7.08 105-5.35 105/7.08 105=0.384 m Answer: 384 mm


57

A5 Catchment with retention reservoir. June. A = 200 ha; Ares = 2.0 ha; Qout = 0.045 m3/s; P =176 mm; Qin = 95 103 m3;E = 2.8 mm/day -------------- a) Runoff coefficient? a = Q/P = Qin/P = 95 103/(0.176*200*104) = 0.27 b) June -91. P = 220 mm./ Qin = ? Assume (*) that the runoff coefficient is generally valid for June.

Qin = a P = 0.27*0.220*200104 = 119 103 (m3) (*) The assumption is open for discussion. c) Hres. ? Water budget : (Qin+PAres) - (Qout+EAres) = Hres Ares

Hres= (Qin-Qout)/Ares +P - E Hres = (119 103- 0.045*3600*24*30)/2.0 104 + 0.220 - 30*2.810-3 Hres = 0.118 + 0.220 - 0.084 = + 0.254 (m)

Answer : a) a=0.27; b) Qin = 119103 m3; c) Hres = + 254 mm A6 a) According to Vattenfring i Sverige; Qmean=287 dm3/s = 0.287 m3/s A=52 km2, Q(1 year)/A=(0.287*3600*24*365)/52 106 = 0.174 m/year = 174 mm/year According to Vattenfring i Sverige; Qmean = 5.5 l/(s km2) b) Runoff coefficient (a) Annual precipitation = 700 mm (from Appendix) a=Q/P=174/700 = 0.25


58

Section B B1 Basic equation: E = -dp/(dV/V) Our case: dV/V = -0.01 E = 2.17109 Pa

dp = 2.17109 0.01 = 2.17107 Pa Answer: Extra pressure needed is 21.7 MPa B2 F F F

Vh

AV

hA

tot= + = +

1 2 1 22 2

/ /

Since 1 = 2 2 Ftot = A 2V/h (22 + 2) With V = 0.3 m/s, Ftot = 29 N/m2, h = 0.06 m 2 = (290.06)/ (20.33) = 0.97 Answer: 2 = 0.97 Pas, 1 = 1.94 Pas B3 Since the plate is moving with a constant velocity, the net force on the plate must be equal to zero. Therefore the gravitational force component of the plate, G, acting along the sloping surface must be equal to the frictional force, F. Thus we have

AhVAFG ===

where h (=1.25 mm) is the thickness of the oil film, A (= 1 m2) is the area of the plate, V (= 0.2 m/s) is the plate velocity, and is the viscosity of the oil. G = (5/13)250 = 96.2 N = 0.60 Pas

Answer: = 0.60 Pas B4 Force balance (constant lift velocity net force on plate equal to zero):

F = Flift +Fbuoyancy Fweight Ffriction = 0

Fbuoyancy = Vplate = 9.819501.51.50.0016 = 33.6 N Fweight = 45 N

Ffriction = A = 2(U/h)A = 22.4[0.06/((0.025-0.0016)/2)]1.51.5 = 55.4 N

Flift = 55.4 + 45 33.6 = 66.8 N Answer: 66.8 N


59

B5 Start with the torque on an infinitesimal area dA at the radial distance r from the centre of rotation.

( )dT dF r dA r rh

rdr r= = = ( )

2 (1)

The expression (r/h) is the velocity gradient since V = 0 at the wall and V = Vdisc = r near the disc. We assume linear velocity variation. (1) dT = 2 (/h) r3 dr

T dTh

dd= =

( )

/

214 20

2 4

Answer : Tdh

=

32

4

B6 Power = P Linear motion: P = Force velocity Rotation: P = Torque angular velocity

T dT dF d dA dBBB

= = = ( / ) ( / )2 2000

(1)

= =constdh

( / )2 (2)

dA d db= (3) Everything constant in axial direction integration can be substituted by multiplication by B Eqs 1, 2, 3

( )( )T B dh

d d Bh

d= =

( / )/

22

43 (4)

P = T and (4) P = /4 0.72(1/0.2310-3) 0.363 (200 1/60 2)2 = 5.03104 Answer: Power lost is 50.3 kW


60

B7 Vertical force balance between surface tension force and weight of water column gives 2 rcos = ghr2 h = (2cos)/(gr) = surface tension for water = 0.073 N/m our case: h = (20.0731) /( g 0.0005) = 0.0298 Answer: = Maximum rise = 30 mm B8 Vertical force balance between surface tension force and weight of water column gives (cos = 1)

h = (2)/(d) = surface tension for water = 0.073 N/m Our case: h = (20.073) /( g 0.001) = 0.0148 Answer: Maximum rise = 15 mm B9 pi -po = (1/R1 + 1/R2)

Special case of spherical bubble: R1 = R2=R (where R is the radius of the bubble). N:B a bubble has two surfaces. pi -po = (22)/R

pi-po = 20 Pa, R = 50/2 = 25 mm = 0.125 N/m

Answer: The tension in the soap film is 0.125 N/m


61

Section C C1 a) Arithmetic mean, use only gauges within the catchment, P = 20 mm. b) The area influenced by each gauge is determined. The area for gauge A is outside the catchment. The area for B is 4*3 km and the area for gauge C is 1*3 km. The areal precipitation is calculated as a weighted average. P = (12*20+3*30)/15 = 22 mm. C2 a) Convert from l/(s ha) to mm/h 1 l/(s ha) = 0.001/10000*1000*3600 = 0.36 mm/h Use the idf curve to find the rainfall intensities for the different return periods T = 1 year gives i = 1538/(10+6.9)+10,7 = 101.7 l/(s ha) or 101.7*0.36 = 37 mm/h T = 2 years gives i = 46 mm/h; T = 5 years gives i = 61 mm/h; T = 10 years gives i = 75 mm/h; b)

c)T = 5 years and Tr = 5, 30, and 60 min gives i = 82, 33, and 21 mm/h C3 a) For the arithmetic averaging method, only the gages that are within the watershed are used, in this case gages B and D. (17.1+15.3)/2=16.2 mm b) The watershed is divided using the Thiessens polygon method according to the figure below.

0

10

20

30

40

50

60

70

80

0 2 4 6 8 10 12

inte

nsity

(mm

/h)

Return period (Y)


62

The area of each polygon within the watershed relative to the total area of the watershed is calculated. Gage P [mm] relative area, Ar P*Ar A 13 0.06 0.78 B 17.1 0.24 4.10 C 22.9 0.07 1.60 D 15.3 0.46 7.04 E 14.1 0.03 0.42 F 12.6 0.09 1.13 G 11.6 0.05 0.58 Sum 15.66 The average rainfall is given by the sum of the P*Ar column; 15.7 mm C4 Diameter of tube 0.046 m Area of tube 0.00166 m2

depth (cm) Weight (g) Density (kg/m3) 91 299 197.7 87 86 84 92 107 320 180 103 100 98 90 87 290 200.6

average depth average density 93.18181818 192.7

Water equivalent = 0.9318*192.7 = 180 mm


63

C5 a) 10 mm/10 min = 0.01/(10*60)*103 * 104 = 167 l/(s ha) IDF- curve gives return time (T) around 5 years The probability for a more intense rainfall during a given year is thus 1/5 b) Flow = A*i = 2500*0.01/(10*60) = 0.042 m3/s C6 a) Elake = k*Epan k = 123/156 = 0.79 b) Elake = 0.79*6.7 = 5.3 mm C7 From Appendix: Ra = 8.7 (linear interpolation 10.4+(8.5-10.4)*(59-50)/(50-60)) Ta4 = 12.9 ea = 9.21, ed=ea*RH = 9.21*0.65 = 5.99 / = 1.26 Convert m/s to miles/day: u2=2*3600*24/1609 = 107.4 (Equation no. in Shaw) RI(1-r)=0.95*8.7*(0.18+0.55*0.4) = 3.31 (11.14) Ro = 12.9*(0.56-0.09*(5.99)0.5)*(0.1+0.9*0.4) = 2.02 (11.15) H = 3.31-2.02 = 1.29 (11.13) Ea = 0.35(0.5+107.4/100)(9.21-5.99) = 1.77 (11.16) Eo = (1.26*1.29+1.77)/(1.26+1) = 1.50 mm (11.12) Answer: a) 3.31/0.95 = 3.48, b) 2.02, c) 1.29, d) 1.77, e) 1.5 mm per day. C8 Air at T = 28 C, RH = 70 % a) Use enclosed table es =28.32 mmHg (=37.67 mbar = 3.77 kPa) b) Saturation deficit = es -e = (1- 70/100) 28.32 = 8.50 mmHg (= 11.31 mbar = 1.13 kPa) c) e = 70/100 28.32 = 19.82 mmHg = 26.37 mbar = 2.64 kPa d) Dew point temperature. Use the table to find at which temperature es = 19.82 mmHg Td = 22.0 C C9 From Appendix: Ra = 16 (linear interpolation 16.1+(16.1-15.6)*(52-50)/(50-60)) Ta4 = 14.4 ea =15.49, ed=ea*RH = 15.49*0.5 = 7.745 / = 1.99


64

Convert m/s to miles/day: u2=1.2*3600*24/1609 = 64.44 (Equation no. in Shaw) RI(1-r)=0.95*16*(0.18+0.55*0.5) = 6.916 (11.14) Ro = 14.4*(0.56-0.09*(7.745)0.5)*(0.1+0.9*0.5) = 2.451 (11.15) H = 6.916-2.451 = 4.4645 (11.13) Ea = 0.35(0.5+64.44/100)(15.49-7.745) = 3.10 (11.16) Eo = (1.99*4.46+3.1)/(1.99+1) = 4.0 mm (11.12) Answer: The potential evaporation is 4.0 mm per day. C10

a) Calculate average values of T, RH, etc, use the averages of the

measurements at 7 13 and 19. For example T = 17,3 C

Lund is approximately on lat. 56. Ra from table n/N can be estimated from cloudiness read from the Appendix cloudiness =5.1 gives n/N = 1-(5.1/8)=0.3625

From Appendix: Ra = 15.85 Ta4 = 14.26 ea =14.81, ed=ea*RH = 14.81*0.74 = 10.96 / = 1.92 Convert m/s to miles/day: u2=3*3600*24/1609 = 161.1 (Equation no. in Shaw) RI(1-r)=0.95*15.85*(0.18+0.55*0.362) = 5.71 (11.14) Ro = 14.26*(0.56-0.09*(10.96)0.5)*(0.1+0.9*0.362) = 1.59 (11.15) H = 5.71-1.59 = 4.12 (11.13) Ea = 0.35(0.5+161.1/100)(14.81-10.96) = 2.84 (11.16) Eo = (1.92*4.12+2.84)/(1.92+1) = 3.68 mm (11.12) For July Epot = 3.68*31=114 mm

The calculated value is lower than the average value (130 mm) during 1961-1978. July 1980 was unusually cold and rainy, so this seems to be reasonable.

b) Flow = (0.114*8E6)/(31*24*3600)= 0.341 m3/s

c) Assume 230 000 inhabitants, consumption 250 l/day and person gives a flow of (230000*250)/(24*3600)=0.665 m3/s The water consumption is of the same order of magnitude as the evaporation.


65

C11 "Evaporated" equals difference in water level plus rain. "Corrected" results from multiplication by Pan coefficient.

Pan coeff= 0.70

Day Rain (mm) Water level Evaporated Corrected

1 0 203.2 2 5.8 205.9 3.1 2.2 3 14.2 217.5 2.6 1.8 4 1.3 214.3 4.5 3.1 5 0.3 208.3 6.3 4.4 6 0 201.7 6.6 4.6 7 0.5 196.7 5.5 3.9 8 0.3 192.4 4.6 3.2 9 0 187.6 4.8 3.3

10 0 182.8 4.8 3.3 C12 Water budget Qin1 + Qin2 + P Qut E = dS/dt Convert to mm Qin1 = Qin2 = 0.14*3600*24*31/3400000 = 0.1103 m = 110.3 mm Qut = 0.24*3600*24*31/3400000 = 0.1891 m = 189.1 mm dS/dt = 18.97 19.08 = -0.11 m = -110 mm (negative since S decreases) This gives E = 110.3 + 110.3 + 3.8 189.1 + 110 = 145.3 mm Answer a) the evaporation was 145 mm b) decrease, c) increase, d) decrease


66

Section D D1 a) -index can in this case be calculated as (P-Q)/t = (55-30)/12 = 2.08 mm/h b) The loss during each period is the lowest value of 1) the rainfall or 2) *t, where t is the length of the period. Period 1, rainfall = 20 mm, *t = 2.08*6 = 12.5 mm, Period 2, rainfall = 10 mm, *t = 2.08*6 = 12.5 mm. Thus the loss during period 1 was 12.5 mm and during period 2 10 mm. Total loss was 22.5 mm. D3 a) The added water during each time period can be used as an estimate of the infiltration velocity at the middle of that time period Area = 0.031416 m2

"mid" time Added water Added water infiltration velocity [minutes] ml mm mm/min

5 105 3.342245989 0.3342246 15 69 2.196333079 0.21963331 25 50 1.591545709 0.15915457 35 39 1.241405653 0.12414057 45 33 1.050420168 0.10504202 55 30 0.954927426 0.09549274 65 28 0.891265597 0.08912656 75 27 0.859434683 0.08594347 extrapolate to get infiltration capacity at t=0 85 27 0.859434683 0.08594347

The parameters f0 and fc can be determined graphically from the figure, f0 = 0.41, fc = 0.086 mm/min

00.05

0.10.15

0.20.25

0.30.35

0.4

0 20 40 60 80 100

Infil

trat

ion

capa

city

[m

m/m

in]

time [min]


67

By rearranging the Horton equation, the k value can be calculated for each time period, an average value of k = 0.062 min-1 b) 15 mm of rain during 30 minutes gives a rainfall intensity of 0.5 mm/min. Thus, the rainfall intensity is always larger than the infiltration capacity. Use the integral of Hortons equation to calculate the infiltrated water: F(t) = 0.086*30+(0.41-0.086)/0.062*(1-exp(-0.062*30)) = 6.99 mm The effective rainfall is then 15 7 = 8 mm D4 Time inf. Capacity accumulated infiltration [h] [mm/h] [mm] 0 25 0 0.125 18.43307485 2.688975048 0.25 13.91969761 4.693434131 0.375 10.81770181 6.227432728 0.5 8.685733363 7.438088879 0.625 7.220454304 8.426515232 0.75 6.213383716 9.262205428 0.875 5.521234898 9.992921701 1 5.045528436 10.65149052

a)


68

D5 Area = 2.26 km2; runoff volume = 5.6 104 m3 Calculate -index For the entire 12 h period: Runoff = 5.6 104/2.26 106 = 0.02478 m = 24.78 mm Total rainfall = 2*7.1+3*11.7+2*5.6+3*3.6+2*1.5 = 74.3 mm Losses = 74.3 24.78 = 49.52 mm Average loss rate = 49.52/12 = 4.13 mm/h But during 7-12 h the rainfall intensity was lower than 4.13 mm/h, this means that the loss rate is smaller than -index. For 0-7 h Runoff = 24.78 mm Total rainfall = 2*7.1+3*11.7+2*5.6 = 60.5 mm Losses = 60.5 24.78 = 35.72 mm Average loss rate = 35.72/7 = 5.10 mm/h The rainfall intensity is always larger, thus -index = 5.10 mm/h D6 a) Bulk density = dry weight/total volume = 0.352/0.0002 = 1760 kg/m3 b) Volume of soil particles = dry weight/particle density = 0.352/2600 = 0.000135 m3 c) Volume of pores = total volume - volume of soil particles = 0.0002 0.000135 = 0.000065 m3 d) Porosity = volume of pores/total volume = 0.000065/0.0002 = 0.323 m3 m-3 e) Gravimetric water content = weight water/weight dry soil = (0.401-0.352)/0.352 = 0.139 g/g f) Volumetric water content = volume of water/volume of soil = (0.401-0.352)/0.2 = 0.245 m3 m-3 D7 Sample, D = 0.03 m, L = 0.05 m gives volume = 3.53 10-5 Weight wet = 0.0649 kg; weight dry = 0.05396 kg; volume of drained water = 5.3 10-6 m3; particle density 2650 kg/m3 Rootzone 1 m; wilting point 0.1; Area 1 104 Assume water density 1000 kg/m3 a) dry bulk density = weight dry/volume = 0.05396/3.53 10-5 = 1528.6 kg/m3 Porosity = 1 - dry bulk density/particle density = 1 1528.6/2650 = 0.423 Initial water content = initial volume of water/volume = (0.0649-0.05396)/(1000*3.53 10-5) = 0.31 m3 m-3 b) Effective porosity = porosity-field capacity = volume of drained water/volume = 5.3 10-6/3.53 10-5 = 0.15 Field capacity = porosity- effective porosity = 0.423-0.15 = 0.273 c) Available water = rootzone*(field capacity wilting point) = 1*(0.273-0.10) = 0.173 m = 173 mm


69

d) Potential evaporation from appendix; May June = 108+132+130 = 370 mm; Rainfall 50 mm. Lowest water content = wilting point + 25% of available water = 0.1 + 0.25*0.173 = 0.14325 m3 m-3; water content in May = field capacity = 0.273 m3 m-3 Thus, in May we have water stored in the rootzone = 1*(0.273-0.14325) = 0.12975 m Irrigation = potential evaporation rainfall stored water = 370-50-129.75 = 190.25 mm/m2, totally 0.19025*1 104 = 1903 m3/ha D8 Amount of water = 65.42 - 56.14 = 9.28 g Moisture content (% of dry weight) = 9.28 / 56.14 = 0.165 (= 16.5 %) Moisture content (% volume) = 9.28 / (5 32/4) = 9.28 / 35.34 = 0.263 (= 26.3 %) Air volume = Total volume - water volume - particle volume = 35.34 - 9.28 - 56.14/2.65 = 4.87 (cm3) Porosity = total pore volume / total volume = (9.28 + 4.87) / 35.34 = 0.401 (=40.1 %) Answer: is 26 % by volume or 16 % by weight. Porosity is 40 %. D9 Sample, D = 0.1 m, L = 0.3 m gives volume = 2.356 10-3 m3 volume of drained water = 365 10-6 m3; porosity = 43% Rootzone 0.5 m; wilting point 0.1; Area 10 104 Assume water density 1000 kg/m3; particle density 2650 kg/m3 a) Effective porosity = volume of drained water/sample volume = 365 10-6/2.356 10-3 = 0.155 m3 m-3 Field capacity = porosity- effective porosity = 0.43 0.155 = 0.275 m3 m-3 b) Available water = rootzone*(field capacity wilting point) = 0.5*(0.275-0.1) = 0.0875 m c) Potential evaporation from appendix; May June = 108+132+130 = 370 mm; Rainfall 50 mm. Lowest water content = wilting point + 25% of available water = 0.1 + 0.25*(0.275-0.1) = 0.14377 m3 m-3; water content in May = field capacity = 0.275 m3 m-3 Thus, in May, the water stored in the rootzone = 0.5*(0.275-0.14377) = 0.06565 m Irrigation = potential evaporation rainfall stored water = 370-50-65.65 = 254.35 mm/m2 Totally 0.25435*10 104 25500 m3


70

D10 Rainfall = 125 mm; evaporation = 50 mm; depth to groundwater = 1.25 m; field capacity = 0.25; wilting point = 0.05 a) infiltration = rainfall-evaporation = 125-50=75 mm Since the soil is at field capacity all infiltrated water will drain to the groundwater, i.e, 75 mm. b) infiltration = rainfall-evaporation = 125-50=75 mm In order to drain water to the groundwater the water content has to be higher than the field capacity. The infiltrated water will increase the water content from 0.05 m3 m-3 to 0.05 + 0.075/1.25 = 0.11 m3 m-3. This is lower than the field capacity, thus no water will reach the groundwater. c) In b) the groundwater table will not change. In a) the water table will rise. The infiltrated water will occupy the effective porosity = s-FC = 0.5-0.25 = 0.25. Total rise =0.075 m /0.25 = 0.3 m.


71

Section E E1 When the groundwater level is lowered by h the corresponding amount of water being extracted is calculated as Sh, where S is the storage coefficient. For an unconfined aquifer this is equal to the effective porosity (or specific yield). For our case:

Volume = Sh area Or 1.2108 = S 3 200106 S = 0.20 Answer: The storage coefficient S = 0.20 E2 Darcy's law: v=Q/A=KdH/dx In this case dH/dx = (pressure at the top of the column-pressure at the bottom of the column)/length of column A = area = 0.35^2*/4=0.09621 m2 Series 1 experiment Flow v =Q/A length of column Pressure dH/dx K number [l/min] [m/s] [m] [mvp] [m/s]

1 3.6 0.000624 0.58 1.11 1.913793 0.0003259 2 7.65 0.001325 0.58 2.36 4.068966 0.0003257 3 12 0.002079 0.58 4 6.896552 0.0003014 4 14.28 0.002474 0.58 4.9 8.448276 0.0002928 5 15.2 0.002633 0.58 5.02 8.655172 0.0003042 6 21.8 0.003776 0.58 7.63 13.15517 0.0002871 7 23.41 0.004055 0.58 8.13 14.01724 0.0002893 8 24.5 0.004244 0.58 8.58 14.7931 0.0002869 9 27.8 0.004816 0.58 9.86 17 0.0002833

10 29.4 0.005093 0.58 10.89 18.77586 0.0002713 average 0.0002968 Series 2 experiment Flow v =Q/A length of column Pressure dH/dx K number [l/min] [m/s] [m] [mvp] [m/s]

1 2.66 0.000461 1.14 2.6 2.280702 0.000202 2 4.28 0.000741 1.14 4.7 4.122807 0.0001798 3 6.26 0.001084 1.14 7.71 6.763158 0.0001603 4 8.6 0.00149 1.14 10.34 9.070175 0.0001643 5 8.9 0.001542 1.14 10.75 9.429825 0.0001635 6 10.4 0.001802 1.14 12.34 10.82456 0.0001664

average 0.0001727


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Series 3 experiment Flow v =Q/A length of column Pressure dH/dx K number [l/min] [m/s] [m] [mvp] [m/s]

1 2.13 0.000369 1.71 2.57 1.502924 0.0002455 2 3.9 0.000676 1.71 5.09 2.976608 0.000227 3 7.25 0.001256 1.71 9.46 5.532164 0.000227 4 8.55 0.001481 1.71 12.35 7.222222 0.0002051

average 0.0002261 Series 4 experiment Flow v =Q/A length of column Pressure dH/dx K number [l/min] [m/s] [m] [mvp] [m/s]

1 5.25 0.000909 1.7 6.98 4.105882 0.0002215 2 7 0.001213 1.7 9.95 5.852941 0.0002072 3 10.3 0.001784 1.7 13.93 8.194118 0.0002177

average 0.0002155 You can also plot Q/A vs. dH/dx and fit a straight line for calculation of K.

This is an example of the procedure for series 1, K=0.000285 m/s. The K values for the other three series using this method are 1.66, 2.15, and 2.15 E-4 m/s. E3 Well A: h = +10.00 Well B : h = +9.47 Well C : h = +9.46

A

--- show Approximate equipot. lines B C


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a) Alt. 1) Use linear interpolation to find equipotential lines of head +9.50 and +9.75. Now, draw a line perpendicular to the equipotential lines. The latter line gives the flow direction. Measure the distance along this line between the +9.50 and +9.75 it comes out as 10.0 m. Slope = S = (9.75 9.50 )/ 10.0 = 0.025 Alt. 2) Use a co-ordinate system (x/y) with x-axis through the wells with lower h-value.(B and C). This gives

Sx = (9.47-9.46) / 25 = 4.00 10-4 For Sy use the third well (A) with h= + 10.00. A second point with the same x-coordinate on the line connecting B and C has h = + 9.465 (through linear interpolation). This gives

Sy = (10.00 9.465 ) / (2531/2/2) = 2.47110-2 = 0.025 Resulting slope S = (Sx2 + Sy2 ) 1/2 = 2.47110-2 b) Since you have information about the conditions inside the triangle you should keep the tracer moving inside that triangle as far as possible. Since the direction of flow is known from a) to be essentially from A directed to a point more or less in the middle between B and C.

Injection in well A New observation well should be located 25 from A in the flow direction.

c) Observed water velocity is 25m/100h Use Darcy eq.

V = K dH/dx BUT real water velocity is

Vreal = Vdarcy / porosity With porosity = 0.35 (given) and dH/dx = 0.025 (from a)) 0.35 25/(1003600) = K 0.025 K = 9.710-4 (m/s) Answer: Hydraulic conductivity K = 1 10-3 m/s E4 Use linear interpolation between the known values of the groundwater table level. This gives: On the line connecting A and B +12 is 21 m from B and +13 is 64 m from B. On the line connecting B and C +12 is 50 m from C. On the line connecting A and C +13 is 38 m from C. With these values, equipotential lines can be drawn. A line perpendicular to the equipotential lines shows the flow direction. It turns out to be essentially parallel with AB and in the direction from A towards B. The Darcy velocity is


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V = K dH/dx Our case:

V = 1.0 10-6 1/43 = 2.32 10-8 (m/s) The real water velocity in the pores is

Vreal = V/por where Vreal = the average water velocity in the pores por = effective porosity (In some literature, por = porosity)

In reality there is some velocity variation in the pores. If we assume that the flow behaves like laminar pipe flow then

Vmax = 2 Vaverage Vreal (maximum) = 22.3210-8 / 0.20 = 2.3210-7 (m/s) = 7.3 (m/year) 100 m can be not be covered in shorter time than 100 /7.3 14 years Answer: The flow direction is from A to B (c.f. fig.). The shortest time for a substance to travel 100 m is 14 years. E5 Steady flow to a well in an unconfined aquifer.

=

2

122

21 ln r

rKQhh

a) Q = 0.004 (m3/s)

K= 1.210-4 (m/s)

The groundwater level in the well is given hw = 4 m at r = rw (= ?)

The drawdown curve must be at or below the bottom of the excavation floor. The most critical point(s) is/are at the corner(s) of the site. h2 = 7 m at r2 = 5 21/2 = 7.07 (m)

=

wr07.7ln

102.1004.047 4

22

rw = 0.32 Answer: Radius of well should be rw = 0.32


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E6 Steady flow to a well in a confined aquifer.

=

w

owo r

rKb

Qhh ln2

Q = ? K = 15 (m/day) B = 15 (m) ro = 0.5 1600 (m) rw = 0.50.3 = 0.15 (m)

)/(49415.0

800ln15152

3 3 daymQ =

=

or Q = 5.7 l/s

Amswer : Q = 5.7 l/s E7 Steady flow to a well in a confined aquifer.

=

1

212 ln2 r

rKb

Qhh

In our case

h2-h1 = (ho-2)-(ho-8) = 8-2 = 6 (m) Q = 0.2 (m3/min) Kb = T = ? r2 = 1000 (m) r1 = 100 (m)

=

1001000ln

260/2.06T

T = 2 10-4 (m2/s) K = T/b = T/20 = 1.010-5 (m/s) Answer: Hydraulic conductivity K = 1.0 10-5 m/s, transmissivity T = 2.0 10-4 (m2/s)


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dxdhKbhQ =

hdhQKbdx =

=2

1

2

1

h

h

x

xhdh

QKbdx

( )212212 2 hhQKbxx =

( )3642

10035

=

Q

dxdhAKQ =

E8 Darcys