Storm Drainage Design Project - Assignment Example

April 18, 2009 Storm Drainage Design Project The graph for the discharge of water in rivers is called a hydrograph. Usually, A line graph is used to present discharge in relation with time. Rainfall is also plotted with time using the bar graph. There are factors that control the shapes of a hydrograph. Weyman. (1975) concluded that the different shapes are shown and the main components are labelled accordingly. Peak rainfall and peak discharge are two different graphs. Lag time is the term for the difference betseen the peak rainfall from the peak discharge. There is less chance of flooding if the lag time is great. A short lag time will indicate that water had already reached the river channel at a fast rate. The rise in the discharge shown in a graph is called the rising limb, whereas the decrease in the discharge is called the falling limb. Large areas of basins normally receive more precipitation than the small ones and therefore they have a larger runoff. A larger size of runoff would mean that there is longer lag time as water has a longer distance to reach the river trunk. In most cases, the shape of the basin is elongated and produces a lower peak flow and longer lag time than a circular basin with the same size (Gillesania, 2006). The effects of the slopes are very important. The channel flow becomes fall faster down a steep slope thus, producing a steeper rising limb and a shorter lag time. The rock's permeability would mean rapid filtration and little overland flow. This will produce a shallow rising limb. Considering the soil as an important factor, infiltration is greater on thick soil even if clay would act as impermeable layer. Longer lag time occurs when there are more infiltration and more shallow rising limb. Concrete and tarmac form impermeable surface that creates s steep rising limb and a shorter time lag. Afforestation intercepts the precipitation, thereby creating a shallow rising limb and a longer time lag. A rapid overland flow is observed with a higher drainage density. The temperature and precipitation is being considered. The production of rapid overland flow and steep rising limb by short intense rainfall is due to precipitation and temperature. Tidal condition is the high spring tides that block the normal exit for the water, extending the length of time the river from the river basin takes to return to its original flow. The volume of water reaching the river from the surface run off is the overland flow, and the through flow is the volume of water reaching the river through the soil and its underlying rock layer Cynon River hygrograph Time respect to discharge Cynon River hydrograph Rainfall with respect to time The data of Cynon river study was plotted having the results above. The study was done within a period of 4 days (96hours) continuously every hour, taking the height reading. All the readings are different from one another. We can see from the readings that there is a steady flow from the start of the reading up to its reading in the 42th hour. The water began to rise after the reading of the 42th hour. This is where the rising limb takes place. The time between the rise of water and the time the water reaches the peak flow is known to be the basin lag time. As it reached the peak on the 57th hour, the water ha now reached the peak discharge and it is now beginning to fall down. From the point, when the water height starts to fall down, the process is called recession limb. After the4 recession limb, discharge of the water will normalize. The storm flow is the total of the overland flow and the through flow. The overland flow is the flow at which water rises above the through flow, and the through flow is the water that rises above the base flow. Channel Design The discharge in the is Q m3. Apply the Manning's formula to design a suitable breadth b of a channel with depth d v = R2/3S1/ 2 where: v = velocity, m/s n R = hydraulic radius, m S = channel bed slope, m/m n = Manning's coefficient of roughness A = db where: A = area, m2 b = breadth, m d = depth, m Q = Av Given Data Q = 1.5 m3/s n = 0.018 S = 1 in 3000 = 0.0003 d = 0.6 m Required = width of base b of the open channel Discharge Q of the river into the open channel Design of water pump to discharge water from the river to the open channel Computations: A = db = (0.6)(b) A = 0.6b Wetted Perimeter = 2d + b = 2(0.6) + b = 1.2+ b Hydraulic radius R = areawetted perimeter = 0.6 b1.2+b v = R2/3S1/2 n Q = Av 1.5 = 0.6b0.61.2+b230.0005120.018 1.5(0.017) = 0.6b0.41.2+b23(0.0221) 0.027 = 0.6b 0.61.2+b2/30.0221 0.0270.02210.6 = b0.61.2+b23 0.0270.0134 = b0.61.2+b23 2.0149 = b0.61.2+b23 2.01493 = b0.6b1.2+b2/3 2.63153 = b30.6b1.2+b2 8.1504 = b30.6b1.2+b2 8.1504 = 0.36b51.44 + 2.4b +b2 8.1504(1.44 + 2.4b + b2) = 0.36b5 11.7365 + 19.5609b + 8.1504b2 = 0.36b5 11.7365 + 19.5609b + 8.1504b2 - 0.36b5 b = 3.4515 m. In the computation of the river flow volume (discharge), we must consider the width of theriver to be 15 meters with depth d Computations; Q = Av where: A = cross-sectional area v = velocity = 4.0 m/s A = bd b = 15 m. A = 15(d) R = areawetted perimeter R = 15d15 + 2d v = R2/3S1/2n v = 15d(15+2d)2/30.00051/20.018 4.0 = 15d(15+2d)2/30.00051/20.018 4.0(0.018) = 15d15+2d2/30.00051/2 0.072 = 15d15+2d2/30.0221 0.0720.0221 = 15d15+2d2/3 3.2579 = 15d15+2d2/3 (3.2579)3 = (15d)2(15 + 2d)2 34.5790 = (15d)2(225 + 60d + 4d2) 34.5790(225 + 60d + 4d2) = 225d2 7,780.275 + 2,074.74d +138.316d2 = 225d2 7,780.275 + 2,074.74d + 138.31d2 - 225d2 = 0 7,780.275 + 2,074.74d - 86.684d2 = 0 By quadratic equation; solve for the value of d, d = -bb2-4ac2a where a = -86.684 b = 2,074.74 c = 7,780.275 d = -2,074.74(2,074.74)2-4-86.684(7,780.275)2(-86.684) d = -2,074.744,304,546.0676 + 2,697,701.4324-173.368 d = -2,074.747,002,247.5000-173.368 d = -2,074.742,646.1760-173.368 d = 3.29 m. = depth of the river Solving for the discharge Q Q = Av Q = 15(3.29) (4) Q = 197.40 m3/s = discharge of water in the river With the result of discharge Q which is 197.40m3/s of the river, compared to the discharge Q of the channel which is 1.5m3/s, it is safe to say that we can draw water from the river without the river drying up. The river will be able to supply the open channel with the required volume of water needed to fill the necessary amount of water needed for the reservoir to distribute water to the end consumers. In order for the channel to have a steady supply of water coming from the river, a design for a pump to be used is necessary. (Young and Freedmen, 2000), stated that The design of the pump will depend on the discharge of water in the open channel. In the computations done regarding the discharge of water in the channel with respect to the discharge of water in the river, the pump to be used must have a discharge equal to the discharge of water in the channel to avoid overflowing or shortage of water supply in the canal since the water from the river that will pass through the channel will go to a reservoir storage. Computations for the design of a water pump: HP = Q x H3960 where Q = discharge H = total head 3960 = constant H = v22g +d Q = Av A = bd A = 3.4515 (.6) A = 2.0709 m2 Q = Av 1.5 = 2.0709 x v v = 1.52..0709 v = 0.7241 m/s H = v22g + d where H = total head v = velocity H = 0,724122(9.81) + 0.6 g = 9.81 = gravitational constant d = depth = 0.524319.62 + 0.6 HP = design load 3960 = constant value H = 0.0267 + 0.6 1.5m3/s = 23775.484712233gallons/min 0.6267 m = 2.0561023622047 feet H = 0.6267 HP = Q x H3960 HP = 23775.484712233 x 2.05610236220473960 HP = 12.344654110956 horsepower The design load of the pump that was computed is very safe and would satisfy the design of the open channel. Since the design load of the pump is to draw water from the river to the open channel is 12.34 horsepower, a pump with a 12.5 horsepower capacity to 15 horsepower pump is needed. The pump will be able to draw water from the river and lift it up to the elevation of the open channel, thus filling in the required volume of water in the channel. Therefore, the computed pump workload design is safe. The pump can be installed anywhere between 3.0 meters to 12 meters above the river bed. The pump to be used must have the capacity to lift water from the river and fill up the open channel without overflowing. The pump must be able to fill the channel with the a cross-section that that had a design for the required capacity of 2.0709m2. Conclusion: It is essential that the stream hydrograph that is taken from a drainage-area be separated into ground water flow and the other components of storm flow must be accurate as the data would permit. W.G Hoyt stated that the peaks occurring in rapid succession that surface run-off may drain out does the task of a ground water hydrograph difficult and having uncertain results. The improvements of the methods and development of new and better methods for the differentiating of ground water run-off and surface run off had been pointed out as one of the deficiencies. In some researches, precipitation and other measurements that are important to watershed studies are being taken from drainage areas supporting continuous stream-flow. References: Dr. Tim Stott, Flood Hydrographs, Fluvial Geomorphology, Learning and Research Technology University of Bristol, April 19,2009. . University of Bristol, April 14, 2009, . Flooding, BBC - GCSE BITESIZE - Flooding, BBC April 12, 2009,. Pump Equation and Formula Calculation, 2007, AJ Designs, April 10,2009, . Gillesania, Diego Inocencio T. 2006, Engineering Formula Series, Civil Engineering, Diego Inocencio T. Gillesania, Manila, Philippines. King, Wisler, and Woodburn, 1988, Hydraulics, John Wiley and Sons, Inc. New York. Waugh D. 1995. Geography: An integrated Approach, Walton-on-Thames, Nelson. Chapter 3 Drainage Basins and Rivers, 48-52. Weyman DR. 1975. Runoff Processes and Streamflow modelling, London, Oxford University Press, 54 pp. Young and Freedman, 2000, University Physics, Addison-Wesley Publishing Company, Inc. Singapore Read More