Hydraulics Assignment: Flow Calculations & Water Distribution

NAME AND ID: Abhijith Ramachandran Nair Sreeja (# 10324876 , section-61) ENVR 91- INFRASTRUCTURE MANAGEMENT HYDRAULICS ASSIGNMENT  TO BE DONE INDIVIDUALLY OR IN GROUPS OF TWO (MAX)  DUE Wednesday, February 21 st IN D2L (11:59 PM) Answer the following 6 questions- Total Marks 60 1. Flow Calculations A. Calculate the flow (in ML/day) through 800 mm diameter pipe that has an average velocity of 1.2 m/s (3 marks) B. The pipe then moves through a reduction from 800 mm to 650 mm, what is the new velocity in m/s ? (3 marks) C. The pipe is then split into two pipes- 1 is 250 mm the other is 200 mm in diameter, what is the velocity (in m/s) through each of these pipes, assuming 60% of flow goes through the larger pipe? (3 marks) D. Draw a diagram. (2 marks) a) Diameter of the pipe=800 mm = 0.8 m Radius of the pipe=0.8/2=0.4 m velocity=1.2 m/s Area of the pipe (A)= A=πr 2 = 3.14× (0.4m) 2 = 3.14×0.16m 2 =0.5024m 2 Flow rate (Q) = A × V = 0.5024m 2 × 1.2m/s = 0.6032m 3 /s conversion factors, 1m 3 =1000L ENVR 91- Infrastructure Management

1 L=10 −6 ML 1 day=86400 s QML/day = 0.6032m 3 /s × 1000 L 1 m 3 × 1 ML 10 6 L × 86400 s 1 day = = 52.11 ML/day b) So, the flow rate through the 800 mm diameter pipe with an average velocity of 1.2 m/s is 52.11 ML/day. Diameter 2, d 2 = 650mm = 0.65m radius 2, r 2 = 20.65m = 0.325m ≈ 0.33m For the new pipe (pipe 2), the cross-sectional area A 2 =π × (0.33m) 2 =π × (0.33m) 2 flow rate, Q = A 2 × V 2 51840m 3 /day × 1 day 86400 s = π× (0.33m) 2 × V 2 V 2 = 1.8m/s The new velocity of the flow in the pipe after the reduction from 800 mm to 650 mm is 1.8m/s. c) Given Flow rate Q=52.1 ML/day Diameter of the larger pipe (Pipe 3), d 3 =250 mm=0.25 m Diameter of the smaller pipe (Pipe 4), d 4 =200 mm=0.2 m Flow through Pipe 3 (Q3): 60%×52.1 ML/day=31.26 ML/day Flow through Pipe 4 (Q4): ENVR 91- Infrastructure Management

40%×52.1 ML/day=20.84 ML/day Q=VA Pipe 3, V 3 = Q 3 / A 3 = 31.26 × 10 6 L / day 1 ML × 1 m 3 1000 L × 1 day 24 hour × 1 hour 60 min × 1 min 60 s ÷ π×(0.125m) 2 =7.37 m/s For Pipe 4, V 4 = Q 4 / A 4 = 20.84 × 10 6 L / day 1 ML × 1 m 3 1000 L × 1 day 24 hour × 1 hour 60 min × 1 min 60 s ÷ π×(0.1m) 2 =7.68m/s The velocity through Pipe 3 is 7.37 m/s, and through Pipe 4 is approximately 7.68 m/s d) ENVR 91- Infrastructure Management

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2. The water entering the distribution system from the high lift pumps at the water treatment plant is at a pressure of 840 kPa. The water tower is 32 m high (height of water). What is the elevation difference between the water treatment plant and the base of the tower (in m) ? (5 marks) State your assumptions (2 marks). Given, Pressure of high lift pump, Pa = 840kPa=84m Height of the tower =32m Pressure at water tower, P b = ρ ×g×h =1000 kg/m 3 × 9.8 m/s 2 × 32 m = 313600 Pa conversion factors, 1 kPa = 1000 Pa 1 meter = 10 kPa ENVR 91- Infrastructure Management

P b = 313600 Pa × (1 meter / 10 kPa) = 31.36 meters head loss = 0, Pa = Pb + Elevation – HL Elevation = Pa – Pb Elevation = 84 m - 31.36 m =52.64 meters The elevation difference between the water treatment plant and the base of the tower is 52.64 meters. Assumptions The diameter of the pipe used for water distribution is 12 inches. No head loss occurs at bends in the pipe, indicating smooth flow conditions. The contribution of pressure from the water head at the treatment plant is considered negligible.  The pressure discharged at the tower is equal to the pressure generated by the pump.  The flow rate of water through the system is constant at 9500 liters per minute (LPM). The pressure at the tower (PT) is maintained at a minimum of 140 kPa. =14m ENVR 91- Infrastructure Management

3. A cylindrical tank is 10 m in diameter and a pressure gauge at the bottom reads 750 kPa. Calculate the height of liquid in the tank if the liquid inside is: A. a) water (SG = 1) (4 marks) B. b) oil (SG = 0.85) (6 marks) Given: Diameter of the tank (d) = 10m Pressure (P) = 750kPa Gravity (g) = 10m/s 2 a) For water (SG = 1): Density of water (ρ) = 1000 kg/m 3 Using the formula: h = P ρ× g = 750000 kg m − 1 s − 2 1000 kgm − 3 × 10 ms − 2 = 750000 10000 = 75m b) For oil (SG = 0.85) Density of oil (ρ) = 850 kg/m 3 Using the formula: ENVR 91- Infrastructure Management

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h = P ρ× g = 750000 kgm − 1 s − 2 850 kgm − 3 × 10 m s − 2 = 750000 8500 = 88.23m Therefore, the height of the liquid in the tank for water is 75m and for oil is 88.23m. ENVR 91- Infrastructure Management

4. For the system below, what size of pump (in meters of head) is needed to ensure water is going to be pumped into the reservoir (8 marks)? Assume: Elevation differential from pump to tank inlet= 75 m Total length of pipe = 3650 m Equivalent length of pipe for elbows = 120 m Pipe diameter = 250 mm Flow = 8,200 LPM PVC schedule 40 pipe Use tables from formula book Assumptions A pipe with a diameter of 250mm and a flow rate of 8200 LPM incurs a head loss of 2.6/100, while the pressure at point C, representing the collection tank, is assumed to be 0 The calculation proceeds as follows: The pressure discharged (PB) is calculated as: PB=Pc+Elevation+Head Loss =75m+2.6/100×(120m+3650m) = 75m+98.02m=173.02m Then, the pressure at the pump (Ppump) is found by subtracting the pressure at point A from PB: Ppump=PB−PA =173.02m−4m=169.02m Therefore, the pressure at the pump is 169.02 meters. ENVR 91- Infrastructure Management 4 M

5. For the system below (bullets and sketch) A. What is the minimum pressure for the booster pump (in m) to meet a minimum pressure of 380 kPa in the system (5 marks) B. A new pump is installed that now boosts the pressure by 850 kPa, what is the new pressure at C (in m) and how much do I need to reduce the pressure at the PSV (in m) to get a pressure of 450 kPa at point E (7 marks) C. State your assumptions (2 marks)  Pressure at A is 450 kPa  B to C is 1.95 km of 12-inch PVC piping with 12 valves  Flow leaving pump = 10,000 LPM  C to D is 3.5 km of 10-inch PVC piping with 36 valves  Flow entering PSV station is 9,500 LPM  All valves are equivalent to 15 m of pipe (each valve) a) At point A, the pressure of 450 kPa corresponds to a 45-meter head, while the distance between points B and C is 1950 meters, with a friction head loss of 1.6 ft/100 ft for a 12-inch pipe carrying 10000 L/min; additionally, there are 12 valves equivalent to 15 meters of pipe each, and the elevation is 47 meters. ENVR 91- Infrastructure Management C P B A Booster Pumping Station 77 M E D P P 47M P P PSV

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The calculation proceeds as follows: PC=PB−elevation−head loss 38 m=PB−47 m−1.6/100(12×15 m+1950 m) PB=38 m+47 m+34.08 m PB=119.08 m This corresponds to 1190.8 kPa. Then, to calculate the pressure at the booster pump: P Booster Pump =P B −P A =119.08 m−45.0 m =74.08 m=74.08 m Therefore, the pressure at the booster pump is 74.08 meters. b) After installing a new booster pump with a pressure of 850 kPa, equivalent to an 85-meter head, and considering that the pressure at point A is 450 kPa or 45 meters head: The pressure at point B (PB) can be found by subtracting the head at point A from the head at the booster pump: P Pump =P B - P A PB=85 m+45 m=130 m This corresponds to 1300 kPa. Then, to calculate the pressure at point C (PC): PC=PB−elevation−head loss =130 m−47 m−1.6/100(1950 m+12×15 m) =83 m−34.08 m =48.92 m To find the pressure at the pressure reducing valve (PRV), we first need the pressure at point D (PD): ENVR 91- Infrastructure Management

PD=PC+elevation−head loss = 48.92 m+77 m−2.6/100 (3500 m+36×15 m) =125.92 m−105.04 m =20.88 m Then, the pressure at the pressure reducing valve (PPSV) is: PPSV=PE−PD =45 m−20.88 m =24.12 m In summary, the pressure at the pressure reducing valve is 24.12 meters. C) The assumptions are: A head loss of 1.6 is considered for a 12-inch pipe with a flow rate of 10000 LPM. A head loss of 2.6 is considered for a 10-inch pipe with a flow rate of 9500 LPM. ENVR 91- Infrastructure Management

6. The Town of Spooksville is installing a new watermain. The contractor has submitted their disinfection proposal for you to review. The watermain is 300 mm in diameter and 1950 m long and the contractor is using a 15% bleach solution for disinfecting the pipe (starting at the beginning) A. Draw a sketch (2 marks) B. How long before the bleach solution makes its way to the end of the pipe if the flow leaving through the end bleeder is 500 L/min (4 marks) C. The contractor’s plan says they will add 120 L of the bleach solution. Comment on this proposal assuming the final concentration for disinfecting the pipe according to regulations is 25 mg/L (4 marks) a) ENVR 91- Infrastructure Management

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b) The diameter of the pipe is 300 mm, which is equivalent to 0.3 meters. The distance of the pipe is 1950 meters. Therefore, the volume of the pipe is calculated as follows: Volume of pipe = Distance of pipe × Cross-sectional Area of Pipe = 1950 m × π × (0.3/2) 2 = 137.84 m 3 Given that the flow of solution at the end of the bleeder is 500 liters per minute: Flow = Volume / Time = 500 L/min = (137.84 m 3 ) / Time Hence, Time = (137.84 × 1000 L) / (500 L/min) = 275.68 min ≈ 4.595 hours, which is approximately 4.6 hours. C)  Initial concentration C1 of bleach = 15% Cl2 or 150000 mg/L Cl2  Volume V1 of bleach = 120 L  Final desired concentration C2 = 25 mg/L  V2 = 137840 L To calculate the volume of bleach needed to achieve the desired concentration: C2= C1×V1/V2 = 150000mg/L×120L/137840L = 130.9mg/L Since the calculated concentration (130.9 mg/L) exceeds the desired concentration (25 mg/L), the contractor needs to dilute the bleach to achieve the desired concentration. To calculate the volume of bleach needed for the desired concentration: V1= C2×V2/C1 = 25mg/L×137840L/150000mg/L = 22.97L ENVR 91- Infrastructure Management

Therefore, the contractor needs to reduce the volume of bleach by 22.97 L to achieve the desired concentration of 25 mg/L in the system. In summary, the contractor needs to decrease the volume of bleach from 120 L to 22.97 L to achieve the desired concentration. ENVR 91- Infrastructure Management

Hydraulics[1](1)

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