Introduction
Fluid flow through pipe involves head loss due to friction. The energy needs to push the water from one point to another increase if the frictional surface area increases. In the case of pumps the energy needs to be supplied to water is increased if the length of the tube and datum height increases. Prior to the purchasing of the pump, it is important to calculate the head and power requirement of the pump based on the scenario. Without any prior information or analysis purchasing a pump may not be sufficient to supply water to the required height. Sometimes the power needed for the pump is much more than the actual power needed to push water. To find a suitable pump for the pumping system one must need to find the head requirement for a particular discharge. In this project, a pump is selected to push water to a height of 45m. There are four different pumps is available to purchase, but a suitable pump for this operation so that the power needed to operate the motor must be minimum. The pipe system has four elbow joints and one gate valve. The suction side has a 75-meter length pipe and the delivery side has a 125m length pipe. There are three different pipe diameter is available to select provided the design must have minimum power usage. The objectives of this project are to find a suitable pump and piping system to transport water to a height. The design involves suitable pipe selection as well as pump selection from the given options. The system needs to operate 350 days per year and for 16 hours a day. The average discharge in each month is shown in the table below. A suitable design must be developed and selected such that the requirement of water discharge for every month must be fulfilled.
Table 1: Average water consumption in each month
| Month | Avg Daily consumption (L/s) |
| January | 101 |
| February | 110 |
| March | 150 |
| April | 170 |
| May | 145 |
| June | 200 |
| July | 240 |
| August | 250 |
| September | 205 |
| October | 160 |
| November | 145 |
| December | 110 |
Theory
In the case of fluid flow over surface friction developed between the surface and fluid. This friction drags each fluid layer near the surface. The velocity near the surface will be less than the free stream velocity. Flow-through pipe involves the same phenomena so that the friction reduces the pressure along the flow field. The pressure loss or head loss in the pipe is calculated from Hazen Williams equation V = k * C * (D/4)0.63 * S0.54 . ‘k’ is 0.85 unit conversion factor, ‘C’ is the Hazen Williams constant, ‘D’ is the hydraulic diameter and ‘S’ is the head loss per meter. Another loss that can take place in the pipe flow is the minor loss factor. Minor loss is calculated as = * /2g
For a pipe, the power requirement increases with a decrease in the diameter as the decrease in diameter induces more pressure loss. At the same time power needed to pump through a pipe also increases with the velocity. Each pump head available for various flow rates will be provided by the manufacturer. While selecting a pump the system curve must be matched with the characteristic curve.
Analysis
From the average monthly flow rate, the maximum flow rate is found to be 250L/s. For the pump selection, the discharge must be a minimum of 250L/s if only one pump is selected. Another wise pump with parallel formation is possible to reach the discharge requirement. General analysis of the piping system and the system curve is shown below.
Figure 1: Head required for each pipe diameter
Figure 2:Power required for each pipe diameter
Pump 1 analysis
The characteristic curve from the manufacturer for pump 1 is shown below. The graph is plotted with the actual head available for each flow rate. The maximum flow rate obtained is 45L/s and the maximum head available is 55 m. No single pump can be used for the requirement as the minimum requirement for a single pump is 250L/s. Another option can be used to provide more height is a series arrangement of pump and parallel arrangement of the pump is needed for more flow rate conditions. To reach the flow rate of 250L/s pump 1 needs 10 units in the parallel condition each with 25L/s. The cost analysis for 25L/s pump is shown below
Table 2: Pump 1 cost analysis
| Pump 1 (25L/s) | 0.25 m | 0.3 m | 0.35 m | 0.25 m | 0.3 m | 0.35 m | 0.25 m | 0.3 m | 0.35 m | ||
| RPM | Head | Horse Power | No of pumps in series | Power requirement | Cost ($) | ||||||
| 3250 | 25 | 23 | 3 | 3 | 2 | 690 | 690 | 460 | 23659.66 | 24339.663 | 18019.66 |
| 3550 | 32 | 30 | 3 | 2 | 2 | 900 | 600 | 600 | 23666.25 | 17346.25 | 18026.25 |
| 3850 | 40 | 40 | 2 | 2 | 2 | 800 | 800 | 800 | 16675.66 | 17355.66 | 18035.66 |
| 4050 | 43 | 46 | 2 | 2 | 2 | 920 | 920 | 920 | 16681.31 | 17361.306 | 18041.31 |
| 4350 | 50 | 49 | 2 | 2 | 1 | 980 | 980 | 490 | 16684.13 | 17364.129 | 11044.13 |
The cost analysis shows that the pump1 can deliver 25L/s with 10 units in parallel and one unit in series can be selected. The cost needed for this purpose is 11044$ and is obtained by using the pipe diameter of 0.35m. Increasing the pipe diameter or the impeller RMP increases the cost of the pump. The pump selected is 40% efficient there for the motor power needed will be 490/0.43 = 1140Horse power. The total running time is 5600 hours. The total energy consumption will be 1140*746*5600*3600/3600000*0.1 = 476246$
Figure 3: pump 1 Characteristic curve
Pump 2 analysis
The characteristic curve from the manufacturer for pump 1 is shown below. The graph is plotted with the actual head available for each flow rate. The maximum flow rate obtained is 88L/s and the maximum head available is 53 m. No single pump can be used for the requirement as the minimum requirement for a single pump is 250L/s. Another option can be used to provide more height is a series arrangement of pump and parallel arrangement of the pump is needed for more flow rate conditions. To reach the flow rate of 250L/s pump 2 needs 5 units in the parallel condition with 50L/s each. The cost analysis for the 50L/s pump is shown below.
Table 3: Pump 2 cost analysis
| Pump 2 (50L/s) | 0.25 m | 0.3 m | 0.35 m | 0.25 m | 0.3 m | 0.35 m | 0.25 m | 0.3 m | 0.35 m | ||
| RPM | Head | Horse Power | No of pumps in series | Power requirement | Cost ($) | ||||||
| 3250 | 23 | 35 | 3 | 3 | 3 | 525 | 525 | 525 | 13920.96 | 14600.955 | 15280.96 |
| 3550 | 30 | 40 | 3 | 2 | 2 | 600 | 400 | 400 | 13925.66 | 10855.66 | 11535.66 |
| 3850 | 35 | 46 | 2 | 2 | 2 | 460 | 460 | 460 | 10181.31 | 10861.306 | 11541.31 |
| 4050 | 42 | 55 | 2 | 2 | 2 | 550 | 550 | 550 | 10189.78 | 10869.775 | 11549.78 |
| 4350 | 58 | 68 | 2 | 1 | 1 | 680 | 340 | 340 | 10202.01 | 7132.008 | 7812.008 |
The cost analysis shows that the pump2 can deliver 50L/s with 5 units in parallel and one unit in series can be selected. The cost needed for this purpose is 7812$ and is obtained by using the pipe diameter of 0.35m. Increasing the pipe diameter or the impeller RPM increases the cost of the pump. The pump selected is 54% efficient there for the motor power needed will be 340/0.54 = 630Horse power. The total running time is 5600 hours. The total energy consumption will be 630*746*5600*3600/3600000*0.1 = 263034$
Figure 4: Pump 2 characteristic curve
Pump 3 analysis
The characteristic curve from the manufacturer for pump 3 is shown below. The graph is plotted with the actual head available for each flow rate. The maximum flow rate obtained is 172L/s and the maximum head available is 53 m. No single pump can be used for the requirement as the minimum requirement for a single pump is 250L/s. Another option can be used to provide more height is a series arrangement of pump and parallel arrangement of the pump is needed for more flow rate conditions. To reach the flow rate of 250L/s pump 3 needs 3 units in the parallel condition with 80L/s each. The cost analysis for the 80L/s pump is shown below.
The cost analysis shows that the pump3 can deliver 80L/s with 3 units in parallel and one unit in series can be selected. The cost needed for this purpose is 7265$ and is obtained by using the pipe diameter of 0.35m. Increasing the pipe diameter or the impeller RPM increases the cost of the pump. The pump selected is 60% efficient there for the motor power needed will be 375/.6 = 625Horse power. The total running time is 5600 hours. The total energy consumption will be 625*746*5600*3600/3600000*0.1 = 261100$
Table 4:Pump 3 cost analysis
| Pump 2 (80L/s) | 0.25 m | 0.3 m | 0.35 m | 0.25 m | 0.3 m | 0.35 m | 0.25 m | 0.3 m | 0.35 m | ||
| RPM | Head | Horse Power | No of pumps in series | Power requirement | Cost ($) | ||||||
| 3250 | 25 | 55 | 3 | 3 | 2 | 495 | 495 | 330 | 12139.78 | 12819.775 | 10349.78 |
| 3550 | 33 | 75 | 3 | 2 | 2 | 675 | 450 | 450 | 12158.6 | 9688.595 | 10368.6 |
| 3850 | 38 | 92 | 2 | 2 | 2 | 552 | 552 | 552 | 9024.592 | 9704.592 | 10384.59 |
| 4050 | 45 | 110 | 2 | 2 | 2 | 660 | 660 | 660 | 9041.53 | 9721.53 | 10401.53 |
| 4350 | 50 | 125 | 2 | 2 | 1 | 750 | 750 | 375 | 9055.645 | 9735.645 | 7265.645 |
Figure 5: Pump 3 characteristic curve
Pump 4 analysis
The characteristic curve from the manufacturer for pump 4 is shown below. The graph is plotted with the actual head available for each flow rate. The maximum flow rate obtained is 225L/s and the maximum head available is 52 m. No single pump can be used for the requirement as the minimum requirement for a single pump is 250L/s. Another option can be used to provide more height is a series arrangement of pump and parallel arrangement of the pump is needed for more flow rate conditions. To reach the flow rate of 250L/s pump 4 needs 2 units in the parallel condition with 125L/s each. The cost analysis for the 125L/s pump is shown below.
The cost analysis shows that the pump4 can deliver 125L/s with 2 units in parallel and 3 units in series can be selected. The cost needed for this purpose is 8115$ and is obtained by using the pipe diameter of 0.35m. Increasing the pipe diameter or the impeller RPM increases the cost of the pump. The pump selected is 62% efficient there for the motor power needed will be 500/.62 =806 power. The total running time is 5600 hours. The total energy consumption will be 625*746*5600*3600/3600000*0.1 = 336714$
| Pump 2 (125L/s) | 0.25 m | 0.3 m | 0.35 m | 0.25 m | 0.3 m | 0.35 m | 0.25 m | 0.3 m | 0.35 m | ||
| RPM | Head | Horse Power | No of pumps in series | Power requirement | Cost ($) | ||||||
| 3250 | 26 | 55 | 3 | 3 | 2 | 330 | 330 | 220 | 8689.775 | 9369.775 | 8049.775 |
| 3550 | 30 | 75 | 3 | 2 | 2 | 450 | 300 | 300 | 8708.595 | 7388.595 | 8068.595 |
| 3850 | 38 | 92 | 2 | 2 | 2 | 368 | 368 | 368 | 6724.592 | 7404.592 | 8084.592 |
| 4050 | 44 | 110 | 2 | 2 | 2 | 440 | 440 | 440 | 6741.53 | 7421.53 | 8101.53 |
| 4350 | 48 | 125 | 2 | 2 | 2 | 500 | 500 | 500 | 6755.645 | 7435.645 | 8115.645 |
Pump selection
Based on the pump selection in each pump type, one pump is finally selected. The total cost for the pump 3 is less compared to the other selections. Both the operating and construction cost of the pump 3 with 80L/s flow rate is selected for the purpose. The total number of pumps is 3 all are connected in parallel to achieve the maximum flow rate. The three pumps are parallel arrangements that are capable fo delivering the required flow rate at any tie in the year.
| Operating cost ($) | Construction cost ($) | Total cost ($) | |
| Pump 1 | 476246 | 11044.129 | 487290.129 |
| Pump 2 | 263034 | 7812.008 | 270846.008 |
| Pump 3 | 261100 | 7265.645 | 268365.645 |
| Pump 4 | 336714 | 8115.645 | 344829.645 |
Figure 6:Pump 4 characteristic curve
Tank analysis
Design of tank needed for storing the pumped water is calculated below
| Month | Avg Daily consumption (L/s) | Totla volume needed per day (m3/s) | 8 hours | Supply | 8 hours | Remaining | Further consumed in 8 hours | left |
| January | 101 | 0.101 | 2908.8 | 0.25 | 7200 | 4291.2 | 2908.8 | 1382.4 |
| February | 110 | 0.11 | 3168 | 0.25 | 7200 | 4032 | 3168 | 864 |
| March | 150 | 0.15 | 4320 | 0.25 | 7200 | 2880 | 4320 | -1440 |
| April | 170 | 0.17 | 4896 | 0.25 | 7200 | 2304 | 4896 | -2592 |
| May | 145 | 0.145 | 4176 | 0.25 | 7200 | 3024 | 4176 | -1152 |
| June | 200 | 0.2 | 5760 | 0.25 | 7200 | 1440 | 5760 | -4320 |
| July | 240 | 0.24 | 6912 | 0.25 | 7200 | 288 | 6912 | -6624 |
| August | 250 | 0.25 | 7200 | 0.25 | 7200 | 0 | 7200 | -7200 |
| September | 205 | 0.205 | 5904 | 0.25 | 7200 | 1296 | 5904 | -4608 |
| October | 160 | 0.16 | 4608 | 0.25 | 7200 | 2592 | 4608 | -2016 |
| November | 145 | 0.145 | 4176 | 0.25 | 7200 | 3024 | 4176 | -1152 |
| December | 110 | 0.11 | 3168 | 0.25 | 7200 | 4032 | 3168 | 864 |
| Tank size needed in the starting (m3) | -27993.6 |
The tank size needed is 28000m3. The design is a cylindrical shape with 20 m height and 43 m diameter. The material must be designed with steel to withstand the water pressure of 20m height.
Conclusion
The pump is suitable for the operation is pump 3 with an impeller with rpm 4350. The running cost and construction cost is lesser than other pump choices. The total cost is 268365.645$. The tank size needed for the same pump working 8hours per day is 28000m3.
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