Classroom Computer Exercise—Pipe Networks answer Q from a to g Obtain or write computer software that is appropriate for

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Classroom Computer Exercise Pipe Networks Answer Q From A To G Obtain Or Write Computer Software That Is Appropriate For 1 (198.26 KiB) Viewed 42 times

Classroom Computer Exercise—Pipe Networks answer Q from a to g Obtain or write computer software that is appropriate for solving pipe networks. Examples are EPANET (public domain from the U.S. Environmental Protection Agency), WaterCAD and WaterGEMS (proprietary from Haestad Methods-Bentley), and KYPipe (proprietary from KYPipe LLC). Or you can write your own spreadsheet program. Answer the following ques- tions by performing a computer analysis of the pipe network described in Example 4.8 and its modifications, provide answer with EPANET screenshoot solution (a) Before using the computer software, what data do you anticipate the software will need to analyze the pipe network in Example 4.8? (b) Now use the computer software to analyze Example 4.8. Enter the data requested by the software and perform a network of analysis. Compare flow rates generated by the computer model to those in the example problem. Why are the solutions not exactly the same? (Note: Some computer models require a pipe material and then assign an "f" value based on the pipe material and the Reynolds number. You may have to "manipulate the model to get it to match the "f" values in Example 4.8.) (c) What would happen to the flow rate in pipe EF if the friction factor was reduced? What would happen to the pressure at F? Write down your answers, and then reduce "f" in EF from 0.021 to 0.014 and perform a new network analysis. List the original and new flow rate in pipe EF and the original and new pressure at node F. (Hint: The program may not let you change the friction factor directly but may allow a change in the pipe material or roughness value. You may have to assume complete turbulence on the Moody diagram and back into the roughness value or pipe material you need to reduce the friction factor in pipe EF to 0.014.) Once you have completed the analysis, restore pipe EF to its original friction factor and proceed to the next question. (d) What would happen to the flow rate in pipe HF if the diameter was doubled? What would happen to the pressure at F? Estimate the magnitude of these changes and write them down. Now double the diameter and analyze the network. Did you rea- son correctly? List the original and new flow rate in pipe HF and the original and new pressure at node F. Now restore HF to its original size and proceed to the next question (e) What would happen to the pressure at F if the demand for water at that point increased by 50 liters/sec ? Estimate the magnitude of these changes and write them down. Now increase the demand for water at F and perform a new network analysis. Did you reason correctly? List the original and new pressure at node F. Now restore the demand at node F to its original value and proceed to the next question. (f) What would happen to the flow rate in pipe EF if a new pipe were added to the system from node G to halfway between nodes A and D? What would happen to the pressure at F? Add this new pipe with the same characteristics as pipe DE and perform a new network analysis. Did you reason correctly? Now restore the network to its original configuration. (g) Perform any other changes that your instructor requests.
Example 4.8 A water-supply distribution system for an industrial Park is schematically shown in Figure 4.9 (a). The demands on the system are currently at junctions C, G, and F with flow rates given in liters per second. Water enters the system at junction A from a water storage tank on a hill. The water surface elevation in the tank is 50 m above the elevation of point A in the industrial park. All the junctions have the same elevation as point A. All pipes are aged ductile iron (e = 0.26 mm) with lengths and diameters provided in the table below. Calculate the flow rate in each pipe. Also determine if the pressure at junction F will be high enough to satisfy the customer there. The required pressure is 185 kPa. A table of pipe and system geometry is a convenient way to organize the available information and make some preliminary calculations. The table below has been set up for that purpose. The first column identifies all of the pipes in the network. Column 2 contains flow rates for each pipe, which were estimated to initiate the Hardy-Cross algorithm. These estimated flow rates are shown in brackets on the system Q=300 А [200] B [80] C [120] 2 [30] G 1 Q = 50 H [100] [0] E [20] 3 D [50] 0 = 100 F [100] [100] Q = 150 (a) Q = 300 A 205 B 80 c 125 30 G 2 33 95 1 O= 50 H 8 E Q = 100 3 63 D F 95 87 Q = 150 (b)
108 Pipelines and Pipe Networks Chap. 4 schematic in Figure 4.9 (a). Note that mass balance was maintained at each junction. Flow direction in the table is indicated using the junction letters that define the pipe. For example, flow in pipe AB is from junc- tion A to junction B. Friction factors (column 6) are found assuming complete turbulence and read from the Moody diagram using e/D or, alternatively, from Equation 3.23. The “K” coefficient (column 7) is used later in the procedure to obtain the head loss in each pipe according to Equation 4.12. Pipe Flow (m/sec) Length (m) Diameter (m) elD f K(sec/m) 300 250 350 125 AB AD BC BG GH CH DE GE EF HF 350 0.20 0.10 0.08 0.12 0.02 0.03 0.10 0.00 0.10 0.05 0.30 0.25 0.20 0.20 0.20 0.20 0.20 0.15 0.20 0.15 125 300 125 350 125 0.00087 0.00104 0.00130 0.00130 0.00130 0.00130 0.00130 0.00173 0.00130 0.00173 0.019 0.020 0.021 0.021 0.021 0.021 0.021 0.022 0.021 0.022 194 423 1,900 678 1,900 678 1,630 2.990 1,900 2,990 Solution The Hardy-Cross method utilizes a relaxation technique (method of successive approximations). Using the estimated flow rates, head losses are found in each pipe by using Equation 4.12, one loop at a time. Equation 4.17a is then used to determine a flow correction and thus improve the flow estimate. The same procedure is applied to all the remaining loops, and then the cycle repeats itself. The process is ended when the flow corrections become acceptably small. At this point, conservation of mass is satisfied for each junction, and the loss of head around each loop is the same for counterclockwise and clockwise flow (conservation of energy). We will proceed through the calculations by using a series of tables; explanations will follow each table. Having said all that, let us commence with loop 1. Loop Pipe Q (m/sec) K (sec?/m) hệ (m) hy/Q (sec/m?) New 0(m/sec) 1 AB BG GE AD DE 0.200 0.120 0.000 (0.100) (0.100) 194 678 2.990 423 1,630 7.76 9.76 0.00 (4.23) (16.3) 38.8 81.3 0.0 (42.3) (163.0) 0.205 0.125 0.005 (0.095) (0.095) The flows listed in column 3 are the original estimates. Flows in loop 1 that are counterclockwise are placed in parentheses. Head losses are computed from: hy = KQ The flow correction is found using Equation 4.17a. Σhic - Σhic (7.76 +9.76) - (4.23 + 16.3) AQ -0.005 m/sec 2[z(hc/Qc) + 3(hce/Qcc)] 2[(38.8 + 81.3) + (42.3 + 163.0)] The negative sign on the flow adjustment indicates that counterclockwise head losses dominate (Ehfcc > Ehfe). Therefore, the flow correction of 0.005 m/sec is applied in the clockwise direction [column 7 (“New C")]. This will help to equalize the losses in the next iteration. Now we continue with loop 2.
Sec. 4.4 Pipe Networks 109 Loop Pipe (m/sec) K sec/m) h, (m) // (sec/m?) New Q (m/sec) 2 BC CH BG GH 0.080 0.030 (0.125) (0.020) 1,900 678 678 1,900 12.2 0.61 (10.6) (0.76) 152.5 20.3 (84.8) (38.0) 0.078 0.028 (0.127) (0.022) Because pipe BG is shared by loops 1 and 2, the revised flow from the calculation in loop 1 is used here. Note that in loop 1 the flow in BG is clockwise; in loop 2, the flow is counterclockwise. The flow correction is found to be Ehre - She (12.2 +0.61) - (10.6 + 0.76) +0.002 m/sec 2[2(hyc/2) + 2(hyc/2)] 2[(152.5 + 20.3) + (84.8 + 38.0)] Clockwise flow dominates the loss so the correction of 0.002 m²/sec is added in the counterclockwise direction. We complete the first iteration by correcting flows in loop 3. AQ m Loop Pipe 0(m/sec) 3 GH HF GE 0.022 0.050 (0.005) (0.100) K (sec?/m”) h(m) h[/Q (sec/m?) New Q (m/sec) 1.900 0.92 41.8 0.035 2,990 7.48 149.6 0.063 2.990 (0.07) (14.0) 0.008 1,900 (19.0) (190.0) (0.087) = Σhic - Σhics (0.92 + 7.48) - (0.07 + 19.0) ΔΩ: 2[3 (h/c/Qc) + 3(hjee/Qcc)] 2[(41.8 + 149.6) + (14.0 + 190.0)] = -0.013 m/sec Counterclockwise head losses dominate so the flow correction is added in the clockwise direction. Note that this is a large enough correction to reverse the flow direction in GE; it will be labeled EG the next time. We now begin the second iteration with loop 1. Loop Pipe Q(m/sec) K (sec/m) h; (m) hy/Q (sec/m?) New Q (m/sec) 1 AB BG AD DE EG 0.205 0.127 (0.095) (0.095) (0.008) 194 678 423 1,630 2.990 8.15 10.9 (3.82) (14.7) (0.19) 39.8 85.8 (40.2) (154.7) (23.8) 0.205 0.127 (0.095) (0.095) (0.008) Σhic - Σhicc (8.15 + 10.9) - (3.82 + 14.7 + 0.19) AQ 2[z(h;/Qc) + 3 (hyce/Oce)] 2[(39.8 + 85.8) + (40.2 + 154.7 + 23.8)] +0.000 m/sec The correction is very small (< 0.0005 m/sec). We continue on to loop 2. = Loop Pipe 2 BC CH BG GH 0 m/sec) 0.078 0.028 (0.127) (0.035) K (sec/m) nf (m) 15/0 (sec/m?) New Q (m/sec) 1.900 11.6 148.7 0.080 678 0.53 18.9 0.030 678 (10.9) (85.8) (0.125) 1.900 (2.33) (66.6) (0.033)
110 Pipelines and Pipe Networks Chap 4 -0.002 m/sec Σhic - Σhic (11.6 + 0.53) - (10.9 + 2.33) AQ 2[(hc/Qc) + 3 (hyce/Qcc)] 2[(148.7 + 18.9) + (85.8 +66.6)] Once again the correction appears to be acceptably small. Finally, we check loop 3. Loop Pipe Q (m/sec) K (sec?/m") hf (m) hy/sec/m?) New 0(m/sec) 3 GH HF EG EF 0.033 0.063 0.008 (0.087) 1,900 2,990 2.990 1,900 2.07 11.9 0.19 (14.4) 62.7 188.9 23.8 (165.5) 0.033 0.063 0.008 (0.087) + + Σhic - Σhree (2.07 + 11.9 +0.19) - (14.4) AQ 2[z(hyd/.) + 3 (hyed/Qcc)] 2(62.7 + 188.9 + 23.8) + (165.5)] = -0.000 m/sec Because the correction is very small on all three loops, the flow rates are accepted and the process is ended. The final flows appear in Figure 4.9(b). The following table below summarizes information about the pipe system. Final head losses are determined by using final flows and Equation 4.12. The head loss is converted to a pressure drop in the last column (AP = yhy). Pipe Q (L/sec) Length (m) Diameter (cm) hj (m) AP (kPa) 300 250 AB AD BC BG GH CH DE EG EF HF 205 95 80 125 33 30 95 8 87 63 350 125 350 125 300 125 350 125 30 25 20 20 20 20 20 15 20 ū8888888 8.2 3.8 12.2 10.6 2.1 0.6 14.7 0.2 14.4 11.9 80.3 37.2 119.4 103.8 20.6 5.9 143.9 2.0 141.0 116.5 Now we determine whether or not the pressure at junction F is large enough to satisfy the customer at that location. We will use an energy balance to do this. But first, because the water surface elevation in the tank is 50 m above junction A, the pressure there is P = yh = (9790 N/m?)(50 m) = 489.5 kPa The pressure at junction F may now be determined by subtracting the pressure drops in pipes AD, DE, and EF or any alternative route from A to F. (In this case, all the junctions are at the same elevation and the vari- ations in the velocity heads are negligible, so an energy balance only involves pressure heads and friction head losses. Therefore, Pp = PA - APAD – APDE - APEF = 489.5 – 37.2 – 143.9 – 141.0 = 167.4 kPa Because the pressure is less than 185 kPa, the industrial customer is not likely to be satisfied. Minor losses in the system were not accounted for, so the pressure is likely to be even lower when the system is running at the full demand specified. It is left to the student to suggest modifications to the system to accommodate the customer (Problem 4.4.1). Also note that the pressure at F could have been determined by subtracting head losses from the total head at A and converting the pressure head to pressure.
The procedure outlined above for the Hardy-Cross method is valid if all the inflows entering the network are known. In practice, this occurs when there is only one source of inflow. In this case, the inflow rate is equal to the sum of the known withdrawal rates at all the junctions. However, if the flow is supplied to the network from two or more sources, as shown in Figure 4.10 (a), the inflow rates entering the network will not be known a priori. Therefore, we need to add inflow path calculations to the procedure. The number of inflow paths to be considered is equal to the number of inflow sources minus one. In Figure 4.10 (a), there are two reservoirs supplying flow to the network. Therefore, only one inflow path needs to be considered. This can be any path connecting the two reservoirs. For exam- ple, one can choose inflow path ABCDG in Figure 4.10 (a). There are several other possibilities like ABFEDG, GDCFBA, and so on. The results will not be affected by the choice of the inflow path. Once an inflow path is selected, the path calculations are carried out in a manner similar to the loop calculations. Using the subscript p (path) to denote the flows in the same direction as the inflow path followed and cp (counterpath) to denote the flows in the opposite direction, the dis- charge correction 4Q is calculated as (Ehfp - Ehfcp) + Hd - H. AQ = (4.18a) hip Σ- + Σ" Qp where H, and H, are the total heads at the beginning (upstream) point of the inflow path and at the end (downstream) point of the path. For path ABCDG in Figure 4.10a, H, = HA and Ha = H. A positive value of AQ indicates that losses in the direction of the path are dominat- ing. Therefore, the correction would be applied in the counterpath direction. In other words, the flows along the inflow path direction would be decreased in absolute value, and those in the opposite direction would be increased by AQ. Equation 4.18a can be used in conjunction with the Manning equation (3.28), as well as the Darcy-Weisbach equation (3.16). However, when the Hazen-Williams equation (3.27) is used, the equation is reformulated as (Ehfp - Ehfcp) + Hd - H. ΔΩ (4.18b) 1.85| Σ + Σ- Op hfcp z ( zakoni + 2 lcp hip hrer ( 2 lcp HG HA G А Qc c D B L h OF F E Figure 4.10 (a) Pipe network with two reservoir sources