Find: QBE gal min 2,000 200 A F B 100 500 Pipe AB BC CD DE EF FA BE C *All flows in [gpm] B 100 500 Pipe AB BC CD DE EF FA BE C 120 L[ft] 400 600 18 d [in] 12 8 E 400 QBE 3,600 C D Find the flowrate through pipe BE, in gallons per minute. [pause] In this problem, --- 2,000 600 A) 2,340 C) 2,620 B) 2,480 D) 2,760
Find: QBE gal min 2,000 200 A F B 100 500 Pipe AB BC CD DE EF FA BE C *All flows in [gpm] B 100 500 Pipe AB BC CD DE EF FA BE C 120 L[ft] 400 600 18 d [in] 12 8 E 400 QBE 3,600 C D 4,400 gallons per minute flow through a pipe network consisting of --- 2,000 600 A) 2,340 C) 2,620 B) 2,480 D) 2,760
Find: QBE gal min 2,000 200 A F B 100 500 Pipe AB BC CD DE EF FA BE C *All flows in [gpm] B 100 500 Pipe AB BC CD DE EF FA BE C 120 L[ft] 400 600 18 d [in] 12 8 E 400 QBE 3,600 C D Supply nodes A, B and C, and demand nodes, D, E and F, and the pipe network, which connect these nodes. 2,000 600 A) 2,340 C) 2,620 B) 2,480 D) 2,760
Find: QBE gal min 2,000 200 A F B 100 500 Pipe AB BC CD DE EF FA BE C *All flows in [gpm] B 100 500 Pipe AB BC CD DE EF FA BE C 120 L[ft] 400 600 18 d [in] 12 8 E 400 QBE 3,600 C D All flowrates listed in this problem are in gallons per minute. 2,000 600 A) 2,340 C) 2,620 B) 2,480 D) 2,760
Find: QBE gal min 2,000 200 A F B 100 500 Pipe AB BC CD DE EF FA BE C *All flows in [gpm] B 100 500 Pipe AB BC CD DE EF FA BE C 120 L[ft] 400 600 18 d [in] 12 8 E 400 QBE 3,600 C D The nodes are connected by pipes, whose roughness, lengths and diameters are provided. [pause] Network flows are governed by ---- 2,000 600 A) 2,340 C) 2,620 B) 2,480 D) 2,760
Find: QBE gal min 2,000 200 A F B 0 = Σ Q node i E 400 QBE 3,600 C D *All flows in [gpm] B n 0 = Σ Q node i E 400 i=1 QBE 3,600 C D the conservation of mass at each node, 2,000 600
Find: QBE gal min 2,000 200 A F B 0 = Σ Q node i E 400 QBE 3,600 *All flows in [gpm] B n 0 = Σ Q node i E 400 i=1 QBE n 3,600 0 = Σ h pipe i i=1 C D and the conservation of energy, around each loop. 2,000 600
Find: QBE gal min 2,000 200 A F B 0 = Σ Q node i E 400 QBE 3,600 *All flows in [gpm] B n 0 = Σ Q node i E 400 i=1 QBE n 3,600 0 = Σ h pipe i i=1 C D Our strategy will be to initially guess the flowrates through the network, such that the first equation is satisfied, --- 2,000 600
? Find: QBE gal min 2,000 200 A F B 0 = Σ Q node i E 400 QBE 3,600 *All flows in [gpm] B n 0 = Σ Q node i E 400 i=1 QBE n 3,600 0 = Σ h pipe i ? i=1 C D and then solve for the headloss around each loop. Depending on the calculated headloss, the flowrates will be revised as necessary, and after enough iterations, --- 2,000 600
Find: QBE gal min 2,000 200 A F B 0 = Σ Q node i E 400 QBE 3,600 *All flows in [gpm] B n 0 = Σ Q node i E 400 i=1 QBE n 3,600 0 = Σ h pipe i i=1 C D of adjusting the flowrates, the sum of the headlosses around each individual loop will equal zero. 2,000 600
Find: QBE gal min h= 2,000 200 A F B 0 = Σ Q node i E 400 QBE 3,600 *All flows in [gpm] B n 0 = Σ Q node i E 400 i=1 QBE n 3,600 0 = Σ h pipe i i=1 C D Using the given data, we’ll want to develop an expression for the headloss through each pipe for a given flowrate. 2,000 600 10.44 * L * Q1.85 h= C1.85 * d4.87
Find: QBE gal min h= 2,000 200 A F B 0 = Σ Q node i E 400 QBE 3,600 *All flows in [gpm] B n 0 = Σ Q node i E 400 i=1 QBE n 3,600 0 = Σ h pipe i i=1 C D [ft] In this equation, the headloss is in units of feet, --- 2,000 600 10.44 * L * Q1.85 h= C1.85 * d4.87
Find: QBE gal min h= 2,000 200 A F B 0 = Σ Q node i E 400 QBE 3,600 *All flows in [gpm] B n 0 = Σ Q node i E 400 i=1 QBE n 3,600 0 = Σ h pipe i i=1 C D [ft] the length of the pipe is in units of feet, --- [ft] 2,000 600 10.44 * L * Q1.85 h= C1.85 * d4.87
Find: QBE gal min h= 2,000 200 A F B 0 = Σ Q node i E 400 QBE 3,600 *All flows in [gpm] B n 0 = Σ Q node i E 400 i=1 QBE n 3,600 0 = Σ h pipe i i=1 C D [ft] gal the flowrate, is in gallons per minute, the friction coefficient --- [ft] 2,000 min 600 10.44 * L * Q1.85 h= C1.85 * d4.87
Find: QBE gal min h= 2,000 200 A F B 0 = Σ Q node i E 400 QBE 3,600 *All flows in [gpm] B n 0 = Σ Q node i E 400 i=1 QBE n 3,600 0 = Σ h pipe i i=1 C D [ft] gal is unitless, and the diameter is in inches. [pause] [ft] 2,000 min 600 10.44 * L * Q1.85 h= C1.85 * d4.87 [in]
Find: QBE gal min h= 2,000 200 A F 100 500 Pipe AB BC CD DE EF FA BE C 120 L[ft] 400 600 18 d [in] 12 8 B E 400 3,600 C D Returning to our given data, we notice we’ve been given ---- 2,000 600 10.44 * L * Q1.85 h= C1.85 * d4.87
Find: QBE gal min h= 2,000 200 A F 100 500 Pipe AB BC CD DE EF FA BE C 120 L[ft] 400 600 18 d [in] 12 8 B E 400 3,600 C D the friction coefficient, the length and the diameter. For convenience, we’ll define the variable K, equal to --- 2,000 600 10.44 * L * Q1.85 h= C1.85 * d4.87
Find: QBE gal min K h= 2,000 200 A F 100 500 Pipe AB BC CD DE EF FA BE 120 L[ft] 400 600 18 d [in] 12 8 B E 400 3,600 C D 10.44 times the length, divided by the friction coefficient to the 1.85 power divided by the diameter raised to the 4.87 power. K 2,000 600 10.44 * L * Q1.85 h= C1.85 * d4.87
Find: QBE gal min K h= 2,000 200 A F 100 500 Pipe AB BC CD DE EF FA BE 120 L[ft] 400 600 18 d [in] 12 8 B E 400 3,600 C D Plugging in the known values for L, C and d, the values of K --- K 2,000 600 10.44 * L * Q1.85 h= C1.85 * d4.87
Find: QBE gal min K h= 2,000 200 A F Pipe K B E AB 3.30*10-6 400 BC 3.57*10-5 3,600 CD 4.17*10-5 DE 2.38*10-5 C D are calculated. [pause] The pipe network has two loops, which we’ll define as a --- EF 3.57*10-5 K 2,000 600 FA 4.13*10-6 10.44 * L * Q1.85 BE 8.03*10-7 h= C1.85 * d4.87
Find: QBE gal min 1 2 K h= 2,000 200 A F Pipe K B E AB 3.30*10-6 400 BC 3.57*10-5 3,600 2 CD 4.17*10-5 DE 2.38*10-5 C D Loop 1 and Loop 2. When calculating the headloss around a loop, --- EF 3.57*10-5 K 2,000 600 FA 4.13*10-6 10.44 * L * Q1.85 BE 8.03*10-7 h= C1.85 * d4.87
Find: QBE gal min 1 2 K h= 2,000 200 A F Pipe K B E AB 3.30*10-6 400 BC 3.57*10-5 3,600 2 CD 4.17*10-5 DE 2.38*10-5 C D we’ll sum the headlosses in clockwise direction. [pause] Now it time to guess the initial flowrates --- EF 3.57*10-5 K 2,000 600 FA 4.13*10-6 10.44 * L * Q1.85 BE 8.03*10-7 h= C1.85 * d4.87
Find: QBE gal min 1 2 K h= 2,000 200 A F Pipe K B E AB 3.30*10-6 400 BC 3.57*10-5 3,600 2 CD 4.17*10-5 DE 2.38*10-5 C D through each pipe in the network, using our best judgment, --- EF 3.57*10-5 K 2,000 600 FA 4.13*10-6 10.44 * L * Q1.85 BE 8.03*10-7 h= C1.85 * d4.87
Find: QBE gal min 2,000 200 A F 8.03*10-7 Pipe AB BC CD DE EF FA BE K 3.30*10-6 3.57*10-5 4.17*10-5 2.38*10-5 4.13*10-6 B E 400 3,600 C D and without violating the conservation of mass. [pause] To do so, it would be prudent to review the possible solutions, --- 2,000 600 n 0 = Σ Q node i i=1
2,000 200 gal min Find: QBE A F 2,340 2,480 2,620 2,760 B E 400 3,600 C D for the correct flowrate through pipe BE. For the first iteration, we’ll guess ---- 2,000 600 n 0 = Σ Q node i i=1
Find: QBE gal min 2,000 200 A F 2,340 2,480 2,620 2,760 B E 400 guess 3,600 C D 2,620, gallons per minute, through pipe BE. [pause] After that, we’ll guess --- 2,000 600 n 0 = Σ Q node i i=1
2,000 200 gal min 780 Find: QBE A F 1,220 580 2,340 2,480 2,620 2,760 B E 400 guess 2,620 3,600 1,000 400 C D the flowrates for the remaining pipes. The headloss through each pipe, --- 1,000 2,000 600 n 0 = Σ Q node i i=1
2,000 200 gal min 780 Find: QBE A F 1,220 580 2,340 2,480 2,620 2,760 B E 400 guess 2,620 3,600 1,000 400 C D is equal to K times the flowrate, in gallons per minute, raised to the 1.85 power. Using our values of K --- 1,000 2,000 600 h = K * Q1.85
Find: QBE gal min 2,000 200 780 A F 1,220 580 Pipe K B E AB 3.30*10-6 400 BC 3.57*10-5 2,620 3,600 1,000 CD 4.17*10-5 400 DE 2.38*10-5 C D we previously calculated, We’ll add a columns for flow and --- EF 3.57*10-5 1,000 2,000 600 FA 4.13*10-6 BE 8.03*10-7 h = K * Q1.85
Find: QBE gal min Pipe K Q h [ft] AB 3.30*10-6 1,220 1,220 BC 3.57*10-5 1,000 1,000 CD 4.17*10-5 1,000 1,000 DE 2.38*10-5 400 -400 headloss. Before calculating the headloss through each pipe, we’ll look at each loop individually. EF 3.57*10-5 580 580 FA 4.13*10-6 780 780 BE 8.03*10-7 2,620 2620
Find: QBE gal min Loop 1 2,000 200 780 A F 1,220 580 B E 400 2,620 3,600 Loop 1 Pipe K Q h [ft] AB For Loop 1, we notice pipes AF and FE flow in clockwise direction, and pipes --- 3.30*10-6 1,220 1,220 EF 3.57*10-5 580 1,000 FA 4.13*10-6 780 1,000 BE 8.03*10-7 2,620 -400
Find: QBE gal min CW flow CCW flow Loop 1 2,000 200 780 A F 1,220 580 400 2,620 3,600 Loop 1 Pipe AB K 3.30*10-6 Q 1,220 h [ft] 1,000 -400 8.03*10-7 EF FA BE 3.57*10-5 4.13*10-6 2,620 580 780 AB and BE flow in the counter-clockwise direction. Therefore, the headloss when traveling through these pipes is, ---
Find: QBE - + + - gal min CW flow CCW flow Loop 1 2,000 200 780 A F 1,220 CW flow 580 B CCW flow E 400 2,620 3,600 Loop 1 Pipe K Q h [ft] - AB positive through the clockwise pipes and negative through the counterclockwise pipes. 3.30*10-6 1,220 + EF 3.57*10-5 580 + FA 4.13*10-6 780 - BE 8.03*10-7 2,620
Find: QBE gal min Loop 1 2,000 200 780 A F 1,220 580 B E 400 2,620 3,600 Loop 1 h = K * Q1.85 Pipe K Q h [ft] AB The headlosses are calculated, and the sum of these headlosses equals --- 3.30*10-6 1,220 -1.69 EF 3.57*10-5 580 4.62 FA 4.13*10-6 780 0.92 BE 8.03*10-7 2,620 -1.69
Find: QBE Σ hi=2.16 [ft] gal min Loop 1 2,000 200 780 A F 1,220 580 B 400 2,620 3,600 Loop 1 h = K * Q1.85 Pipe K Q h [ft] AB 2.16 feet. This means, that for the assumed flowrates, --- 3.30*10-6 1,220 -1.69 EF 3.57*10-5 580 4.62 FA 4.13*10-6 780 0.92 BE 8.03*10-7 2,620 -1.69
Find: QBE Σ hi=2.16 [ft] gal min Loop 1 2,000 200 780 A F 1,220 580 B 400 2,620 3,600 Loop 1 h = K * Q1.85 Pipe K Q h [ft] AB the flow through route AFE incurs 2.16 more feet of head, than the flow, through route ABE. 3.30*10-6 1,220 -1.69 EF 3.57*10-5 580 4.62 FA 4.13*10-6 780 0.92 BE 8.03*10-7 2,620 -1.69
Find: QBE Σ hi=2.16 [ft] gal min Loop 1 2,000 200 780 A F 1,220 580 B 400 2,620 3,600 Loop 1 h = K * Q1.85 Pipe K Q h [ft] AB In order to satisfy the conservation of energy equation, where --- 3.30*10-6 1,220 -1.69 EF 3.57*10-5 580 4.62 FA 4.13*10-6 780 0.92 BE 8.03*10-7 2,620 -1.69
Find: QBE Σ hi=2.16 [ft] gal min Loop 1 2,000 200 780 A F 1,220 580 B 0 = Σ h link i E 400 i=1 2,620 3,600 Loop 1 h = K * Q1.85 Pipe K Q h [ft] AB the sum of the headlosses around the loop equals zero, --- 3.30*10-6 1,220 -1.69 EF 3.57*10-5 580 4.62 FA 4.13*10-6 780 0.92 BE 8.03*10-7 2,620 -1.69
Find: QBE Σ hi=2.16 [ft] gal min Loop 1 2,000 200 780 A F 1,220 580 B 0 = Σ h link i E 400 i=1 2,620 3,600 Loop 1 h = K * Q1.85 Pipe K Q h [ft] AB we would have to increase the flow in the counter-clockwise direction --- 3.30*10-6 1,220 -1.69 EF 3.57*10-5 580 4.62 FA 4.13*10-6 780 0.92 BE 8.03*10-7 2,620 -1.69
Find: QBE Σ hi=2.16 [ft] gal min Loop 1 2,000 200 780 A F 1,220 580 +Q 0 = Σ h link i E 400 i=1 2,620 3,600 Loop 1 h = K * Q1.85 Pipe K Q h [ft] AB a certain amount, for all pipes in the loop. Before we do that, we’ll look at Loop 2. 3.30*10-6 1,220 -1.69 EF 3.57*10-5 580 4.62 FA 4.13*10-6 780 0.92 BE 8.03*10-7 2,620 -1.69
Find: QBE gal min Loop 2 B 2,620 E 400 3,600 1,000 400 C D 1,000 2,000 Pipe K Q h [ft] BC Loop 2 is calculated the same way as loop 1. The clockwise headlosses are positive, and --- 3.57*10-5 1,000 CD 4.17*10-5 1,000 DE 2.38*10-5 400 BE 8.03*10-7 2,620
Find: QBE + - - + gal min CW flow CCW flow Loop 2 B 2,620 E 400 3,600 1,000 400 CW flow C D CCW flow 1,000 2,000 600 Loop 2 Pipe K Q h [ft] + BC and the counter-clockwise headlosses are negative. The headlosses are calculated --- --- 3.57*10-5 1,000 - CD 4.17*10-5 1,000 - DE 2.38*10-5 400 + BE 8.03*10-7 2,620
Find: QBE Σ hi=-1.98 [ft] gal min Loop 2 B 2,620 E 400 3,600 1,000 400 C D 1,000 2,000 600 Loop 2 h = K * Q1.85 Pipe K Q h [ft] BC and summed, and we learn that, for the given flowrates, --- 3.57*10-5 1,000 12.66 CD 4.17*10-5 1,000 -14.78 DE 2.38*10-5 400 -1.55 BE 8.03*10-7 2,620 1.69
Find: QBE Σ hi=-1.98 [ft] gal min Loop 2 B 2,620 E 400 3,600 1,000 400 C D 1,000 2,000 600 Loop 2 h = K * Q1.85 Pipe K Q h [ft] BC there is 1.98 feet more headloss incurred along route CDE, than along route --- 3.57*10-5 1,000 12.66 CD 4.17*10-5 1,000 -14.78 DE 2.38*10-5 400 -1.55 BE 8.03*10-7 2,620 1.69
Find: QBE Σ hi=-1.98 [ft] gal min Loop 2 B 2,620 E 400 3,600 1,000 400 C D 1,000 2,000 600 Loop 2 h = K * Q1.85 Pipe K Q h [ft] BC CBE. Therefore, to equalize the headlosses, --- 3.57*10-5 1,000 12.66 CD 4.17*10-5 1,000 -14.78 DE 2.38*10-5 400 -1.55 BE 8.03*10-7 2,620 1.69
Find: QBE Σ hi=-1.98 [ft] gal min Loop 2 B 2,620 E 400 3,600 1,000 +Q C D 1,000 2,000 600 Loop 2 h = K * Q1.85 Pipe K Q h [ft] BC the flowrates should be increased by a certain amount in the clockwise direction, for all pipes in the loop. [pause] If we look at our entire network again, ---- 3.57*10-5 1,000 12.66 CD 4.17*10-5 1,000 -14.78 DE 2.38*10-5 400 -1.55 BE 8.03*10-7 2,620 1.69
2,000 200 gal min 780 Find: QBE A F 1,220 580 1 2,340 2,480 2,620 2,760 B E 400 guess 2,620 3,600 1,000 2 400 C D we recall that Loop 1 should increase flows in the --- 1,000 2,000 600
2,000 200 gal min 780 Find: QBE A F 1,220 580 +Q 2,340 2,480 2,620 2,760 B E 400 guess 2,620 3,600 1,000 +Q 400 C D counter-clockwise direction, and Loop 2 should increase flows in the clockwise direction. 1,000 2,000 600
2,000 200 gal min 780 Find: QBE A F 1,220 580 +Q 2,340 2,480 2,620 2,760 B E 400 guess 2,620 3,600 1,000 +Q 400 C D Both of these changes would result in a higher flowrate in pipe BE. 1,000 2,000 600
2,000 200 gal min 780 Find: QBE A F 1,220 580 +Q 2,340 2,480 2,620 2,760 B E 400 guess 2,620 3,600 1,000 +Q 400 C D Since our guess for the flowrate through pipe BE was 2,620 gallons per minute, and calculations from the first iteration indicate we increase the flowrates, through this pipe, --- 1,000 2,000 600
2,000 200 gal min 780 Find: QBE A F 1,220 580 +Q 2,340 2,480 2,620 2,760 B E 400 3,600 guess 1,000 +Q 400 C D the flowrate through pipe BE must be 2,760, gallons per minute. The remaining pipe flows are guessed and adjusted until --- 1,000 2,000 600
2,000 200 gal min 670 Find: QBE A F 1,330 470 2,340 2,480 2,620 2,760 B E 400 2,760 3,600 1,030 370 C D n the conservation of mass and the conservation energy equations are both satisfied. 0 = Σ Q node i 970 2,000 600 i=1 n 0 = Σ h pipe i i=1
2,000 200 gal min 670 Find: QBE A F 1,330 470 2,340 2,480 2,620 2,760 B E 400 2,760 3,600 1,030 370 C D n Since the flowrate through pipe BE equals 2,760 gallons per minute, --- 0 = Σ Q node i 970 2,000 600 i=1 n 0 = Σ h pipe i i=1
Find: QBE gal min AnswerD 2,000 200 670 A F 1,330 470 2,340 2,480 2,620 2,760 B E 400 2,760 3,600 1,030 370 C D n the answer is D. 0 = Σ Q node i 970 2,000 600 i=1 n 0 = Σ h pipe i AnswerD i=1
Find: QC [L/s] 3.4 34 340 3,400 Δh=20 [m] Tank pipe A 1 pipe B Tank 2 pipe C (all pipes) pipe D pipe d [cm] L [m] 3.4 34 340 3,400 A 50 200 B 20 400 C 40 400 D 50 500
? Index σ’v = Σ γ d γT=100 [lb/ft3] +γclay dclay 1 Find: σ’v at d = 30 feet (1+wc)*γw wc+(1/SG) σ’v = Σ γ d d Sand 10 ft γT=100 [lb/ft3] 100 [lb/ft3] 10 [ft] 20 ft Clay = γsand dsand +γclay dclay A W S V [ft3] W [lb] 40 ft text wc = 37% ? Δh 20 [ft] (5 [cm])2 * π/4