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 Find the flowrate through pipe C, in liters per second. [pause] In this problem, --- A 50 200 B 20 400 C 40 400 D 50 500

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 water flows from Tank 1 to Tank 2 through a pipe network, consisting to pipes A, B, C and D. The diameters and lengths --- A 50 200 B 20 400 C 40 400 D 50 500

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 of all 4 pipes are provided in the data table. The total headloss --- A 50 200 B 20 400 C 40 400 D 50 500

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 between the tanks, and the friction coefficient for all pipe, is provided. To begin, --- A 50 200 B 20 400 C 40 400 D 50 500

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 we’ll make some general observations about this particular pipe network, --- A 50 200 B 20 400 C 40 400 D 50 500

Find: QC [L/s] Δh=20 [m] Tank pipe A 1 pipe B Tank 2 f=0.03 pipe C (all pipes) pipe D pipe d [cm] L [m] Guiding Equations which will guide us, as we solve the problem. [pause] From the conservation of mass, we note that --- A 50 200 B 20 400 C 40 400 D 50 500

Find: QC [L/s] Δh=20 [m] Tank pipe A 1 pipe B Tank 2 f=0.03 pipe C (all pipes) pipe D pipe d [cm] L [m] Guiding Equations the flowrate through pipe A, must equal the flowrates through through pipes B and C, added together, which must also equal the flowrate though pipe D. A 50 200 QA = QB+QC = QD B 20 400 C 40 400 D 50 500

Find: QC [L/s] Δh=20 [m] Tank pipe A 1 pipe B Tank 2 f=0.03 pipe C (all pipes) pipe D pipe d [cm] L [m] Guiding Equations We’ll call this our first guiding equation. [pause] Since pipes B and C are parallel, --- A 50 200 QA = QB+QC = QD (1) B 20 400 C 40 400 D 50 500

Find: QC [L/s] Δh=20 [m] Tank pipe A 1 pipe B Tank 2 f=0.03 pipe C (all pipes) pipe D pipe d [cm] L [m] Guiding Equations the headlosses through these two pipes are equal, and this will be our second --- A 50 200 QA = QB+QC = QD (1) B 20 400 hL,B = hL,C C 40 400 D 50 500

Find: QC [L/s] Δh=20 [m] Tank pipe A 1 pipe B Tank 2 f=0.03 pipe C (all pipes) pipe D pipe d [cm] L [m] Guiding Equations guiding equation. [pause] Lastly, we know the total head loss between the two tanks equals 20 meters, --- A 50 200 QA = QB+QC = QD (1) B 20 400 hL,B = hL,C (2) C 40 400 D 50 500

Find: QC [L/s] Δh=20 [m] Tank pipe A 1 pipe B Tank 2 f=0.03 pipe C (all pipes) pipe D pipe d [cm] L [m] Guiding Equations So 20 meters equals the headloss through pipes A, C and D, will be our third guiding equation. [pause] Since the problem provides the --- A 50 200 QA = QB+QC = QD (1) B 20 400 hL,B = hL,C (2) C 40 400 20 [m] = hL,A + hL,C + hL,D D 50 500 (3)

Find: QC [L/s] Δh=20 [m] Tank pipe A 1 pipe B Tank 2 f=0.03 pipe C (all pipes) pipe D pipe d [cm] L [m] Guiding Equations friction factor, f, we’ll first solve for the --- A 50 200 QA = QB+QC = QD (1) B 20 400 hL,B = hL,C (2) C 40 400 20 [m] = hL,A + hL,C + hL,D D 50 500 (3)

Find: QC [L/s] Δh=20 [m] Tank pipe A 1 pipe B Tank 2 f=0.03 pipe C (all pipes) pipe D pipe d [cm] L [m] Guiding Equations flowrates through each pipe. A 50 200 QA = QB+QC = QD (1) B 20 400 hL,B = hL,C (2) C 40 400 20 [m] = hL,A + hL,C + hL,D D 50 500 (3)

Find: QC [L/s] hL=f * Δh=20 [m] Tank pipe A 1 pipe B Tank 2 f=0.03 pipe C (all pipes) pipe D pipe d [cm] L [m] Guiding Equations The flowrate divided by the area is substituted --- A 50 200 QA = QB+QC = QD (1) B 20 400 L * V2 C 40 400 hL=f * d * 2 * g D 50 500

Find: QC [L/s] hL=f * Δh=20 [m] Tank pipe A 1 pipe B Tank 2 f=0.03 pipe C (all pipes) pipe D pipe d [cm] L [m] Guiding Equations into the headloss equation, for the velocity. [pause] A 50 200 QA = QB+QC = QD (1) B 20 400 L * V2 Q C 40 400 hL=f * d * 2 * g A D 50 500

Find: QC [L/s] hL=f hL=f * * Δh=20 [m] Tank pipe A 1 pipe B Tank 2 pipe C (all pipes) pipe D L * Q2 pipe d [cm] L [m] hL=f If we revise this equation, --- * d * 2 * g * A2 A 50 200 B 20 400 L * V2 Q C 40 400 hL=f * d * 2 * g A D 50 500

Find: QC [L/s] hL=f hL=f * * Δh=20 [m] Tank pipe A 1 pipe B Tank 2 pipe C (all pipes) pipe D L * Q2 pipe d [cm] L [m] hL=f and then solve for the flowrate, Q, we can calcualte --- * d * 2 * g * A2 A 50 200 B 20 400 L * V2 Q C 40 400 hL=f * d * 2 * g A D 50 500

Find: QC [L/s] hL=f hL * d * 2 * g * A2 * Q= Δh=20 [m] Tank pipe A 1 pipe B Tank 2 f=0.03 pipe C (all pipes) pipe D L * Q2 pipe d [cm] L [m] hL=f the flowrate, for any pipe. For example, if we want the flowrate pipe A, --- * d * 2 * g * A2 A 50 200 hL * d * 2 * g * A2 B 20 400 Q= f * L C 40 400 D 50 500

Find: QC [L/s] hL=f hL,A * dA * 2 * g * A2 * Q= Δh=20 [m] Tank pipe A 1 pipe B Tank 2 f=0.03 pipe C (all pipes) pipe D L * Q2 pipe d [cm] L [m] hL=f we’ll substituted in the headloss, diameter, area, and length, unique to pipe A. [pause] This brings up the issue of units. * d * 2 * g * A2 A 50 200 hL,A * dA * 2 * g * A2 B 20 400 A Q= f * LA C 40 400 D 50 500

Find: QC [L/s] hL * d * 2 * g * A2 Q= Δh=20 [m] Tank pipe A 1 pipe B pipe C (all pipes) pipe D pipe d [cm] L [m] For this problem, we’ll use the following units, --- A 50 200 hL * d * 2 * g * A2 B 20 400 Q= f * L C 40 400 D 50 500

Find: QC [L/s] hL * d * 2 * g * A2 Q= Δh=20 [m] Tank pipe A 1 pipe B pipe C (all pipes) pipe D pipe d [cm] L [m] [m] head loss in meters, diamter in --- A 50 200 hL * d * 2 * g * A2 B 20 400 Q= f * L C 40 400 D 50 500

Find: QC [L/s] hL * d * 2 * g * A2 Q= Δh=20 [m] Tank pipe A 1 pipe B pipe C (all pipes) pipe D m pipe d [cm] L [m] [m] [m] s2 meters, acceleration in meters per second squared, area in -- A 50 200 hL * d * 2 * g * A2 B 20 400 Q= f * L C 40 400 D 50 500

Find: QC [L/s] hL * d * 2 * g * A2 Q= Δh=20 [m] Tank pipe A 1 pipe B pipe C (all pipes) pipe D m pipe d [cm] L [m] [m] [m] s2 [m2] meters squared, length in meters. And the flow rate in --- A 50 200 hL * d * 2 * g * A2 B 20 400 Q= f * L C 40 400 [m] D 50 500

Find: QC [L/s] hL * d * 2 * g * A2 Q= Δh=20 [m] Tank pipe A 1 pipe B pipe C (all pipes) pipe D m pipe d [cm] L [m] [m] [m] s2 [m2] meters cubed per second. For ease of calculations, let’s expand the data table to include --- A 50 200 hL * d * 2 * g * A2 B 20 400 Q= f * L C 40 400 m3 [m] D 50 500 s

Find: QC [L/s] hL * d * 2 * g * A2 Q= [m] [m] [m4] f * L [m] f=0.03 pipe d [cm] d [m] A2 [m4] L [m] a column for diameter, in meters, and a column for area squared, in meters to the 4th power. A 50 200 B 20 400 C 40 400 D 50 500

Find: QC [L/s] hL * d * 2 * g * A2 Q= [m] [m] [m4] f * L [m] f=0.03 pipe d [cm] d [m] A2 [m4] L [m] These values are calculated based on the given diameters, for each pipe. We’ll remove the first --- A 50 0.50 0.03853 200 B 20 0.20 9.872*10-4 400 C 40 0.40 0.01580 400 D 50 0.50 0.03853 500

Find: QC [L/s] hL * d * 2 * g * A2 Q= [m] [m] [m4] f * L [m] f=0.03 pipe d [m] A2 [m4] L [m] diameter column, and solve for the flowrate, --- A 0.50 0.03853 200 B 0.20 9.872*10-4 400 C 0.40 0.01580 400 D 0.50 0.03853 500

Find: QC [L/s] hL,A * dA * 2 * g * A 2 QA= f * LA f=0.03 pipe d [m] A2 [m4] L [m] Q, for pipe A. [pause] Substituting in the known values for --- A 0.50 0.03853 200 B 0.20 9.872*10-4 400 C 0.40 0.01580 400 D 0.50 0.03853 500

Find: QC [L/s] hL,A * dA * 2 * g * A 2 QA= 9.81 f * LA f=0.03 pipe m 9.81 s2 hL,A * dA * 2 * g * A 2 A QA= f * LA f=0.03 pipe d [m] A2 [m4] L [m] diameter, gravitational acceleration, area squared, friction coefficient, and length. The flowrate through pipe A equals, --- A 0.50 0.03853 200 B 0.20 9.872*10-4 400 C 0.40 0.01580 400 D 0.50 0.03853 500

Find: QC [L/s] hL,A * dA * 2 * g * A 2 QA= =0.2510 * hL,A 9.81 f * LA m 9.81 s2 hL,A * dA * 2 * g * A 2 A QA= =0.2510 * hL,A f * LA f=0.03 pipe d [m] A2 [m4] L [m] 0.2510 times the square root of the headloss through pipe A. [pause] If we calculate --- A 0.50 0.03853 200 B 0.20 9.872*10-4 400 C 0.40 0.01580 400 D 0.50 0.03853 500

Find: QC [L/s] hL,B * dB * 2 * g * A 2 hL,C * dB * 2 * g * A 2 QA=0.2510 * hL,A QB= f * LB hL,C * dB * 2 * g * A 2 hL,D * dB * 2 * g * A 2 C D QC= QD= f * LC f * LD pipe d [m] A2 [m4] L [m] the flowrate through the other three pipes, in the same manner, --- A 0.50 0.03853 200 B 0.20 9.872*10-4 400 C 0.40 0.01580 400 D 0.50 0.03853 500

Find: QC [L/s] QA=0.2510 * hL,A QB=0.0180 * hL,B QC=0.1017 * hL,C QD=0.1587 * hL,D pipe d [m] A2 [m4] L [m] we’ll find they are all a function of the square root of the headloss, through that particular pipe. [pause] To find all flowrates and all headlosses, --- A 0.50 0.03853 200 B 0.20 9.872*10-4 400 C 0.40 0.01580 400 D 0.50 0.03853 500

Find: QC [L/s] QA=0.2510 * hL,A QB=0.0180 * hL,B QC=0.1017 * hL,C 20 [m] = hL,A + hL,C + hL,D (3) QB=0.0180 * hL,B QC=0.1017 * hL,C QD=0.1587 * hL,D pipe d [m] A2 [m4] L [m] we’ll turn to our third guiding equation. [pause] If we represent the headloss --- A 0.50 0.03853 200 B 0.20 9.872*10-4 400 C 0.40 0.01580 400 D 0.50 0.03853 500

Find: QC [L/s] ƒ(hL,A ) ƒ(hL,A ) QA=0.2510 * hL,A QB=0.0180 * hL,B 20 [m] = hL,A + hL,C + hL,D (3) QB=0.0180 * hL,B QC=0.1017 * hL,C QD=0.1587 * hL,D pipe d [m] A2 [m4] L [m] across pipes C and D, as a function of the headloss across pipe A, --- A 0.50 0.03853 200 B 0.20 9.872*10-4 400 C 0.40 0.01580 400 D 0.50 0.03853 500

Find: QC [L/s] ƒ(hL,A ) ƒ(hL,A ) QA=0.2510 * hL,A QB=0.0180 * hL,B 20 [m] = hL,A + hL,C + hL,D (3) QB=0.0180 * hL,B QC=0.1017 * hL,C QD=0.1587 * hL,D we could solve for the headloss across pipe A, explicitly. The headloss across pipe D, --- Tank Δh=20 [m] pipe A 1 pipe B Tank 2 pipe C pipe D

Find: QC [L/s] ƒ(hL,A ) ƒ(hL,A ) QA=0.2510 * hL,A QB=0.0180 * hL,B 20 [m] = hL,A + hL,C + hL,D (3) QB=0.0180 * hL,B QC=0.1017 * hL,C QD=0.1587 * hL,D as a function of the headloss across pipe A, can be computed since --- Tank Δh=20 [m] pipe A 1 pipe B Tank 2 pipe C pipe D

Find: QC [L/s] ƒ(hL,A ) ƒ(hL,A ) QA=0.2510 * hL,A QB=0.0180 * hL,B 20 [m] = hL,A + hL,C + hL,D (3) QA = QB+QC = QD (1) QB=0.0180 * hL,B QC=0.1017 * hL,C QD=0.1587 * hL,D the flowrate through pipe A equals the flowrate through pipe D. [pause] Substituting in the flowrates, --- Tank Δh=20 [m] pipe A 1 pipe B Tank 2 pipe C pipe D

Find: QC [L/s] ƒ(hL,A ) ƒ(hL,A ) QA=0.2510 * hL,A QB=0.0180 * hL,B 20 [m] = hL,A + hL,C + hL,D (3) QA = QB+QC = QD (1) QB=0.0180 * hL,B QC=0.1017 * hL,C 0.2510* hL,A=0.1587* hL,D QD=0.1587 * hL,D for pipes A and D, we find that --- Tank Δh=20 [m] pipe A 1 pipe B Tank 2 pipe C pipe D

Find: QC [L/s] ƒ(hL,A ) ƒ(hL,A ) hL,D=2.5 * hL,A QA=0.2510 * hL,A 20 [m] = hL,A + hL,C + hL,D (3) QA = QB+QC = QD (1) QB=0.0180 * hL,B QC=0.1017 * hL,C 0.2510* hL,A=0.1587* hL,D QD=0.1587 * hL,D hL,D=2.5 * hL,A the headloss across pipe D equals 2.5 times the headloss across pipe pipe A. Tank Δh=20 [m] pipe A 1 pipe B Tank 2 pipe C pipe D

Find: QC [L/s] ƒ(hL,A ) 2.5 * hL,A QA=0.2510 * hL,A QB=0.0180 * hL,B 20 [m] = hL,A + hL,C + hL,D (3) QB=0.0180 * hL,B QC=0.1017 * hL,C QD=0.1587 * hL,D Next we’ll solve for the headloss across --- Tank Δh=20 [m] pipe A 1 pipe B Tank 2 pipe C pipe D

Find: QC [L/s] ƒ(hL,A ) 2.5 * hL,A QA=0.2510 * hL,A QB=0.0180 * hL,B 20 [m] = hL,A + hL,C + hL,D (3) QB=0.0180 * hL,B QC=0.1017 * hL,C QD=0.1587 * hL,D pipe C in terms of the headloss across pipe A. To do so, --- Tank Δh=20 [m] pipe A 1 pipe B Tank 2 pipe C pipe D

Find: QC [L/s] ƒ(hL,A ) 2.5 * hL,A QA=0.2510 * hL,A QB=0.0180 * hL,B 20 [m] = hL,A + hL,C + hL,D (3) QA = QB+QC = QD (1) QB=0.0180 * hL,B QC=0.1017 * hL,C QD=0.1587 * hL,D we note the flowrate through pipe A equals the flowrate across pipes B and C, combined. And we also know --- Tank Δh=20 [m] pipe A 1 pipe B Tank 2 pipe C pipe D

Find: QC [L/s] ƒ(hL,A ) 2.5 * hL,A QA=0.2510 * hL,A QB=0.0180 * hL,B 20 [m] = hL,A + hL,C + hL,D (3) QA = QB+QC = QD (1) QB=0.0180 * hL,B hL,B = hL,C (2) QC=0.1017 * hL,C QD=0.1587 * hL,D the headloss across pipe B equals the headloss across pipe C. This means, --- Tank Δh=20 [m] pipe A 1 pipe B Tank 2 pipe C pipe D

Find: QC [L/s] ƒ(hL,A ) 2.5 * hL,A QA=0.2510 * hL,A QB=0.0180 * hL,C 20 [m] = hL,A + hL,C + hL,D (3) QA = QB+QC = QD (1) QB=0.0180 * hL,C hL,B = hL,C (2) QC=0.1017 * hL,C QD=0.1587 * hL,D we can substitute headloss C, in for headloss B. Tank Δh=20 [m] pipe A 1 pipe B Tank 2 pipe C pipe D

Find: QC [L/s] ƒ(hL,A ) 2.5 * hL,A QA=0.2510 * hL,A QB=0.0180 * hL,C 20 [m] = hL,A + hL,C + hL,D (3) QA = QB+QC = QD (1) QB=0.0180 * hL,C hL,B = hL,C (2) QC=0.1017 * hL,C QB + QC =0.0180 * hL,C + 0.1017 * hL,C and the flowrate, through pipes B plus C, equals, --- Tank Δh=20 [m] pipe A 1 pipe B Tank 2 pipe C pipe D

Find: QC [L/s] ƒ(hL,A ) 2.5 * hL,A QA=0.2510 * hL,A QB=0.0180 * hL,C 20 [m] = hL,A + hL,C + hL,D (3) QA = QB+QC = QD (1) QB=0.0180 * hL,C hL,B = hL,C (2) QC=0.1017 * hL,C QB + QC =0.0180 * hL,C + 0.1017 * hL,C QB + QC =0.1197 * hL,C 0.1197 times the square root of the headloss across pipe C. Tank Δh=20 [m] pipe A 1 pipe B Tank 2 pipe C pipe D

Find: QC [L/s] ƒ(hL,A ) 2.5 * hL,A QA=0.2510 * hL,A QB=0.0180 * hL,C 20 [m] = hL,A + hL,C + hL,D (3) QA = QB+QC = QD (1) QB=0.0180 * hL,C QC=0.1017 * hL,C QB + QC =0.1197 * hL,C Plugging in the flowrates to equation 1, --- Tank Δh=20 [m] pipe A 1 pipe B Tank 2 pipe C pipe D

Find: QC [L/s] ƒ(hL,A ) 2.5 * hL,A QA=0.2510 * hL,A QB=0.0180 * hL,C 20 [m] = hL,A + hL,C + hL,D (3) QA = QB+QC = QD (1) QB=0.0180 * hL,C 0.2510 * hL,A =0.1197* hL,C QC=0.1017 * hL,C QB + QC =0.1197 * hL,C the headloss across pipe C is equal to --- Tank Δh=20 [m] pipe A 1 pipe B Tank 2 pipe C pipe D

Find: QC [L/s] ƒ(hL,A ) hL,C=4.397 * hL,A 2.5 * hL,A QA=0.2510 * hL,A 20 [m] = hL,A + hL,C + hL,D (3) QA = QB+QC = QD (1) QB=0.0180 * hL,C 0.2510 * hL,A =0.1197* hL,C QC=0.1017 * hL,C hL,C=4.397 * hL,A QB + QC =0.1197 * hL,C 4.397 times the headloss across pipe A. [pause] Now, equation 3 --- Tank Δh=20 [m] pipe A 1 pipe B Tank 2 pipe C pipe D

Find: QC [L/s] hL,C=4.397 * hL,A 2.5 * hL,A 4.397 * hL,A QA=0.2510 * hL,A 20 [m] = hL,A + hL,C + hL,D (3) QA = QB+QC = QD (1) QB=0.0180 * hL,C 0.2510 * hL,A =0.1197* hL,C QC=0.1017 * hL,C hL,C=4.397 * hL,A QB + QC =0.1197 * hL,C reduces to 1 unknown variable, the headloss across pipe A. Tank Δh=20 [m] pipe A 1 pipe B Tank 2 pipe C pipe D

Find: QC [L/s] 20 [m] = hL,A + 4.397 * hL,A+ 2.5 * hL,A 2.5 * hL,A 20 [m] = hL,A + hL,C + hL,D (3) 20 [m] = hL,A + 4.397 * hL,A+ 2.5 * hL,A [pause] The headloss across pipe A equals --- Tank Δh=20 [m] pipe A 1 pipe B Tank 2 pipe C pipe D

Find: QC [L/s] 20 [m] = hL,A + 4.397 * hL,A+ 2.5 * hL,A 20 [m] = hL,A + hL,C + hL,D (3) 20 [m] = hL,A + 4.397 * hL,A+ 2.5 * hL,A hL,A = 2.53 [m] 2.53 meters. [pause] This makes --- Tank Δh=20 [m] pipe A 1 pipe B Tank 2 pipe C pipe D

Find: QC [L/s] hL,B,hL,C =4.397 * hL,A hL,D=2.5 * hL,A 20 [m] = hL,A + hL,C + hL,D (3) 20 [m] = hL,A + 4.397 * hL,A+ 2.5 * hL,A hL,A = 2.53 [m] hL,B,hL,C =4.397 * hL,A hL,D=2.5 * hL,A the headloss through pipes B, C and D equal to --- Tank Δh=20 [m] pipe A 1 pipe B Tank 2 pipe C pipe D

Find: QC [L/s] hL,B,hL,C =4.397 * hL,A hL,B,hL,C =11.14 [m] 20 [m] = hL,A + hL,C + hL,D (3) 20 [m] = hL,A + 4.397 * hL,A+ 2.5 * hL,A hL,A = 2.53 [m] hL,B,hL,C =4.397 * hL,A hL,B,hL,C =11.14 [m] hL,D=2.5 * hL,A hL,D=6.33 [m] 11.14 meters, 11.14 meters, and 6.33 meters, respectively. Substituting these headloss --- Tank Δh=20 [m] pipe A 1 pipe B Tank 2 pipe C pipe D

Find: QC [L/s] hL,B,hL,C =11.14 [m] hL,D=6.33 [m] hL,A = 2.53 [m] QA=0.2510 * hL,A QB=0.0180 * hL,B QC=0.1017 * hL,C hL,A = 2.53 [m] hL,B,hL,C =11.14 [m] QD=0.1587 * hL,D hL,D=6.33 [m] values into our flowrate equations, we learn, the flowrates through pipes A, B, C and D are --- Tank Δh=20 [m] pipe A 1 pipe B Tank 2 pipe C pipe D

Find: QC [L/s] QA=0.2510 * 2.53 QB=0.0180 * 11.14 QC=0.1017 * 11.14 m3 s QA,QD=0.399 QB=0.0180 * 11.14 m3 s QB=0.060 QC=0.1017 * 11.14 m3 s QC=0.339 QD=0.1587 * 6.33 0.399 meters cubed per second, 0.060 meters cubed per second, 0.339 meter cubed per second, and 0.399 meters cubed per second, respectively. Tank Δh=20 [m] pipe A 1 pipe B Tank 2 pipe C pipe D

Find: QC [L/s] QA=0.2510 * 2.53 QB=0.0180 * 11.14 QC=0.1017 * 11.14 m3 QA,QD=0.399 s QB=0.0180 * 11.14 m3 QB=0.060 s QC=0.1017 * 11.14 m3 s QC=0.339 QD=0.1587 * 6.33 Since the question asks to find the flowrate through pipe C, Tank Δh=20 [m] pipe A 1 pipe B Tank 2 pipe C pipe D

Find: QC [L/s] QA=0.2510 * 2.53 QB=0.0180 * 11.14 QC=0.1017 * 11.14 m3 QA,QD=0.399 s QB=0.0180 * 11.14 m3 QB=0.060 s QC=0.1017 * 11.14 m3 s QC=0.339 QD=0.1587 * 6.33 L 1,000 m3 in liters per second, we multiply by 1,000, and the flowrate through pipe C equals, --- Tank Δh=20 [m] pipe A 1 pipe B Tank 2 pipe C pipe D

Find: QC [L/s] QA=0.2510 * 2.53 QB=0.0180 * 11.14 QC=0.1017 * 11.14 m3 QA,QD=0.399 s QB=0.0180 * 11.14 m3 QB=0.060 s QC=0.1017 * 11.14 m3 QC=0.339 s QD=0.1587 * 6.33 L QC=339 [L/s] 1,000 m3 339 liters per second. [pause] Tank Δh=20 [m] pipe A 1 pipe B Tank 2 pipe C pipe D

Find: QC [L/s] 3.4 34 340 3,400 QA,QD=0.399 QB=0.060 QC=0.339 m3 QA,QD=0.399 s m3 QB=0.060 s m3 QC=0.339 s L QC=339 [L/s] 1,000 m3 When reviewing the possible solutions, --- Tank Δh=20 [m] pipe A 1 pipe B Tank 2 pipe C pipe D

Find: QC [L/s] AnswerC 3.4 34 340 3,400 QA,QD=0.399 QB=0.060 QC=0.339 m3 QA,QD=0.399 s m3 QB=0.060 s m3 QC=0.339 s AnswerC L QC=339 [L/s] 1,000 m3 The answer is C. Tank Δh=20 [m] pipe A 1 pipe B Tank 2 pipe C pipe D

? 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