Find: Q gal min 1,600 1,800 2,000 2,200 Δh pipe entrance fresh water h

Find: Q gal min 1,600 1,800 2,000 2,200 Δh pipe entrance fresh water h
#2 Δh min #1 pipe entrance fresh (sharp edge) water 1,600 1,800 2,000 2,200 h o 90 elbow check (standard) pump o valve T=60 F curve Find the flowrate, in gallons per minute. [pause] In this problem, --- Q Δh=150 [ft] pipe length diameter material suction 80 [ft] 8 [in] steel discharge 640 [ft] 8 [in] steel

#2 Δh min #1 pipe entrance fresh (sharp edge) water 1,600 1,800 2,000 2,200 h o 90 elbow check (standard) pump o valve T=60 F curve fresh water is pumped from reservoir 1 to reservoir 2. Q Δh=150 [ft] discharge suction pipe 640 [ft] 80 [ft] length 8 [in] diameter steel material

#2 Δh min #1 pipe entrance fresh (sharp edge) water 1,600 1,800 2,000 2,200 h o 90 elbow check (standard) pump o valve T=60 F curve Both reservoirs are open to the atmosphere and the elevation difference of their water surfaces are given. Q Δh=150 [ft] discharge suction pipe 640 [ft] 80 [ft] length 8 [in] diameter steel material

#2 Δh min #1 pipe entrance fresh (sharp edge) water 1,600 1,800 2,000 2,200 h o 90 elbow check (standard) pump o valve T=60 F curve Properties of the suction and discharge piping are provided, as well as the pump curve, --- Q Δh=150 [ft] discharge suction pipe 640 [ft] 80 [ft] length 8 [in] diameter steel material

Find: Q gal min Q h [ft] Δh fresh water 1,400 260 h 1,600 220 1,800
#2 Δh min Q gal min h [ft] #1 fresh water 1,400 260 h 1,600 220 1,800 195 o 90 elbow 2,000 165 check (standard) pump o 2,200 145 valve T=60 F curve which has an operating range between 1,400 and 2,400 gallons per minute. [pause] Taking a closer look at the graph, --- Q 2,400 130 Δh=150 [ft] discharge suction pipe 640 [ft] 80 [ft] length 8 [in] diameter steel material

Find: Q gal min Q h [ft] Δh fresh water 1,400 260 1,600 220 h 1,800
#2 Δh min Q gal min h [ft] #1 fresh water 1,400 260 1,600 220 h 1,800 195 2,000 165 2,200 145 the pump curve is plotted against the --- 2,400 130 pump curve Q

#2 Δh min Q gal min h [ft] #1 fresh water 1,400 260 1,600 220 h 1,800 195 2,000 165 system 2,200 145 system curve, and the pump will operate at the flowrate --- curve 2,400 130 pump curve Q

#2 Δh min Q gal min h [ft] #1 fresh water 1,400 260 1,600 220 h 1,800 195 2,000 165 system 2,200 145 where these two lines intersect. [pause] The problem statement provides --- curve 2,400 130 pump curve Q

#2 Δh min Q gal min h [ft] #1 fresh water 1,400 260 1,600 220 h 1,800 195 2,000 165 system 2,200 145 data points for the pump curve, and points on the system curve ---- curve 2,400 130 pump curve Q

Find: Q gal min Q h [ft] Δh fresh water 260 1,400 220 1,600 195 1,800
#2 Δh min Q gal min h [ft] #1 fresh water 260 1,400 220 1,600 195 1,800 165 2,000 145 2,200 130 2,400 h system can be calculated based on the headloss seen by the system, at various flowrates. The possible solutions --- curve pump curve Q

Find: Q gal min ? ? ? ? 1,600 1,800 2,000 2,200 Q h [ft] Δh fresh
#2 Δh min ? 1,600 1,800 2,000 2,200 Q gal min h [ft] #1 fresh ? water 1,400 260 ? 1,600 220 ? 1,800 195 2,000 165 system 2,200 145 are the flowrates 1,600 to 2,200 gallons per minute. Since the pump curve minus the system curve is monotonic, it would be prudent to --- curve 2,400 130 pump curve Q

#2 Δh min ? 1,600 1,800 2,000 2,200 Q gal min h [ft] #1 fresh ? water 1,400 260 ? 1,600 220 ? 1,800 195 2,000 165 system 2,200 145 first evaluate one of the two middle flowrates, 1,800 gallons per minute or 2,000 gallons per minute. [pause] If we evaluate --- curve 2,400 130 pump curve Q

#2 Δh min ? 1,600 1,800 2,000 2,200 Q gal min h [ft] #1 fresh ? water 1,400 260 ? 1,600 220 ? 1,800 195 2,000 165 system 2,200 145 2,000 gallons per minute, and the headloss in the system equals --- curve 2,400 130 pump curve Q

#2 Δh min ? 1,600 1,800 2,000 2,200 Q gal min h [ft] #1 fresh ? water 1,400 260 ? 1,600 220 ? 1,800 195 2,000 165 system 2,200 145 165 feet, then 2,000 gallons per minute is the correct answer. curve 2,400 130 pump curve Q

#2 Δh min ? 1,600 1,800 2,000 2,200 Q gal min h [ft] #1 fresh ? water 1,400 260 ? 1,600 220 ? 1,800 195 2,000 165 system 2,200 145 However, if the system head at 2,000 gallons per minute, is greater than 165 feet, --- curve 2,400 130 pump curve Q

#2 Δh min ? 1,600 1,800 2,000 2,200 Q gal min h [ft] #1 fresh ? water 1,400 260 ? 1,600 220 ? 1,800 195 2,000 165 system 2,200 145 then the actual flowrate is less than 2,000 gallons per minute, and if the system head at 2,000 gallons per minute, is less than 165 feet, --- curve 2,400 130 pump curve Q

#2 Δh min ? 1,600 1,800 2,000 2,200 Q gal min h [ft] #1 fresh ? water 1,400 260 ? 1,600 220 ? 1,800 195 2,000 165 system 2,200 145 then the actual flowrate is more than the 2,000 gallons per minute. curve 2,400 130 pump curve Q

#2 Δh min ? 1,600 1,800 2,000 2,200 Q gal min h [ft] #1 fresh ? water 1,400 260 ? 1,600 220 ? 1,800 195 2,000 165 system 2,200 145 Now it’s time to solve for the system head, at the flowrate of 2,000 gallons per minute. curve 2,400 130 pump curve Q

Find: Q + + - + + ρ*g ρ*g gal min 2 +hL h = P2 v2 P1 v1 2 y2 y1 Δh
#2 Δh min #1 Δh=150 [ft] fresh gal min water assume Q=2,000 2 P2 v2 P1 v1 2 + y2 + - + y1 + h = +hL ρ*g ρ*g 2*g 2*g The system head, for a given flowrate, equals ---

Find: Q + + - + + ρ*g ρ*g gal min 2 +hL h = P2 v2 P1 v1 2 y2 y1 Δh
#2 Δh min #1 Δh=150 [ft] fresh gal min water assume Q=2,000 2 P2 v2 P1 v1 2 + y2 + - + y1 + h = +hL ρ*g ρ*g 2*g 2*g the total head at reservoir 2, minus --- hT,2

Find: Q + + - + + ρ*g ρ*g gal min 2 2 +hL h = P2 v2 P1 v1 y2 y1 Δh
#2 Δh min #1 Δh=150 [ft] fresh gal min water assume Q=2,000 2 2 P2 v2 P1 v1 + y2 + - + y1 + h = +hL ρ*g ρ*g 2*g 2*g the total head at reservoir 1, plus --- hT,2 hT,1

#2 Δh min #1 Δh=150 [ft] fresh gal min water assume Q=2,000 2 2 P2 v2 P1 v1 + y2 + - + y1 + h = +hL ρ*g ρ*g 2*g 2*g the headloss through the system, between the two reservoirs. hT,2 hT,1

#2 Δh min #1 Δh=150 [ft] fresh gal min water assume Q=2,000 2 2 P2 v2 P1 v1 + y2 + - + y1 + h = +hL ρ*g ρ*g 2*g 2*g Since both reservoirs are open to the atmosphere and static, the pressure head and --- hT,2 hT,1

#2 Δh min #1 Δh=150 [ft] fresh gal min water assume Q=2,000 2 2 P2 v2 P1 v1 + y2 + - + y1 + h = +hL ρ*g ρ*g 2*g 2*g velocity head values are zero, and the equation reduces down to --- hT,2 hT,1

Find: Q + + - + + ρ*g ρ*g gal min 2 2 +hL h = h =y2-y1+hL P2 v2 P1 v1
#2 Δh min #1 Δh=150 [ft] fresh gal min water assume Q=2,000 2 2 P2 v2 P1 v1 + y2 + - + y1 + h = +hL ρ*g ρ*g 2*g 2*g the change in elevation head plus the headloss through the system. [pause] The change in elevation head equals, ---- hT,2 hT,1 h =y2-y1+hL

Find: Q + + - + + ρ*g ρ*g gal min 2 2 +hL h = h =y2-y1+hL P2 v2 P1 v1
#2 Δh min #1 Δh=150 [ft] fresh gal min water assume Q=2,000 2 2 P2 v2 P1 v1 + y2 + - + y1 + h = +hL ρ*g ρ*g 2*g 2*g delta h, 150 feet, and the headloss through the system equals --- hT,2 hT,1 h =y2-y1+hL

Find: Q + + - + + ρ*g ρ*g gal min 2 2 + +hL h = h =y2-y1+hL hL=f P2 v2
#2 Δh min #1 Δh=150 [ft] fresh gal min water assume Q=2,000 2 2 P2 v2 P1 v1 + y2 + - + y1 + h = +hL ρ*g ρ*g 2*g 2*g the major losses, along the wall of the pipeline, plus hT,2 hT,1 L V2 V2 h =y2-y1+hL hL=f + ΣK * * * d 2*g 2*g major loss

Find: Q + + - + + ρ*g ρ*g gal min 2 2 + +hL h = h =y2-y1+hL hL=f P2 v2
#2 Δh min #1 Δh=150 [ft] fresh gal min water assume Q=2,000 2 2 P2 v2 P1 v1 + y2 + - + y1 + h = +hL ρ*g ρ*g 2*g 2*g the minor losses, caused by the fittings, valve and entrance. [pause] hT,2 hT,1 L V2 V2 h =y2-y1+hL hL=f + ΣK * * * d 2*g 2*g major loss minor loss

Find: Q gal min + hL=f Δh assume Q=2,000 fresh water L V2 V2 ΣK d 2*g
#2 Δh min gal min #1 assume Q=2,000 fresh water L V2 V2 hL=f + ΣK * * * d 2*g 2*g In the headloss equation, the problem statement provides the diameter as ---

Find: Q gal min + =0.667[ft] hL=f Δh assume Q=2,000 fresh water L V2
#2 Δh min gal min #1 assume Q=2,000 fresh water L V2 V2 hL=f + ΣK * * * d 2*g 2*g 1 in ft =0.667[ft] * 8 inches, or feet, through both suction and discharge pipes. We also know the total --- 12 discharge suction pipe 640 [ft] 80 [ft] length 8 [in] diameter steel material

#2 Δh min gal min #1 assume Q=2,000 fresh water L V2 V2 hL=f + ΣK * * * d 2*g 2*g 1 ft =0.667[ft] * Σ=720 [ft] length of pipe is 720 feet, as well as the gravitational acceleration, --- 12 in discharge suction pipe 640 [ft] 80 [ft] length 8 [in] diameter steel material

#2 Δh min gal min #1 assume Q=2,000 fresh water L V2 V2 hL=f + ΣK * * * d 2*g 2*g 1 ft ft =0.667[ft] 32.2 * Σ=720 [ft] g, equal to 32.2 feet per second squared. [pause] But we don’t yet know --- 12 in s2 discharge suction pipe 640 [ft] 80 [ft] length 8 [in] diameter steel material

#2 Δh min gal min #1 assume Q=2,000 fresh water L V2 V2 hL=f + ΣK * * * d 2*g 2*g friction the friction factor, f, --- factor s2 ft g=32.2 d=0.667[ft] L=720 [ft]

#2 Δh min gal min #1 assume Q=2,000 fresh water L V2 V2 hL=f + ΣK * * * d 2*g 2*g ft friction velocity s the velocity, v, in feet per second, or --- factor s2 ft g=32.2 d=0.667[ft] L=720 [ft]

#2 Δh min gal min #1 assume Q=2,000 fresh water L V2 V2 hL=f + ΣK * * * d 2*g 2*g minor headloss ft coefficient(s) friction velocity s the sum of the minor headloss coefficients, sigma K. factor s2 ft g=32.2 d=0.667[ft] L=720 [ft]

#2 Δh min gal min #1 assume Q=2,000 fresh water L V2 V2 hL=f + ΣK * * * d 2*g 2*g minor headloss ft coefficient(s) friction velocity s We’ll first solve for the velocity. factor s2 ft g=32.2 d=0.667[ft] L=720 [ft]

#2 Δh min gal min #1 assume Q=2,000 fresh water L V2 V2 hL=f + ΣK * * * d 2*g 2*g Q flowrate V= area A The velocity equals the flowrate divided by the area, where the assumed flowrate of --- s2 ft g=32.2 d=0.667[ft] L=720 [ft]

#2 Δh min gal #1 assume Q=2,000 fresh min water L V2 V2 hL=f + ΣK * * * d 2*g 2*g 1 ft3 1 min * * gal s Q flowrate 7.48 60 V= area A 2,000 gallons per minute is converted to cubic feet per second, and the s2 ft g=32.2 d=0.667[ft] L=720 [ft]

Find: Q gal min + π hL=f Δh assume Q=2,000 fresh water L V2 V2 ΣK d
#2 Δh min gal #1 assume Q=2,000 fresh min water L V2 V2 hL=f + ΣK * * * d 2*g 2*g 1 ft3 1 min * * gal s Q flowrate 7.48 60 V= π area A area is calculated from the diameter. *(0.667 [ft])2 s2 ft 4 g=32.2 d=0.667[ft] L=720 [ft]

#2 Δh min gal #1 assume Q=2,000 fresh min water L V2 V2 hL=f + ΣK * * * d 2*g 2*g 1 ft3 1 min * * gal s Q flowrate 7.48 60 V= π area A The velocity of flow equals, --- *(0.667 [ft])2 s2 ft 4 g=32.2 ft3 4.456 s d=0.667[ft] V= 0.349 [ft2] L=720 [ft]

#2 Δh min gal #1 assume Q=2,000 fresh min water L V2 V2 hL=f + ΣK * * * d 2*g 2*g 1 ft3 1 min * * gal s Q flowrate 7.48 60 V= π area A 12.77 feet per second. [pause] *(0.667 [ft])2 s2 ft 4 g=32.2 ft3 4.456 s ft d=0.667[ft] V= =12.77 s 0.349 [ft2] L=720 [ft]

#2 Δh min gal min #1 assume Q=2,000 fresh water L V2 V2 hL=f + ΣK * * * d 2*g 2*g minor headloss ft coefficient(s) friction velocity s With the velocity solved, we’ll next determine the --- factor s2 ft g=32.2 ft3 4.456 s ft d=0.667[ft] V= =12.77 s 0.349 [ft2] L=720 [ft]

#2 Δh min gal min #1 assume Q=2,000 fresh water L V2 V2 hL=f + ΣK * * * d 2*g 2*g minor headloss ft coefficient(s) friction velocity s sum of the minor headloss coefficients. factor s2 ft g=32.2 d=0.667[ft] s ft V=12.77 L=720 [ft]

Find: Q gal min + hL=f Δh pipe entrance fresh water L V2 V2 ΣK d 2*g
#2 Δh min #1 pipe entrance fresh (sharp edge) water L V2 V2 hL=f + ΣK * * * d 2*g 2*g 90 elbow check (standard) valve The problem statement identified the system as having a sharp edge pipe entrance, a check valve, and 2 standard 90 degree elbows. s2 ft g=32.2 d=0.667[ft] s ft V=12.77 L=720 [ft]

#2 Δh min #1 pipe entrance fresh (sharp edge) water L V2 V2 hL=f + ΣK * * * d 2*g 2*g 90 elbow check valve (standard) item qty K qty * K The headloss coefficients for these three items are looked up and --- entrance 1 0.50 0.50 check valve 1 2.30 2.30 o 90 elbow 2 0.90 1.80

#2 Δh min #1 pipe entrance fresh (sharp edge) water L V2 V2 hL=f + ΣK * * * d 2*g 2*g 90 elbow check valve (standard) item qty K qty * K added to the table. Since there are 2 elbows fittings, we’ll multiply the quantity of each item by it’s K value, --- entrance 1 0.50 0.50 check valve 1 2.30 2.30 o 90 elbow 2 0.90 1.80

#2 Δh min #1 pipe entrance fresh (sharp edge) water L V2 V2 hL=f + ΣK * * * d 2*g 2*g 90 elbow check valve (standard) item qty K qty * K and find the sum. --- entrance 1 0.50 0.50 check valve 1 2.30 2.30 o 90 elbow 2 0.90 1.80

#2 Δh min #1 pipe entrance fresh (sharp edge) water L V2 V2 hL=f + ΣK * * * d 2*g 2*g 90 elbow check valve (standard) item qty K qty * K which is [pause] entrance 1 0.50 0.50 check valve 1 2.30 2.30 o 90 elbow 2 0.90 1.80 ΣK= 4.60

Find: Q gal min + hL=f Δh fresh water L V2 V2 ΣK d 2*g 2*g
#2 Δh min #1 fresh water L V2 V2 hL=f + ΣK * * * d 2*g 2*g minor headloss ft coefficient(s) friction velocity s The last variable to solve for is the --- factor s2 ft g=32.2 ΣK= 4.60 d=0.667[ft] s ft V=12.77 L=720 [ft]

Find: Q gal min + hL=f Δh fresh water L V2 V2 ΣK d 2*g 2*g
#2 Δh min #1 fresh water L V2 V2 hL=f + ΣK * * * d 2*g 2*g minor headloss ft coefficient(s) friction velocity s friction factor, f. [pause] The friction factor is determined using the --- factor s2 ft g=32.2 ΣK= 4.60 d=0.667[ft] s ft V=12.77 L=720 [ft]

Find: Q gal min + hL=f f f=ƒ(Re, ) Re ε ε Δh T=60 F d=0.667[ft] fresh
#2 Δh min o #1 T=60 F d=0.667[ft] fresh V=12.77[ft/s] steel water L V2 V2 hL=f + ΣK * * * d 2*g 2*g f moody ε d f=ƒ(Re, ) diagram relative Reynolds number, the relative roughness, and the Moody Diagram. Reynolds number equals --- roughness ε d Reynolds number Re

Find: Q gal min + hL=f f f=ƒ(Re, ) Re= Re ε ε Δh T=60 F d=0.667[ft]
#2 Δh min o #1 T=60 F d=0.667[ft] fresh V=12.77[ft/s] steel water L V2 V2 hL=f + ΣK * * * d 2*g 2*g f moody ε f=ƒ(Re, ) diagram d V*d the velocity times the diameter, divided by the kinematic viscosity. Re= ε d Re

#2 Δh min o #1 T=60 F d=0.667[ft] fresh V=12.77[ft/s] steel water L V2 V2 hL=f + ΣK * * * d 2*g 2*g f moody ε f=ƒ(Re, ) diagram d V*d We’ve already solved for the velocity and diameter, And the kinematic viscosity --- Re= ε d Re

#2 Δh min o #1 T=60 F d=0.667[ft] fresh V=12.77[ft/s] steel water L V2 V2 hL=f + ΣK * * * d 2*g 2*g f moody ε f=ƒ(Re, ) diagram d V*d for fresh water can be looked up in a table, based on it’s temperature. For 60 degrees Fahrenheit, --- Re= ε d Re

#2 Δh min o #1 T=60 F d=0.667[ft] fresh V=12.77[ft/s] steel water L V2 V2 hL=f + ΣK * * * d 2*g 2*g f moody ε f=ƒ(Re, ) diagram d V*d the kinematic viscosity equals times 10 to the –5th, feet squared per second. The Reynolds number equals --- Re= ε d ft2 1.214*10-5 s Re

Find: Q gal min + hL=f f f=ƒ(Re, ) Re= Re=7.02*105 Re ε ε Δh T=60 F
#2 Δh min o #1 T=60 F d=0.667[ft] fresh V=12.77[ft/s] steel water L V2 V2 hL=f + ΣK * * * d 2*g 2*g f moody ε f=ƒ(Re, ) diagram d V*d 7.02 times 10 to the 5th power. [pause] The relative roughness equals --- Re= ε d ft2 1.214*10-5 s Re=7.02*105 Re

Find: Q gal min + hL=f f f=ƒ(Re, ) Re ε ε ε Δh T=60 F d=0.667[ft]
#2 Δh min o #1 T=60 F d=0.667[ft] fresh V=12.77[ft/s] steel water L V2 V2 hL=f + ΣK * * * d 2*g 2*g f moody ε f=ƒ(Re, ) diagram d roughness ε the roughness distance, epsilon, divided by the inside diameter of the pipe, d. Based on the pipe material, --- d ε diameter d Re Re=7.02*105

#2 Δh min o #1 T=60 F d=0.667[ft] fresh V=12.77[ft/s] steel water L V2 V2 hL=f + ΣK * * * d 2*g 2*g f moody ε f=ƒ(Re, ) diagram d roughness ε the roughness can be looked up from a table of values. For steel, the roughness equals --- d ε diameter d Re Re=7.02*105

#2 Δh min o #1 T=60 F d=0.667[ft] fresh V=12.77[ft/s] steel water L V2 V2 hL=f + ΣK * * * d 2*g 2*g f moody ε f=ƒ(Re, ) 2*10-4 [ft] diagram d roughness ε 2 times 10 to the –4, feet. [pause] d ε diameter d Re Re=7.02*105

#2 Δh min o #1 T=60 F d=0.667[ft] fresh V=12.77[ft/s] steel water L V2 V2 hL=f + ΣK * * * d 2*g 2*g f moody ε f=ƒ(Re, ) 2*10-4 [ft] diagram d roughness ε Knowing the diameter already, the relative roughness computes to --- d ε diameter d Re Re=7.02*105

Find: Q gal min + hL=f f f=ƒ(Re, ) Re ε ε ε ε Δh T=60 F d=0.667[ft]
#2 Δh min o #1 T=60 F d=0.667[ft] fresh V=12.77[ft/s] steel water L V2 V2 hL=f + ΣK * * * d 2*g 2*g f moody ε f=ƒ(Re, ) 2*10-4 [ft] diagram d roughness ε 3 times 10 to the –4. [pause] d ε diameter d ε = 3*10-4 d Re Re=7.02*105

Find: Q gal min + hL=f f f=ƒ(Re, ) Re ε ε ε ε Δh T=60 F d=0.667[ft]
#2 Δh min o #1 T=60 F d=0.667[ft] fresh V=12.77[ft/s] steel water L V2 V2 hL=f + ΣK * * * d 2*g 2*g f moody ε f=ƒ(Re, ) 2*10-4 [ft] diagram d roughness ε With knowledge of Reynolds’s number and the relative roughness, --- d ε diameter d ε = 3*10-4 d Re Re=7.02*105

Find: Q gal min + hL=f f f=ƒ(Re, ) Re ε ε ε f=0.016 ε Δh T=60 F
#2 Δh min o #1 T=60 F d=0.667[ft] fresh V=12.77[ft/s] steel water L V2 V2 hL=f + ΣK * * * d 2*g 2*g f moody ε f=ƒ(Re, ) 2*10-4 [ft] diagram d roughness ε the friction factor equals [pause] d ε diameter f=0.016 d ε = 3*10-4 d Re Re=7.02*105

#2 Δh min gal min #1 assume Q=2,000 fresh water L V2 V2 hL=f + ΣK * * * d 2*g 2*g friction With the friction factor solved, all variables on the right hand side of the equation are known and plugged in. factor s2 ft g=32.2 f=0.016 ΣK= 4.60 d=0.667[ft] s ft V=12.77 L=720 [ft]

#2 Δh min gal min #1 assume Q=2,000 fresh water L V2 V2 hL=f + ΣK * * * d 2*g 2*g The headloss between reservoirs 1 and 2, at a flowrate of 2,000 gallons per minute --- ft g=32.2 f=0.016 s2 ΣK= 4.60 d=0.667[ft] ft V=12.77 L=720 [ft] s

Find: Q gal min + hL=f hL=55.4 [ft] Δh assume Q=2,000 fresh water L V2
#2 Δh min gal min #1 assume Q=2,000 fresh water L V2 V2 hL=f + ΣK * * * d 2*g 2*g equals, 55.4 feet. [pause] By knowing this headloss, --- ft g=32.2 f=0.016 hL=55.4 [ft] s2 ΣK= 4.60 d=0.667[ft] ft V=12.77 L=720 [ft] s

Find: Q gal min + hL=f h =y2-y1+hL hL=55.4 [ft] Δh assume Q=2,000
#2 Δh min gal min #1 assume Q=2,000 fresh water L V2 V2 hL=f + ΣK * * * d 2*g 2*g h =y2-y1+hL the total head in the system, while operating at 2,000 gallons per minute equals, --- hL=55.4 [ft] Δh=150 [ft]

#2 Δh min gal min #1 assume Q=2,000 fresh water L V2 V2 hL=f + ΣK * * * d 2*g 2*g h =y2-y1+hL the 150 feet of elevation head, plus, --- hL=55.4 [ft] Δh=150 [ft]

#2 Δh min gal min #1 assume Q=2,000 fresh water L V2 V2 hL=f + ΣK * * * d 2*g 2*g h =y2-y1+hL the 55.4 feet of headloss due to friction, which equals, --- hL=55.4 [ft] Δh=150 [ft]

Find: Q gal min + hL=f h =y2-y1+hL h =205.4 [ft] hL=55.4 [ft] Δh
#2 Δh min gal min #1 assume Q=2,000 fresh water L V2 V2 hL=f + ΣK * * * d 2*g 2*g h =y2-y1+hL 205.4 feet. hL=55.4 [ft] h =205.4 [ft] Δh=150 [ft]

Find: Q gal min h =205.4 [ft] Δh assume Q=2,000 fresh water h gal min
#2 Δh min gal #1 assume Q=2,000 fresh min water h =205.4 [ft] h So far we have 1 data point on the system curve, --- Q

Find: Q gal min h =205.4 [ft] Δh assume Q=2,000 fresh water h system
#2 Δh min gal #1 assume Q=2,000 fresh min water h =205.4 [ft] h system at the flowrate of 2,000 gallons per minute and the head of feet. From the pump curve data, ---- curve Q

Find: Q gal min h =205.4 [ft] Q h [ft] Δh assume Q=2,000 fresh water
#2 Δh min gal #1 assume Q=2,000 fresh min water h =205.4 [ft] system 260 1,400 220 1,600 195 1,800 165 2,000 145 2,200 130 2,400 Q h [ft] gal min h curve point pump system we notice that at a flowrate of 2,000 gallons per minute, curve curve data Q

#2 Δh min gal #1 assume Q=2,000 fresh min water h =205.4 [ft] system 260 1,400 220 1,600 195 1,800 165 2,000 145 2,200 130 2,400 Q h [ft] gal min h curve point pump system the headloss is only 165 feet. Therefore, the pump curve passes below --- curve curve data Q

#2 Δh min gal #1 assume Q=2,000 fresh min water h =205.4 [ft] system 260 1,400 220 1,600 195 1,800 165 2,000 145 2,200 130 2,400 Q h [ft] gal min h curve point pump system the data point on the system curve. This means, the actual flowrate is --- curve curve data pump curve Q

#2 Δh min gal #1 assume Q=2,000 fresh min water h =205.4 [ft] system 260 1,400 220 1,600 195 1,800 165 2,000 145 2,200 130 2,400 Q h [ft] gal min h curve point pump system less than 2,000 gallons per minute, ruling out solutions C and D. Solutions A and B were, --- curve curve data pump curve Q

#2 Δh min gal #1 assume Q=2,000 fresh min water h =205.4 [ft] system 260 1,400 220 1,600 195 1,800 165 2,000 145 2,200 130 2,400 Q h [ft] gal min h curve point pump system 1,600 and 1,800 gallons per minute. However, when considering our data point on the system curve, --- curve curve data pump curve Q

#2 Δh min gal #1 assume Q=2,000 fresh min water h =205.4 [ft] system 260 1,400 220 1,600 195 1,800 165 2,000 145 2,200 130 2,400 Q h [ft] gal min h curve point pump system we can rule out the 1,600 gallons per minute solution because, as flowrate decreases from 2,000 gallons per minute to 1,600 gallons per minute, we would also expect the head to drop from feet to a lower head value, --- curve curve data pump curve Q

#2 Δh min gal #1 assume Q=2,000 fresh min water h =205.4 [ft] system 260 1,400 220 1,600 195 1,800 165 2,000 145 2,200 130 2,400 Q h [ft] gal min h curve point pump system as would happen when if the pump were to operate at 1,800 gallons per minute. curve curve data pump curve Q

Find: Q gal min h =205.4 [ft] Q h [ft] 1,600 1,800 2,000 2,200 Δh
#2 Δh min gal #1 assume Q=2,000 fresh min water h =205.4 [ft] 260 1,400 220 1,600 195 1,800 165 2,000 145 2,200 130 2,400 Q h [ft] gal min 1,600 1,800 2,000 2,200 system When reviewing the possible solutions, --- curve pump curve Q

Find: Q gal min AnswerB h =205.4 [ft] Q h [ft] 1,600 1,800 2,000
#2 Δh min gal #1 assume Q=2,000 fresh min water h =205.4 [ft] 260 1,400 220 1,600 195 1,800 165 2,000 145 2,200 130 2,400 Q h [ft] gal min AnswerB 1,600 1,800 2,000 2,200 system the answer is B. curve pump curve Q

? 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

Find: Q gal min 1,600 1,800 2,000 2,200 Δh pipe entrance fresh water h

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Find: Q gal min 1,600 1,800 2,000 2,200 Δh pipe entrance fresh water h

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Presentation on theme: "Find: Q gal min 1,600 1,800 2,000 2,200 Δh pipe entrance fresh water h"— Presentation transcript:

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