Find: max L [ft] 470 1,330 1,780 2,220 Pmin=50 [psi] hP=130 [ft] tank

Find: max L [ft] 470 1,330 1,780 2,220 Pmin=50 [psi] hP=130 [ft] tank
elevation=220 [ft] community Qmax water L pipe d= 8 [in] 470 1,330 1,780 2,220 Find the maximum length, L, in feet. [pause] In this problem, --- C= 100 gal min Qmax=500

elevation=220 [ft] community Qmax water L pipe d= 8 [in] 470 1,330 1,780 2,220 a 8-inch diameter water pipe conveys water from a tank, to a residential community, across a distance L. C= 100 gal min Qmax=500

elevation=220 [ft] community Qmax water L pipe d= 8 [in] 470 1,330 1,780 2,220 At an elevation of 220 feet, the tank provides a pressure head of 130 feet. C= 100 gal min Qmax=500

elevation=220 [ft] community Qmax water L pipe d= 8 [in] 470 1,330 1,780 2,220 At an elevation of 220 feet, the community requires, a minimum of 50 pounds per square inch, of service pressure. C= 100 gal min Qmax=500

elevation=220 [ft] community Qmax water L pipe d= 8 [in] 470 1,330 1,780 2,220 The pipe roughness, and max flowrate in the pipe, are also given. [pause] C= 100 gal min Qmax=500

elevation=220 [ft] community Qmax water L pipe d= 8 [in] 470 1,330 1,780 2,220 Since this problem deals with pressure loss in a pipe, --- C= 100 gal min Qmax=500

elevation=220 [ft] community Qmax water L pipe d= 8 [in] 470 1,330 1,780 2,220 and provides the Hazen-Williams roughness coefficient, C, --- C= 100 gal min Qmax=500

Find: max L [ft] hL= hP=130 [ft] Pmin=50 [psi] tank elevation=220 [ft]
community Qmax water pipe L d= 8 [in] C= 100 we’ll begin with the Hazen-William’s equation for headloss through a pipe, where --- gal min 10.44 * L * Q1.85 Qmax=500 hL= C1.85 * d4.87

community Qmax water pipe L d= 8 [in] headloss [ft] C= 100 the headloss, in feet, equals, --- gal min 10.44 * L * Q1.85 Qmax=500 hL= C1.85 * d4.87

community Qmax water pipe L d= 8 [in] headloss [ft] length [ft] C= 100 flowrate 10.44 * L * Q1.85 10.44, times the length in feet, times the flowrate, in gallons per minute, raised to the 1.85 power, --- gal min gal Qmax=500 hL= C1.85 * d4.87 min

community Qmax water pipe L d= 8 [in] headloss [ft] length [ft] C= 100 flowrate all divided by the quantity, the roughness coefficient, raised to the 1.85 power, times the diameter, in inches, raised to the 4.87 power. gal min 10.44 * L * Q1.85 gal Qmax=500 hL= C1.85 * d4.87 min roughness diameter [in] coefficient

community Qmax water pipe L d= 8 [in] headloss [ft] length [ft] C= 100 flowrate Keep in mind, this equation has different variations, --- gal min 10.44 * L * Q1.85 gal Qmax=500 hL= C1.85 * d4.87 min roughness diameter [in] coefficient

community Qmax water pipe L d= 8 [in] headloss [ft] length [ft] C= 100 flowrate based on the units for each variable. gal min 10.44 * L * Q1.85 gal Qmax=500 hL= C1.85 * d4.87 min roughness diameter [in] coefficient

community Qmax water pipe L d= 8 [in] length [ft] C= 100 gal min 10.44 * L * Q1.85 If we solve for the length, L, we notice the problem statement already provides --- Qmax=500 hL= C1.85 * d4.87 hL * C1.85 * d4.87 L = 10.44 * Q1.85

community Qmax water pipe L d= 8 [in] length [ft] C= 100 gal 10.44 * L * Q1.85 3 of the 4 variables on the right hand side of the equation. Qmax=500 hL= min C1.85 * d4.87 hL * C1.85 * d4.87 L = 10.44 * Q1.85

Find: max L [ft] ? hL= hP=130 [ft] Pmin=50 [psi] tank
elevation=220 [ft] community Qmax water pipe L d= 8 [in] length [ft] C= 100 gal min 10.44 * L * Q1.85 The value for headloss is currently unknown. [pause] The headloss in the pipe, equals, --- Qmax=500 hL= C1.85 * d4.87 ? hL * C1.85 * d4.87 L = 10.44 * Q1.85

Find: max L [ft] hP=130 [ft] Pmin=50 [psi] tank elevation=220 [ft]
community Qmax water pipe L hL =hT,tank-hT,community the total head at the tank, minus, the total head at the community. [pause] Writing out Bernoulli’s equation, ---

Find: max L [ft] + + - + + ρ*g ρ*g Ptank Pcom com ytank ycom
Pmin=50 [psi] hP=130 [ft] tank elevation=220 [ft] community Qmax water pipe L hL =hT,tank-hT,community we can simplify by noticing the the elevation is --- Ptank Pcom v2 v2 tank com hL = + ytank + - + ycom + ρ*g ρ*g 2*g 2*g

Pmin=50 [psi] hP=130 [ft] tank elevation=220 [ft] community Qmax water pipe L hL =hT,tank-hT,community 220 feet for both ends of the pipe, --- Ptank Pcom v2 v2 tank com hL = + ytank + - + ycom + ρ*g ρ*g 2*g 2*g

Pmin=50 [psi] hP=130 [ft] tank elevation=220 [ft] community Qmax water pipe L hL =hT,tank-hT,community so the elevation terms are equal, and cancel out. [pause] Since the velocity is equal to the … Ptank Pcom v2 v2 tank com hL = + ytank + - + ycom + ρ*g ρ*g 2*g 2*g

Pmin=50 [psi] hP=130 [ft] tank elevation=220 [ft] community Qmax water pipe L hL =hT,tank-hT,community flowrate in the pipe, divided by the area of the pipe, --- Ptank Pcom v2 v2 tank com hL = + ytank + - + ycom + ρ*g ρ*g 2*g 2*g Q v= A

Find: max L [ft] + + - + + π ρ*g ρ*g Ptank Pcom com ytank ycom * 4 D2
Pmin=50 [psi] hP=130 [ft] tank elevation=220 [ft] community Qmax water pipe L hL =hT,tank-hT,community both of which are constant throughout the length of the pipe, the velocity is also constant, --- Ptank Pcom v2 v2 tank com hL = + ytank + - + ycom + ρ*g ρ*g 2*g 2*g π Q v= D2 * A 4

Find: max L [ft] + + - + + π ρ*g ρ*g Ptank Pcom com ytank ycom
Pmin=50 [psi] hP=130 [ft] tank elevation=220 [ft] community Qmax water pipe L hL =hT,tank-hT,community and v tank equals v community and the velocity terms --- Ptank Pcom v2 v2 tank com hL = + ytank + - + ycom + ρ*g ρ*g 2*g 2*g π Q vtank = vcom v= D2 * A 4

Find: max L [ft] + + - + + π ρ*g ρ*g Ptank Pcom com ytank ycom
Pmin=50 [psi] hP=130 [ft] tank elevation=220 [ft] community Qmax water pipe L hL =hT,tank-hT,community cancel out of the equation. [pause] The headloss in the pipe is now equal to, --- Ptank Pcom v2 v2 tank com hL = + ytank + - + ycom + ρ*g ρ*g 2*g 2*g π Q vtank = vcom v= D2 * A 4

Find: max L [ft] + + - + + - ρ*g ρ*g ρ*g ρ*g Ptank Pcom com ytank ycom
Pmin=50 [psi] hP=130 [ft] tank elevation=220 [ft] community Qmax water pipe L hL =hT,tank-hT,community the pressure head at the tank minus the pressure head at the community. Ptank Pcom v2 v2 tank com hL = + ytank + - + ycom + ρ*g ρ*g 2*g 2*g Ptank Pcom - hL = ρ*g ρ*g

Pmin=50 [psi] hP=130 [ft] tank elevation=220 [ft] community Qmax water pipe L hL =hT,tank-hT,community The pressure head at the tank, is given as, 130 feet, and the minimum pressure at the community --- Ptank Pcom v2 v2 tank com hL = + ytank + - + ycom + ρ*g ρ*g 2*g 2*g Ptank Pcom - hL = ρ*g ρ*g

Pmin=50 [psi] hP=130 [ft] tank elevation=220 [ft] community Qmax water pipe L hL =hT,tank-hT,community as 50, pounds per square inch. Ptank Pcom v2 v2 tank com hL = + ytank + - + ycom + ρ*g ρ*g 2*g 2*g Ptank Pcom - hL = ρ*g ρ*g

Find: max L [ft] ρ*g Pcom hP=130 [ft] Pmin=50 [psi] tank
elevation=220 [ft] community Qmax water pipe L Carefully solving this second term, --- Pcom hL =130 [ft]- ρ*g

Find: max L [ft] ρ*g 2 Pcom hP=130 [ft] Pmin=50 [psi] tank
elevation=220 [ft] community Qmax water pipe L in 2 12 lbm*ft and canceling units wherever possible, * lbf in2 ft 32.2 50 * lbf*s2 Pcom lbm ft 62.4 hL =130 [ft]- 32.2 * ρ*g ft3 s2

Find: max L [ft] ρ*g 2 Pcom hP=130 [ft] Pmin=50 [psi] tank
elevation=220 [ft] community Qmax water pipe L in 2 12 lbm*ft The minimum pressure head at the community equals --- * lbf in2 ft 32.2 50 * lbf*s2 Pcom lbm ft 62.4 hL =130 [ft]- 32.2 * ρ*g ft3 s2

Find: max L [ft] ρ*g 2 Pcom Pmin=50 [psi] hP=130 [ft] tank
elevation=220 [ft] community Qmax water pipe L Pcom ρ*g in 2 12 lbm*ft 115.4 feet, which makes the total head loss across the pipe equal to --- * lbf ft 32.2 50 * in2 lbf*s2 lbm ft 62.4 32.2 hL =130[ft]-115.4[ft] * ft3 s2

Find: max L [ft] ρ*g 2 Pcom Pmin=50 [psi] hP=130 [ft] tank
elevation=220 [ft] community Qmax water pipe L Pcom ρ*g in 2 12 lbm*ft 14.6 feet. [pause] With the head loss calculated, --- * lbf ft 32.2 50 * in2 lbf*s2 hL =14.6[ft] lbm ft 62.4 32.2 hL =130[ft]-115.4[ft] * ft3 s2

community Qmax water pipe L d= 8 [in] hL * C1.85 * d4.87 L = C= 100 all the needed variables are solved for, to determine the maximum length, L, --- 10.44 * Q1.85 gal min Qmax=500 hL =14.6[ft]

community Qmax water pipe L d= 8 [in] hL * C1.85 * d4.87 L = C= 100 the 8-inch pipe can extend and still maintain the minimum pressure of 50 psi is equal to, --- 10.44 * Q1.85 gal Qmax=500 min hL =14.6[ft]

Find: max L [ft] L =1,781 [ft] hP=130 [ft] Pmin=50 [psi] tank
elevation=220 [ft] community Qmax water pipe L d= 8 [in] hL * C1.85 * d4.87 L = C= 100 1,781 feet. 10.44 * Q1.85 gal Qmax=500 min L =1,781 [ft] hL =14.6[ft]

Find: max L [ft] L =1,781 [ft] 470 1,330 1,780 2,220 hP=130 [ft]
Pmin=50 [psi] hP=130 [ft] tank elevation=220 [ft] community Qmax water pipe L d= 8 [in] hL * C1.85 * d4.87 470 1,330 1,780 2,220 L = C= 100 When reviewing the possible solutions, --- 10.44 * Q1.85 gal Qmax=500 min L =1,781 [ft] hL =14.6[ft]

Find: max L [ft] L =1,781 [ft] AnswerC 470 1,330 1,780 2,220
Pmin=50 [psi] hP=130 [ft] tank elevation=220 [ft] community Qmax water pipe L d= 8 [in] hL * C1.85 * d4.87 470 1,330 1,780 2,220 L = C= 100 The answer is C. 10.44 * Q1.85 gal Qmax=500 min L =1,781 [ft] hL =14.6[ft] AnswerC

Find: hL [ft] ε=1*10-4 [ft] 4 8 12 16 D Q L D= 6 [in] Q=500
gal min 500 gallons per minute flows through a 6-inch diameter, 500-foot long pipe. Q=500 L = 500 [ft] ε=1*10-4 [ft] T= 50 F o

Find: hL [ft] ε=1*10-4 [ft] 4 8 12 16 D Q L D= 6 [in] Q=500
gal min The temperature of the water, and the pipe roughness are also provided. Q=500 L = 500 [ft] ε=1*10-4 [ft] T= 50 F o

Find: hL [ft] ε=1*10-4 [ft] hL=f * * D= 6 [in] Q=500 L = 500 [ft] D
gal min D= 6 [in] Q=500 L = 500 [ft] D T= 50 F o Q ε=1*10-4 [ft] L A headloss equation for flow through a pressurized pipe is, --- L V2 hL=f * * D 2*g

gal min D= 6 [in] Q=500 L = 500 [ft] D T= 50 F o Q ε=1*10-4 [ft] L headloss the headloss, equals, L V2 hL=f * * D 2*g

gal min D= 6 [in] Q=500 L = 500 [ft] D T= 50 F o Q ε=1*10-4 [ft] L headloss the friction factor, --- L V2 hL=f * * D 2*g friction factor

gal min D= 6 [in] Q=500 L = 500 [ft] D T= 50 F o Q ε=1*10-4 [ft] L headloss length times the length divided by the diameter, --- L V2 hL=f * * D 2*g friction diameter factor

gal min D= 6 [in] Q=500 L = 500 [ft] D T= 50 F o Q ε=1*10-4 [ft] L headloss length velocity times the velocity squared, divided by 2 times the gravitational acceleration. L V2 hL=f * * D 2*g gravitational acceleration friction diameter factor

gal min D= 6 [in] Q=500 L = 500 [ft] D T= 50 F o Q ε=1*10-4 [ft] L headloss length velocity [pause] The problem statement provides --- L V2 hL=f * * D 2*g gravitational acceleration friction diameter factor

gal min D= 6 [in] Q=500 L = 500 [ft] D T= 50 F o Q ε=1*10-4 [ft] L headloss length velocity the length, and the diameter, and we also know --- L V2 hL=f * * D 2*g gravitational acceleration friction diameter factor

gal min D= 6 [in] Q=500 L = 500 [ft] D T= 50 F o Q ε=1*10-4 [ft] L headloss length velocity the gravitational acceleration. [pause] So we’ll need to determine the --- ft s2 L V2 32.2 hL=f * * D 2*g gravitational acceleration friction diameter factor

gal min D= 6 [in] Q=500 L = 500 [ft] D T= 50 F o Q ε=1*10-4 [ft] L headloss length velocity friction factor, and the velocity. [pause] The average velocity of flow, v, --- ft s2 L V2 32.2 hL=f * * D 2*g gravitational acceleration friction diameter factor

gal min D= 6 [in] Q=500 L = 500 [ft] D T= 50 F o Q ε=1*10-4 [ft] L headloss length Q velocity, V= equals the flowrate divided by the area. A L V2 hL=f * * D 2*g friction diameter factor

gal D= 6 [in] Q=500 min L = 500 [ft] D T= 50 F o Q ε=1*10-4 [ft] L headloss length Q velocity, V= The problem statement provides the flowrate, Q, A L V2 hL=f * * D 2*g friction diameter factor

gal D= 6 [in] Q=500 min L = 500 [ft] D T= 50 F o ε=1*10-4 [ft] 1 min s L 60 headloss 1 length ft3 Q velocity, V= 7.48 gal which we’ll convert to cubic feet per second. [pause] A L V2 hL=f * * D 2*g friction diameter factor

Find: hL [ft] π ε=1*10-4 [ft] hL=f * * * 4 D2 D= 6 [in] Q=500
gal D= 6 [in] Q=500 min L = 500 [ft] D T= 50 F o ε=1*10-4 [ft] 1 min L 60 s headloss 1 length ft3 Q velocity, V= 7.48 gal The area is a function of diameter. A L V2 π hL=f * * D2 D 2*g * 4 friction diameter factor

Find: hL [ft] π ε=1*10-4 [ft] hL=f * * * 4 D2 1 D= 6 [in] 12 Q=500
gal D= 6 [in] 12 Q=500 min L = 500 [ft] D T= 50 F o ε=1*10-4 [ft] 1 min L 60 s headloss 1 length ft3 Q velocity, V= 7.48 gal The diameter is converted to feet, and then substituted in. After solving for Q, and A, we compute --- A L V2 π hL=f * * D2 D 2*g * 4 friction diameter factor

gal D= 6 [in] 12 in Q=500 min L = 500 [ft] D T= 50 F o ε=1*10-4 [ft] 1 min L 60 s headloss 1 length ft3 Q velocity, V= 7.48 gal 1.114 cubic feet per second divided by square feet, and the velocity through the pipe equals, --- A L V2 ft3 s π hL=f * * 1.114 D 2*g * D2 V= 4 [ft2] friction factor diameter

gal D= 6 [in] 12 in Q=500 min L = 500 [ft] D T= 50 F o ε=1*10-4 [ft] 1 min L 60 s headloss 1 length ft3 Q velocity, V= 7.48 gal 5.673 feet per second. [pause] A L V2 ft3 π hL=f * * 1.114 s D2 D 2*g * V= 4 [ft2] friction factor diameter V = [ft/s]

gal D= 6 [in] 12 in Q=500 min L = 500 [ft] D T= 50 F o ε=1*10-4 [ft] 1 min L 60 s headloss 1 length ft3 Q velocity, V= 7.48 gal With the velocity calculated, --- A L V2 ft3 π hL=f * * 1.114 s D2 D 2*g * V= 4 [ft2] friction factor diameter V = [ft/s]

gal D= 6 [in] 12 in Q=500 min L = 500 [ft] D T= 50 F o ε=1*10-4 [ft] 1 min L 60 s headloss 1 length ft3 Q velocity, V= 7.48 gal the last remaining variable is the friction factor, f. A L V2 ft3 π hL=f * * 1.114 s D2 D 2*g * V= 4 [ft2] friction factor diameter V = [ft/s]

Find: hL [ft] ε=1*10-4 [ft] ε ƒ( ,Re) hL=f * * D= 6 [in]
gal min D= 6 [in] V = [ft/s] Q=500 L = 500 [ft] D T= 50 F o Q ε=1*10-4 [ft] L headloss ε D ƒ( ,Re) The friction factor is a function of --- L V2 hL=f * * D 2*g friction factor

gal min D= 6 [in] V = [ft/s] Q=500 L = 500 [ft] D T= 50 F o Q ε=1*10-4 [ft] L headloss ε ƒ( ,Re) Reynolds D Number the Reynolds Number, and L V2 hL=f * * D 2*g friction factor

gal min D= 6 [in] V = [ft/s] Q=500 L = 500 [ft] D T= 50 F o Q ε=1*10-4 [ft] L headloss ε ƒ( ,Re) Reynolds D Number the relative roughness. [pause] The relative roughness equals the --- L V2 relative hL=f * * D 2*g roughness friction factor

Find: hL [ft] ε ε=1*10-4 [ft] ε ƒ( ,Re) hL=f * * D= 6 [in]
gal min D= 6 [in] V = [ft/s] Q=500 L = 500 [ft] D T= 50 F o Q ε=1*10-4 [ft] L headloss ε ƒ( ,Re) Reynolds D Number roughness roughness length on the inside of the pipe, epsilon, divided by the inside diameter of the pipe, D. Since both of these values have been given, --- L V2 relative ε hL=f * * D 2*g roughness D friction diameter factor

gal min D= 6 [in] V = [ft/s] Q=500 L = 500 [ft] D T= 50 F o Q ε=1*10-4 [ft] L headloss ε ƒ( ,Re) Reynolds D Number roughness they are substituted into the equation, and the relative roughness equals, --- L V2 relative ε hL=f * * D 2*g roughness D 1 ft in friction diameter factor 12

gal min D= 6 [in] V = [ft/s] Q=500 L = 500 [ft] D T= 50 F o Q ε=1*10-4 [ft] L headloss ε ƒ( ,Re) Reynolds D Number roughness 5 times 10 to the –5. [pause] L V2 relative ε hL=f * * =5*10-5 D 2*g roughness D 1 friction ft diameter factor 12 in

Find: hL [ft] ε=1*10-4 [ft] ε ƒ( ,Re) hL=f Re= * * D= 6 [in]
gal min D= 6 [in] V = [ft/s] Q=500 L = 500 [ft] D T= 50 F o Q ε=1*10-4 [ft] L headloss ε ƒ( ,Re) Reynolds D Number The Reynolds Number equals the velocity times the diameter, --- L V2 V*D hL=f Re= * * 5*10-5 D 2*g friction factor

gal min D= 6 [in] V = [ft/s] Q=500 L = 500 [ft] D T= 50 F o Q ε=1*10-4 [ft] L headloss ε ƒ( ,Re) Reynolds D Number divided by the kinematic viscosity. L V2 V*D hL=f Re= * * 5*10-5 D 2*g friction kinematic factor viscosity

gal min D= 6 [in] V = [ft/s] Q=500 L = 500 [ft] D T= 50 F o Q ε=1*10-4 [ft] L headloss ε ƒ( ,Re) Reynolds D Number The kinematic viscosity is a function of the water temperature, --- L V2 V*D hL=f Re= * * 5*10-5 D 2*g friction kinematic factor viscosity

Find: hL [ft] ε=1*10-4 [ft] ε ƒ( ,Re) Re= D= 6 [in] V = 5.673 [ft/s]
gal min D= 6 [in] V = [ft/s] Q=500 L = 500 [ft] D T= 50 F o Q ε=1*10-4 [ft] L ε ƒ( ,Re) Reynolds D Number T [ F] o and can be looked up in a table. Since the temperature is --- [ft2/s] V*D Re= 40 o 1.664*10-5 50 o 1.410*10-5 kinematic 60 o 1.217*10-5 viscosity

Find: hL [ft] ε=1*10-4 [ft] ε ƒ( ,Re) Re= … D= 6 [in] V = 5.673 [ft/s]
gal min D= 6 [in] V = [ft/s] Q=500 L = 500 [ft] D T= 50 F o Q ε=1*10-4 [ft] L ε ƒ( ,Re) Reynolds D Number T [ F] o 50 degrees Fahrenheit, the kinematic viscosity, --- [ft2/s] V*D Re= 40 o 1.664*10-5 50 o 1.410*10-5 kinematic 60 o 1.217*10-5 … viscosity

gal min D= 6 [in] V = [ft/s] Q=500 L = 500 [ft] D T= 50 F o Q ε=1*10-4 [ft] L ε ƒ( ,Re) Reynolds D Number T [ F] o is times 10 to the –5, feet squared per second. [ft2/s] V*D Re= 40 o 1.664*10-5 50 o 1.410*10-5 kinematic 60 o 1.217*10-5 … viscosity

gal min D= 6 [in] V = [ft/s] Q=500 L = 500 [ft] D T= 50 F o Q ε=1*10-4 [ft] L ε ƒ( ,Re) Reynolds D Number T [ F] o After the velocity and diameter are plugged in, the Reynolds Number equals, --- [ft2/s] V*D Re= 40 o 1.664*10-5 50 o 1.410*10-5 kinematic 60 o 1.217*10-5 … viscosity

gal min D= 6 [in] V = [ft/s] Q=500 L = 500 [ft] D T= 50 F o Q ε=1*10-4 [ft] L ε ƒ( ,Re) Reynolds D =2.012*105 Number T [ F] o 2.012 times 10 to the 5th. [pause] With the relative roughness, --- [ft2/s] V*D Re= 40 o 1.664*10-5 50 o 1.410*10-5 kinematic 60 o 1.217*10-5 … viscosity

Find: hL [ft] ε=1*10-4 [ft] ε f=ƒ( ,Re) D= 6 [in] V = 5.673 [ft/s]
gal min D= 6 [in] V = [ft/s] Q=500 L = 500 [ft] D T= 50 F o Q ε=1*10-4 [ft] L ε f=ƒ( ,Re) D 2.012*105 and Reynolds Number. We’ll last turn to the Moody Diagram --- 5*10-5

Find: hL [ft] f Re ε=1*10-4 [ft] ε f=ƒ( ,Re) ε D= 6 [in]
gal min D= 6 [in] V = [ft/s] Q=500 L = 500 [ft] D T= 50 F o Q ε=1*10-4 [ft] L ε f=ƒ( ,Re) D 2.012*105 f to determine the friction factor. The Moody Diagram relates the --- 5*10-5 ε D Re

Find: hL [ft] f Re ε=1*10-4 [ft] ε f=ƒ( ,Re) ε D= 6 [in]
gal min D= 6 [in] V = [ft/s] Q=500 L = 500 [ft] D T= 50 F o Q ε=1*10-4 [ft] L ε f=ƒ( ,Re) D 2.012*105 f friction factor to the Reynolds Number, and the relative roughness. Where the left part of the diagram --- 5*10-5 ε D Re

Find: hL [ft] f Re ε=1*10-4 [ft] ε f=ƒ( ,Re) turbulent flow laminar ε
gal min D= 6 [in] V = [ft/s] Q=500 L = 500 [ft] D T= 50 F o Q ε=1*10-4 [ft] L ε f=ƒ( ,Re) D 2.012*105 turbulent f flow is used for laminar flow, and the rest of the diagram, for turbulent flow. Since the Reynolds Number in this problem is more than a few thousand, --- 5*10-5 laminar ε flow D Re

Find: hL [ft] f Re ε=1*10-4 [ft] ε f=ƒ( ,Re) turbulent flow laminar ε
gal min D= 6 [in] V = [ft/s] Q=500 L = 500 [ft] D T= 50 F o Q ε=1*10-4 [ft] L ε f=ƒ( ,Re) D 2.012*105 turbulent f flow the flow is turbulent. In the diagram, each curved line represents a --- 5*10-5 laminar ε flow D Re

Find: hL [ft] f Re ε=1*10-4 [ft] ε f=ƒ( ,Re) ε ε ε ε D= 6 [in]
gal min D= 6 [in] V = [ft/s] Q=500 L = 500 [ft] D T= 50 F o Q ε=1*10-4 [ft] L ε f=ƒ( ,Re) D 2.012*105 turbulent flow f ε constant relative roughness value, where the bottom-most line is considered --- = 5*10-5 D laminar ε = flow D ε = D ε = D Re

gal min D= 6 [in] V = [ft/s] Q=500 L = 500 [ft] D T= 50 F o Q ε=1*10-4 [ft] L ε f=ƒ( ,Re) D 2.012*105 turbulent flow f ε a “smooth” pipe. [pause] To find the friction factor, --- = 5*10-5 D laminar ε = flow D ε = D “smooth” ε = pipe D Re

gal min D= 6 [in] V = [ft/s] Q=500 L = 500 [ft] D T= 50 F o Q ε=1*10-4 [ft] L ε f=ƒ( ,Re) D 2.012*105 turbulent flow f ε we’ll find the Reynolds Number on the horizontal axis, then trace up --- = 5*10-5 D laminar ε = flow D ε = D ε = D Re

gal min D= 6 [in] V = [ft/s] Q=500 L = 500 [ft] D T= 50 F o Q ε=1*10-4 [ft] L ε f=ƒ( ,Re) D 2.012*105 turbulent flow f ε to the relative roughness contour equal to 5 times 10 to the –5. = 5*10-5 D laminar ε = flow D ε = D ε = D Re

gal min D= 6 [in] V = [ft/s] Q=500 L = 500 [ft] D T= 50 F o Q ε=1*10-4 [ft] L ε f=ƒ( ,Re) D 2.012*105 turbulent flow f ε Then trace left to find the vertical axis, to find the value of f. For this problem, the friction factor is most nearly, ---- = 5*10-5 D laminar ε = flow D ε = D ε = D Re

Find: hL [ft] f Re ε=1*10-4 [ft] ε f=ƒ( ,Re) ε ε ε f=0.016 ε D= 6 [in]
gal min D= 6 [in] V = [ft/s] Q=500 L = 500 [ft] D T= 50 F o Q ε=1*10-4 [ft] L ε f=ƒ( ,Re) D 2.012*105 turbulent flow f ε If reading the Moody Diagram is difficult, --- = 5*10-5 D laminar ε = flow D ε f=0.016 = D ε = D Re

Find: hL [ft] ε=1*10-4 [ft] f( = 5*10-5) ε f=ƒ( ,Re) ε Re f=0.016
gal min D= 6 [in] V = [ft/s] Q=500 L = 500 [ft] D T= 50 F o Q ε=1*10-4 [ft] L ε f=ƒ( ,Re) D 2.012*105 ε these values can also be looked up, in published tables. Re f( = 5*10-5) 5*10-5 D 1.5*105 0.0169 f=0.016 2.0*105 0.0160 2.5*105 0.0154

Find: hL [ft] ε=1*10-4 [ft] f( = 5*10-5) ε f=ƒ( ,Re) ε Re f=0.016
gal min D= 6 [in] V = [ft/s] Q=500 L = 500 [ft] D T= 50 F o Q ε=1*10-4 [ft] L ε f=ƒ( ,Re) D 2.012*105 ε [pause] By knowing the friction factor, --- Re f( = 5*10-5) 5*10-5 D 1.5*105 0.0169 f=0.016 2.0*105 0.0160 2.5*105 0.0154

Find: hL [ft] ε=1*10-4 [ft] hL=f * * D= 6 [in] V = 5.673 [ft/s] Q=500
gal min D= 6 [in] V = [ft/s] Q=500 L = 500 [ft] D T= 50 F o Q ε=1*10-4 [ft] L f=0.016 length velocity the headloss through the pipe can be calculated. L V2 hL=f * * D 2*g gravitational acceleration friction diameter factor

Find: hL [ft] ε=1*10-4 [ft] hL=f * * D= 6 [in] V = 5.673 [ft/s] Q=500
gal min D= 6 [in] V = [ft/s] Q=500 L = 500 [ft] D T= 50 F o Q ε=1*10-4 [ft] L f=0.016 1 ft in 12 The variables are substituted into the equation, --- L V2 hL=f ft s2 * * 32.2 D 2*g

Find: hL [ft] ε=1*10-4 [ft] hL=f hL=8.0 [ft] * * D= 6 [in]
gal min D= 6 [in] V = [ft/s] Q=500 L = 500 [ft] D T= 50 F o Q ε=1*10-4 [ft] L f=0.016 1 ft 12 in and the head loss equals 8.0 feet. L V2 hL=f ft * * 32.2 D 2*g s2 hL=8.0 [ft]

Find: hL [ft] ε=1*10-4 [ft] hL=f hL=8.0 [ft] 4 8 12 16 * * D= 6 [in]
gal min D= 6 [in] V = [ft/s] Q=500 L = 500 [ft] D T= 50 F o Q ε=1*10-4 [ft] L f=0.016 4 8 12 16 1 ft 12 in Looking over the possible solutions, --- L V2 hL=f ft * * 32.2 D 2*g s2 hL=8.0 [ft]

Find: hL [ft] ε=1*10-4 [ft] AnswerB hL=f hL=8.0 [ft] 4 8 12 16 * *
gal min D= 6 [in] V = [ft/s] Q=500 L = 500 [ft] D T= 50 F o Q ε=1*10-4 [ft] L f=0.016 4 8 12 16 1 ft 12 in the answer is B. L V2 hL=f ft * * 32.2 D 2*g s2 hL=8.0 [ft] AnswerB

? 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: max L [ft] 470 1,330 1,780 2,220 Pmin=50 [psi] hP=130 [ft] tank

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Find: max L [ft] 470 1,330 1,780 2,220 Pmin=50 [psi] hP=130 [ft] tank

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Presentation on theme: "Find: max L [ft] 470 1,330 1,780 2,220 Pmin=50 [psi] hP=130 [ft] tank"— Presentation transcript:

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