Find: Phome [psi] 40 C) 60 B) 50 D) 70 ft s v=5 C=100

Find: Phome [psi] 40 C) 60 B) 50 D) 70 ft s v=5 C=100
Elevtank=540 [ft] home Elevhome=380 [ft] L=1,000 [ft] d=4 [in] o Find the pressure at the home, in pounds per square inch. [pause] In this problem, --- gate 90 elbow meter (standard) valves (rotary) 40 C) 60 B) 50 D) 70

Elevtank=540 [ft] home Elevhome=380 [ft] L=1,000 [ft] d=4 [in] o water is stored in an elevated tank, with a water surface elevation, of 540 feet. gate 90 elbow meter (standard) valves (rotary) 40 C) 60 B) 50 D) 70

Elevtank=540 [ft] home Elevhome=380 [ft] L=1,000 [ft] d=4 [in] o The elevation at the home, where the pressure is tested, is 380 feet. gate 90 elbow meter (standard) valves (rotary) 40 C) 60 B) 50 D) 70

Elevtank=540 [ft] home Elevhome=380 [ft] L=1,000 [ft] d=4 [in] o A 4-inch diameter pipe conveys water from the tank to the home, --- gate 90 elbow meter (standard) valves (rotary) 40 C) 60 B) 50 D) 70

Elevtank=540 [ft] home Elevhome=380 [ft] L=1,000 [ft] d=4 [in] o and various pipe appurtenances and flow parameters, are also labeled. [pause] To find the pressure, at the home, --- gate 90 elbow meter (standard) valves (rotary) 40 C) 60 B) 50 D) 70

Find: Phome [psi] + + - + + ρ*g ρ*g Ptank Phome home ytank yhome ft s
v=5 C=100 Elevtank=540 [ft] home Elevhome=380 [ft] L=1,000 [ft] d=4 [in] Ptank Phome v2 v2 we’ll begin with Bernoulli’s equation, which states, --- tank home hL = + ytank + - + + yhome ρ*g ρ*g 2*g 2*g

Find: Phome [psi] + + - + + ρ*g ρ*g Ptank Phome home ytank yhome ft s
v=5 C=100 Elevtank=540 [ft] home Elevhome=380 [ft] L=1,000 [ft] d=4 [in] Ptank Phome v2 v2 the headloss, between the tank, and the home, equals --- tank home hL = + ytank + - + + yhome ρ*g ρ*g 2*g 2*g headloss [ft]

Find: Phome [psi] + + - + + ρ*g ρ*g hT,tank Ptank Phome home ytank
ft s v=5 C=100 Elevtank=540 [ft] home Elevhome=380 [ft] L=1,000 [ft] d=4 [in] Ptank Phome v2 v2 the total head at the tank, minus, the total head, --- tank home hL = + ytank + - + + yhome ρ*g ρ*g 2*g 2*g headloss [ft] hT,tank

Find: Phome [psi] + + - + + ρ*g ρ*g hT,tank hT,home Ptank Phome home
ft s v=5 C=100 Elevtank=540 [ft] home Elevhome=380 [ft] L=1,000 [ft] d=4 [in] Ptank Phome v2 v2 at the home. [pause] Since the problem asks to find --- tank home hL = + ytank + - + + yhome ρ*g ρ*g 2*g 2*g headloss [ft] hT,tank hT,home

Find: Phome [psi] + + - + + ρ*g ρ*g hT,tank hT,home Ptank Phome home
ft v=5 C=100 s Elevtank=540 [ft] home Elevhome=380 [ft] L=1,000 [ft] d=4 [in] Ptank Phome v2 v2 the pressure at the home, we’ll isolate the pressure term, P sub home. [pause] tank home hL = + ytank + - + + yhome ρ*g ρ*g 2*g 2*g headloss [ft] hT,tank hT,home

Find: Phome [psi] + + - + + + + + - ρ*g ρ*g ρ*g Ptank Phome home ytank
ft v=5 C=100 s Elevtank=540 [ft] home Elevhome=380 [ft] L=1,000 [ft] d=4 [in] Ptank Phome v2 v2 Since the water tank is not pressurized, --- tank home hL = + ytank + - + + yhome ρ*g ρ*g 2*g 2*g Ptank v2 v2 home tank + Phome=ρ*g* + ytank + - yhome -hL ρ*g 2*g 2*g

ft v=5 C=100 s Elevtank=540 [ft] home Elevhome=380 [ft] L=1,000 [ft] d=4 [in] Ptank Phome v2 v2 the pressure head at the tank equals zero. tank home hL = + ytank + - + + yhome ρ*g ρ*g 2*g 2*g Ptank v2 v2 home tank + Phome=ρ*g* + ytank + - yhome -hL ρ*g 2*g 2*g

ft v=5 C=100 s Elevtank=540 [ft] home Elevhome=380 [ft] L=1,000 [ft] d=4 [in] Ptank Phome v2 v2 And since the velocity of water in the tank is zero, --- tank home hL = + ytank + - + + yhome ρ*g ρ*g 2*g 2*g Ptank v2 v2 home tank + Phome=ρ*g* + ytank + - yhome -hL ρ*g 2*g 2*g

ft v=5 C=100 s Elevtank=540 [ft] home Elevhome=380 [ft] L=1,000 [ft] d=4 [in] Ptank Phome v2 v2 the velocity head at the tank equals zero. [pause] Now the equation simplifies a bit. tank home hL = + ytank + - + + yhome ρ*g ρ*g 2*g 2*g Ptank v2 v2 home tank + Phome=ρ*g* + ytank + - yhome -hL ρ*g 2*g 2*g

Find: Phome [psi] - + + + - ρ*g home Phome=ρ*g* ytank - yhome -hL
ft v=5 C=100 s Elevtank=540 [ft] home Elevhome=380 [ft] L=1,000 [ft] d=4 [in] v2 At this point, the only unknown variable is the headloss, --- home Phome=ρ*g* ytank - yhome - -hL 2*g Ptank v2 v2 home tank + Phome=ρ*g* + ytank + - yhome -hL ρ*g 2*g 2*g

Find: Phome [psi] - + + + - ρ*g home Phome=ρ*g* ytank - yhome -hL
ft v=5 C=100 s Elevtank=540 [ft] home Elevhome=380 [ft] L=1,000 [ft] d=4 [in] v2 h sub L. [pause] The headloss--- home Phome=ρ*g* ytank - yhome - -hL 2*g Ptank v2 v2 home tank + Phome=ρ*g* + ytank + - yhome -hL ρ*g 2*g 2*g

Find: Phome [psi] + ΣK * ft v=5 C=100 s Elevtank=540 [ft] home
Elevhome=380 [ft] L=1,000 [ft] d=4 [in] is a combination of the major losses --- major minor losses losses 3.022 * L * v1.85 v2 hL= + ΣK * C1.85 * d1.17 2 * g

Elevhome=380 [ft] L=1,000 [ft] d=4 [in] along the pipeline wall, plus the minor losses, --- major minor losses losses 3.022 * L * v1.85 v2 hL= + ΣK * C1.85 * d1.17 2 * g

Elevhome=380 [ft] L=1,000 [ft] d=4 [in] caused by fittings, valves, meters and other appurtenances. For this version of the --- major minor losses losses 3.022 * L * v1.85 v2 hL= + ΣK * C1.85 * d1.17 2 * g

Elevhome=380 [ft] L=1,000 [ft] d=4 [in] Hazen-Williams major headloss equation, the headloss is in feet, the length of the pipe --- major minor losses losses 3.022 * L * v1.85 v2 hL= + ΣK * C1.85 * d1.17 2 * g

Elevhome=380 [ft] L=1,000 [ft] d=4 [in] headloss [ft] is in feet, the velocity is in feet per second, --- ft velocity s length [ft] 3.022 * L * v1.85 v2 hL= + ΣK * C1.85 * d1.17 2 * g

Elevhome=380 [ft] L=1,000 [ft] d=4 [in] headloss [ft] the friction coefficient is unitless, and the diameter, is in feet. For the minor headloss variables, sigma K represents --- ft velocity s length [ft] 3.022 * L * v1.85 v2 hL= + ΣK * C1.85 * d1.17 2 * g diameter [ft]

Elevhome=380 [ft] L=1,000 [ft] d=4 [in] headloss [ft] minor headloss the sum of the minor headloss coefficients, and the v and g variables represent the --- ft velocity coefficients s length [ft] 3.022 * L * v1.85 v2 hL= + ΣK * C1.85 * d1.17 2 * g diameter [ft]

Elevhome=380 [ft] L=1,000 [ft] d=4 [in] headloss [ft] minor headloss velocity, and gravitational acceleration, in feet per second, and feet per second squared, respectively. The only unknown variable in this equation --- ft velocity coefficients s length [ft] ft gravitational acceleration s2 3.022 * L * v1.85 v2 hL= + ΣK * C1.85 * d1.17 2 * g diameter [ft]

Elevhome=380 [ft] L=1,000 [ft] d=4 [in] headloss [ft] minor headloss is the sum of the minor headloss coefficients, sigma K. [pause] ft velocity coefficients s length [ft] ft gravitational acceleration s2 3.022 * L * v1.85 v2 hL= + ΣK * C1.85 * d1.17 2 * g diameter [ft]

Find: Phome [psi] ft s v=5 C=100 Elevtank=540 [ft] home
Elevhome=380 [ft] L=1,000 [ft] d=4 [in] o The problem statement labels the different pipe appurtenances, and the the loss coefficients for each item --- gate 90 elbow meter (standard) valves (rotary)

Elevhome=380 [ft] L=1,000 [ft] d=4 [in] o meter 90 elbow valve can be looked up in a standard table of values. The value of K equals --- (rotary) Item K qty K*qty o 90 elbow (std) K gate valve K meter (rotary) K

Elevhome=380 [ft] L=1,000 [ft] d=4 [in] o meter 90 elbow valve 0.9 for the standard 90 degree elbow, --- (rotary) Item K qty K*qty o 90 elbow (std) 0.90 gate valve K meter (rotary) K

Elevhome=380 [ft] L=1,000 [ft] d=4 [in] o meter 90 elbow valve 0.19 for the fully-open gate valve, and --- (rotary) Item K qty K*qty o 90 elbow (std) 0.90 gate valve 0.19 meter (rotary) K

Elevhome=380 [ft] L=1,000 [ft] d=4 [in] o meter 90 elbow valve 10.0 for the rotary meter. [pause] The quantities of each of these appurtenances --- (rotary) Item K qty K*qty o 90 elbow (std) 0.90 gate valve 0.19 meter (rotary) 10.0

Elevhome=380 [ft] L=1,000 [ft] d=4 [in] o meter 90 elbow valve is added to the table, and then multiplied out. (rotary) Item K qty K*qty o 90 elbow (std) 0.90 2 gate valve 0.19 3 meter (rotary) 10.0 1

Elevhome=380 [ft] L=1,000 [ft] d=4 [in] o meter 90 elbow valve [pause] The sum of these products equals --- (rotary) Item K qty K*qty o 90 elbow (std) 0.90 2 1.80 gate valve 0.19 3 0.57 meter (rotary) 10.0 1 10.0

Find: Phome [psi] ΣK*qty= 12.37 ft s v=5 C=100 Elevtank=540 [ft] home
Elevhome=380 [ft] L=1,000 [ft] d=4 [in] o meter 90 elbow valve 12.37, and is used for the variable of sigma K, in the minor headloss calculation. (rotary) Item K qty K*qty ΣK*qty= o 90 elbow (std) 0.90 2 1.80 12.37 gate valve 0.19 3 0.57 meter (rotary) 10.0 1 10.0

Find: Phome [psi] + ΣK=12.37 ΣK * ft v=5 C=100 s Elevtank=540 [ft]
Elevhome=380 [ft] L=1,000 [ft] d=4 [in] ΣK=12.37 Returning to the equation for headloss, --- 3.022 * L * v1.85 v2 hL= + ΣK * C1.85 * d1.17 2 * g

Elevhome=380 [ft] L=1,000 [ft] d=4 [in] ΣK=12.37 the known variables are substituted into the equation, and the headloss between the tank and the home equals --- 3.022 * L * v1.85 v2 hL= + ΣK * ft g=32.2 C1.85 * d1.17 2 * g s2 ft 1 * in 12

Elevhome=380 [ft] L=1,000 [ft] d=4 [in] ΣK=12.37 47.6 feet. [pause] By knowing the headloss, we can return to our equation for --- 3.022 * L * v1.85 v2 hL= + ΣK * ft g=32.2 C1.85 * d1.17 2 * g s2 hL=42.8 [ft] [ft] ft 1 * hL=47.6 [ft] in 12

Find: Phome [psi] - Phome=ρ*g* home ytank - yhome -hL ft v=5 C=100 s
Elevtank=540 [ft] home Elevhome=380 [ft] L=1,000 [ft] d=4 [in] hL=47.6 [ft] v2 Phome=ρ*g* home the pressure, at the home. ytank - yhome - -hL 2*g

Find: Phome [psi] - Phome=ρ*g* home ytank - yhome -hL ft v=5 C=100 s
Elevtank=540 [ft] home Elevhome=380 [ft] L=1,000 [ft] d=4 [in] hL=47.6 [ft] ft v2 32.2 Phome=ρ*g* home The elevations, velocity head and head loss are plugged in, and then multiplied by the --- ytank - yhome - -hL s2 2*g

Find: Phome [psi] - Phome=ρ*g* home ytank - yhome -hL Phome=62.4 ft
v=5 C=100 s Elevtank=540 [ft] home Elevhome=380 [ft] L=1,000 [ft] d=4 [in] hL=47.6 [ft] ft v2 32.2 Phome=ρ*g* unit weight of water, and converted to the correct units such that the --- ytank - yhome - home -hL s2 2*g 2 lbm ft 1 lbf*s2 1 ft Phome=62.4 *32.2 * * ft3 s2 32.2 lbm*ft in 12 *112.0 [ft]

Find: Phome [psi] - Phome=ρ*g* home ytank - yhome -hL Phome=62.4
ft v=5 C=100 s Elevtank=540 [ft] home Elevhome=380 [ft] L=1,000 [ft] d=4 [in] hL=47.6 [ft] ft v2 32.2 Phome=ρ*g* home pressure at the home equals 48.5, pounds per square inch. [pause] ytank - yhome - -hL s2 2*g 2 lbm ft 1 lbf*s2 1 ft Phome=62.4 *32.2 * * ft3 s2 lbm*ft in 32.2 12 Phome=48.5 [psi] *112.0 [ft]

Find: Phome [psi] - 40 50 Phome=ρ*g* home ytank - yhome 60 -hL 70
ft v=5 C=100 s Elevtank=540 [ft] home Elevhome=380 [ft] L=1,000 [ft] d=4 [in] hL=47.6 [ft] 40 50 60 70 ft v2 32.2 Phome=ρ*g* home When reviewing the possible solutions, --- ytank - yhome - -hL s2 2*g 2 lbm ft 1 lbf*s2 1 ft Phome=62.4 *32.2 * * ft3 s2 lbm*ft in 32.2 12 Phome=48.5 [psi] *112.0 [ft]

Find: Phome [psi] - AnswerB 40 50 Phome=ρ*g* 60 home ytank - yhome
ft v=5 C=100 s Elevtank=540 [ft] home Elevhome=380 [ft] L=1,000 [ft] d=4 [in] hL=47.6 [ft] 40 50 60 70 ft v2 32.2 Phome=ρ*g* the answer is B. ytank - yhome - home -hL s2 2*g 2 lbm ft 1 lbf*s2 1 ft Phome=62.4 *32.2 * * ft3 s2 32.2 lbm*ft in 12 Phome=48.5 [psi] AnswerB *112.0 [ft]

Find: Q gal min ΣK*qty 1,600 1,800 2,000 2,200 Δh pipe entrance fresh
#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 ΣK*qty discharge 640 [ft] 8 [in] steel

? 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: Phome [psi] 40 C) 60 B) 50 D) 70 ft s v=5 C=100

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Find: Phome [psi] 40 C) 60 B) 50 D) 70 ft s v=5 C=100

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Presentation on theme: "Find: Phome [psi] 40 C) 60 B) 50 D) 70 ft s v=5 C=100"— Presentation transcript:

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