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Find: hT [ft] @ Section 3 754.4 754.8 755.2 755.6 1 2 3 4 Section
Station WSEL Part Area P [mile] [ft] [ft2] [ft] 3 377.86 753.07 M 45,950 1,440 O 1,990 167 ft3 s nM= 0.025 nO= 0.050 Q=431,000 1 2 3 4 Station [mile] flow direction 754.4 754.8 755.2 755.6 Find the total head at Section 3, in feet. [pause] In this problem, ---
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Find: hT [ft] @ Section 3 754.4 754.8 755.2 755.6 1 2 3 4 Section
Station WSEL Part Area P [mile] [ft] [ft2] [ft] 3 377.86 753.07 M 45,950 1,440 O 1,990 167 ft3 s nM= 0.025 nO= 0.050 Q=431,000 1 2 3 4 Station [mile] flow direction 754.4 754.8 755.2 755.6 the river station and water surface elevation for a given cross-section are provided, as well as the ---
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Find: hT [ft] @ Section 3 754.4 754.8 755.2 755.6 1 2 3 4 Section
Station WSEL Part Area P [mile] [ft] [ft2] [ft] 3 377.86 753.07 M 45,950 1,440 O 1,990 167 ft3 s nM= 0.025 nO= 0.050 Q=431,000 1 2 3 4 Station [mile] flow direction 754.4 754.8 755.2 755.6 area and wetted perimeter, divided into the ---
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Find: hT [ft] @ Section 3 overbank main 754.4 754.8 755.2 755.6 1 2 3
Station WSEL Part Area P [mile] [ft] [ft2] [ft] 3 377.86 753.07 M 45,950 1,440 overbank main O 1,990 167 ft3 s nM= 0.025 nO= 0.050 Q=431,000 1 2 3 4 Station [mile] flow direction 754.4 754.8 755.2 755.6 main part of the channel, M, and the overbank part of the channel, O.
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Find: hT [ft] @ Section 3 overbank main 754.4 754.8 755.2 755.6 1 2 3
Station WSEL Part Area P [mile] [ft] [ft2] [ft] 3 377.86 753.07 M 45,950 1,440 overbank main O 1,990 167 ft3 s nM= 0.025 nO= 0.050 Q=431,000 1 2 3 4 Station [mile] flow direction 754.4 754.8 755.2 755.6 The problem also provides the roughness coefficient for these two channel parts, as well as the constant flowrate.
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Find: hT [ft] @ Section 3 overbank main 754.4 754.8 755.2 755.6 1 2 3
Station WSEL Part Area P [mile] [ft] [ft2] [ft] 3 377.86 753.07 M 45,950 1,440 overbank main O 1,990 167 ft3 s nM= 0.025 nO= 0.050 Q=431,000 1 2 3 4 flow direction 754.4 754.8 755.2 755.6 From the conservation of energy, ---
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Find: hT [ft] @ Section 3 1 2 3 4 Station [mile] 377.70 377.78 377.86
377.94 1 2 3 4 Station [mile] flow direction the total head in the river decreases as the river flows from section 4, to section 1.
![Find: hT Section Station [mile]](http://slideplayer.com./slide/16356259/95/images/7/Find%3A+hT+Section+Station+%5Bmile%5D.jpg "Station [mile] flow direction. the total head in the river decreases as the river flows from section 4, to section 1.")
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Find: hT [ft] @ Section 3 1 2 3 4 Station [mile] 377.70 377.78 377.86
377.94 1 2 3 4 Station [mile] flow direction The problem asks to find the total head at section 3, ---
![Find: hT Section Station [mile]](http://slideplayer.com./slide/16356259/95/images/8/Find%3A+hT+Section+Station+%5Bmile%5D.jpg "Station [mile] flow direction. The problem asks to find the total head at section 3, ---")
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Find: hT [ft] @ Section 3 + + ρ*g hT = 1 2 3 4 y P α*v2 Station [mile]
flow direction where the total head equals the pressure head, plus the elevation head, plus the velocity head. 377.70 377.78 377.86 377.94 Station [mile] P α*v2 hT = + y + ρ*g 2*g
![Find: hT Section ρ*g hT = y P α*v2 Station [mile]](http://slideplayer.com./slide/16356259/95/images/9/Find%3A+hT+Section+%CF%81%2Ag+hT+%3D+y+P+%CE%B1%2Av2+Station+%5Bmile%5D.jpg "flow direction. where the total head equals the pressure head, plus the elevation head, plus the velocity head Station [mile] P. α*v2. hT = + y. + ρ*g. 2*g.")
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Find: hT [ft] @ Section 3 + + ρ*g hT = 1 2 3 4 y P α*v2 Station [mile]
flow direction The pressure head plus the elevation head is equal to the water surface elevation, --- 377.70 377.78 377.86 377.94 Station [mile] P α*v2 hT = + y + ρ*g 2*g
![Find: hT Section ρ*g hT = y P α*v2 Station [mile]](http://slideplayer.com./slide/16356259/95/images/10/Find%3A+hT+Section+%CF%81%2Ag+hT+%3D+y+P+%CE%B1%2Av2+Station+%5Bmile%5D.jpg "flow direction. The pressure head plus the elevation head is equal to the water surface elevation, Station [mile] P. α*v2. hT = + y. + ρ*g. 2*g.")
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Find: hT [ft] @ Section 3 + + ρ*g hT = 1 2 3 4 y P α*v2 Station [mile]
flow direction which is provided in the problem statement, as feet. 377.70 377.78 377.86 377.94 Station [mile] P α*v2 hT = + y + ρ*g 2*g
![Find: hT Section ρ*g hT = y P α*v2 Station [mile]](http://slideplayer.com./slide/16356259/95/images/11/Find%3A+hT+Section+%CF%81%2Ag+hT+%3D+y+P+%CE%B1%2Av2+Station+%5Bmile%5D.jpg "flow direction. which is provided in the problem statement, as feet Station [mile] P. α*v2. hT = + y. + ρ*g. 2*g.")
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Find: hT [ft] @ Section 3 + + ρ*g hT,3 = 1 2 3 4 y3 P3 α3*v3
flow direction Therefore, this values can be plugged into the equation, in place of pressure head plus elevation head, at section 4. Now all we have left to solve for --- 377.70 377.78 377.86 377.94 Station [mile] P3 α3*v3 2 hT,3 = + y3 + ρ*g 2*g
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Find: hT [ft] @ Section 3 + + ρ*g hT,3 = 1 2 3 4 y3 P3 α3*v3
flow direction is the velocity head. [pause] The velocity head is the difference between the --- 377.70 377.78 377.86 377.94 Station [mile] P3 α3*v3 2 hT,3 = + y3 + ρ*g 2*g
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Find: hT [ft] @ Section 3 + + ρ*g hT,3 = 1 2 3 4 y3 α3*v3 P3 α3*v3
hydraulic grade line α3*v3 2 2*g flow direction hydraulic grade line, represented by the water surface elevation, and the 377.70 377.78 377.86 377.94 Station [mile] P3 α3*v3 2 hT,3 = + y3 + ρ*g 2*g
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Find: hT [ft] @ Section 3 + + ρ*g hT,3 = 1 3 4 y3 α3*v3 P3 α3*v3
energy 3 4 hydraulic grade line grade line α3*v3 2 2*g flow direction energy grade line, represented by the dashed line. 377.70 377.78 377.86 377.94 Station [mile] P3 α3*v3 2 hT,3 = + y3 + ρ*g 2*g
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Find: hT [ft] @ Section 3 + + ρ*g hT,3 = 1 3 4 y3 α3*v3 P3 α3*v3
energy 3 4 hydraulic grade line grade line α3*v3 2 2*g flow direction The velocity head is composed of 2 unknown terms, --- 377.70 377.78 377.86 377.94 Station [mile] P3 α3*v3 2 hT,3 = + y3 + ρ*g 2*g
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Find: hT [ft] @ Section 3 + + ρ*g hT,3 = 1 3 4 y3 α3*v3 P3 α3*v3
energy 3 4 hydraulic grade line grade line α3*v3 2 2*g flow direction the average velocity, v, and --- 377.70 377.78 377.86 377.94 Station [mile] P3 α3*v3 2 average velocity hT,3 = + y3 + ρ*g 2*g
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Find: hT [ft] @ Section 3 + + ρ*g hT,3 = 1 3 4 y3 α3*v3 P3 α3*v3
energy 3 4 hydraulic grade line grade line α3*v3 2 2*g flow direction and an energy coefficient to account for the non-uniform velocity distribution in the river, alpha, we’ll refer to this alpha as the velocity coefficient. 377.70 377.78 377.86 377.94 Station [mile] P3 α3*v3 2 average velocity hT,3 = + y3 + ρ*g 2*g velocity coefficient
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Find: hT [ft] @ Section 3 + + ρ*g hT,3 = 1 3 4 y3 α3*v3 P3 α3*v3
energy 3 4 hydraulic grade line grade line α3*v3 2 2*g flow direction We’ll first solve for the average velocity. The average velocity at section 3 equals, --- 377.70 377.78 377.86 377.94 Station [mile] P3 α3*v3 2 average velocity hT,3 = + y3 + ρ*g 2*g velocity coefficient
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Find: hT [ft] @ Section 3 + + ρ*g hT,3 = y3 P3 α3*v3 Q v3= A3
the flowrate, Q, divided by the area at section 3, A sub 3. The problem statement provides the flowrate as --- A3 P3 α3*v3 2 average velocity hT,3 = + y3 + ρ*g 2*g velocity coefficient
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Find: hT [ft] @ Section 3 + + ρ*g hT,3 = y3 P3 α3*v3 Q=431,000 Q v3=
431,000 cubic feet per second, and if we recall the data table, --- A3 P3 α3*v3 2 average velocity hT,3 = + y3 + ρ*g 2*g velocity coefficient
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Find: hT [ft] @ Section 3 + + ρ*g hT,3 = y3 P3 α3*v3 Section Station
WSEL Part Area P [mile] [ft] [ft2] [ft] 3 377.86 753.07 M 45,950 1,440 O 1,990 167 ft3 Q=431,000 s Q v3= we notice the areas have been provided, for --- A3 P3 α3*v3 2 average velocity hT,3 = + y3 + ρ*g 2*g velocity coefficient
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Find: hT [ft] @ Section 3 + + ρ*g hT,3 = y3 P3 α3*v3 Section Station
WSEL Part Area P [mile] [ft] [ft2] [ft] 3 377.86 753.07 M 45,950 1,440 O 1,990 167 ft3 Q=431,000 s Q v3= the main channel, and for the overbank part of the channel. Adding these areas together, ---- A3 P3 α3*v3 2 average velocity hT,3 = + y3 + ρ*g 2*g velocity coefficient
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Find: hT [ft] @ Section 3 + + ρ*g + hT,3 = y3 P3 α3*v3 Section Station
WSEL Part Area P [mile] [ft] [ft2] [ft] 3 377.86 753.07 M 45,950 1,440 + O 1,990 167 ft3 47,940 [ft2] Q=431,000 s Q v3= the total area at section 3 equals, 47,940 feet squared. [pause] Plugging this value into the velocity equation, --- A3 P3 α3*v3 2 average velocity hT,3 = + y3 + ρ*g 2*g velocity coefficient
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Find: hT [ft] @ Section 3 + + ρ*g + hT,3 = y3 P3 α3*v3 Section Station
WSEL Part Area P [mile] [ft] [ft2] [ft] 3 377.86 753.07 M 45,950 1,440 + O 1,990 167 ft3 47,940 [ft2] Q=431,000 s ft Q v3=8.99 v3= The velocity at section 3 equals 8.99 feet per second. [pause] s A3 P3 α3*v3 2 average velocity hT,3 = + y3 + ρ*g 2*g velocity coefficient
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Find: hT [ft] @ Section 3 + + ρ*g + hT,3 = y3 P3 α3*v3 Section Station
WSEL Part Area P [mile] [ft] [ft2] [ft] 3 377.86 753.07 M 45,950 1,440 + O 1,990 167 ft3 47,940 [ft2] Q=431,000 s ft Q v3=8.99 v3= With the velocity solved, we next calculate --- s A3 P3 α3*v3 2 average velocity hT,3 = + y3 + ρ*g 2*g velocity coefficient
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Find: hT [ft] @ Section 3 + + ρ*g Σ(K3/A2) α = hT,3 = y3 Σ K3 Σ A2 P3
Station WSEL Part Area P [mile] [ft] [ft2] [ft] 3 377.86 753.07 M 45,950 1,440 O 1,990 167 Σ(K3/A2) α = Σ K3 the velocity coefficient which equals, the sum of the --- Σ A2 A3=47,940 [ft2] P3 α3*v3 2 average velocity hT,3 = + y3 + ρ*g 2*g velocity coefficient
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Find: hT [ft] @ Section 3 + + ρ*g Σ(K3/A2) α = hT,3 = y3 Σ K3 Σ A2 P3
Station WSEL Part Area P [mile] [ft] [ft2] [ft] 3 377.86 753.07 M 45,950 1,440 O 1,990 167 Σ(K3/A2) α = Σ K3 conveyances cubed divided by the area squared, all divided by the quotient of the --- Σ A2 A3=47,940 [ft2] P3 α3*v3 2 average velocity hT,3 = + y3 + ρ*g 2*g velocity coefficient
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Find: hT [ft] @ Section 3 + + ρ*g Σ(K3/A2) α = hT,3 = y3 Σ K3 Σ A2 P3
Station WSEL Part Area P [mile] [ft] [ft2] [ft] 3 377.86 753.07 M 45,950 1,440 O 1,990 167 Σ(K3/A2) α = Σ K3 sum of the conveyance cubed divided by the sum of the area squared, where the summation of --- Σ A2 A3=47,940 [ft2] P3 α3*v3 2 average velocity hT,3 = + y3 + ρ*g 2*g velocity coefficient
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Find: hT [ft] @ Section 3 + + ρ*g Σ(K3/A2) α = hT,3 = y3 Σ K3 Σ A2 P3
Station WSEL Part Area P [mile] [ft] [ft2] [ft] 3 377.86 753.07 M 45,950 1,440 O 1,990 167 Σ(K3/A2) α = Σ K3 conveyances and areas refer to the main channel, and the overbank part of the channel. Σ A2 A3=47,940 [ft2] P3 α3*v3 2 average velocity hT,3 = + y3 + ρ*g 2*g velocity coefficient
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Find: hT [ft] @ Section 3 + + ρ*g Σ(K3 /A3 ) α3 = hT,3 = y3 Σ K3 Σ A3
Station WSEL Part Area P [mile] [ft] [ft2] [ft] 3 377.86 753.07 M 45,950 1,440 O 1,990 167 Σ(K3 /A3 ) 3 2 α3 = Σ K3 3 When referring to section 3, the subscript 3 is added to each term. Since we already solved for the area, --- Σ A3 2 A3=47,940 [ft2] P3 α3*v3 2 average velocity hT,3 = + y3 + ρ*g 2*g velocity coefficient
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Find: hT [ft] @ Section 3 + + ρ*g Σ(K3 /A3 ) α3 = hT,3 = y3 Σ K3 Σ A3
Station WSEL Part Area P [mile] [ft] [ft2] [ft] 3 377.86 753.07 M 45,950 1,440 O 1,990 167 Σ(K3 /A3 ) 3 2 α3 = Σ K3 3 we’ll next have to solve for the conveyance, --- Σ A3 2 A3=47,940 [ft2] P3 α3*v3 2 average velocity hT,3 = + y3 + ρ*g 2*g velocity coefficient
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Find: hT [ft] @ Section 3 + + ρ*g Σ(K3 /A3 ) α3 = hT,3 = y3 Σ K3 Σ A3
Station WSEL Part Area P [mile] [ft] [ft2] [ft] 3 377.86 753.07 M 45,950 1,440 O 1,990 167 Σ(K3 /A3 ) 3 2 1.49 K= α3 = A R2/3 * * n Σ K3 3 K sub 3, which equals 1.49 divided by the --- Σ A3 2 A3=47,940 [ft2] P3 α3*v3 2 average velocity hT,3 = + y3 + ρ*g 2*g velocity coefficient
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Find: hT [ft] @ Section 3 + + ρ*g Σ(K3 /A3 ) α3 = hT,3 = y3 Σ K3 Σ A3
Station WSEL Part Area P [mile] [ft] [ft2] [ft] 3 377.86 753.07 M 45,950 1,440 O 1,990 167 Σ(K3 /A3 ) 3 2 1.49 K= α3 = A R2/3 * * n Σ K3 3 roughness coefficient, n, times ---- Σ A3 2 roughness coefficient A3=47,940 [ft2] P3 α3*v3 2 average velocity hT,3 = + y3 + ρ*g 2*g velocity coefficient
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Find: hT [ft] @ Section 3 + + ρ*g Σ(K3 /A3 ) α3 = hT,3 = y3 Σ K3 Σ A3
Station WSEL Part Area P [mile] [ft] [ft2] [ft] 3 377.86 753.07 M 45,950 1,440 O 1,990 167 area [ft2] Σ(K3 /A3 ) 3 2 1.49 K= α3 = A R2/3 * * n Σ K3 3 the area, A, in feet squared, times --- Σ A3 2 roughness coefficient A3=47,940 [ft2] P3 α3*v3 2 average velocity hT,3 = + y3 + ρ*g 2*g velocity coefficient
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Find: hT [ft] @ Section 3 + + ρ*g Σ(K3 /A3 ) α3 = hT,3 = y3 Σ K3 Σ A3
Station WSEL Part Area P [mile] [ft] [ft2] [ft] 3 377.86 753.07 M 45,950 1,440 O 1,990 167 area [ft2] Σ(K3 /A3 ) 3 2 1.49 K= α3 = A R2/3 * * n Σ K3 3 hydraulic radius [ft] the hydraulic radius, R, in feet, raised to the 2/3s power. [pause] Returning to the data table, --- Σ A3 2 roughness coefficient A3=47,940 [ft2] P3 α3*v3 2 average velocity hT,3 = + y3 + ρ*g 2*g velocity coefficient
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Find: hT [ft] @ Section 3 + + ρ*g Σ(K3 /A3 ) α3 = hT,3 = y3 Σ K3 Σ A3
Station WSEL Part Area P [mile] [ft] [ft2] [ft] 3 377.86 753.07 M 45,950 1,440 O 1,990 167 area [ft2] Σ(K3 /A3 ) 3 2 1.49 K= α3 = A R2/3 * * n Σ K3 3 hydraulic radius [ft] we’ll remove the columns for Section, Station, water surface elevation, --- Σ A3 2 roughness coefficient A3=47,940 [ft2] P3 α3*v3 2 average velocity hT,3 = + y3 + ρ*g 2*g velocity coefficient
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Find: hT [ft] @ Section 3 + + ρ*g hT,3 = y3 area [ft2] 1.49 K= A R2/3
Part Area P [ft2] [ft] M 45,950 1,440 O 1,990 167 area [ft2] 1.49 K= A R2/3 * * and shift the other three columns over. n hydraulic radius [ft] roughness P3 α3*v3 2 average velocity hT,3 = + y3 + ρ*g 2*g velocity coefficient
![Find: hT Section ρ*g hT,3 = y3 area [ft2] 1.49 K= A R2/3](http://slideplayer.com./slide/16356259/95/images/38/Find%3A+hT+Section+%CF%81%2Ag+hT%2C3+%3D+y3+area+%5Bft2%5D+1.49+K%3D+A+R2%2F3.jpg "Part. Area. P. [ft2] [ft] M. 45,950. 1,440. O. 1, area [ft2] K= A. R2/3. * * and shift the other three columns over. n. hydraulic. radius [ft] roughness. P3. α3*v3. 2. average velocity. hT,3 = + y3. + ρ*g. 2*g. velocity coefficient.")
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Find: hT [ft] @ Section 3 + + ρ*g hT,3 = y3 Section 3 area [ft2] 1.49
Part Area P [ft2] [ft] M 45,950 1,440 O 1,990 167 area [ft2] 1.49 K= A R2/3 * * [pause] The equation for conveyance includes the hydraulic radius, R, --- n hydraulic radius [ft] roughness P3 α3*v3 2 average velocity hT,3 = + y3 + ρ*g 2*g velocity coefficient
![Find: hT Section ρ*g hT,3 = y3 Section 3 area [ft2] 1.49](http://slideplayer.com./slide/16356259/95/images/39/Find%3A+hT+Section+%CF%81%2Ag+hT%2C3+%3D+y3+Section+3+area+%5Bft2%5D+1.49.jpg "Part. Area. P. [ft2] [ft] M. 45,950. 1,440. O. 1, area [ft2] K= A. R2/3. * * [pause] The equation for conveyance includes the hydraulic radius, R, --- n. hydraulic. radius [ft] roughness. P3. α3*v3. 2. average velocity. hT,3 = + y3. + ρ*g. 2*g. velocity coefficient.")
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Find: hT [ft] @ Section 3 + + ρ*g hT,3 = y3 Section 3 area [ft2] 1.49
Part Area P R [ft2] [ft] [ft] M 45,950 1,440 31.91 O 1,990 167 11.92 area [ft2] 1.49 A K= R = A R2/3 * * which equals the area, divided by the wetted perimeter, which equals, --- n P hydraulic radius [ft] roughness P3 α3*v3 2 average velocity hT,3 = + y3 + ρ*g 2*g velocity coefficient
![Find: hT Section ρ*g hT,3 = y3 Section 3 area [ft2] 1.49](http://slideplayer.com./slide/16356259/95/images/40/Find%3A+hT+Section+%CF%81%2Ag+hT%2C3+%3D+y3+Section+3+area+%5Bft2%5D+1.49.jpg "Part. Area. P. R. [ft2] [ft] [ft] M. 45,950. 1, O. 1, area [ft2] A. K= R = A. R2/3. * * which equals the area, divided by the wetted perimeter, which equals, --- n. P. hydraulic. radius [ft] roughness. P3. α3*v3. 2. average velocity. hT,3 = + y3. + ρ*g. 2*g. velocity coefficient.")
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Find: hT [ft] @ Section 3 + + ρ*g hT,3 = y3 Section 3 area [ft2] 1.49
Part Area P R [ft2] [ft] [ft] M 45,950 1,440 31.91 O 1,990 167 11.92 area [ft2] 1.49 A K= R = A R2/3 * * 31.91 feet, and feet, for the main channel, and overbank channel, respectively. [pause] The problem statement also n P hydraulic radius [ft] roughness P3 α3*v3 2 average velocity hT,3 = + y3 + ρ*g 2*g velocity coefficient
![Find: hT Section ρ*g hT,3 = y3 Section 3 area [ft2] 1.49](http://slideplayer.com./slide/16356259/95/images/41/Find%3A+hT+Section+%CF%81%2Ag+hT%2C3+%3D+y3+Section+3+area+%5Bft2%5D+1.49.jpg "Part. Area. P. R. [ft2] [ft] [ft] M. 45,950. 1, O. 1, area [ft2] A. K= R = A. R2/3. * * feet, and feet, for the main channel, and overbank channel, respectively. [pause] The problem statement also. n. P. hydraulic. radius [ft] roughness. P3. α3*v3. 2. average velocity. hT,3 = + y3. + ρ*g. 2*g. velocity coefficient.")
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Find: hT [ft] @ Section 3 + + ρ*g hT,3 = y3 Section 3 area [ft2] 1.49
Part Area P R n [ft2] [ft] [ft] M 45,950 1,440 31.91 0.025 O 1,990 167 11.92 0.050 area [ft2] 1.49 K= A R2/3 * * provided the the roughness coefficients, n, of and 0.050, for these two channel parts. n hydraulic radius [ft] roughness P3 α3*v3 2 average velocity hT,3 = + y3 + ρ*g 2*g velocity coefficient
![Find: hT Section ρ*g hT,3 = y3 Section 3 area [ft2] 1.49](http://slideplayer.com./slide/16356259/95/images/42/Find%3A+hT+Section+%CF%81%2Ag+hT%2C3+%3D+y3+Section+3+area+%5Bft2%5D+1.49.jpg "Part. Area. P. R. n. [ft2] [ft] [ft] M. 45,950. 1, O. 1, area [ft2] K= A. R2/3. * * provided the the roughness coefficients, n, of and 0.050, for these two channel parts. n. hydraulic. radius [ft] roughness. P3. α3*v3. 2. average velocity. hT,3 = + y3. + ρ*g. 2*g. velocity coefficient.")
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Find: hT [ft] @ Section 3 + + ρ*g hT,3 = y3 Section 3 area [ft2] 1.49
Part Area P R n K [ft2] [ft] [ft] M 45,950 1,440 31.91 0.025 2.76*107 O 1,990 167 11.92 0.050 3.09*105 area [ft2] 1.49 K= A R2/3 * * Finally we can calculate the conveyance, K, and we’ll disregarding the units. The sum of the conveyances equals, --- n hydraulic radius [ft] roughness P3 α3*v3 2 average velocity hT,3 = + y3 + ρ*g 2*g velocity coefficient
![Find: hT Section ρ*g hT,3 = y3 Section 3 area [ft2] 1.49](http://slideplayer.com./slide/16356259/95/images/43/Find%3A+hT+Section+%CF%81%2Ag+hT%2C3+%3D+y3+Section+3+area+%5Bft2%5D+1.49.jpg "Part. Area. P. R. n. K. [ft2] [ft] [ft] M. 45,950. 1, *107. O. 1, *105. area [ft2] K= A. R2/3. * * Finally we can calculate the conveyance, K, and we’ll disregarding the units. The sum of the conveyances equals, --- n. hydraulic. radius [ft] roughness. P3. α3*v3. 2. average velocity. hT,3 = + y3. + ρ*g. 2*g. velocity coefficient.")
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Find: hT [ft] @ Section 3 + + ρ*g + hT,3 = y3 Section 3 Σ K3=2.79*107
Part Area P R n K [ft2] [ft] [ft] M 45,950 1,440 31.91 0.025 2.76*107 O 1,990 167 11.92 0.050 + 3.09*105 Σ K3=2.79*107 area [ft2] 1.49 K= A R2/3 * * 2.79 times 10 to the 7. [pause] By knowing the sum of the conveyances and ---- n hydraulic radius [ft] roughness P3 α3*v3 2 average velocity hT,3 = + y3 + ρ*g 2*g velocity coefficient
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Find: hT [ft] @ Section 3 + + ρ*g + hT,3 = y3 Section 3 Σ K3=2.79*107
Part Area P R n K [ft2] [ft] [ft] M 45,950 1,440 31.91 0.025 2.76*107 O 1,990 167 11.92 0.050 + 3.09*105 Σ K3=2.79*107 Σ A3=47,940 the sum of the areas, we can calculate --- P3 α3*v3 2 average velocity hT,3 = + y3 + ρ*g 2*g velocity coefficient
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Find: hT [ft] @ Section 3 + + ρ*g Σ(K3 /A3 ) α3 = hT,3 = y3 Section 3
Part Area P R n K [ft2] [ft] [ft] M 45,950 1,440 31.91 0.025 2.76*107 O 1,990 167 11.92 0.050 3.09*105 Σ K3=2.79*107 Σ(K3 /A3 ) 3 2 α3 = Σ A3=47,940 Σ K3 3 the velocity coefficient, alpha. [pause] Σ A3 2 P3 α3*v3 2 average velocity hT,3 = + y3 + ρ*g 2*g velocity coefficient
47
Find: hT [ft] @ Section 3 + + = ρ*g Σ(K3 /A3 ) + α3 = hT,3 = y3
Part Area P R n K [ft2] [ft] [ft] M 45,950 1,440 31.91 0.025 2.76*107 O 1,990 167 11.92 0.050 3.09*105 Σ(K3 /A3 ) 3 2 (2.76*107)3 (3.09*105)3 + α3 = = (45,950)2 (1,990)2 Σ K3 3 Plugging in the the appropriate values, alpha equals --- (2.79*107)3/(47,940)2 Σ A3 2 P3 α3*v3 2 average velocity hT,3 = + y3 + ρ*g 2*g velocity coefficient
48
Find: hT [ft] @ Section 3 + + = ρ*g Σ(K3 /A3 ) + α3 = hT,3 = y3
Part Area P R n K [ft2] [ft] [ft] M 45,950 1,440 31.91 0.025 2.76*107 O 1,990 167 11.92 0.050 3.09*105 Σ(K3 /A3 ) 3 2 (2.76*107)3 (3.09*105)3 + α3 = = (45,950)2 (1,990)2 =1.054 Σ K3 3 [pause] Returning to Bernoulli’s equation, --- (2.79*107)3/(47,940)2 Σ A3 2 P3 α3*v3 2 average velocity hT,3 = + y3 + ρ*g 2*g velocity coefficient
49
Find: hT [ft] @ Section 3 + + ρ*g α3 =1.054 hT,3 = y3 Section 3
Σ K3=2.79*107 Σ A3=47,940 Part Area P R n K [ft2] [ft] [ft] M 45,950 1,440 31.91 0.025 2.76*107 O 1,990 167 11.92 0.050 3.09*105 α3 =1.054 WSEL3= [ft] ft v3=8.99 g=32.174 ft we can solve for the total head at section 3 by plugging in the values for the ---- s s2 P3 α3*v3 2 hT,3 = + y3 + ρ*g 2*g
50
Find: hT [ft] @ Section 3 + + ρ*g α3 =1.054 hT,3 = y3 Section 3
Σ K3=2.79*107 Σ A3=47,940 Part Area P R n K [ft2] [ft] [ft] M 45,950 1,440 31.91 0.025 2.76*107 O 1,990 167 11.92 0.050 3.09*105 α3 =1.054 WSEL3= [ft] ft v3=8.99 g=32.174 ft the water surface elevation, --- s s2 P3 α3*v3 2 hT,3 = + y3 + ρ*g 2*g
51
Find: hT [ft] @ Section 3 + + ρ*g α3 =1.054 hT,3 = y3 Section 3
Σ K3=2.79*107 Σ A3=47,940 Part Area P R n K [ft2] [ft] [ft] M 45,950 1,440 31.91 0.025 2.76*107 O 1,990 167 11.92 0.050 3.09*105 α3 =1.054 WSEL3= [ft] ft v3=8.99 g=32.174 ft the velocity, the velocity coefficient, --- s s2 P3 α3*v3 2 hT,3 = + y3 + ρ*g 2*g
52
Find: hT [ft] @ Section 3 + + ρ*g α3 =1.054 hT,3 = y3 Section 3
Σ K3=2.79*107 Σ A3=47,940 Part Area P R n K [ft2] [ft] [ft] M 45,950 1,440 31.91 0.025 2.76*107 O 1,990 167 11.92 0.050 3.09*105 α3 =1.054 WSEL3= [ft] ft v3=8.99 g=32.174 ft and the gravitational acceleration. The total head at section 3 equals, ---- s s2 P3 α3*v3 2 hT,3 = + y3 + ρ*g 2*g
53
Find: hT [ft] @ Section 3 + + ρ*g α3 =1.054 hT,3 = y3 Section 3
Σ K3=2.79*107 Σ A3=47,940 Part Area P R n K [ft2] [ft] [ft] M 45,950 1,440 31.91 0.025 2.76*107 O 1,990 167 11.92 0.050 3.09*105 α3 =1.054 WSEL3= [ft] ft v3=8.99 g=32.174 ft feet. [pause] s s2 P3 α3*v3 2 hT,3 = + y3 + hT,3= [ft] ρ*g 2*g
54
Find: hT [ft] @ Section 3 + + ρ*g α3 =1.054 hT,3 = y3 Section 3
Σ K3=2.79*107 Σ A3=47,940 Part Area P R n K [ft2] [ft] [ft] M 45,950 1,440 31.91 0.025 2.76*107 O 1,990 167 11.92 0.050 3.09*105 754.4 754.8 755.2 755.6 α3 =1.054 WSEL3= [ft] ft v3=8.99 g=32.174 ft When reviewing the possible solutions, --- s s2 P3 α3*v3 2 hT,3 = + y3 + hT,3= [ft] ρ*g 2*g
55
Find: hT [ft] @ Section 3 + + ρ*g α3 =1.054 hT,3 = y3 Section 3
Σ K3=2.79*107 Σ A3=47,940 Part Area P R n K [ft2] [ft] [ft] M 45,950 1,440 31.91 0.025 2.76*107 O 1,990 167 11.92 0.050 3.09*105 754.4 754.8 755.2 755.6 α3 =1.054 WSEL3= [ft] ft v3=8.99 g=32.174 ft the answer is A. s s2 P3 α3*v3 2 hT,3 = + y3 + hT,3= [ft] ρ*g 2*g AnswerA
56
? 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
![Index σ’v = Σ γ d γT=100 [lb/ft3] +γclay dclay 1](http://slideplayer.com./slide/16356259/95/images/56/Index+%CF%83%E2%80%99v+%3D+%CE%A3+%CE%B3+d+%CE%B3T%3D100+%5Blb%2Fft3%5D+%2B%CE%B3clay+dclay+1.jpg "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.")
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