Find: hL [m] 0.63 0.84 1.02 1.31 rectangular d channel b b=3 [m] hydraulic jump Find the headloss, in meters, across the hydraulic jump. [pause] In this problem, --- d2 Q d1 d1=0.5 [m] d2=2.0 [m]

Find: hL [m] 0.63 0.84 1.02 1.31 rectangular d channel b b=3 [m] hydraulic jump a hydraulic jump occurs in a rectangular channel. d2 Q d1 d1=0.5 [m] d2=2.0 [m]

Find: hL [m] 0.63 0.84 1.02 1.31 rectangular d channel b b=3 [m] hydraulic jump The channel width is provided, as well as the initial and final water depths. [pause] d2 Q d1 d1=0.5 [m] d2=2.0 [m]

Find: hL [m] + + ρ*g P y hT= v2 d 2*g b b=3 [m] hydraulic jump d2 Q d1 Bernoulli’s equation states the total head equals, --- d2 Q d1 d1=0.5 [m] d2=2.0 [m]

Find: hL [m] + + ρ*g P y hT= v2 d 2*g b pressure b=3 [m] head hydraulic jump the pressure head, --- d2 Q d1 d1=0.5 [m] d2=2.0 [m]

Find: hL [m] + + ρ*g P y hT= v2 d 2*g b pressure b=3 [m] head hydraulic elevation head jump plus the elevation head, --- d2 Q d1 d1=0.5 [m] d2=2.0 [m]

Find: hL [m] + + ρ*g P y hT= v2 d 2*g b pressure pressure velocity b=3 [m] head head head hydraulic elevation head jump plus the velocity head. d2 Q d1 d1=0.5 [m] d2=2.0 [m]

Find: hL [m] + + ρ*g P y hT= v2 d 2*g b pressure pressure velocity b=3 [m] head head head hydraulic elevation head jump Since open channel flow is not pressurized flow, --- d2 Q d1 d1=0.5 [m] d2=2.0 [m]

Find: hL [m] + + ρ*g P y hT= v2 d 2*g b pressure pressure velocity P v2 + y + hT= d ρ*g 2*g b pressure pressure velocity b=3 [m] head head head hydraulic elevation head jump the pressure head for this problem, is zero, and the equation simplifies. d2 Q d1 d1=0.5 [m] d2=2.0 [m]

Find: hL [m] + + + ρ*g P y hT= hT= y v2 d 2*g v2 b 2*g b=3 [m] P v2 + y + hT= d ρ*g 2*g v2 b hT= y + 2*g b=3 [m] hydraulic jump The headloss across the hydraulic jump, is equal to --- d2 Q d1 d1=0.5 [m] d2=2.0 [m]

Find: hL [m] + + + ρ*g P y hT= hT= y hL= h1 – h2 v2 d 2*g v2 b 2*g P v2 + y + hT= d ρ*g 2*g v2 b hT= y + 2*g b=3 [m] hydraulic hL= h1 – h2 jump the initial total head, minus the final total head, which can be written out in terms of the --- d2 Q d1 d1=0.5 [m] d2=2.0 [m]

Find: hL [m] + + + + - + ρ*g P y hT= hT= y hL= h1 – h2 hL= y1 y2 v2 d P v2 + y + hT= d ρ*g 2*g v2 b hT= y + 2*g b=3 [m] hydraulic hL= h1 – h2 jump initial and final elevation and velocity heads. [pause] The elevation heads, v1 v2 d2 2 2 hL= y1 + - y2 + Q 2*g 2*g d1 d1=0.5 [m] d2=2.0 [m]

Find: hL [m] + + + + - + ρ*g P y hT= hT= y hL= h1 – h2 hL= y1 y2 v2 d P v2 + y + hT= d ρ*g 2*g v2 b hT= y + 2*g b=3 [m] hydraulic hL= h1 – h2 jump y1 and y2, are simply the water depths given in the problem statement, d1 and d2. v1 v2 d2 2 2 hL= y1 + - y2 + Q 2*g 2*g d1 d1=0.5 [m] d2=2.0 [m]

Find: hL [m] + + + + - + ρ*g P y hT= hT= y hL= h1 – h2 hL= y1 y2 v2 d P v2 + y + hT= d ρ*g 2*g v2 b hT= y + 2*g b=3 [m] hydraulic hL= h1 – h2 jump The value of g, is a constant, but we don’t know the velocities, --- v1 v2 d2 2 hL= y1 + - 2 y2 + Q 2*g 2*g d1 d1=0.5 [m] m 9.81 d2=2.0 [m] s2

Find: hL [m] + + + + - + ρ*g ? ? P y hT= hT= y hL= h1 – h2 hL= y1 y2 P v2 + y + hT= d ρ*g 2*g v2 b hT= y + 2*g b=3 [m] ? ? hydraulic hL= h1 – h2 jump v1 and v2. [pause] Velocities 1 and 2 equal, --- v1 v2 d2 2 2 hL= y1 + - y2 + Q 2*g 2*g d1 d1=0.5 [m] m 9.81 d2=2.0 [m] s2

Find: hL [m] + - + ? ? v1= v2= hL= y1 y2 Q A1 d Q A2 b b=3 [m] hydraulic jump the flowrate, divided by areas 1 and 2, respectively, --- v1 v2 d2 2 hL= y1 + - 2 y2 + Q 2*g 2*g d1 d1=0.5 [m] m 9.81 d2=2.0 [m] s2

Find: hL [m] + - + ? ? v1= = v2= = hL= y1 y2 Q Q A1 b*d1 d Q Q A2 b*d2 b=3 [m] ? ? hydraulic jump where the areas 1 and 2, equal the channel base times the depths 1 and 2, respectively. v1 v2 d2 2 2 hL= y1 + - y2 + Q 2*g 2*g d1 d1=0.5 [m] m 9.81 d2=2.0 [m] s2

Find: hL [m] + - + ? ? v1= = v2= = hL= y1 y2 Q Q A1 b*d1 d Q Q A2 b*d2 b=3 [m] ? ? hydraulic jump Since we already know the base width and the depths, we need to solve for the flowrate, --- v1 v2 d2 2 2 hL= y1 + - y2 + Q 2*g 2*g d1 d1=0.5 [m] m 9.81 d2=2.0 [m] s2

Find: hL [m] + - + ? ? ? ? v1= = v2= = hL= y1 y2 Q Q A1 b*d1 d Q Q A2 b=3 [m] ? ? hydraulic jump Q. [pause] An equation which relates the conjugate depths across --- v1 v2 d2 2 hL= y1 + - 2 y2 + Q 2*g 2*g d1 d1=0.5 [m] m 9.81 d2=2.0 [m] s2

Find: hL [m] ? ? v1= = v2= = d2 1 = 1 + 8 * Fr1 -1 d1 2 Q Q A1 b*d1 d b=3 [m] d2 1 = * 1 + 8 * Fr1 -1 2 hydraulic d1 2 jump a hydraulic jump, is, d2 divided by d1, equals, 1/2 times the quantity, the square root of 1 plus eight times the upstream froude number squared, minus 1. For a rectangular channel, the Froude number equals --- d2 Q d1 d1=0.5 [m] d2=2.0 [m]

Find: hL [m] ? v1= = v v2= = Fr= d2 1 = 1 + 8 * Fr1 -1 d1 2 Q Q A1 b*d1 d Q Q v v2= = Fr= A2 b*d2 g*d b b=3 [m] d2 1 = * 1 + 8 * Fr1 -1 2 hydraulic d1 2 jump the velocity, divided by the square root of the gravitational acceleration times the depth, where the velocity equals, --- d2 Q d1 d1=0.5 [m] d2=2.0 [m]

Find: hL [m] ? v1= = v= v v2= = Fr= d2 1 = 1 + 8 * Fr1 -1 d1 2 Q Q Q A1 b*d1 A d Q Q v v2= = Fr= A2 b*d2 g*d b b=3 [m] d2 1 = * 1 + 8 * Fr1 -1 2 hydraulic d1 2 jump the flowrate, divided by the area, and the area equals, --- d2 Q d1 d1=0.5 [m] d2=2.0 [m]

Find: hL [m] ? v1= = v= v v2= = Fr= d2 1 = 1 + 8 * Fr1 -1 d1 2 A=b*d Q g*d b b=3 [m] d2 1 = * 1 + 8 * Fr1 -1 2 hydraulic d1 2 jump the base width times the depth. [pause] After these substitutions are made, the hydraulic jump equation --- d2 Q d1 d1=0.5 [m] d2=2.0 [m]

Find: hL [m] ? v1= = v= v v2= = Fr= d2 1 = 1 + 8 * Fr1 -1 d1 2 d2 1 = A=b*d Q Q ? Q v1= = v= A1 b*d1 A d Q Q v v2= = Fr= A2 b*d2 g*d b b=3 [m] d2 1 = * 1 + 8 * Fr1 -1 2 hydraulic d1 2 jump is effectively 1 equation and 1 unknown, and that unknown value is our flowrate, Q. [pause] After a few lines of algebra, --- d2 d2 1 Q 2 Q = 1 + 8 * -1 d1 * d1 2 g*b2*d1 3 d1=0.5 [m] d2=2.0 [m]

Find: hL [m] ? +1 -1 v1= = v= v v2= = Fr= d2 1 = 1 + 8 * Fr1 -1 d1 2 A=b*d Q Q ? Q v1= = v= A1 b*d1 A d Q Q v v2= = Fr= A2 b*d2 g*d b b=3 [m] d2 1 = * 1 + 8 * Fr1 -1 2 hydraulic d1 2 jump the flowrate is simply a function of the gravitational acceleration, and the given data. d2 d2 1 Q 2 Q = 1 + 8 * -1 d1 * d1 2 g*b2*d1 3 d1=0.5 [m] 2 d2 g*b2*d1 * 2 3 Q= +1 -1 * d2=2.0 [m] d1 8

Find: hL [m] +1 -1 v1= = v2= = 2 d2 d1 Q Q A1 b*d1 d Q Q A2 b*d2 b b=3 [m] hydraulic jump m Plugging in the values for g, b, d1 and d2, the flowrate computes to, --- 9.81 d2 s2 Q d1 d1=0.5 [m] 2 d2 g*b2*d1 * 2 3 Q= +1 -1 * d2=2.0 [m] d1 8

Find: hL [m] +1 -1 v1= = v2= = Q=10.50 2 d2 d1 Q Q A1 b*d1 d Q Q A2 b=3 [m] m3 Q=10.50 s hydraulic jump m 10.50 meters cubed per second. [pause] To solve for the upstream and downstream velocities, --- 9.81 d2 s2 Q d1 d1=0.5 [m] 2 d2 g*b2*d1 * 2 3 Q= +1 -1 * d2=2.0 [m] d1 8

Find: hL [m] +1 -1 v1= = v2= = Q=10.50 2 d2 d1 Q Q A1 b*d1 d Q Q A2 b=3 [m] m3 Q=10.50 s hydraulic jump m the flowrate, base width, and depths are substituted in, and we get --- 9.81 d2 s2 Q d1 d1=0.5 [m] 2 d2 g*b2*d1 * 2 3 Q= +1 -1 * d2=2.0 [m] d1 8

Find: hL [m] +1 -1 v1= = =7.00 v2= = Q=10.50 2 d2 d1 Q Q A1 b*d1 d Q Q s A1 b*d1 d Q Q v2= = A2 b*d2 b b=3 [m] m3 Q=10.50 s hydraulic jump m velocity 1 equals 7 meters per second, and --- 9.81 d2 s2 Q d1 d1=0.5 [m] 2 d2 g*b2*d1 * 2 3 Q= +1 -1 * d2=2.0 [m] d1 8

Find: hL [m] +1 -1 v1= = =7.00 v2= = =1.75 Q=10.50 2 d2 d1 Q Q A1 b*d1 s A1 b*d1 d Q Q m v2= = =1.75 s A2 b*d2 b b=3 [m] m3 Q=10.50 s hydraulic jump m velocity 2 equals 1.75 meters per second. [pause] By knowing our velocities, --- 9.81 d2 s2 Q d1 d1=0.5 [m] 2 d2 g*b2*d1 * 2 3 Q= +1 -1 * d2=2.0 [m] d1 8

Find: hL [m] + - + ? ? v1=7.00 v2=1.75 hL= y1 y2 d b b=3 [m] hydraulic s d m v2=1.75 s b b=3 [m] ? ? hydraulic jump we can return to our equation, for headloss across the hydraulic jump, --- v1 v2 d2 2 hL= y1 + - 2 y2 + Q 2*g 2*g d1 d1=0.5 [m] m 9.81 d2=2.0 [m] s2

Find: hL [m] - + + v1=7.00 v2=1.75 hL= y1 y2 d b b=3 [m] hydraulic s d m v2=1.75 s b b=3 [m] hydraulic jump plug in the values for velocity, --- v1 v2 d2 2 - 2 hL= y1 + y2 + Q 2*g 2*g d1 d1=0.5 [m] m 9.81 d2=2.0 [m] s2

Find: hL [m] - + + v1=7.00 v2=1.75 hL= y1 y2 d b b=3 [m] hydraulic s d m v2=1.75 s b b=3 [m] hydraulic jump plug in the values for acceleration and the depths, and the headloss computes to --- v1 v2 d2 2 - 2 hL= y1 + y2 + Q 2*g 2*g d1 d1=0.5 [m] m 9.81 d2=2.0 [m] s2

Find: hL [m] + - + hL= 0.841 [m] v1=7.00 v2=1.75 hL= y1 y2 d b b=3 [m] s d m v2=1.75 s b hL= 0.841 [m] b=3 [m] hydraulic jump 0.841 meters. v1 v2 d2 2 hL= y1 + - 2 y2 + Q 2*g 2*g d1 d1=0.5 [m] m 9.81 d2=2.0 [m] s2

Find: hL [m] + - + hL= 0.841 [m] v1=7.00 v2=1.75 0.63 0.84 1.02 1.31 s d m v2=1.75 0.63 0.84 1.02 1.31 s b hL= 0.841 [m] b=3 [m] hydraulic jump Looking over the possible solutions, --- v1 v2 d2 2 hL= y1 + - 2 y2 + Q 2*g 2*g d1 d1=0.5 [m] m 9.81 d2=2.0 [m] s2

Find: hL [m] + - + hL= 0.841 [m] AnswerB v1=7.00 v2=1.75 0.63 0.84 1.02 1.31 s b hL= 0.841 [m] b=3 [m] AnswerB hydraulic jump the answer is B. v1 v2 d2 2 2 hL= y1 + - y2 + Q 2*g 2*g d1 d1=0.5 [m] m 9.81 d2=2.0 [m] s2

? 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