Find: V [ft/s] xL xR b b=5 [ft] xL=3 xR=3 ft3 s

Find: V [ft/s] 4.6 5.5 6.4 7.3 1 1 xL xR b b=5 [ft] xL=3 xR=3 ft3 s
Find the velocity, v, in feet per second. [pause] In this problem, --- ft3 s Q=100 S=0.01 n=0.020

water flows in a trapezoidal channel at 100 cubic feet per second. ft3 s Q=100 S=0.01 n=0.020

The base width and side slopes of the channel are given, --- ft3 s Q=100 S=0.01 n=0.020

as well as the channel slope and roughness coefficient. ft3 s Q=100 S=0.01 n=0.020

Find: V [ft/s] Q V = A 1 1 xL xR b b=5 [ft] xL=3 xR=3 ft3 s Q=100
The velocity of flow in the channel equals --- ft3 s Q=100 S=0.01 n=0.020

Find: V [ft/s] Q V = A 1 1 xL xR b b=5 [ft] flowrate xL=3 xR=3 ft3 s
the flowrate, Q, divided by the --- ft3 s Q=100 S=0.01 n=0.020

Find: V [ft/s] Q V = A 1 1 xL xR b b=5 [ft] flowrate xL=3 area xR=3
area, A. Since the problem statement provides the flowrate, --- ft3 s Q=100 S=0.01 n=0.020

Find: V [ft/s] ? Q V = A 1 1 xL xR b b=5 [ft] flowrate xL=3 area xR=3
all we need to determine is the area. [pause] From Manning’s equation, --- ft3 Q=100 s S=0.01 n=0.020

Find: V [ft/s] ? Q V = A Q= A * R2/3 * S 1/2 1 1 xL xR b b=5 [ft]
flowrate V = xL=3 A area xR=3 ? we can relate the area, in feet squared, to the flowrate, --- ft3 area [ft2] Q=100 s k S=0.01 Q= A * R2/3 * S 1/2 * n n=0.020

flowrate V = xL=3 A area xR=3 ? flowrate In cubic feet per second, the roughness, and the slope. [pause] For English units, k equals ---- ft3 area [ft2] Q=100 ft3 s s k S=0.01 Q= A * R2/3 * S 1/2 * n n=0.020 slope roughness

flowrate V = xL=3 A area xR=3 ? flowrate [pause] At this point, there are 2 unknown variables, --- ft3 area [ft2] Q=100 ft3 s s k S=0.01 Q= A * R2/3 * S 1/2 * n n=0.020 slope roughness 1.49

flowrate V = xL=3 A area xR=3 ? flowrate area, A, and the hydraulic radius, R. However these two variables can be related --- ft3 hydraulic area [ft2] Q=100 ft3 s radius [ft] s k S=0.01 Q= A * R2/3 * S 1/2 * n n=0.020 slope roughness 1.49

flowrate V = xL=3 A area xR=3 ? flowrate based on the given, channel geometry. ft3 hydraulic area [ft2] Q=100 ft3 s radius [ft] s k S=0.01 Q= A * R2/3 * S 1/2 * n n=0.020 slope roughness 1.49

Find: V [ft/s] Q= A * R2/3 * S 1/2 1 1 xL xR b b=5 [ft] xL=3 xR=3
flowrate If we isolate the unknown variables, we have --- ft3 hydraulic area [ft2] Q=100 ft3 s radius [ft] s k S=0.01 Q= A * R2/3 * S 1/2 * n n=0.020 slope roughness 1.49

Find: V [ft/s] A * R2/3= Q= A * R2/3 * S 1/2 1 1 xL xR b b=5 [ft]
Q * n A * R2/3= xL=3 k * S 1/2 xR=3 flowrate The area times the hydraulic radius to the 2/3s power equals the flowrate times the roughness all divided by k times the square root of the channel’s slope. ft3 s hydraulic area [ft2] Q=100 ft3 radius [ft] s k S=0.01 Q= A * R2/3 * S 1/2 * n n=0.020 slope roughness 1.49

Find: V [ft/s] A * R2/3= 1 1 xL xR b 1.49 b=5 [ft] Q * n xL=3
k * S 1/2 xR=3 plugging in the known variables, and temporarily dropping the units, the right hand side of the equation calculates to --- ft3 Q=100 s S=0.01 n=0.020

Find: V [ft/s] A * R2/3= A * R2/3 = 13.42 1 1 xL xR b 1.49 b=5 [ft]
Q * n A * R2/3= xL=3 k * S 1/2 xR=3 [pause] Since the hydraulic radius equal the --- ft3 A * R2/3 = 13.42 Q=100 s S=0.01 n=0.020

Find: V [ft/s] A * R2/3= A * R2/3 = 13.42 A R = P 1 1 xL xR b 1.49
b=5 [ft] Q * n A * R2/3= xL=3 k * S 1/2 xR=3 area divided by the perimeter, --- ft3 A * R2/3 = 13.42 A Q=100 R = s P S=0.01 n=0.020

Find: V [ft/s] A * R2/3= A * R2/3 = 13.42 A R = P 1 1 xL xR b 1.49
b=5 [ft] Q * n A * R2/3= xL=3 k * S 1/2 xR=3 R can be substituted for, and we get --- ft3 A * R2/3 = 13.42 A Q=100 R = s P S=0.01 n=0.020

Find: V [ft/s] A * R2/3= A * R2/3 = 13.42 A R = P A5/3 =13.42 P2/3 1 1
xL xR b 1.49 b=5 [ft] Q * n A * R2/3= xL=3 k * S 1/2 xR=3 A to the 5/3s, over P to the 2/3s, equals [pause] At this point it would be typical to develop an expression for area, in terms of the wetted perimeter, --- ft3 A * R2/3 = 13.42 A Q=100 R = s P A5/3 S=0.01 =13.42 P2/3 n=0.020

Find: V [ft/s] A * R2/3= A * R2/3 = 13.42 A R = P A5/3 =13.42 P2/3
xL xR b 1.49 b=5 [ft] Q * n A * R2/3= xL=3 k * S 1/2 xR=3 or an expression for the wetted perimeter, in terms of the area, so we have just 1 unknown variable to solve for. However, it would easier if we --- ft3 A * R2/3 = 13.42 A Q=100 R = s P A5/3 S=0.01 =13.42 P2/3 n=0.020 A=ƒ1(P) or P=ƒ2(A)

Find: V [ft/s] d A * R2/3= A * R2/3 = 13.42 A R = P A5/3 =13.42 no
xL xR b 1.49 b=5 [ft] Q * n A * R2/3= xL=3 k * S 1/2 xR=3 define a variable for the depth of water in the channel, d, and develop expressions for both --- ft3 A * R2/3 = 13.42 A Q=100 R = s P A5/3 S=0.01 =13.42 no P2/3 n=0.020 A=ƒ1(P) or P=ƒ2(A)

Find: V [ft/s] d A=ƒA(d) and P=ƒP(d) A * R2/3= A * R2/3 = 13.42 A R =
xL xR b 1.49 b=5 [ft] Q * n A * R2/3= xL=3 k * S 1/2 xR=3 the area and wetted perimeter, in terms of d, substitute those expressions --- ft3 A * R2/3 = 13.42 A Q=100 R = s P A5/3 S=0.01 =13.42 no P2/3 n=0.020 A=ƒ1(P) or P=ƒ2(A)

Find: V [ft/s] d A=ƒA(d) and P=ƒP(d) A * R2/3= A * R2/3 = 13.42
xL xR b 1.49 b=5 [ft] Q * n A * R2/3= xL=3 k * S 1/2 xR=3 back into the previous equation, solve for the depth directly, then solve for the area, --- ft3 A * R2/3 = 13.42 Q=100 s ƒA(d)5/3 A5/3 S=0.01 =13.42= P2/3 ƒP(d)2/3 n=0.020

Find: V [ft/s] d A=ƒA(d) and P=ƒP(d) A * R2/3= A * R2/3 = 13.42
xL xR b 1.49 b=5 [ft] Q * n A * R2/3= xL=3 k * S 1/2 xR=3 and then solve for the velocity. ft3 A * R2/3 = 13.42 Q=100 s ƒA(d)5/3 A5/3 S=0.01 =13.42= P2/3 ƒP(d)2/3 A=ƒ(d) n=0.020

Find: V [ft/s] d A=ƒA(d) and P=ƒP(d) A * R2/3= A * R2/3 = 13.42 V=ƒ(A)
xL xR b 1.49 b=5 [ft] Q * n A * R2/3= xL=3 k * S 1/2 xR=3 [pause] We’ll begin by first --- ft3 A * R2/3 = 13.42 Q=100 s V=ƒ(A) ƒA(d)5/3 A5/3 S=0.01 =13.42= P2/3 ƒP(d)2/3 A=ƒ(d) n=0.020

Find: V [ft/s] d A=ƒA(d) and P=ƒP(d) A * R2/3= A * R2/3 = 13.42 V=ƒ(A)
xL xR b 1.49 b=5 [ft] Q * n A * R2/3= xL=3 k * S 1/2 xR=3 developing an expression for the area, in terms of the depth. ft3 A * R2/3 = 13.42 Q=100 s V=ƒ(A) ƒA(d)5/3 A5/3 S=0.01 =13.42= P2/3 ƒP(d)2/3 A=ƒ(d) n=0.020

Find: V [ft/s] d A=ƒA(d) and P=ƒP(d) 1 1 xR xL b xL=3 xR=3 b=5 [ft]
The cross-sectional area can be divided into 3 parts, --- ft3 Q=100 s S=0.01 n=0.020

Find: V [ft/s] A1 A2 A3 d A=ƒA(d) and P=ƒP(d) 1 1 xR xL b xL=3 xR=3
b=5 [ft] A1, A2, and A3. [pause] From the diagram, it is clear --- ft3 Q=100 s S=0.01 n=0.020

Find: V [ft/s] A1 A2 A3 d A2 = b * d A=ƒA(d) and P=ƒP(d) 1 1 xR xL b
b=5 [ft] area 2 equals the base width, b, times the depth, d. [pause] Areas 1 and 3 are both right triangles, with top legs --- ft3 Q=100 s S=0.01 n=0.020

Find: V [ft/s] A1 A2 A3 d A2 = b * d A=ƒA(d) and P=ƒP(d) d*xL d*xR 1 1
b=5 [ft] equal to d times x sub L and d times x sub R, for areas 1 and 3, respectively. This makes area 1 equal to --- ft3 Q=100 s S=0.01 n=0.020

Find: V [ft/s] A1 A2 A3 d A2 = b * d A1 = xL * d2 A=ƒA(d) and P=ƒP(d)
d*xL d*xR A1 A2 A3 d 1 1 xR xL b xL=3 xR=3 A2 = b * d b=5 [ft] 1 A1 = 1/2 times x sub L times d squared, and area 3 equals, --- xL * d2 ft3 * 2 Q=100 s S=0.01 n=0.020

Find: V [ft/s] A1 A2 A3 d A2 = b * d A1 = xL * d2 A3 = xR * d2
A=ƒA(d) and P=ƒP(d) d*xL d*xR A1 A2 A3 d 1 1 xR xL b xL=3 xR=3 A2 = b * d b=5 [ft] 1 A1 = 1/2 times x sub R times d squared, which makes the total area equal to, --- xL * d2 ft3 * 2 Q=100 s 1 A3 = xR * d2 S=0.01 * 2 n=0.020

Find: V [ft/s] A1 A2 A3 d A2 = b * d A1 = xL * d2 A3 = xR * d2
A=ƒA(d) and P=ƒP(d) d*xL d*xR A1 A2 A3 d 1 1 xR xL b xL=3 xR=3 A2 = b * d b=5 [ft] 1 A1 = the base width times the depth, plus, 1/2 depth squared plus the quantity x sub L plus x sub R. [pause] Similarly, the wetted perimeters is --- xL * d2 ft3 * 2 Q=100 s 1 A3 = xR * d2 S=0.01 * 2 1 n=0.020 A = b*d + *d2 *(xL+xR) 2

Find: V [ft/s] d P1 P3 P2 A=ƒA(d) and P=ƒP(d) d*xL d*xR b xL=3 xR=3
b=5 [ft] the sum of 3 side lengths, P1, P2 and P3. ft3 Q=100 s S=0.01 n=0.020

Find: V [ft/s] d P1 P3 P2 P2 = b A=ƒA(d) and P=ƒP(d) d*xL d*xR b xL=3
b=5 [ft] From observation, P2 equals the base width, b, and P1 and --- ft3 Q=100 s S=0.01 n=0.020

Find: V [ft/s] d P1 P3 P2 P2 = b A=ƒA(d) and P=ƒP(d) d*xL d*xR b xL=3
b=5 [ft] P3 are the hypotenuses of right triangles, so P1 equals the square root of --- ft3 Q=100 s S=0.01 n=0.020

Find: V [ft/s] d P1 P3 P2 P2 = b P1 = d2 + (xL*d)2 A=ƒA(d) and P=ƒP(d)
d*xR d P1 P3 b P2 xL=3 xR=3 P2 = b b=5 [ft] P1 = d2 + (xL*d)2 d squared plus the quantity x sub L times d, squared, and P3 equals, --- ft3 Q=100 s S=0.01 n=0.020

Find: V [ft/s] d P1 P3 P2 P2 = b P1 = d2 + (xL*d)2 P3 = d2 + (xR*d)2
A=ƒA(d) and P=ƒP(d) d*xL d*xR d P1 P3 b P2 xL=3 xR=3 P2 = b b=5 [ft] P1 = d2 + (xL*d)2 the square root of d squared plus the quantity x sub R times d, squared, which makes the wetted perimeter equal to --- ft3 Q=100 s P3 = d2 + (xR*d)2 S=0.01 n=0.020

A=ƒA(d) and P=ƒP(d) d*xL d*xR d P1 P3 b P2 xL=3 xR=3 P2 = b b=5 [ft] P1 = d2 + (xL*d)2 the sum of P1, P2, and P3. With an expression for --- ft3 Q=100 s P3 = d2 + (xR*d)2 S=0.01 P =b+ d* xL2+1 + d* xR2+1 n=0.020

A=ƒA(d) and P=ƒP(d) d*xL d*xR d P1 P3 b P2 xL=3 xR=3 P2 = b b=5 [ft] P1 = d2 + (xL*d)2 the wetted permitted, in terms of the depth, d, --- ft3 Q=100 s P3 = d2 + (xR*d)2 S=0.01 P =b+ d* xL2+1 + d* xR2+1 n=0.020

Find: V [ft/s] A1 A2 A3 d P1 P3 P2 A = b*d + P =b+ d* xL2+1 + d* xR2+1
A=ƒA(d) and P=ƒP(d) d*xL d*xR A1 A2 A3 d P1 P3 b P2 xL=3 xR=3 b=5 [ft] and an expression for the area in terms of d, --- ft3 Q=100 s 1 A = b*d + *d2 *(xL+xR) 2 S=0.01 P =b+ d* xL2+1 + d* xR2+1 n=0.020

A=ƒA(d) and P=ƒP(d) d*xL d*xR A1 A2 A3 d P1 P3 b P2 xL=3 xR=3 A5/3 =13.42 b=5 [ft] P2/3 these values are plugged into --- ft3 Q=100 s 1 A = b*d + *d2 *(xL+xR) 2 S=0.01 P =b+ d* xL2+1 + d* xR2+1 n=0.020

A=ƒA(d) and P=ƒP(d) d*xL d*xR A1 A2 A3 d P1 P3 b P2 xL=3 xR=3 A5/3 =13.42 b=5 [ft] P2/3 our partially solved Manning’s equation, for A and P. ft3 Q=100 s 1 A = b*d + *d2 *(xL+xR) 2 S=0.01 P =b+ d* xL2+1 + d* xR2+1 n=0.020

Find: V [ft/s] A1 A2 A3 d P1 P3 P2 b*d + b+ d* xL2+1 + d* xR2+1
A=ƒA(d) and P=ƒP(d) d*xL d*xR A1 A2 A3 d P1 P3 b P2 A5/3 b= 5 [ft] =13.42 P2/3 xL=3 [pause] Plugging in the known values for --- xR=3 1 5/3 b*d + *d2 *(xL+xR) 2 =13.42 b+ d* xL2+1 + d* xR2+1 2/3

Find: V [ft/s] A1 A2 A3 d P1 P3 P2 b*d + b+ d* xL2+1 + d* xR2+1
A=ƒA(d) and P=ƒP(d) d*xL d*xR A1 A2 A3 d P1 P3 b P2 A5/3 b= 5 [ft] =13.42 P2/3 xL=3 base width, x sub L, and x sub R, and --- xR=3 1 5/3 b*d + *d2 *(xL+xR) 2 =13.42 b+ d* xL2+1 + d* xR2+1 2/3

Find: V [ft/s] d b*d + b+ d* xL2+1 + d* xR2+1 5*d+3*d2 -13.42 = 0
Q=100 d 1 1 n=0.020 xL xR S=0.01 b 5/3 5*d+3*d2 b= 5 [ft] = 0 2/3 5+2*d* 10 xL=3 subtracting from both sides, we now have one equation and --- xR=3 1 5/3 b*d + *d2 *(xL+xR) 2 =13.42 b+ d* xL2+1 + d* xR2+1 2/3

Find: V [ft/s] d 5*d+3*d2 -13.42 = 0 5+2*d* 10 ft3 s Q=100 1 1 n=0.020
xR=3 xL xR xL=3 S=0.01 b b= 5 [ft] 5/3 5*d+3*d2 = 0 2/3 5+2*d* 10 1 unknown variable, the depth d. At this point we could start guessing values of d and ---

Find: V [ft/s] d 5*d+3*d2 -13.42 = 0 d LHS 5+2*d* 10 d1 LHS1 ft3 s
Q=100 d 1 1 n=0.020 xR=3 xL xR xL=3 S=0.01 b b= 5 [ft] 5/3 5*d+3*d2 = 0 d LHS 2/3 5+2*d* 10 solve for the left hand side of the equation, --- d1 LHS1

Find: V [ft/s] d 5*d+3*d2 -13.42 = 0 d LHS 5+2*d* 10 d1 LHS1 d2 LHS2
Q=100 d 1 1 n=0.020 xR=3 xL xR xL=3 S=0.01 b b= 5 [ft] 5/3 5*d+3*d2 = 0 d LHS 2/3 5+2*d* 10 until it equals 0. d1 LHS1 d2 LHS2

Find: V [ft/s] d 5*d+3*d2 -13.42 = 0 d LHS 5+2*d* 10 d1 LHS1 d2 LHS2
Q=100 d 1 1 n=0.020 xR=3 xL xR xL=3 S=0.01 b b= 5 [ft] 5/3 5*d+3*d2 = 0 d LHS 2/3 5+2*d* 10 Unfortunately, this relatively inefficient search algorithm took me 11 iterations, find the depth to 4 significant figures. [pause] So instead, --- d1 LHS1 d2 LHS2 d3

Find: V [ft/s] f(dn) d f(dn) f’(dn) dn+1 = dn- f’(dn) d1 f(d1) f’(d1)
5/3 5*d+3*d2 = 0 d LHS 2/3 5+2*d* 10 we’ll employ Newton’s method for solving roots by making an initial guess, d sub 1, --- d1 LHS1 d2 LHS2 d3

5/3 5*d+3*d2 = 0 d LHS 2/3 5+2*d* 10 solving the function at d1, --- d1 LHS1 d2 LHS2 d3

5/3 5*d+3*d2 = 0 d LHS 2/3 5+2*d* 10 solving the derivative of the function at d1, --- d1 LHS1 d2 LHS2 d3

5/3 5*d+3*d2 = 0 d LHS 2/3 5+2*d* 10 and then using the equation to approximate the second depth value. Graphically, --- d1 LHS1 d2 LHS2 d3

5/3 5*d+3*d2 f(d)= -13.42 2/3 5+2*d* 10 if the function for f of d, is continuously differentiable, which it is, ---

5/3 5*d+3*d2 f(d)= -13.42 2/3 5+2*d* 10 then successive approximations of d --- dn+1 dn

5/3 5*d+3*d2 f(d)= -13.42 2/3 5+2*d* 10 intelligently consider the previous approximation of d, --- dn+1 dn

5/3 5*d+3*d2 f(d)= -13.42 2/3 5+2*d* 10 the value of the function at d, --- dn+1 dn

5/3 5*d+3*d2 f(d)= -13.42 2/3 5+2*d* 10 and the derivative of the function evaluated at d. [pause] dn+1 dn

Find: V [ft/s] = f(dn) d f(dn) f’(dn) dn+1 = dn- f’(dn) d1 f(d1)
δ a a’*b-a*b’ d2 = δd b b2 5/3 5*d+3*d2 f(d)= -13.42 2/3 5+2*d* 10 Applying the quotient rule from calculus, we can solve for the --- dn+1 dn

Find: V [ft/s] = = - f(dn) d f(dn) f’(dn) dn+1 = dn- f’(dn) d1 δ a
a’*b-a*b’ = δd b b2 5/3 5*d+3*d2 f(d)= -13.42 2/3 5+2*d* 10 derivative of our function, with respect to d. Now to begin the algorithm, --- δf(d) 5 (5*d+3*d2)2/3 * (5+6*d) = * δd 3 (5+2*d* 10)2/3 4 (5*d+3*d2)2/3 * 10*d - * 3 (5+2*d* 10)2/3 *d

Find: V [ft/s] = - f(dn) d f(dn) f’(dn) dn+1=dn- f’(dn) d1 5*d+3*d2
5/3 5*d+3*d2 f(d)= -13.42 2/3 5+2*d* 10 we’ll choose an initial depth, d1, --- δf(d) 5 (5*d+3*d2)2/3 * (5+6*d) = * δd 3 (5+2*d* 10)2/3 4 (5*d+3*d2)2/3 * 10*d - * 3 (5+2*d* 10)2/3 *d

Find: V [ft/s] = - f(dn) d f(dn) f’(dn) dn+1=dn- f’(dn) 3.000 5*d+3*d2
5/3 5*d+3*d2 f(d)= -13.42 2/3 5+2*d* 10 3 feet. Then we’ll solve for the value of the function and the --- δf(d) 5 (5*d+3*d2)2/3 * (5+6*d) = * δd 3 (5+2*d* 10)2/3 4 (5*d+3*d2)2/3 * 10*d - * 3 (5+2*d* 10)2/3 *d

Find: V [ft/s] = - f(dn) d f(dn) f’(dn) dn+1=dn- f’(dn) 3.000 5*d+3*d2
5/3 5*d+3*d2 f(d)= -13.42 2/3 5+2*d* 10 value of the derivative of the function, for d equal to 3, --- δf(d) 5 (5*d+3*d2)2/3 * (5+6*d) = * δd 3 (5+2*d* 10)2/3 4 (5*d+3*d2)2/3 * 10*d - * 3 (5+2*d* 10)2/3 *d

Find: V [ft/s] = - f(dn) d f(dn) f’(dn) dn+1=dn- f’(dn) 3.000 47.62
44.97 5/3 5*d+3*d2 f(d)= -13.42 =47.62 2/3 5+2*d* 10 which equals 47.62, and 44.97, respectively. [pause] Our second approximation for depth --- δf(d) 5 (5*d+3*d2)2/3 * (5+6*d) = * δd 3 (5+2*d* 10)2/3 4 (5*d+3*d2)2/3 * 10*d - =44.97 * 3 (5+2*d* 10)2/3 *d

44.97 5/3 5*d+3*d2 f(d)= -13.42 =47.62 2/3 5+2*d* 10 of water in the channel is the previous approximation, 3, --- δf(d) 5 (5*d+3*d2)2/3 * (5+6*d) = * δd 3 (5+2*d* 10)2/3 4 (5*d+3*d2)2/3 * 10*d - =44.97 * 3 (5+2*d* 10)2/3 *d

44.97 5/3 5*d+3*d2 f(d)= -13.42 =47.62 2/3 5+2*d* 10 minus divided by 44.97, which equals, --- δf(d) 5 (5*d+3*d2)2/3 * (5+6*d) = * δd 3 (5+2*d* 10)2/3 4 (5*d+3*d2)2/3 * 10*d - =44.97 * 3 (5+2*d* 10)2/3 *d

44.97 d2=1.941 [ft] 5/3 5*d+3*d2 f(d)= -13.42 =47.62 2/3 5+2*d* 10 1.941 feet. [pause] δf(d) 5 (5*d+3*d2)2/3 * (5+6*d) = * δd 3 (5+2*d* 10)2/3 4 (5*d+3*d2)2/3 * 10*d - =44.97 * 3 (5+2*d* 10)2/3 *d

44.97 1.941 47.62 44.97 d2=1.941 [ft] 5/3 5*d+3*d2 f(d)= -13.42 2/3 5+2*d* 10 Adding this value to the table, and solving for --- δf(d) 5 (5*d+3*d2)2/3 * (5+6*d) = * δd 3 (5+2*d* 10)2/3 4 (5*d+3*d2)2/3 * 10*d - * 3 (5+2*d* 10)2/3 *d

44.97 1.941 47.62 44.97 d2=1.941 [ft] 5/3 5*d+3*d2 f(d)= -13.42 2/3 5+2*d* 10 the function and the derivative again, using this second depth, we get, --- δf(d) 5 (5*d+3*d2)2/3 * (5+6*d) = * δd 3 (5+2*d* 10)2/3 4 (5*d+3*d2)2/3 * 10*d (5*d+3*d2)2/3 * 10*d - * 3 (5+2*d* 10)2/3 *d (5+2*d* 10)2/3 *d

44.97 1.941 10.51 25.77 5/3 5*d+3*d2 f(d)= -13.42 =10.51 2/3 5+2*d* 10 10.51 and Plugging these values in, --- δf(d) 5 (5*d+3*d2)2/3 * (5+6*d) = * δd 3 (5+2*d* 10)2/3 4 (5*d+3*d2)2/3 * 10*d (5*d+3*d2)2/3 * 10*d - =25.77 * 3 (5+2*d* 10)2/3 *d (5+2*d* 10)2/3 *d

44.97 1.941 10.51 25.77 5/3 5*d+3*d2 f(d)= -13.42 =10.51 2/3 5+2*d* 10 our third approximation for the depth is --- δf(d) 5 (5*d+3*d2)2/3 * (5+6*d) = * δd 3 (5+2*d* 10)2/3 4 (5*d+3*d2)2/3 * 10*d (5*d+3*d2)2/3 * 10*d - =25.77 * 3 (5+2*d* 10)2/3 *d (5+2*d* 10)2/3 *d

44.97 1.941 10.51 25.77 d3=1.533 [ft] 5/3 5*d+3*d2 f(d)= -13.42 =10.51 2/3 5+2*d* 10 1.533 feet. [pause] δf(d) 5 (5*d+3*d2)2/3 * (5+6*d) = * δd 3 (5+2*d* 10)2/3 4 (5*d+3*d2)2/3 * 10*d (5*d+3*d2)2/3 * 10*d - =25.77 * 3 (5+2*d* 10)2/3 *d (5+2*d* 10)2/3 *d

44.97 1.941 10.51 25.77 d3=1.533 [ft] 1.533 10.51 25.77 5/3 5*d+3*d2 f(d)= -13.42 =10.51 2/3 5+2*d* 10 Continuing with this algorithm, we reduce the --- δf(d) 5 (5*d+3*d2)2/3 * (5+6*d) = * δd 3 (5+2*d* 10)2/3 4 (5*d+3*d2)2/3 * 10*d (5*d+3*d2)2/3 * 10*d - =25.77 * 3 (5+2*d* 10)2/3 *d (5+2*d* 10)2/3 *d

44.97 1.941 10.51 25.77 1.533 1.308 19.46 5*d+3*d 1.466 1.308 19.46 f(d)= 2/3 5+2*d* 10 number of iterations needed --- δf(d) 5 (5*d+3*d2)2/3 * (5+6*d) = * δd 3 (5+2*d* 10)2/3 4 (5*d+3*d2)2/3 * 10*d (5*d+3*d2)2/3 * 10*d - =25.77 * 3 (5+2*d* 10)2/3 *d (5+2*d* 10)2/3 *d

44.97 1.941 10.51 25.77 1.533 1.308 19.46 5*d+3*d 1.466 0.037 18.48 f(d)= 5+2*d* 10 1.464 0.037 18.48 to find the correct depth. δf(d) 5 (5*d+3*d2)2/3 * (5+6*d) = * δd 3 (5+2*d* 10)2/3 4 (5*d+3*d2)2/3 * 10*d (5*d+3*d2)2/3 * 10*d - =25.77 * 3 (5+2*d* 10)2/3 *d (5+2*d* 10)2/3 *d

44.97 1.941 10.51 25.77 1.533 1.308 19.46 5*d+3*d 1.466 0.037 18.48 f(d)= 5+2*d* 10 1.464 0.001 18.46 [pause] By knowing the depth of water --- δf(d) 5 (5*d+3*d2)2/3 * (5+6*d) = * δd 3 (5+2*d* 10)2/3 4 (5*d+3*d2)2/3 * 10*d (5*d+3*d2)2/3 * 10*d - =25.77 * 3 (5+2*d* 10)2/3 *d (5+2*d* 10)2/3 *d

Find: V [ft/s] f(dn) d f(dn) f’(dn) dn+1=dn- f’(dn) 3.000 47.62 44.97
d=1.464 [ft] 1.941 10.51 25.77 1.533 1.308 19.46 1.466 0.037 18.48 1.464 0.001 18.46 in the channel is feet, ---

Find: V [ft/s] A = b*d+ f(dn) d f(dn) f’(dn) dn+1=dn- f’(dn) 1.464
0.001 18.46 d=1.464 [ft] b=5 [ft] 1 xR=3 A = b*d+ *d2 *(xL+xR) 2 xL=3 ft3 s Q=100 we can plug this value into our equation for area,

0.001 18.46 d=1.464 [ft] b=5 [ft] 1 xR=3 A = b*d+ *d2 *(xL+xR) 2 xL=3 ft3 s Q=100 in addition to the other variables, and the area equals, ---

0.001 18.46 d=1.464 [ft] b=5 [ft] 1 xR=3 A = b*d+ *d2 *(xL+xR) 2 xL=3 A=13.75 [ft2] ft3 s Q=100 13.75 feet squared. [pause] Now we can solve for the ---

0.001 18.46 d=1.464 [ft] b=5 [ft] 1 xR=3 A = b*d+ *d2 *(xL+xR) 2 xL=3 A=13.75 [ft2] ft3 s Q=100 velocity, which equals flowrate --- Q V= A

0.001 18.46 d=1.464 [ft] b=5 [ft] 1 xR=3 A = b*d+ *d2 *(xL+xR) 2 xL=3 A=13.75 [ft2] ft3 Q=100 divided by the area, and equals, --- s Q V= A

0.001 18.46 d=1.464 [ft] b=5 [ft] 1 xR=3 A = b*d+ *d2 *(xL+xR) 2 xL=3 A=13.75 [ft2] ft3 Q=100 7.27 feet per second. s Q V= A ft V=7.27 s

0.001 18.46 d=1.464 [ft] b=5 [ft] 1 xR=3 A = b*d+ *d2 *(xL+xR) 4.6 5.5 6.4 7.3 2 xL=3 A=13.75 [ft2] ft3 Q=100 Reviewing the possible solutions, --- s Q V= A ft V=7.27 s

Find: V [ft/s] A = b*d+ AnswerD f(dn) d f(dn) f’(dn) dn+1=dn- f’(dn)
1.464 0.001 18.46 d=1.464 [ft] b=5 [ft] 1 xR=3 A = b*d+ *d2 *(xL+xR) 4.6 5.5 6.4 7.3 2 xL=3 A=13.75 [ft2] ft3 Q=100 the answer is D. s Q V= A ft V=7.27 AnswerD s

? 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: V [ft/s] xL xR b b=5 [ft] xL=3 xR=3 ft3 s

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Find: V [ft/s] xL xR b b=5 [ft] xL=3 xR=3 ft3 s

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Presentation on theme: "Find: V [ft/s] xL xR b b=5 [ft] xL=3 xR=3 ft3 s"— Presentation transcript:

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