ASSIGNMENT 2 (Due September 13, 2006 )

[Originally Due September 6, 2006]

Reading: Chapter 2

Questions: (Due September 6, 2006 )

Problems: Chapter 2

Prob 2, 4, 5, 6

Problem 2

2. Calculate the critical depth for a discharge of 10 m3/s (350 ft3/s) in the following channels:

a. Rectangular channel; b = 3.0 m (9.8 ft)

A=by=3.0y ; T=b=3.0 ;

D=A/T=3.0y/3.0=1.0y ; u=Q/A=10/{3.0y} ;

${u^2}/{g} = D ={100}/{88.2y^2} = 1.0y$ $right 88.2y^3 = 100 right y=root{3}{100/88.2}=$ 1.0427

y_c=1.04 m

b. Triangular channel; z = 0.5 m (1.6 ft)

A=zy^2=0.5y^2 ; T=2zy=y ;

$D=A/T={0.5y^2} / {y}=0.5y$ ; $u=Q/A=10/{0.5y^2}=20/{y^2}$ ;

${u^2}/{g} = D =400/{9.8y^4}=0.5y$ right y^5 = 800/9.8 right y= 2.411975

y_c=2.41 m

c. Trapezoidal channel; b = 3.0 m (9.8 ft), z = 1.5

A=(b+zy)y=(3.0+1.5y)y ; T=b+2zy=3.0+3.0y ;

D=A/T={(3.0+1.5y)y} / {3.0+3.0y} ; u=Q/A=10/{(3.0+1.5y)y} ;

${u^2}/{g} = D = {10^2}/{9.8{(3.0+1.5y)^2}y^2} = {(3.0+1.5y)y} / {3.0+3.0y}$

$right {300/9.8}+{300/9.8}y = (3.0+1.5y)^3y^3 =$ (3.0+1.5y)(3.0+1.5y)(3.0+1.5y)y^3= y^3(3.0+1.5y)(9+9y+2.25y^2) = y^3(27+13.5y+27y+13.5y^2+6.75y^2+3.375y^3) right 3.375y^6 + 20.25y^5 + 40.5y^4 +27y^3 +0y^2 -30.612y -30.612 = 0

From Polynomial Web Solver:

For the real Polynomial:
+3.375x^6+20.25x^5+40.5x^4+27x^3-30.612x-30.612
The Solutions are:
X1=(-2.572344199612621+i0.7414903876526688)
X2=(-2.572344199612621-i0.7414903876526688)
X3=0.8919105836287637
X4=(-0.3203415425738132-i1.086154696110772)
X5=(-0.3203415425738132+i1.086154696110772)
X6=-1.106539099255896
Time used: 203 msec.

Only one of these makes sense.

y_c=0.89 m

d. Circular channel; D = 3.0 m (9.8 ft).

$A={d^2/8}(theta - sin theta)$ ¹⁾; $T=2 sqrt{y(d_0 - y)}$ ;

$D=A/T={{d^2/8}(theta - sin theta)}/{2 sqrt{y(d_0 - y)}} =3$ ; $u=Q/A=80/{d^2(theta - sin theta)}$ ;

${u^2}/{g} = D =3.0={{theta - sin theta}/{8 sin(0.5 theta)}} = 6400/{9.8 d^4 (theta - sin theta)^2}$

This is no fun. Yea for Straub! (Assume α=1)

$psi = alpha {{Q^2}/g} = 100/9.8$

$y_c approx (1.01/{{d_0}^0.26})psi^0.25 = (1.01/{3.0^0.26})(100/9.8)^0.25$ approx 0.759*1.7873 approx 1.3566

$0.02 le {y_c/d_0} le 0.85$ right 0.02 le {1.3566/3} le 0.85 right 0.02 le 0.4522 le 0.85 ✔

y_c≈1.4 m

Problem 4

4. Water is flowing at a velocity of 4.6 m/s (15 ft/s) and at a depth of 0.75 m (2.5 ft) in a channel of rectangular section. Find the change in the absolute depth of flow produced by

u_1=4.6 ; d_1=0.75 ; z_1=0 ; b_1=b_2=b ; g=9.8 ;

a. A smooth upward step of 0.15 m (0.49 ft)

z_2=0.15

$u_1=Q/A_1=Q/{b_1 d_1} right Q=u_1 b_1 d_1;$ $u_2 = Q/A_2 = Q/{b_2 d_2} = {u_1 b_1 d_1}/{b_2 d_2}$

$H = z_1 + d_1 + {{u_1}^2}/{2g} = z_2 + d_2 + {{u_2}^2}/{2g}=$ $z_2 + d_2 + {({u_1 b_1 d_1}/{b_2 d_2})^2} {1/{2g}}$ $right 2gz_1 + 2gd_1 + {{u_1}^2} = 2gz_2 + 2gd_2 + {({u_1 b_1 d_1}/{b_2 d_2})^2}$ $right 2g{z_1} {b_2}^2 {d_2}^2 + 2gd_1 {b_2}^2 {d_2}^2 + {u_1}^2 {b_2}^2 {d_2}^2 - 2g{z_2} {b_2}^2 {d_2}^2 - 2g {b_2}^2 {d_2}^3 - {u_1}^2 {b_1}^2 {d_1}^2 =0$ = $0 + 14.7 b^2 {d_2}^2 + 21.16 b^2 {d_2}^2 - 2.94 b^2 {d_2}^2 - 19.6 b^2 {d_2}^3 - 11.9025 b^2$ = $14.7{d_2}^2 + 21.16{d_2}^2 - 2.94{d_2}^2 - 19.6{d_2}^3 - 11.9025$ = $-19.6{d_2}^3 +32.92{d_2}^2 -11.9025=0$

For the real Polynomial:
-19.6x^3+32.92x^2-11.9025
The Solutions are:
X1=1.3428032719405436
X2=-0.5248570581005619
X3=0.861645622894712
Time used: 15 msec.

The depth starts at 0.75 m, so since 0.86 and 1.34 are on opposite sides of the energy curve, 0.75 must be on the same side as 0.86.

d₂=0.86 m

b. A smooth downward step of 0.15 m (0.49 ft)

z_2=-0.15

$2g{z_1} {b_2}^2 {d_2}^2 + 2gd_1 {b_2}^2 {d_2}^2 + {u_1}^2 {b_2}^2 {d_2}^2 - 2g{z_2} {b_2}^2 {d_2}^2 - 2g {b_2}^2 {d_2}^3 - {u_1}^2 {b_1}^2 {d_1}^2 =0$ = $0 + 14.7 b^2 {d_2}^2 + 21.16 b^2 {d_2}^2 + 2.94 b^2 {d_2}^2 - 19.6 b^2 {d_2}^3 - 11.9025 b^2$ = $14.7{d_2}^2 + 21.16{d_2}^2 + 2.94{d_2}^2 - 19.6{d_2}^3 - 11.9025$ = $-19.6{d_2}^3 +38.8{d_2}^2 -11.9025=0$

For the real Polynomial:
-19.6x^3+38.8x^2-11.9025
The Solutions are:
X1=1.7900795032749868
X2=-0.4953462607159007
X3=0.6848585941756077
Time used: 15 msec.

From the previous problem, the flow starts at the lower part of the energy curve.

d₂=0.68 m

c. The maximum allowable size of an upward step for the flow to be possible upstream as described.

Solve for z_2 such that z_2+d_2 = d_1, d_2=d_1-z_2=0.75-z_2 right z_2=0.75-d_2

$right 2g{z_1} {b_2}^2 {d_2}^2 + 2gd_1 {b_2}^2 {d_2}^2 + {u_1}^2 {b_2}^2 {d_2}^2 - 2g{z_2} {b_2}^2 {d_2}^2 - 2g {b_2}^2 {d_2}^3 - {u_1}^2 {b_1}^2 {d_1}^2 =0$ = $0 + 14.7 b^2 {d_2}^2 + 21.16 b^2 {d_2}^2 - 19.6{z_2} b^2 {d_2}^2 - 19.6 b^2 {d_2}^3 - 11.9025 b^2$ = $14.7 {d_2}^2 + 21.16 {d_2}^2 - 19.6{z_2} {d_2}^2 - 19.6 {d_2}^3 - 11.9025$ = $35.86 {d_2}^2 - 19.6(0.75-d_2) {d_2}^2 - 19.6 {d_2}^3 - 11.9025$ = $35.86 {d_2}^2 - 14.7 {d_2}^2 +19.6 {d_2}^3 - 19.6 {d_2}^3 - 11.9025$ = $21.16 {d_2}^2 - 11.9025=0$

$right {d_2}^2 =0.5625$ $right {d_2} =0.75$ $right {z_2} =0$

That didn't work well. Suppose the maximum step is the minimum area which must be the most efficient and so flowing at critical depth. This happens where F²=1.

$F^2 = 1 = {u^2}/{gl}$ = ${u^2}/{gL}$ = ${u^2}/{gD}$ = ${u^2 T}/{gA}$ = ${{u_2}^2 b_2}/{g b_2 d_2}$ = ${{u_2}^2}/{g d_2} = 1$ → ${u_2}^2=g d_2$

$H = z_1 + d_1 + {{u_1}^2}/{2g}$ = $z_2 + d_2 + {{u_2}^2}/{2g}$ = $z_2 + d_2 + {g d_2}/{2g}$ = $z_2 + {3/2}d_2$ = 0 + 0.75 + 1.07959 → $z_2=1.82959-{3/2}d_2$

$2g{z_1} {b_2}^2 {d_2}^2 + 2gd_1 {b_2}^2 {d_2}^2 + {u_1}^2 {b_2}^2 {d_2}^2 - 2g{z_2} {b_2}^2 {d_2}^2 - 2g {b_2}^2 {d_2}^3 - {u_1}^2 {b_1}^2 {d_1}^2 =0$ = $0 + 14.7 b^2 {d_2}^2 + 21.16 b^2 {d_2}^2 - 19.6{z_2} b^2 {d_2}^2 - 19.6 b^2 {d_2}^3 - 11.9025 b^2$ = $14.7 {d_2}^2 + 21.16 {d_2}^2 - 19.6{z_2} {d_2}^2 - 19.6 {d_2}^3 - 11.9025$ = $35.86 {d_2}^2 - 19.6(1.82959-{3/2}d_2) {d_2}^2 - 19.6 {d_2}^3 - 11.9025$ = $35.86 {d_2}^2 - 35.86 {d_2}^2 +29.4 {d_2}^3 - 19.6 {d_2}^3 - 11.9025$ = 9.8(d_2}^3 - 11.9025=0

For the real Polynomial:
+9.8x^3-11.9025
The Solutions are:
X1=(-0.5334667745570523+i0.9239915576827066)
X2=(-0.5334667745570523-i0.9239915576827066)
X3=1.0669335491141045
Time used: 16 msec.

$d_2=1.0669 right z_2=1.82959-{3/2}1.0669=0.22924$

Maximum step up: 0.23 m

Problem 5

5. Water is flowing at a velocity of 11 ft/s (3.4 m/s) and at a depth of 11 ft (3.4 m) in a channel of rectangular cross section with a width of 11 ft (3.4 m). Find the changes in depth and absolute water level produced by

g=32.2 ; z_1=z_2=0 ; d_1=11 ; b_1=11 ; u_1=11 ;

$F^2 = {u^2}/{gl}$ = ${u^2}/{gL}$ = ${u^2}/{gD}$ = ${u^2 T}/{gA}$ = ${u^2 b_1}/{g b_1 d_1}$ = ${u^2}/{g d_1}$ = ${11^2}/{354.2}$ =0.3416 < 1 → Subcritical Flow

a. A smooth contraction in width of 10 ft (3.0 m)

Wow! b_2=11-10=1 ;

$2g{z_1} {b_2}^2 {d_2}^2 + 2gd_1 {b_2}^2 {d_2}^2 + {u_1}^2 {b_2}^2 {d_2}^2 - 2g{z_2} {b_2}^2 {d_2}^2 - 2g {b_2}^2 {d_2}^3 - {u_1}^2 {b_1}^2 {d_1}^2 =0$ $0 + 708.4{d_2}^2 + 121{d_2}^2 - 0 - 64.4{d_2}^3 - 1771561 =0$ = $-64.4{d_2}^3 +829.4{d_2}^2 -1771561=0$

For the real Polynomial:
-64.4x^3+829.4x^2-1771561
The Solutions are:
X1=-26.447887345249374
X2=(19.663384666413506+i25.56288849930444)
X3=(19.663384666413506-i25.56288849930444)
Time used: 32 msec.

Flow not possible for 10 ft contraction

b. A smooth expansion in width of 14 ft (4.3 m)

b_2=11+14=25

$2g{z_1} {b_2}^2 {d_2}^2 + 2gd_1 {b_2}^2 {d_2}^2 + {u_1}^2 {b_2}^2 {d_2}^2 - 2g{z_2} {b_2}^2 {d_2}^2 - 2g {b_2}^2 {d_2}^3 - {u_1}^2 {b_1}^2 {d_1}^2 =0$ = $0+442750{d_2}^2 + 75625{d_2}^2 -0- 40250{d_2}^3 - 1771561$ = $-40250{d_2}^3 +518375{d_2}^2 -1771561=0$

For the real Polynomial:
-40250x^3+518375x^2-1771561
The Solutions are:
X1=12.601722064240404
X2=-1.7354255684774667
X3=2.0125854918147033
Time used: 16 msec.

Since the depth will not be increasing, it falls past critical depth.

z₂=d₂=2.0 ft

c. The greatest allowable contraction for the flow to be possible upstream as described.

This will happen when d₁=d₂=z₁=z₂=11 ft, yes? No? How about where F²=1?

F^2 = 1 → ${u_2}^2=g d_2$

$H = z_1 + d_1 + {{u_1}^2}/{2g}$ = $z_2 + d_2 + {{u_2}^2}/{2g}$ = $z_2 + d_2 + {g d_2}/{2g}$ = $0 + {3/2}d_2$ = 0 + 11 + 1.87888 → d_2=8.5859

$2g{z_1} {b_2}^2 {d_2}^2 + 2gd_1 {b_2}^2 {d_2}^2 + {u_1}^2 {b_2}^2 {d_2}^2 - 2g{z_2} {b_2}^2 {d_2}^2 - 2g {b_2}^2 {d_2}^3 - {u_1}^2 {b_1}^2 {d_1}^2 =0$ = $0 + 52222{b_2}^2 + 8920 {b_2}^2 - 0 - 40761{b_2}^2 - 1771561 =0$ → $20381{b_2}^2=1771561$ → ${b_2}^2=86.922$ → ${b_2}=9.323$

b₂=9.3 ft, a 1.7 ft contraction

Problem 6

6. Water is flowing at a velocity of 10 ft/s (3.0 m/s) and a depth of 10 ft (3.0 m) in a channel of rectangular section 10 (f.00 m) wide. There is a smooth upward step of 4 ft (1.2 m) in the channel bed. What expansion in width must take place for the upstream flow to be possible as specified?

g=32.2 z_1=0 ; d_1=10 ; b_1=10 ; u_1=10 ; z_2=4 ;

${F_1}^2 = {u^2}/{g d_1}$ = {100}/{322} =0.3106 < 1 → Subcritical Flow

${F_2}^2 = 1= {{u_2}^2}/{g d_2}$ → ${u_2}^2=g d_2$

$H = z_1 + d_1 + {{u_1}^2}/{2g} = z_2 + d_2 + {g d_2/{2g}$ = $0 + 10 + {10^2}/{64.4} = 4 + {3/2}d_2$ → ${3/2}d_2=7.5528$ → d_2=5.035

$2g{z_1} {b_2}^2 {d_2}^2 + 2gd_1 {b_2}^2 {d_2}^2 + {u_1}^2 {b_2}^2 {d_2}^2 - 2g{z_2} {b_2}^2 {d_2}^2 - 2g {b_2}^2 {d_2}^3 - {u_1}^2 {b_1}^2 {d_1}^2 =0$ = $0 + 16326{b_2}^2 + 2535{b_2}^2 - 6530{b_2}^2 -8220{b_2}^2 - 1000000 =0$ = $4111{b_2}^2 - 1000000 =0$ → ${b_2}^2 = 243.25$ → b_2 = 15.5965