ASSIGNMENT 4 (Due October 4, 2007 )

— David Wagner 2007/10/02 17:23

Reading: Chapter 3; Questions: (Due October 4, 2007 ); Problems: Chapter 3; Prob 4, 10, 14, 15

Problem 4

A rectangular channel 8 ft (2.4 m) wide carries 100 ft³/s (2.8 m³/s) at a depth of 0.5 ft (0.15 m). This channel is connected by a 50 ft (15 m) long straight wall transition to a rectangular channel 10 ft (3.0 m) wide. Determine whether a hydraulic jump occurs in this transition, if the headloss within the transition is h_f ft. (Assume that h_f is uniformly distributed through the transition.) If a hydraulic jump occurs, what is the approximate location of the jump relative to the beginning of the transition section?

a. Downstream depth 4 ft (1.2 m) and h_f = 0 ft.
b. Downstream depth 4 ft (1.2 m) and h_f = 1 ft (0.30 m).
c. Compare the answers to Parts (a) and (b) and discuss any differences.
d. Downstream depth 3.6 ft (1.1 m) and h_f = 1 ft (0.30 m).

g=32.2 ft/s²; γ=62.24 lb/ft³@77ºF; Q=100 ft³/s, y₁ = 0.5 ft, b₁=8 ft, b₂=10 ft.

A_1=8*0.5= 4 ft²; u_1=Q/A=100/4= 25 ft/s; $F_1={u_1}/{sqrt{gy_1}} = 25/{sqrt{32.2*0.5}}=6.23$ , 4.5 < 6.23 < 9 ⇒ Supercritical flow, Real Jump possible; $E_1=y_1+{{{u_1}^2}/{2g}}=0.5+{25^2}/{2*32.2}=0.5+9.7=$ 10.2 ft.

Part (a)

y₂ = 4 ft

A_2=10*4= 40 ft²; u_2=Q/A=100/40= 2.5 ft/s; $F_1={u_2}/{sqrt{gy_2}} = 2.5/{sqrt{32.2*4}}=$ 0.22 < 1 ⇒ Subcritical; $E_2=y_2+{{{u_2}^2}/{2g}}=4+{2.5^2}/{2*32.2}=4+0.1=$ 4.1 ft.

∴Jump occurs.

Gamma_1=1; ${G_1}^2={Gamma_1}^2{F_1}^2={F_1}^2$

$E_L=E_1=y_L+{{Q^2}/{2gb^2 y_L}}=y_L+{10000/{2*32.2*b^2 {y_L}^2}}=10.2$ ;

$M_L=M_H={Q^2}/{g y_L b} + {b{y_L}^2}/2 = {Q^2}/{g y_H b} + {b {y_H}^2}/2=$ ${10000}/{32.2 y_L b} + {b {y_L}^2}/2 = {10000}/{32.2 y_H b} + {b {y_H}^2}/2=$ $310.56/{b y_L} + {b {y_L}^2}/2 = 310.56/{b y_H} + {b{y_H}^2}/2 right$ $621.12/{y_L} + {b^2 {y_L}^2} = 621.12/{y_H} + {b^2 {y_H}^2}$ .

Terrific. What are the formulæ for y_L and y_H needed to fill out the table below?

$E_L=E_1=y_L+{{Q^2}/{2gb^2 y_L}}=y_L+{10000/{2*32.2*b^2 {y_L}^2}}=10.2 right$ ${y_L}^3+{155.28/{b^2}}=10.2{y_L}^2 right$ ${y_L}^3-10.2{y_L}^2+{155.28/{b^2}}=0$ ;

$E_H=E_2=y_H+{{Q^2}/{2gb^2 y_H}}=y_H+{10000/{2*32.2*b^2 {y_H}^2}}=4.1 right$ ${y_H}^3+{155.28/{b^2}}=4.1{y_H}^2 right$ ${y_H}^3-4.1{y_H}^2+{155.28/{b^2}}=0$ ;

Oh, jeepers. These are cubics. Is it necessary to solve two cubics about five times each by hand? Yes.¹⁾

Distance (ft)	Width (ft)	y_L (ft)	y_H (ft)
0	8.0	0.50
10	8.4	0.48
20	8.8	0.45
30	9.2
40	9.6
50	10.0		4.0

Newton-Raphson to the rescue?

${y_L}^3-10.2{y_L}^2+{155.28/{8.4^2}}=0$ ; ${y_L}^3-10.2{y_L}^2+2.2=0$ ;

y_L=0.5 ; ${0.5}^3-10.2*{0.5}^2+2.2=0.125-2.55+2.2=-0.225$ ; $m=3{y_L}^2 -20.4y_L=3*0.5^2-20.4*0.5=0.75-10.2=-9.45$ .

y_L=0.5-0.225/9.45=0.4762 ; ${0.4762}^3-10.2*{0.4762}^2+2.2=0.108-2.313+2.2=-0.005$ ; $m=3{y_L}^2 -20.4y_L=3*0.4762^2-20.4*0.4762=0.6803-9.7145=-9.0342$ .

y_L=0.4762-0.005/9.0342=0.4756 .

It looks like only one iteration is necessary when starting from the known nearby value.

${y_L}^3-10.2{y_L}^2+{155.28/{8.8^2}}=0$ ; ${y_L}^3-10.2{y_L}^2+2.0=0$ ;

y_L=0.4756 ; ${0.4756}^3-10.2*{0.4756}^2+2.0=0.1076-2.3072+2.0=-0.1996$ ; $m=3{y_L}^2 -20.4y_L=3*0.4756^2-20.4*0.4756=0.6786-9.7022=-9.0236$ ;

y_L=0.4756-0.1996/9.0236=0.4535 .

And so on….

${y_H}^3-4.1{y_H}^2+{155.28/{b^2}}=0$ ; $m=3{y_H}^2-8.2{y_H}$ ;

Time for a spreadsheet.

l	b	zero	m	yL	ML	zero	m	yH	MH	deltaM
0	8.0			0.50	78.69	0.23	14.57	3.94	72.04	6.65
10	8.4	-0.22	-9.45	0.48	78.74	0.20	14.78	3.96	75.16	3.58
20	8.8	-0.20	-9.04	0.45	78.78	0.17	14.96	3.97	78.30	0.48
30	9.2	-0.17	-8.64	0.43	78.82	0.15	15.11	3.98	81.46	-2.64
40	9.6	-0.15	-8.29	0.42	78.85	0.08	15.20	3.99	84.63	-5.77
50	10.0	-0.14	-7.96	0.40	78.89			4.00	86.63	-7.74

And finally, the answer to just part a.

20'

For part (b), I think everything is the same except the equations for E, which will decrease an additional 1' over the 50' length.

$E_L=E_1=y_L+{{Q^2}/{2gb^2 y_L}}=y_L+{10000/{2*32.2*b^2 {y_L}^2}}=10.2-{l/50} right$ ${y_L}^3+{155.28/{b^2}}=(10.2-{l/50}){y_L}^2 right$

${y_L}^3-(10.2-{l/50}){y_L}^2+{155.28/{b^2}}=0$ ;

$m=3{y_L}^2 -(20.4-{l/25})y_L$ ;

$E_H=E_2=y_H+{{Q^2}/{2gb^2 y_H}}=y_H+{10000/{2*32.2*b^2 {y_H}^2}}=4.1-{l/50} right$ ${y_H}^3+{155.28/{b^2}}=(4.1-{l/50}){y_H}^2 right$

${y_H}^3-(4.1-{l/50}){y_H}^2+{155.28/{b^2}}=0$ ;

$m=3{y_H}^2-(8.2-{l/25}){y_H}$ ;

l	b	zero	m	yL	ML	zero	m	yH	MH	deltaM
0	8.0			0.50	78.69	-2.32	11.66	3.96	76.16	2.53
10	8.4	-0.17	-9.25	0.48	78.73	-2.10	10.26	3.77	54.95	23.77
20	8.8	-0.15	-8.74	0.46	78.76	-1.74	9.13	3.56	48.50	30.26
30	9.2	-0.13	-8.26	0.45	78.79	-1.80	7.46	3.37	45.87	32.93
40	9.6	-0.11	-7.82	0.43	78.82	-0.24	8.37	3.13	44.34	34.47
50	10.0	-0.10	-7.42	0.42	78.84			3.10	44.24	34.60

0': The jump starts right at the contraction

c. Just a small amount of energy loss gets the jump happening right away.

d. Decrease the 4.1 factor to 3.7

$E_L=E_1=y_L+{{Q^2}/{2gb^2 y_L}}=y_L+{10000/{2*32.2*b^2 {y_L}^2}}=10.2-{l/50} right$ ${y_L}^3+{155.28/{b^2}}=(10.2-{l/50}){y_L}^2 right$

${y_L}^3-(10.2-{l/50}){y_L}^2+{155.28/{b^2}}=0$ ;

$m=3{y_L}^2 -(20.4-{l/25})y_L$ ;

$E_H=E_2=y_H+{{Q^2}/{2gb^2 y_H}}=y_H+{10000/{2*32.2*b^2 {y_H}^2}}=3.7-{l/50} right$ ${y_H}^3+{155.28/{b^2}}=(3.7-{l/50}){y_H}^2 right$

${y_H}^3-(3.7-{l/50}){y_H}^2+{155.28/{b^2}}=0$ ;

$m=3{y_H}^2-(7.4-{l/25}){y_H}$ ;

l	b	zero	m	yL	ML	zero	m	yH	MH	deltaM
0	8.0			0.50	78.69	-1.71	8.57	3.53	47.86	30.83
10	8.4	-0.17	-9.25	0.48	78.73	-1.55	7.28	3.33	45.48	33.25
20	8.8	-0.15	-8.74	0.46	78.76	-1.18	6.40	3.11	44.29	34.47
30	9.2	-0.13	-8.26	0.45	78.79	-1.26	4.77	2.93	43.84	34.95
40	9.6	-0.11	-7.82	0.43	78.82	0.23	6.21	2.66	43.73	35.09
50	10.0	-0.10	-7.42	0.42	78.84			4.00	86.63	-7.78

The Froude number still indicates subcritical at 3.6'.

Now the jump is near the end at about	50'

Problem 10

A trapezoidal channel is 20 ft (6.1 m) wide at the bottom and has side slopes of 1.5 (horizontal to 1 (vertical). For a flow of 500 ft³/s (14 m³/s), calculate the depth sequent to a supercritical depth of 2.0 ft (0.61 m).

b=20 ft; z=1.5; Q=500 ft³/s; y_1=2.0 ft.

A_1=(b+zy_1)y_1=(20+1.5*2)2= 46 ft² $k_1=b/{zy_1}=20/{1.5*2}=$ 6.67 ft; $D_1={A_1}/P_{wet,1}=46/{b+2 sqrt{{y_1}^2+{zy_1}^2}}= =46/{20+2 sqrt{2^2+{1.5*2}^2}}= =46/{20+2 sqrt{13}}= 46/27.21=$ 1.69 ft; ${F_1}^2={Q^2}/{g{A_1}^2 D_1}=250000/{32.2*2116*1.69}=$ 2.17 >1✔; T_1=b+zy_1=20+2*1.5*2= 26 ft;

$A_2=(b+zy_2)y_2=(20+1.5y_2)y_2=20y_2 + 1.5 {y_2}^2$ ; $k_2=b/{zy_2}=20/{1.5 y_2}$

$A_1 k_1 y_1 - A_2 k_2 y_2 = {Q^2/g} ({1/{A_2}} - {1/{A_1}})$

$k_2{{A_2 y_2}/{A_1 y_1}} - k_1 = {F_1}^2 {A_1/{y_1 T_1}}(1-{{A_1}/{A_2}})=$ ${20/{1.5 y_2}} {{A_2 y_2}/{46*2}} - 6.67 = 2.17 {46/{2* 26}}(1-{{46}/{A_2}})=$ ${{A_2}/6.9} - 6.67 = 0.8846-{40.69/{A_2}} right$ $0.15{A_2}^2 - 5.785A_2 + 40.69 =0$

$A_2 = {5.785 pm sqrt{5.785^2-4*0.15*40.69}}/{2*0.15}=19.28 pm {sqrt{33.47-24.41}/0.3}=19.28 pm {3/0.3}=19.28 pm 10$

The larger area has subcritical flow

$A_2=29.28=20y_2 + 1.5 {y_2}^2 right$ $1.5 {y_2}^2 +20 y_2 - 29.28=0 right$ $y_2 = {-20 pm sqrt{20^2+4*1.5*29.28}}/{2*1.5}=-6.67 pm {sqrt{400+175.7}/3}=-6.67 pm 8 right$

1.33 ft

Problem 14

A wave is moving down a wide rectangular channel in the direction of flow. Upstream of the wave, the water velocity is 5 ft/s (1.5 m/s) with a depth of 5 ft (1.5 m). Downstream of the wave, the water velocity is 4 ft/s (1.2 m/s) with a depth of 7 ft (2.1 m). What is the velocity of the wave?

b=5 , y_1=5 , u_1=5 , y_2=7 , u_2=4 .

Q=u_1 A_1 =u_1 b d_1 = 5*5*5= 125 cf/s

$-Delta M =-M_{w}= M_1 - M_2 = {Q^2}/{gA_1} + {y_1/2}A_1 - {Q^2}/{gA_2} - {y_2/2}A_2=$ ${Q^2}/{gby_1} + {y_1/2}by_1 - {Q^2}/{gby_2} - {y_2/2}by_2=$ ${125^2}/{32.2*5*5} + {5*5*5/2} - {125^2}/{32.2*5*7} - {7*7*5/2}=$ 19.4+62.5-13.9-122.5= -54.5 slug-ft/s.

h_w=y_2-y_w=7-5= 2 ft, A_w=b h_w=5*2= 10 ft², Q_w=A_w u_w=10 u_w .

$M_w = 54.5 = {{Q_w}^2}/{g A_w} + {h_w/2}A_w=$ ${100/{32.2*10}} {u_w}^2 - {2/2}10=-0.31{u_w}^2 -10 right$ ${u_w}^2={54.5-10}/0.31 =143.55 right$

12.0 ft/s

Well, that's the long way around. Here's the short path.

c=sqrt{gy}=sqrt{32.2*(7-5)}= 8.025 relative to the downstream water velocity.

u_w=u_2+c=4+8= 12.0 ft/sec✔

Problem 15

A rectangular irrigation channel 10 ft (3.0 m) wide conveys a flow of 500 ft³/s (14 m³/s) at a depth of 10 ft (3.0 m). If the upstream discharge is suddenly increased to 700 ft³/s (20 m³/s), estimate the height and velocity of the resulting surge wave.

10 ft.

Q_1= 700 ft³/s.

Q_2= 500 ft³/s, y_2= 10 ft, $u_2={Q_2}/{A_2} = {500}/{10*10}=$ 5 ft/sec

Find y_1 , u_w

Change the coordinate system to one at a height of y_2=10 ft and moving at a velocity of u_2=5 ft/sec. This makes the problem equivalent to suddenly flooding an empty channel, and the relatively immobile bottom of this empty channel is actually the moving surface of the original flow.

10 ft.

Q_w=Q_1=700 - 500 200 ft³/s; $u_w={Q_w}/A = {Q_w}/{b y_w}=200/{10 y_w}= 20/{y_w}$ .

Note in this empty channel frame of reference, the wave propagates at the same speed the water moves.

$c=u_w=sqrt{g y_w}=20/{y_w} right$ $g y_w = 400/{{y_w}^2} right$ ${y_w}^3 = 400/g right$

$y_w=root{3}{400/32.2}=$ 2.316 ft.

$u_w=20/{y_w}=20/{2.316}=8.636;$ u_2=u_1+u_w=5+8.636= 13.636 ft/sec.

Height of wave	2.3 ft	above downstream water surface
Propagation rate	13.6 ft/sec	relative to a stationary observer

Did it wrong, first.

Q_w=Q_1=700 - 500 200 ft³/s; $q_w={Q_w}/A = {Q_w}/{b y_w}=200/{10 y_w}= 20/{y_w}$ ; $u_w=q_w/y_w=20/{{y_w}^2}$ .

Note the wave propagates at the same speed the water moves.

$c=u_w=sqrt{g y_w}=20/{{y_w}^2} right$ $g y_w = 400/{{y_w}^4} right$ $g = 400/{{y_w}^3} right$

$y_w=root{3}{400/32.2}=$ 2.316 ft.

$u_w=20/{{y_w}^2}=20/{2.316^2}=3.73 right$ u_2=u_1+u_w=5+3.73= 8.73 ft/sec.

Height of wave: 2.316 ft above downstream water surface Propagation rate:8.73 ft/sec relative to a stationary observer

Things tried unsuccessfully:

In this frame the downstream momentum is zero, the wave momentum is the upstream momentum, and the height of the wave is y_1.

Q_2= 0 ft³/s, y_2= 0 ft, u_2= 0 ft/sec M_2= 0 slug-ft/sec

$M_w=M_2=0={{Q_w}^2}/{g b {y_w}^2} + {{y_w}b y_w}/2 ={200^2}/{32.2*10*{y_w}^2} + 10/2={124.22/{{y_w}^2}}+5$

Uh huh. Imaginary depth. Lovely.

Since there are no other constraints, assume the wave propagates at the velocity requiring minimum energy. (This is probably not right since the wave should propagate at such a rate as to dissipate the most energy.)

$M_w=M_1-M_2={{Q_1}^2}/{g b {y_1}^2} + {{y_1}b y_1}/2 - {{Q_2}^2}/{g b {y_2}^2} + {{y_2}b y_2}/2=$

$q={Q_1}/A = {Q_1}/{b y_1}=200/{10 y_1}$

¹⁾ $b^2=155/{10.2 {y_L}^2 - {y_L}^3}$ . ${A_L}^2=b^2{y_L}^2={155{y_L}^2}/{4.1 {y_L}^2 - {y_L}^3}=$ $155/{10.2 - y_L} right$ $A_L=sqrt{155/{10.2 - y_L}}$ . $b^2=155/{4.1 {y_H}^2 - {y_H}^3}$ . ${A_H}^2=b^2{y_H}^2={155{y_H}^2}/{4.1 {y_H}^2 - {y_H}^3}=$ $155/{4.1 - y_H} right$ $A_H=sqrt{155/{4.1 - y_H}}$ . $M_H = M_L = Q^2/{gA_H} + {y_H A_H}/2 = Q^2/{gA_L} + {y_L A_L}/2=$ $10000/{32.2 sqrt{155/{4.1 - y_H}}} + {y_H sqrt{155/{4.1 - y_H}}}/2=$ $10000/{32.2 sqrt{155/{10.2 - y_L}}} + {y_L sqrt{155/{10.2 - y_L}}}/2=$

ASSIGNMENT 4 (Due October 4, 2007 )

Problem 4

Problem 10

Problem 14

Problem 15

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