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CE 4603: Water Resources Engineering — David Wagner 2007/04/15 16:35

Homework #9 Due 4/18

Solve problems 3.26, 3.27, 3.36, 3.37, and 3.40 of Water Resources Engineering by David Chin, 2nd edition, pages 255-257.

3.26

Determine the critical depth for 50 m³/s flowing in a trapezoidal channel with bottom-width 4 m and side slopes of 1.5:1 (H:V). If the depth of flow is 3 m, calculate the Froude number and state whether the flow is subcritical or supercritical. [See example 3.5, page 122 and page 121.]

T=4+3y ; A=3y+1.5y^2 ; Q = 50 ; y=3

{Q^2}/g={A_c^3}/T_c = {50^2}/9.81 = {(3y_c+1.5y_c^2)^3}/{4+3y_c} JpGraph Function Plot

→ yc ≈ 2.25 my > yc → subcritical

Fr^2 = {Q^2T}/{gA^3} = {50^2*(4+3*3)}/{9.81*(3*3+1.5*3^2)^3} = {2500*13}/{9.81*22.5^3} = 32500/111740 = 0.291 → Fr=0.539

Fr < 1 → subcritical

3.27

A rectangular channel 2 m wide carries 3 m³/s of water at a depth of 1.2 m. If an obstruction 40 cm wide is placed in the middle of this channel, find the elevation of the water surface at the constriction. What is the minimum width of the constriction that will not cause a rise in the water surface upstream? [See Example 3.6, page 124.]

Q=3 ; y1=1.2 A_1 = 2*1.2 = 2.4 V_1 = Q/A_1 = 3/2.4 = 1.25

E_1 = y_1 + {{V_1^2}/2g} = 1.2+{{1.25^2}/2*9.81} = 1.2+0.07964 = 1.28

A_2 = (2-0.4)*y_2 = 1.6y_2

E_1 = E_2 = 1.28 = y_2 + {Q^2}/{2gA^2} = y_2 + {3^2}/{2*9.81*(1.6*y_2)^2} = y_2 + 9/{50.227*y_2^2} = y_2 + 0.179/{y_2^2} = 1.28

JpGraph Function Plot

Fr_1 = V_1/sqrt{gy_1} = 1.25/sqrt{9.81*1.2} = 1.25/3.43 = 0.364→ subcritical → y2=max(0.48,1.16)

y2=1.16 m

E_1={3/2}root{3}{{Q^2}/{b^2g}} = 1.28 ={3/2}root{3}{{3^2}/{b^2*9.81}} right 0.8533^3*b^2 = {9/9.81} right b^2 = 1.477 right

b=1.215 m

3.36

Water flows at 36 m³/s in a rectangular channel of width 10 m and a Manning n of 0.030. If the depth of flow at a channel section is 3 m and the slope of the channel is 0.001, classify the water-surface profile. What is the slope of the water surface at the observed section. Would the shape of the water-surface profile be much different if the depth of flow were equal to 2 m? [See Example 3.7, pg 130-133]

n=0.030 ; S0=0.001 ; Q = 36 ; b=10 ; y=3 ; A=by=30 ; P=b+2y=16 ; T=b=10

Q={1/n}{A_n^{5/3}}/{P_n^{2/3}}sqrt{S_0} = 36 = {1/0.30}{(10*y_n)^{5/3}}/{(10+2*y_n)^{2/3}}sqrt{0.001} JpGraph Function Plot

→ yn≈14 m

A_c^3/T_c = Q^2/g = (10*y_c)^3/10 = 36^2/9.81 JpGraph Function Plot

→yc≈1.1 m

yc < y < yn →

M2dy/dx < 0

R=A/P = by/(b+2y) = 10*3/{10+2*3} = 30/16 = 1.875

S_f=({nQ}/AR^{2/3})^2 = ({0.030*36}/{30*1.875^{2/3}})^2 = (1.08/45.617)^2 = 0.00056

Fr=sqrt{{Q^2T}/{gA^3}} = sqrt{{36^2*10}/{9.81*30^3}} = sqrt{12960/264870} = 0.224

dy/dx = -{S_0-S_f}/{1-Fr^2} = -{0.001- 0.00056}/{1-0.224^2} = -0.000439/0.950 =

dy/dx = -0.000463 m/mSame profile (M2) at y=2m.

3.37

Water flows at 30 m³/s in a rectangular channel of width 8 m. The Manning n of the channel is 0.035. Determine the range of channel slopes that would be classified as steep and the range that would be classified as mild.

n=0.035 ; Q=30 ; b=T=8 ; A=by ; P = b+2y ; R=A/P=by/(b+2y)

Steep: yn < yc; Mild: yn > yc → yn=yc divides the two. Call this y=yn=yc

A_c^3/T_c = Q^2/g = {8*y}^3/8 = 30^2/9.81 = 64*y^3 right y=1.1275

A_n^5/P_n^2 = ({Qn/sqrt{s_0}})^3 = (by)^5/(b+2y)^2 = ({30*0.035}/sqrt{S_0})^3 = (8*1.1275)^5/(8+2*1.1275)^2 = (1.05/sqrt{S_0})^3 = 59708/105.17 = 567.76 right 1.05/sqrt{S_0}=8.2805 right S_0 = (1.05/8.2805)^2 = 0.000016

Steep: Slope > 0.000016Mild: Slope < 0.000016

3.40

If 100 m³/s of water flows in a channel 8 m wide at a depth of 0.9 m, calculate the downstream depth required to form a hydraulic jump and the fraction of the initial energy lost in the jump. [See Example 3.8, page 135-136.]

Q=100 ; y1=0.9 ; T=8 ; q=Q/T=12.5

Fr_1=q/{y_1 sqrt{gy_1}}=12.5/{0.9*sqrt{9.81*0.9}}=12.5/2.674=4.674→supercritical✔

y_2/y_1 = {1/2}(sqrt{1+8FR_1^2}-1)={sqrt{1+8*4.674}-1}/2={sqrt{38.39}-1}/2=2.6 right y_2 = 2.6*0.9 =2.34

y_2=2.34 m

Delta E=(y_2-y_1)^3/{4y_1 y_2} = (2.34-0.9)^3/{4*0.9*2.34} = 1.44^3/8.42 =0.355 m

ΔE=0.355 m

Discussion


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