Homework hints, ninth week: 10.2,11,38,48,59,79

10.2 A shallow-water wave 1 cm high propagates into still water of depth 1.1 m. Compute (a) the wave speed; and (b) the induced velocity \delta V.

Solution: Add in \delta y from (10.7) and (10.9):

c = (gy(1+\delta y/y)(1+\delta y/2y))^1/2 = (9.81(1.1)(1+0.01/1.1)(1+0.01/2x1.1))^1/2 = 3.31 m/s

\delta V = c \delta y / (y + \delta y ) = 3.31(0.01)/(1.1 + 0.01) = 0.030 m/s

**10.4 The water channel flow has a free surface in three places. Does it qualify as an open channel flow? Explain. What does the dashed line represent?

Not an open channel flow because there is no flow between the surfaces of the manometer tubes. The dashed line is the hydraulic grade line, representing the frictional head h_f.

**10.5 Water flows rapidly in a channel 15 cm deep. Piercing the surface with a pencil point creates a wedge like wave of included angle 35 degrees. Estimate the velocity V of the water flow.

This is a supercritical flow, analogous to supersonic gas flow. The Froude number serves the same purpose as the Mach number in creating Mach waves. The sine of the angle \theta is 1/Fr, where Fr = V/(gh)^1/2. Sin(17) = (gh)^1/2 / V, g=9.81, h=0.15, V=4.03 m/s.

**10.7 Pebbles dropped successively at the same point into a water channel flow of depth 65 cm create two circular ripples. Estimate the a. Froude no.; b. the stream velocity. The 3m circle is displaced 4m from the 9m circle.

If the drop site is a distance X upstream of the center of the small circle: X=3V/c_o ; large circle: X+4 = 9V/c_o ; solve Fr = V/c_o = 2/3. c_o = sqrt gh = sqrt 9.81 0.65 = 2.53 m/s, so V=2/3 2.53 = 1.68 m/s.

**10.9 Equation 10.10 is for a single disturbance wave. For periodic small amplitude surface waves of wavelength \lambda and period T, inviscid theory predicts a wave propagation speed

c_o^2 =( g \lambda / 2 \pi) tanh (2 \pi y / \lambda)

where y is the water depth and surface tension is neglected. a. Determine if this expression is affected by the Reynolds number, Froude number, or Weber number. Derive the limiting values of this expression for b. y << \lambda and c. y >> \lambda d. For what ratio y/\lambda is the wave speed within 1 % of limit c.?

Because viscosity and surface tension are not in the formula, there can be no Reynolds or Weber number effects. The Froude number is crucial. Solve for Fr = c_o/sqrt g \lambda = sqrt 1/2\pi tanh 2\pi y / \lambda = function (y/lambda).

b. y<<\lambda : tanh x = x if x << 1, so c_o^2 = gy (same as 10.10)

c. y>>\lambda : tanh x = 1 for large x; c_o^2 = g \lambda / 2 \pi (periodic deep water waves)

d. c_o = 0.99 c_o, deep : if tanh 2 \pi y /\lambda = 0.995 = tanh 3 , so y/\lambda = 0.48.

10.11 A rectangular channel is 2 m wide and contains water 3 m deep. If the slope is 0.85 degrees and the lining is corrugated metal, estimate the discharge for uniform flow.

Soln: For corrugated metal, take Manning's n=0.022. Get the hydraulic radius: R_h = A/P = 2(3)/(3+2+3) = 0.75 m ; Q = (1/n)(A R_h^2/3 S_o^1/2 = (1/0.022)(6 0.75^2/3 [tan(0.85)]^1/2 ; or Q = 27 m^3 / s.

**10.15 The finished concrete channel is designed for a flow rate of 6 m^3 /s at a normal depth of 1 m. Determine a. the design slope of the channel and b. the percentage of reduction in flow if the surface is asphalt.

For finished concrete n=0.012. Find the hydraulic radius and then S_0: R_h = A/P = 3/5 = 0.6m Chezy: Q = 6 = 1.0/0.012 (3)(0.6)^2/3 S_0 ^1/2 ; S_0 = 0.00114.

b. Asphalt n=0.016; Q=6 n_1 / n_2 = 4.5 m^3 /s. 25% less.

**10.28 Show that, for any straight, prismatic channel in uniform flow, the average wall shear stress is given by \tau_ave = \rho g R_h S_0.

For a control volume enclosing the fluid prism of length L (partly filled cylinder for example inclined at \theta):

sum F_along flow = W_fluid sin \theta = \tau_ave A_wall , or \tau_ave PL = \rho g A L sine \theta . But sine \theta = S_0 by definition, leaving \tau_ave = \rho g R_h sin \theta.

**10.32 A 2m diameter clay tile sewer pipe runs half full on a slope of 0.25 deg. Compute the normal flow rate in gal/min.

For clay tile n=0.014. For a half full circle; A = 1.57 m^2 , R_h = R/2 = 0.5 m. Q (Chezy) = 4.67 m^3/s = 74000 gal/min.

**10.37 The answer to 10.34 is that the 8 ft wide channel is 29% more efficient than the 4 ft width. Verify this result using the best efficiency concept of 10.3.

From eq 10.26 of the text, a rectangular channel has maximum flow for a given perimeter when width b = 2 y. This is almost exactly the case with our 8 ft wide channel: b=8 ft, y=4.07 ft, b/y = 1.97. The alternative case was way off this ideal shape with b/y = 4.09/9.31=0.43.

10.38 A rectangular channel has b=3m and y=1m. if n and S_0 are the same, what is the diameter of a semicircular channel which will have the same discharge? Compare the two wetted perimeters.

The rectangular channel has A=3m^2 and P=5m. Set the flow rates equal:

Q_rect = 1/n (3m^2)(3/5 m)^2/3 S_0^1/2 = Q_semicircle = 1/n (|pi R^2/2)^2/3 S_0^1/2

so R=1.334m, D_semicircle=2.67m

The semicircle perimeter is P=\pi R = 4.19m, 16% less than the rectangle P=5m.

**10.46 Sellin suggests that a channel for minimum erosion has a half sine wave shape,with h=h_0 sin(\pi z/b). For uniform flow, find the most efficient ratio h_0 / b.

Evaluate the area and perimeter:

A= 2/\pi h_0 b ; P=int[0,b] [1+(dh/dz)^2]^1/2 dz , where dh/dz=\pi A/b cos(\pi z /b)

Evaluate the integral graphically or by curve fitting. Gives h_0 / b = 0.50 to 0.55.

10.48 A wide, clean-earth river has a flow rate q = 150 f^3 / (s ft). What is the critical depth? If the actual depth is 12 ft., what is the Froude number of the river? Compute the critical slope by (a) Manning's formula and (b) the Moody chart.

Soln.: For clean earth, take n = 0.030 and roughness \eps = 0.8 ft. The critical depth is

y_c = (q^2/g)^1/3 = [(150)^2 /32.2]^1/3 = 8.87 ft

If y_actual = 12 ft, Fr = V/V_c = q/y / (g y_c)^1/2 = 150/12 / (32.2(8.87))^1/2 = 0.739

Critical slope:

a. Manning: S_c = g n^2 / \xi y_c^1/3 = 32.2(0.030)^2 / 2.208(8.87)^1/3 = 0.00634

b. Moody: 1/f^1/2 = -2 log [0.8/3.7(4x8.87)] , or f = 0.0509 , S_c = f/8 = 0.00637.

10.59 Uniform water flow in a wide brick channel of slope 0.02 degrees moves over a 10 cm bump. A slight depression in the water surface results. If the minimum depth over the bump is 50 cm, compute a. the velocity over the bump; and b. the flow rate per meter of width.

Soln.: For brickwork, take n = 0.015. Since the water level decreases over the bump, the upstream flow is subcritical. For a 3wide channel, R_h = y/2, and Eq. 10.39 holds:

y^3 - E_2 y_2^2 + q^2 / 2g = 0, q = V_1 y_1 , E_2 = V_1^2/ 2g + y_1 = \delta h , |delta h = 0.1 m, y_2 = 0.5 m.

For uniform flow, q = (1/0.015) y_1 (y_1 / 2)^2/3 (sin 0.02 )^1/2 = 0.785 y_1^5/3 . Solve these two simultaneously for y_1 = 0.608 m. V_1 = 0.563 m/s and q = 0.342 m^3/s m. The upstream flow is subcritical, Fr_1 = 0.23.

**10.68 Modify problem 10.65 to have a supercritical approach condition V_0 = 6 m/s and y_0 = 1m. If you have time for only one case, use h_max = 35 cm (10.66) for which the maximum Froude number is 1.47. If more time is available, it is instructive to examine a complete family of surface profiles for 1 cm < h_max < 52 cm (solution 10.67).

The Froude numbers at the point of maximum bump height are as follows:

h_max	0	10	20	30	35	40	50	52.1
Fr_bump	1.92	1.8	1.68	1.55	1.47	1.39	1.15	1.00

10.79 Show that the Froude number downstream of a hydraulic jump will be given by Fr_2 = 8^1/2 Fr_1 / {[1 + 8 Fr_1^2]^1/2 - 1}^3/2 . Does the formula remain correct if we reverse subscripts 1 and 2? Why?

Soln.: Take the ratio of Froude numbers, use continuity, and eliminate y_2 / y_1 :

Fr_2/Fr_1 = [V_1/(gy_1)^1/2]/ [V_2/(gy_2)^1/2] = [V_1/(y_1)^1/2]/ [V_2/(y_2)^1/2] , but V_1/ V_2 = y_2 / y_1 from continuity, so Fr_2/Fr_1 = ( y_2 / y_1)^3/2 .

From 10.43, y_2 / y_1 = (1/2) {[1 + 8 Fr_1^2]^1/2 - 1} , so rearrange to Fr_2 = 8^1/2 Fr_1 / {[1 + 8 Fr_1^2]^1/2 - 1}^3/2 . The formula is indeed symmetric, you can reverse 1 and 2.