Chegg a Small River With a Flow Rate of 15 Feeds

Problems and Solutions
to
Chapter 19: Flow Measurement

Chapter 19 is worth 45 points (5 points per problem)

Problem 19.1
Determine the discharge of a 4-in. diameter end-cap orifice on a 6-in. diameter pipe if C = 0.7 and the head on the orifice is 12 in .

Solution to 19.1
q = Ca(2gh)**½ = Ca(2gh)^1/2

where:
q = discharge, cu ft/s
C = discharge coefficient = 0.7
a = cross-sectional area of orifice, sq ft = Pi * r² = 3.142 * (2 in./12 in./ft)² = 0.087 sq ft
g = acceleration of gravity = 32.2 ft/s²
h = static head above center of orifice, ft = 12 inches

q = (0.7)(0.087 sq ft)[(2)(32.2 ft/s²)(12 inches)(1 ft/12 inches)]^1/2
=

0.49 cu ft/s

Problem 19.2
Discuss the advantages and disadvantages of impeller meters for flow measurement in pipes.

Solution to 19.2
Advantages: reads both cumulative volume and instantaneous flow rate directly; accurate; east of use; can be used where flumes or weirs impractical.
Disadvantages: sensitive to debris and water hardness build up; needs to be maintained and is difficult to maintain; difficult to relocate; measures velocity at single point in channel; depth of meter placement affects accuracy/readings.

Problem 19.3
Determine the discharge in a trapezoidal concrete-lined canal using the float method. The canal has a 1 ft bottom width and 1:1 side slopes with a flow depth of 1.3 ft. The times for a float to travel 200 feet were 72, 79, 74, 76, and 79 seconds.

Solution to 19.3
q = av
average time = (72+79+74+76+79)/5 = 76 s
cross-sectional area = bd + ed, also area = area of square + 2 * area of outside triangles = (1 ft)(1.3 ft) + [2 * 1/2* (1.3 ft)(1.3 ft) = 2.99 sq ft = 3.0 sq ft

1:1 side slopes, therefore d = e = 1.3 ft
float velocity = 200 ft/76 s = 2.63 ft/s
assume average velocity = 80% of float velocity = 2.63 ft/s x 0.80 = 2.10 ft/s

q = (3.0 ft²)(2.10 ft/s) = [(1 ft)(1.3 ft) + (1.3 ft.)(1.3 ft)](2.63 ft/s)(0.80) = 6.3 cu ft/s

Problem 19.4
Determine the discharge of a stream having a cross-sectional area of 50 sq ft and a wetted perimeter of 16 ft, using the slope-area method. The difference in elevation of the water surface 500 ft apart is 0.3 ft. The channel has straight banks with some weeds.

Solution to 19.4
q = Av
A = 50 sq ft
v= c/n * R^2/3 * S^1/2

where:
C = Constant = 1.49
n = roughness coefficient = 0.035 (see Table 8.2, p. 129)
R = hydraulic radius, ft = A/P = 50 sq ft/16 ft = 3.13 ft
S = channel gradient, ft/ft = 0.3 ft/500 ft = 0.0006 ft/ft

v = (1.49/0.035) (3.13 ft)^2/3 (0.0006 ft/ft)^1/2 = 2.24 ft/s

q = (50 sq ft)(2.24 ft/s) = 112.1 cu ft/s

Problem 19.5
A velocity of 2 ft/s is measured with a current meter placed 0.4 ft above the bottom of a small stream. The estimated cross-sectional area of the stream is 2 sq ft, and the flow depth is 1 ft. What is the estimated flow rate?

Solution to 19.5
Q = AV = (2 sq ft)(2 ft/s) = 4 cu ft/s

Problem 19.6
Determine the flow rate over a suppressed weir if the weir crest is 24 in. long and the head 30 in. upstream is 6 in. Determine the flow rate over a contracted weir having the same dimensions.

Solution to 19.6
a. Q = CLh^3/2

where
C = discharge coefficient
L = length of weir crest, ft = 24 in. = 2 ft
h = head on weir, ft = 6 in. = 0.5 ft
assume C = 3.4 (C is related to h/P. We don't know P, nor can we estimate it from the data given in the problem statement. Therefore, use the recommended h/P of 0.5 for which C=3.4 (see page 352).)

Q = (3.4)(2 ft)(0.5 ft)^3/2 = 2.4 cu ft/s

b. Q = CLh^3/2
assume C = 3.16 (see p. 353)
L = 2 ft
h = 0.5 ft
Q = (3.16)(2 ft)(0.5 ft)^3/2 = 2.23 cu ft/s

Problem 19.7
A trapezoidal weir has crest length of 18 in. Determine the discharge over the weir if the head is 5 in. at a point 2 ft upstream.

Solution to 19.7
Q = CLh^3/2

where
L = 18 in. = 1.5 ft
h = 5 in. = 0.42 ft
assume 4:1 side slope, therefore C = 3.37 (see p. 353)

Q = (3.37)(1.5 ft)(0.42 ft)^3/2 = 1.38 cu ft/s

Problem 19.8
Determine the discharge over a 90-degree V-notch weir if the head is 0.6 ft .

Solution to 19.8
Q = Ch^5/2

where
h = 0.6 ft
assume: C = 2.47 (see p. 354)

Q = (2.47)(0.6 ft)^5/2 = 0.7 cu fts

Problem 19.9
Determine the discharge through a Parshall flume having a throat width of 1.5 ft for H(a) = 1.1 ft. and H(b) = 0.75 ft.

Solution to 19.9

Determine if discharge at H(b) is less than 0.7H(a).

0.75 ft < (0.7)(1.1 ft) = 7.7 is a true statement, therefore, the equation for flow through a Parshall flumes is

Q = 4.1 WH(a)^1.584

where
W = throat width, ft = 1.5 ft
H(a) = head in converging section, ft = 1.1 ft

Q = (4.1)(1.5 ft)(1.1 ft)**1.564 = 7.15 cu ft/s

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Source: https://ag.arizona.edu/classes/ABE404/Assignments/Solution19.htm

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