Design of Canals Questions and Answers

Answer: B

The keyed pair is 20.26 m bottom width and 1.56 m depth.

Lacey's regime equations with silt factor f = 1 give the velocity, area and wetted perimeter:

\[ \begin{aligned} V &= \left(\frac{Qf^{2}}{140}\right)^{1/6} = 0.750\ \text{m/s} \\ A &= Q/V = 33.3\ \text{m}^{2} \\ P &= 4.75\sqrt{Q} = 23.75\ \text{m} \end{aligned} \]

For a side slope of ½ H : 1 V, \(P = b + \sqrt{5}\,d\) and \(A = bd + d^{2}/2\). Solving these two equations gives a depth of about 1.6 m and a bottom width of about 20.2 m, which matches the keyed pair to within the rounding of the regime constants.

The pair with 1.38 m depth gives an area of only about 29 m², too small for the regime velocity. 23.75 m is the whole wetted perimeter mistaken for the bottom width, and 32.78 m is wider still.

Answer: A

The critical depth is about 1.09 m.

At critical flow \(Q^{2}/g = A^{3}/T\), where for this section \(A = (5 + y)y\) and the top width is \(T = 5 + 2y\). The left side is \(20^{2}/9.81 = 40.77\). Trying y = 1.09 m:

\[ \begin{aligned} A &= (6.09)(1.09) = 6.638\ \text{m}^{2} \\ T &= 7.18\ \text{m} \\ A^{3}/T &= 292.5/7.18 = 40.74 \end{aligned} \]

That matches 40.77, so y = 1.09 m. 1.18 m is the critical depth of a 5 m wide rectangular channel, \((q^{2}/g)^{1/3}\) with q = 4 m²/s, which ignores the extra area of the side slopes; at 1.18 m the trapezoidal expression is already about 52.6. 2.12 and 2.62 m are far deeper than either.

Answer: A

In Lacey's theory the velocity does not depend on the depth of water as such.

Lacey's regime equations express the velocity through the silt factor f and the hydraulic mean depth R:

\[ \begin{aligned} V &= \sqrt{\tfrac{2}{5} f R} \\ V &= 10.8 R^{2/3} S^{1/3} \end{aligned} \]

The second, the regime flow equation, brings in the bed slope S. So hydraulic mean depth, slope and silt factor all enter the velocity. The depth of water itself does not appear: R is the area divided by the wetted perimeter, not the flow depth.

Answer: D

The key gives ignored.

Kennedy's theory (1895) gives the critical non-silting, non-scouring velocity as V_o = 0.55 m d^0.64 (with K = 0.55 for his Upper Bari Doab canals), relating velocity to depth only. A recognised limitation of the theory is that it ignores the bed-width to depth ratio: it gives no relation for B/D, so the same velocity can be met by many different channel sections. Lacey's later regime theory addressed this.

Since Kennedy prescribed no B/D range, none of the numerical ranges 'less than 1', '1 to 5' or '5 to 10' belongs to his relation.

Answer: D

Lacey's bed slope is \(\frac{F^{5/3}}{3340\ Q^{1/6}}\).

Lacey's regime theory for alluvial channels gives the slope in terms of the silt factor \(f\) and the discharge \(Q\) in cumecs:

\[ S = \frac{f^{5/3}}{3340\,Q^{1/6}} \]

A coarser silt (larger \(f\)) needs a steeper slope to keep it moving, and a larger channel can carry it at a flatter slope, which is why \(f\) is in the numerator and \(Q\) in the denominator. The exponents 1/3, 4/3 and 2/3 on \(f\), and 1/2 and 1/4 on \(Q\), do not belong to the regime equations.

Answer: A

Chezy's formula is \(V = C \times \sqrt{R \times S}\).

For uniform flow in an open channel, Chezy's equation gives the mean velocity as the Chezy coefficient times the square root of the hydraulic radius times the bed slope:

\[ V = C\sqrt{RS} \]

The hydraulic radius R is the flow area divided by the wetted perimeter, and S is the slope of the channel bed. The square root covers the product RS, not a sum, so \(C\sqrt{R+S}\) is wrong, and the forms without a square root (\(CR + S\), \(C + RS\)) are not dimensionally sensible.