With B = 6D and side slope 1:1, what is the area in terms of D, and what is D?

A = (B + D)D = 7D^2, so D = sqrt(17.143/7) = 1.56 m

How is the bed slope found?

From Manning: S = (V n / R^(2/3))^2 with R = A/P

From Kennedy's V0 = 0.55 D^0.64

From Lacey's S = f^(5/3)/(3340 Q^(1/6))

The Kennedy check gives m = V/V0 = 2.39. What does this mean?

The velocity is far above the usual 0.8-1.2, so it suits only a lined or protected channel

The channel will silt up heavily

The design is ideal for an ordinary earth canal

The side slope must be flatter than 1:1

2077 Chaitra · 8 marks

Kennedy: Q 30, V 1.75 m/s

Short answer

Full supply depth 1.56 m, bed width 9.39 m, bed slope 1 in 1339 (0.000747). Note: critical velocity ratio m=2.39 exceeds typical alluvial range; section suitable only for lined or well-protected channel.

An irrigation channel is to carry a full supply discharge of 30 m³/s at a velocity of 1.75 m/s. The side slopes are to be 1:1. The ratio of full supply depth to bed width is to be 1:6. Assuming the Manning's n as 0.018, calculate the full supply depth, bed width, and bed slope of the channel using Kennedy's method.

Given: $Q = 30\ \text{m}^3/\text{s}$ , $V = 1.75$ m/s, side slope 1H:1V, $D:B = 1:6$ , $n = 0.018$ .

Step 1: Area

A = \frac{Q}{V} = \frac{30}{1.75} = 17.143\ \text{m}^2

Step 2: Depth and bed width

With $B = 6D$ and $z = 1$ :

\begin{aligned} A &= (B + zD)D = (6D + D)D = 7D^2 \\ D &= \sqrt{17.143/7} = 1.565\ \text{m} \approx 1.56\ \text{m} \\ B &= 6D = 9.39\ \text{m} \end{aligned}

Step 3: Bed slope (Manning)

\begin{aligned} A &= (9.39 + 1.56)\times 1.56 = 17.08\ \text{m}^2 \\ P &= B + 2D\sqrt{2} = 9.39 + 2\times 1.56\times 1.4142 = 13.80\ \text{m} \\ R &= \frac{A}{P} = 1.238\ \text{m} \\ S &= \left(\frac{Vn}{R^{2/3}}\right)^2 = \left(\frac{1.75\times 0.018}{1.238^{2/3}}\right)^2 = 0.000747 = \frac{1}{1339} \end{aligned}

Step 4: Kennedy check

Critical velocity $V_0 = 0.55D^{0.64} = 0.55\times 1.56^{0.64} = 0.731$ m/s, so the critical velocity ratio is $m = V/V_0 = 2.39$ . This is much higher than the usual 0.8-1.2, so a velocity of 1.75 m/s suits only a stable boundary (a lined or well-protected channel), not an ordinary alluvial earth canal.

Answer: Full supply depth $\approx 1.56$ m, bed width $\approx 9.39$ m, bed slope $\approx 1$ in 1339 (0.000747).

✓ Answer checked by Claude Opus 5.5 (high effort)

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All Irrigation and Drainage Engineering chapter 4 questions →

Kennedy: Q 30, V 1.75 m/s

Step 1: Area

Step 2: Depth and bed width

Step 3: Bed slope (Manning)

Step 4: Kennedy check

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