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Balancing depth: bed width 25 m

Short answer

Equate cutting area (B + n1*d)*d to filling area 2*h*(b + n2*h). Solve quadratic 1.75d² + 25d - 165 = 0. Balancing depth = 4.91 m.

Calculate the balancing depth of an irrigation canal using the following data: bed width of canal = 25 m; side slope of cutting = 1.75:1 (H:V); side slope of banking = 2.5:1 (H:V); top width of bank = 4 m; height of embankment = 5 m.

Given: B=25B = 25 m, n1=1.75n_1 = 1.75 (cutting), n2=2.5n_2 = 2.5 (banking), b=4b = 4 m, h=5h = 5 m. Both banks are built with slope 2.5:1 on both faces; no berm.

Filling area (two banks):

Af=2h (b+n2h)=2×5×(4+2.5×5)=165 m2A_f = 2h\,(b + n_2 h) = 2\times5\times(4+2.5\times5) = 165\ \text{m}^2

Cutting area: Ac=(25+1.75d) dA_c = (25 + 1.75d)\,d

Equate Ac=AfA_c = A_f:

1.75d2+25d−165=01.75d^2 + 25d - 165 = 0 d=−25+252+4×1.75×1652×1.75=−25+17803.5=4.911 md = \frac{-25+\sqrt{25^2+4\times1.75\times165}}{2\times1.75} = \frac{-25+\sqrt{1780}}{3.5} = 4.911\ \text{m}

Check: Ac=165.00A_c = 165.00 m2^2 = Af=165A_f = 165 m2^2.

Answer: balancing depth = 4.91 m.

✓ Answer checked by Claude Opus 5.5 (high effort)

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