A weir in a horizontal channel is 1 m high and 4 m wide. The water depth upstream is 1.6 m. Estimate the discharge if the weir is (a) sharp-crested and (b) round-nosed with an unfinished concrete broad crest 1.2 m long. Neglect V1^2/(2g).

Question

A weir in a horizontal channel is 1 m high and 4 m wide. The water depth upstream is 1.6 m. Estimate the discharge if the weir is (a) sharp-crested and (b) round-nosed with an unfinished concrete broad crest 1.2 m long. Neglect [latex]V_1^2/(2g)[/latex].

Accepted Answer

Part (a)

We are given Y = 1 m and H + Y ≈ 1.6 m, hence H ≈ 0.6 m. Since H [latex]\ll[/latex] b, we assume that the weir is “wide.” For a sharp crest, Eq. (10.56) applies:

Wide sharp-crested weir: [latex]C_d \approx 0.564 + 0.0846 \frac{H}{Y} for \frac{H}{Y} \leq 2[/latex] (10.56)

[latex]C_d \approx 0.564 + 0.0846 \frac{0.6 m}{1 m} \approx 0.615[/latex]

Then the discharge is given by the basic correlation, Eq. (10.55):

[latex]Q_{weir} = C_db \sqrt{g} \left(H + \frac{V^2_1}{2g} ight)^{3/2} \approx C_db \sqrt{g}H^{3/2}[/latex] (10.55)

[latex]Q = C_db \sqrt{g}H^{3/2} = (0.615)(4 m) \sqrt{(9.81 m/s^2)}(0.6 m)^{3/2} \approx 3.58 m^3/s[/latex]

We check that H/Y = 0.6 < 2.0 for Eq. (10.56) to be valid. From continuity, [latex]V_1 = Q/(by_1) =[/latex] 3.58/[(4.0)(1.6)] = 0.56 m/s, giving a Reynolds number [latex]V_1H/ u \approx 3.4 E5[/latex].

Part (b)

For a round-nosed broad-crested weir, Eq. (10.57) applies. For an unfinished concrete surface, read ε ≈ 2.4 mm from Table 10.1. Then the displacement thickness is

Round-nosed broad-crested weir: [latex]C_d \approx 0.544 \left(1 - \frac{\delta^*/L}{H/L} ight)^{3/2}[/latex] (10.57)

[latex]\frac{\delta^*}{L} \approx 0.001 + 0.2 \sqrt{\epsilon /L} = 0.001 + 0.2 \left(\frac{0.0024 m}{1.2 m} ight)^{1/2} \approx 0.00994[/latex]

Table 10.1 Experimental Values of Manning’s n Factor[latex]^*[/latex]		Average roughness height ε
	n	ft	mm
Artificial lined channels:
Glass	0.010 ± 0.002	0.0011	0.3
Brass	0.011 ± 0.002	0.0019	0.6
Steel, smooth	0.012 ± 0.002	0.0032	1
Painted	0.014 ± 0.003	0.008	2.4
Riveted	0.015 ± 0.002	0.012	3.7
Cast iron	0.013 ± 0.003	0.0051	1.6
Concrete, finished	0.012 ± 0.002	0.0032	1
Unfinished	0.014 ± 0.002	0.008	2.4
Planed wood	0.012 ± 0.002	0.0032	1
Clay tile	0.014 ± 0.003	0.008	2.4
Brickwork	0.015 ± 0.002	0.012	3.7
Asphalt	0.016 ± 0.003	0.018	5.4
Corrugated metal	0.022 ± 0.005	0.12	37
Rubble masonry	0.025 ± 0.005	0.26	80
Excavated earth channels:
Clean	0.022 ± 0.004	0.12	37
Gravelly	0.025 ± 0.005	0.26	80
Weedy	0.030 ± 0.005	0.8	240
Stony, cobbles	0.035 ± 0.010	1.5	500
Natural channels:
Clean and straight	0.030 ± 0.005	0.8	240
Sluggish, deep pools	0.040 ± 0.010	3	900
Major rivers	0.035 ± 0.010	1.5	500
Floodplains:
Pasture, farmland	0.035 ± 0.010	1.5	500
Light brush	0.05 ± 0.02	6	2000
Heavy brush	0.075 ± 0.025	15	5000
Trees	0.15 ± 0.05	?	?

[latex]^*[/latex]A more complete list is given in Ref. 2, pp. 110–113.

Then Eq. (10.57) predicts the discharge coefficient:

[latex]C_d \approx 0.544 \left(1 - \frac{0.00994}{0.6 m/1.2 m} ight)^{3/2} \approx 0.528[/latex]

The estimated flow rate is thus

[latex]Q = C_db \sqrt{g}H^{3/2} = 0.528(4 m) \sqrt{(9.81 m^2/s)}(0.6 m)^{3/2} \approx 3.07 m^3/s[/latex]

Check that H/L = 0.5 < 0.7 as required. The approach Reynolds number is [latex]V_1H/ u \approx 2.9 E5[/latex], just barely below the recommended limit in Eq. (10.57).

Since [latex]V_1[/latex] ≈ 0.5 m/s, [latex]V^2_1/(2g)[/latex] ≈ 0.012 m, so the error in taking total head equal to 0.6 m is about 2 percent. We could correct this for upstream velocity head if desired.

Question 10.11: A weir in a horizontal channel is 1 m high and 4 m wide. The...

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