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## Q. 15.11

The normal depth in a 6 m wide rectangular irrigation channel is 0.5 m. If the volume ﬂowrate is 0.75 m³/s, and the bed slope is 0.002, as shown in Figure 15.36, ﬁnd the head loss and power dissipation per kilometer of channel. What are the average velocity, Froude number, and average wall shear stress in the channel? ## Verified Solution

From SB = tan θ = (h1 − h2)/L, which is Eq. 15.34, the drop in elevation per kilometer of channel is found to be h1 − h2 = SBL = 0.002(1 km)(1000 m/1 km ) = 2 m . The head loss is given by Eq. 15.48b as hL = h1 − h2, so the head loss is hL = 2 m. The power dissipation in a kilometer of channel is calculated by using Eq. 15.48c and found to be

P = ρg(h1 − h2 )Q = 1000 kg/m3(9.81 m/s2)(2 m)(0.75 m3/s)

= 1.47 × 104 (N-m)/s ≈ 15 kW

Next we calculate the average velocity as V = Q/A = (0.75 m3/s)/[(6 m)(0.5 m )] = 0.25 m/s, and Fr = V/$\sqrt{gy_N}= (0.25\ m/s)/\sqrt{(9.81\ m/s^2)(0.5\ m)}$ = 0.11. Note that this ﬂow is subcritical. The average wall shear stress is calculated by Eq. 15.49: $\bar \tau _W$ = ρgRHSB. Using RH = A/P = [(6 m)(0.5 m )]/(6 m + 1 m ) = 0.429 m, we ﬁnd

$\bar \tau _W$ = ρgRHSB = (1000 kg/m3)(9.81 m/s2)(0.429 m)(0.002) = 8.42 N/m2