Question

Water flows at 8.4 m3/s in a trapezoidal channel with a bottom width of 2 m and side slopes of 2:1 (H:V). Over a distance of 100 m, the bottom width expands to 2.5 m, with the side slopes remaining constant at 2:1. If the depth of flow at both of these sections is 1 m and the channel slope is 0.001, calculate the head loss between the sections. What is the power in kilowatts that is dissipated

Answer 1

the head loss between the sections is 0.147m

The power in kilowatts that is dissipated 12.1 kW

Explanation:

First we find the area of the trapezoidal channel at section 1 A₁ using the expression

A₁ = b₁y₁ + m₁y₁²

where b₁ is width of channel at section 1 (2m), y₁ is depth of flow at section 1 (1m), m₁ is side slope of the channel at section 1 (2).

so we substitute in the expression

A₁ = ( 2 × 1 ) + ( 2 × 1²) = 4m²

next we find the area of the trapezoidal channel at section 1 A₂ using the expression.

A₂ = b₂y₂ + m₂y₂²

where b₂ is width of channel at section 2 (2.5m), y₂ is depth of flow at section 2 (1m), m₂ is side slope of the channel at section 2 (2).

so we substitute in the expression

A₂ = ( 2.5 × 1 ) + ( 2 × 1²) = 4.5m²

Next we determine the velocity at section 1 ( V₁) using the expression

V₁= Q/A₁

where Q is the flow rate of the channel (8.4m³/s)

so we substitute

V₁ = 8.4 / 4 = 2.10 m/s

the velocity at section 2 ( V₂) using the expression

V₂= Q/A₂

where Q is the flow rate of the channel (8.4m³/s)

we substitute

V₂ = 8.4 / 4.5 = 1.87 m/s

Now we find the friction head loss per unit length ( Sf) using this expression;

y₁ + v₁²/2g - y₂ - v₂²/2g = L ( Sf - S₀ )

where g is acceleration due to gravity ( 9.81 m/s), L is distance between the inflow and outflow sections of the control volume ( 100m), S₀ is the longitudinal slope of the section( 0.001)

now we substitute

1 + (2.10²/(2×9.81)) - 1 - (1.87²/(2×9.81)) = 100( Sf - 0.001 )

100Sf - 0.1 = 0.0465

100Sf = 0.1465

Sf = 0.00147

Now to calculate the head loss between the sections hL, we say

hL = L × Sf

remember oue L = 100m and Sf = 0.00147

so we substitute

hL = 100 × 0.00147

hL = 0.147m

∴ the head loss between the sections is 0.147m

To find the power in kilowatts that is dissipated we say;

P = YωQhL

where Yω is the specific weight of water ( at 20°C = 9.79kN/m³

so P = 9.79 × 8.4 × 0.147

P = 12.0887 ≈ 12.1 kW

∴ The power in kilowatts that is dissipated 12.1 kW