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

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

answer:

a

explanation:

answer:

(b)

explanation: