gen-define-the-continuity-equation-in-open-channels-and-give-one-c56b

Define the continuity equation in open channels and give one short worked example.

# Continuity Equation in Open Channels ## Definition The **continuity equation** (or equation of continuity) in open channels is based on the principle of conservation of mass. For steady flow, it states that the discharge (flow rate) remains constant along a channel, assuming no water is added or removed. **Mathematical form:** $$Q = A_1V_1 = A_2V_2 = \text{constant}$$ Where: - Q = discharge (m³/s or ft³/s) - A = cross-sectional area of flow (m² or ft²) - V = average velocity (m/s or ft/s) - Subscripts 1 and 2 refer to different sections along the channel --- ## Worked Example **Problem:** A rectangular channel is 3 m wide with water flowing at a depth of 1.2 m and velocity of 2 m/s. The channel narrows to 2 m wide. If the depth in the narrow section is 1.5 m, find the velocity in the narrow section. **Solution:** *Section 1 (wide section):* - Width: b₁ = 3 m - Depth: y₁ = 1.2 m - Velocity: V₁ = 2 m/s - Area: A₁ = b₁ × y₁ = 3 × 1.2 = 3.6 m² *Section 2 (narrow section):* - Width: b₂ = 2 m - Depth: y₂ = 1.5 m - Area: A₂ = b₂ × y₂ = 2 × 1.5 = 3.0 m² *Apply continuity equation:* $$A_1V_1 = A_2V_2$$ $$3.6 \times 2 = 3.0 \times V_2$$ $$V_2 = \frac{7.2}{3.0} = 2.4 \text{ m/s}$$ **Answer:** The velocity in the narrow section is **2.4 m/s**.