The discharge of water, through a retangular channel of width 6m is 18m^3/s when the depth is 2m. Calcullate:
(i) the specific energy of the flowing water,
(ii) critical depth,
(iii) critical velocity
(iv) minimum specific energy

To solve this problem, we need to use concepts from open channel flow, specifically related to **specific energy** and **critical flow** conditions.

### Given:
- Channel width \( b = 6 \, \text{m} \)
- Discharge \( Q = 18 \, \text{m}^3/\text{s} \)
- Depth of flow \( y = 2 \, \text{m} \)

### (i) **Specific Energy (E)**
The specific energy of flowing water in an open channel is the sum of the depth of water and the velocity head. It is given by the formula:

\[
E = y + \frac{v^2}{2g}
\]

Where:
- \( y \) is the depth of flow (2 m),
- \( v \) is the velocity of flow,
- \( g \) is the acceleration due to gravity (\(9.81 \, \text{m/s}^2\)).

First, we need to calculate the velocity \( v \) of the flow using the discharge formula:

\[
Q = A \times v
\]

Where:
- \( Q = 18 \, \text{m}^3/\text{s} \),
- \( A \) is the cross-sectional area of flow (\( A = b \times y \)).

\[
A = 6 \, \text{m} \times 2 \, \text{m} = 12 \, \text{m}^2
\]

Thus,

\[
v = \frac{Q}{A} = \frac{18}{12} = 1.5 \, \text{m/s}
\]

Now, we can calculate the specific energy:

\[
E = y + \frac{v^2}{2g} = 2 + \frac{(1.5)^2}{2 \times 9.81} = 2 + \frac{2.25}{19.62}
\]

\[
E = 2 + 0.115 = 2.115 \, \text{m}
\]

So, the specific energy is **2.115 m**.

### (ii) **Critical Depth (y_c)**
The critical depth in a rectangular channel is given by the formula:

\[
y_c = \left( \frac{Q^2}{g b^2} \right)^{\frac{1}{3}}
\]

Substituting the known values:

\[
y_c = \left( \frac{18^2}{9.81 \times 6^2} \right)^{\frac{1}{3}} = \left( \frac{324}{353.16} \right)^{\frac{1}{3}} = \left( 0.917 \right)^{\frac{1}{3}}
\]

\[
y_c = 0.97 \, \text{m}
\]

So, the critical depth is **0.97 m**.

### (iii) **Critical Velocity (v_c)**
The critical velocity is calculated using the relation:

\[
v_c = \frac{Q}{A_c}
\]

Where \( A_c = b \times y_c \) is the area of flow at critical depth.

\[
A_c = 6 \times 0.97 = 5.82 \, \text{m}^2
\]

Now,

\[
v_c = \frac{18}{5.82} = 3.09 \, \text{m/s}
\]

So, the critical velocity is **3.09 m/s**.

### (iv) **Minimum Specific Energy (E_{min})**
The minimum specific energy occurs at critical flow, and is given by:

\[
E_{min} = \frac{3}{2} \times y_c
\]

\[
E_{min} = \frac{3}{2} \times 0.97 = 1.455 \, \text{m}
\]

Thus, the minimum specific energy is **1.455 m**.

---

### Summary:
1. Specific energy of flowing water: **2.115 m**
2. Critical depth: **0.97 m**
3. Critical velocity: **3.09 m/s**
4. Minimum specific energy: **1.455 m**

Do you want details or have any questions?

Here are some related questions to further explore:
1. How does the specific energy change with varying discharge rates?
2. What happens to the velocity when the channel width increases?
3. How is critical depth affected by different channel shapes?
4. What is the significance of critical flow conditions in channel design?
5. How can we estimate the flow regime (subcritical or supercritical)?

**Tip:** Critical flow represents the condition where the flow velocity is equal to the wave speed, which plays a vital role in understanding transitions between subcritical and supercritical flow.

The discharge of water, through a rectangular channel of width 6m is 18m^3/s when the depth is 2m. Calculate: (i) the specific energy of the flowing water, (ii) critical depth, (iii) critical velocity, (iv) minimum specific energy

This problem involves calculating the specific energy, critical depth, and velocity for a rectangular channel with a width of 6m and a discharge of 18 m^3/s. Learn step-by-step how to apply key fluid mechanics formulas, including specific energy and critical flow conditions. Suitable for undergraduate engineering students.

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