4.14 The physical head of an impoundment power plant is 100 m. When the penstock is fully closed, the pressure at the outtake end is 106 N/m2. When the penstock is fully open, the pressure is reduced to 9 × 105 N/m2. Compute the effective head.

The effective head of a hydropower plant refers to the actual height of water that is converted into usable energy. To compute the effective head, we need to consider the pressure difference when the penstock is fully closed and fully open. 

The relationship between pressure and head can be expressed using the following equation from fluid mechanics:

\[
H = \frac{P}{\rho g}
\]

Where:
- \( H \) is the head (in meters),
- \( P \) is the pressure (in pascals, N/m²),
- \( \rho \) is the density of water (\(1000 \, \text{kg/m}^3\)),
- \( g \) is the gravitational acceleration (\(9.81 \, \text{m/s}^2\)).

### Step 1: Compute the initial head when the penstock is closed
Using the pressure when the penstock is fully closed, \( P_{\text{closed}} = 106 \, \text{N/m}^2 \):

\[
H_{\text{closed}} = \frac{P_{\text{closed}}}{\rho g} = \frac{106}{1000 \times 9.81}
\]

### Step 2: Compute the head when the penstock is open
Using the pressure when the penstock is fully open, \( P_{\text{open}} = 9 \times 10^5 \, \text{N/m}^2 \):

\[
H_{\text{open}} = \frac{P_{\text{open}}}{\rho g} = \frac{9 \times 10^5}{1000 \times 9.81}
\]

### Step 3: Compute the effective head
The effective head is the difference between the physical head (100 m) and the head corresponding to the pressure when the penstock is fully open:

\[
H_{\text{effective}} = 100 - H_{\text{open}}
\]

Let's now compute the values.### Results:

- The head when the penstock is fully closed is approximately \( 0.011 \, \text{m} \).
- The head when the penstock is fully open is approximately \( 91.74 \, \text{m} \).
- The effective head is approximately \( 8.26 \, \text{m} \).

This effective head represents the usable portion of the water's height that contributes to energy generation.

Would you like further clarification or details?

Here are 5 related questions to expand your understanding:
1. What is the significance of the effective head in hydropower generation?
2. How does penstock design affect the pressure and flow in a hydropower plant?
3. How is the efficiency of a hydropower plant influenced by the physical head?
4. What role does water density play in computing the head in hydropower systems?
5. How can we increase the effective head in a given hydropower setup?

**Tip:** Regularly monitor penstock conditions, as improper functioning can reduce the effective head and decrease energy output.

The physical head of an impoundment power plant is 100 m. When the penstock is fully closed, the pressure at the outtake end is 106 N/m². When the penstock is fully open, the pressure is reduced to 9 × 10^5 N/m². Compute the effective head.

This detailed solution explains how to compute the effective head of a hydropower plant using fluid mechanics principles. Given the physical head of 100 m and pressure data (106 N/m² when closed, 9 × 10^5 N/m² when open), we calculate the effective head of approximately 8.26 m. Suitable for undergraduate engineering students.

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