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la altura entre la bomba y la piscina es de 50 metros, además necesito que la perdida de carga, es decir hf se utilice la ecuacion empirica. el diámetro de la tuberia se considera como 150mm

To solve this problem, we need to determine the total head loss (\(hf\)) and the power required by the pump in HP.

### Given Data:
1. **Height between the pump and pool**: 50 meters.
2. **Pipe diameter (\(D\))**: 150 mm = 0.15 m.
3. **Flow rate**: 70 m³ needs to be delivered in 30 minutes, which implies:
   \[
   Q = \frac{70 \text{ m}^3}{30 \times 60 \text{ seconds}} = \frac{70}{1800} \text{ m}^3/\text{s} = 0.03889 \text{ m}^3/\text{s}
   \]
4. **Reynolds number (\(Re\))**: Approximately \(500,000\).
5. **Relative roughness (\(\frac{e}{D}\))**: \(0.001\) as calculated from \(e = 0.15 \text{ mm} = 0.00015 \text{ m}\).

### To Do:
1. **Calculate the Head Loss (\(hf\)) using an empirical equation (Darcy-Weisbach equation)**:
   \[
   hf = f \cdot \frac{L}{D} \cdot \frac{V^2}{2g}
   \]
   where:
   - \(f\) is the friction factor.
   - \(L\) is the length of the pipe.
   - \(V\) is the velocity of water.
   - \(g\) is the acceleration due to gravity (\(9.81 \text{ m/s}^2\)).

2. **Determine the friction factor \(f\)** using the Moody chart or the Colebrook equation:
   \[
   \frac{1}{\sqrt{f}} = -2 \log \left(\frac{\frac{e}{D}}{3.7} + \frac{2.51}{Re \sqrt{f}}\right)
   \]
   Since \(Re = 500,000\) and \(\frac{e}{D} = 0.001\).

3. **Calculate Power required by the pump**:
   \[
   \text{Power} = \rho \cdot g \cdot Q \cdot H
   \]
   where \(H\) includes both static head and head loss due to friction.

Let's start by calculating the velocity \(V\) in the pipe: 

### Velocity Calculation:
\[
V = \frac{Q}{A} = \frac{0.03889 \text{ m}^3/\text{s}}{\pi \left(\frac{0.15}{2}\right)^2} = \frac{0.03889}{0.01767} = 2.20 \text{ m/s}
\]

Now, let's find the friction factor \(f\) using the given Reynolds number and relative roughness. We can then proceed to calculate the head loss (\(hf\)) and the required power.Here's the calculation summary:

1. **Velocity of Water in the Pipe (\(V\))**: \(2.20 \, \text{m/s}\).
2. **Friction Factor (\(f\))**: \(0.0202\) (using the Colebrook equation).
3. **Head Loss (\(hf\))**: \(3.33 \, \text{m}\).

Now, let's calculate the required power for the pump, considering both the static head (50 meters) and the head loss (\(hf\)).

### Total Head (\(H\)):
\[
H = 50 \, \text{m} + 3.33 \, \text{m} = 53.33 \, \text{m}
\]

### Power Required by the Pump:
\[
\text{Power} = \rho \cdot g \cdot Q \cdot H
\]
Assuming water density (\(\rho\)) is \(1000 \, \text{kg/m}^3\), let's calculate the power required by the pump in horsepower (HP).The required power for the pump is approximately **27.28 HP**.

### Summary:
- **Total Head (\(H\))**: 53.33 m
- **Power Required**: 27.28 HP

Would you like more details or have any questions?

### Related Questions:
1. How would the power requirement change if the pipe diameter were different?
2. How does the friction factor affect the overall pump power requirement?
3. What impact does changing the height between the pump and pool have on power requirements?
4. How would varying flow rates influence the head loss and required pump power?
5. What are other methods to reduce head loss in the system?

**Tip:** Regular maintenance and proper sizing of pipes can significantly reduce the friction factor, leading to energy savings and improved pump efficiency.

Learn how to calculate total head loss (hf) and the power required by the pump for a 50-meter height difference between a pump and pool. Includes detailed step-by-step solutions using the Darcy-Weisbach equation and Colebrook equation. Suitable for engineers and professionals in fluid mechanics.

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