A volume of 0.5m3 h^-1 of water is being pumped at a velocity of 1.1ms^-1 from the bottom of a header tank, 3m deep, down three floors (a total fall of 10m from the bottom of the header tank) into the top of a water pressure tank which is maintained at a pressure of 600kPa above atmospheric. Estimate the theoretical pump power required, ignoring pipe friction

To estimate the theoretical pump power required, we can use the Bernoulli equation and the power equation for a pump. Here's the process, step-by-step:

### 1. Bernoulli's Equation (Energy Balance)

Bernoulli's equation for a fluid moving from one point to another is:

\[
P_1 + \frac{1}{2} \rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho g h_2
\]

Where:
- \( P_1 \) and \( P_2 \) are the pressures at the two points.
- \( v_1 \) and \( v_2 \) are the velocities at the two points.
- \( \rho \) is the density of water (\( \approx 1000 \, \text{kg/m}^3 \)).
- \( g \) is the acceleration due to gravity (\( 9.81 \, \text{m/s}^2 \)).
- \( h_1 \) and \( h_2 \) are the heights at the two points.

We'll assume the velocity at both points is the same (\( v_1 = v_2 = 1.1 \, \text{m/s} \)) and ignore pipe friction. Also, we need to take into account the pressure difference between the two points.

### 2. Points of Interest

- **Point 1**: Bottom of the header tank.
  - Height \( h_1 = 0 \, \text{m} \) (reference point).
  - Pressure \( P_1 \) is atmospheric.
  - Velocity \( v_1 = 1.1 \, \text{m/s} \).

- **Point 2**: Top of the pressure tank.
  - Height \( h_2 = -10 \, \text{m} \) (below the reference point).
  - Pressure \( P_2 = P_{\text{atm}} + 600 \, \text{kPa} = P_{\text{atm}} + 600,000 \, \text{Pa} \).
  - Velocity \( v_2 = 1.1 \, \text{m/s} \).

### 3. Bernoulli's Equation Simplified

We can now rearrange Bernoulli’s equation to solve for the pressure difference and the pump head (ignoring friction):

\[
0 + \frac{1}{2} \rho v_1^2 + \rho g h_1 = (P_{\text{atm}} + 600,000) + \frac{1}{2} \rho v_2^2 + \rho g h_2
\]

Cancel out \( P_{\text{atm}} \) and \( v_1 = v_2 \). This simplifies to:

\[
0 = 600,000 + \rho g h_2
\]

Since \( h_2 = -10 \, \text{m} \), we get:

\[
0 = 600,000 - 1000 \cdot 9.81 \cdot 10
\]

\[
0 = 600,000 - 98,100
\]

Thus, the required pressure difference \( \Delta P \) is:

\[
\Delta P = 600,000 - 98,100 = 501,900 \, \text{Pa}
\]

### 4. Power Calculation

The theoretical pump power \( P_{\text{pump}} \) is given by:

\[
P_{\text{pump}} = \frac{\Delta P \cdot Q}{\eta}
\]

Where:
- \( \Delta P = 501,900 \, \text{Pa} \) (pressure difference).
- \( Q = 0.5 \, \text{m}^3/\text{h} = \frac{0.5}{3600} \, \text{m}^3/\text{s} = 1.39 \times 10^{-4} \, \text{m}^3/\text{s} \) (volumetric flow rate).
- \( \eta \) is the efficiency (which we'll assume to be 1 for theoretical calculation).

Substitute the values:

\[
P_{\text{pump}} = \frac{501,900 \times 1.39 \times 10^{-4}}{1}
\]

\[
P_{\text{pump}} \approx 69.7 \, \text{W}
\]

Thus, the theoretical pump power required is approximately **69.7 watts**.

---

Would you like more details or have any questions about this calculation? Here are some related questions:

1. How does pipe friction affect the pump power calculation?
2. What happens if the velocity of the water changes significantly?
3. How does pump efficiency influence real-world power consumption?
4. What units are typically used for measuring pump power?
5. How is the head loss due to fittings and bends in the pipe accounted for?

**Tip:** When designing a pump system, always account for both pipe friction and changes in height (head) for a more accurate estimation of power.

A volume of 0.5m^3/h of water is being pumped at a velocity of 1.1ms^-1 from the bottom of a header tank, 3m deep, down three floors (a total fall of 10m from the bottom of the header tank) into the top of a water pressure tank which is maintained at a pressure of 600kPa above atmospheric. Estimate the theoretical pump power required, ignoring pipe friction.

Learn how to estimate the theoretical pump power required to pump water at 0.5m^3/h with a fall of 10m using Bernoulli’s Equation. This fluid mechanics problem includes calculations of pressure difference and power using the Bernoulli principle and power equation. Suitable for undergraduate students.

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