Question

Calculate the pump head if suction pressure is 4 bar and discharge pressure is 6 bar. Consider density of water as 997 kg/m$^3$

• A. 70 m
• B. 20.45 m
• C. 25 m
• D. 67 m

Hint:

Ensure all units are converted to SI before using the formula: $1 \text{ bar} = 10^5 \text{ Pa}$. A quick estimation is $H \approx \frac{\Delta P \text{ (in bar)} \times 10.2}{\text{Specific gravity}} \approx 2 \times 10.2 = 20.4$ m.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
A pump increases the energy of a fluid to overcome flow resistance and elevate the fluid.
This energy increase is measured as the pump head, which represents the height of a fluid column corresponding to the pressure difference developed by the pump.
The pump head is calculated from the net pressure difference between the suction and discharge sides.
Key Formula or Approach:
The equation relating pressure difference ($\Delta P$) to head ($H$) is:
\[ H = \frac{\Delta P}{\rho \cdot g} \]
Where:
$\Delta P = P_{\text{discharge}} - P_{\text{suction}}$ is the pressure difference (in Pascals).
$\rho$ is the density of the fluid (in kg/m$^3$).
$g$ is the acceleration due to gravity (approximately $9.81 \text{ m/s}^2$).

Step 2: Detailed Explanation:
First, we find the net pressure difference developed by the pump:
The discharge pressure is:
\[ P_{\text{discharge}} = 6 \text{ bar} \]
The suction pressure is:
\[ P_{\text{suction}} = 4 \text{ bar} \]
Therefore, the pressure difference ($\Delta P$) is:
\[ \Delta P = 6 - 4 = 2 \text{ bar} \]
We convert this pressure difference from bar to Pascals (SI units), where $1 \text{ bar} = 10^5 \text{ N/m}^2$:
\[ \Delta P = 2 \times 10^5 \text{ N/m}^2 \]
The density of water is given as:
\[ \rho = 997 \text{ kg/m}^3 \]
The acceleration due to gravity is:
\[ g = 9.81 \text{ m/s}^2 \]
Substituting these values into the head formula:
\[ H = \frac{2 \times 10^5}{997 \cdot 9.81} \]
First, calculate the denominator:
\[ 997 \cdot 9.81 = 9780.57 \]
Now, solve for the head ($H$):
\[ H = \frac{200000}{9780.57} \]
\[ H \approx 20.448 \text{ m} \]
Rounding to two decimal places yields 20.45 meters.
Therefore, the pump head is 20.45 m.

Step 3: Final Answer
The pump head is 20.45 m.