Step 1: Understanding the Concept:
For a non-viscous, incompressible fluid in steady flow through a horizontal pipe, Bernoulli’s Principle applies.
Since the pipe is horizontal, the potential energy per unit volume is constant throughout.
Therefore, the sum of pressure energy and kinetic energy per unit volume remains constant.
Pressure is expressed here in terms of "pressure head," which is the height of a liquid column that would exert that pressure.
Key Formula or Approach:
Bernoulli's Equation for horizontal flow:
\[ P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2 \]
To convert this to "head" (units of meters), divide by \(\rho g\):
\[ \frac{P_1}{\rho g} + \frac{v_1^2}{2g} = \frac{P_2}{\rho g} + \frac{v_2^2}{2g} \]
Let \(h = \frac{P}{\rho g}\) be the pressure head.
\[ h_1 + \frac{v_1^2}{2g} = h_2 + \frac{v_2^2}{2g} \]
Step 2: Detailed Explanation:
Given:
Initial Pressure Head \(h_1 = 2\text{ m}\).
Initial Velocity \(v_1 = 2\text{ m/s}\).
Final Velocity \(v_2 = 3\text{ m/s}\).
Acceleration due to gravity \(g \approx 10\text{ m/s}^2\) (standard approximation for CET).
Substitute the values into the Bernoulli head equation:
\[ 2 + \frac{2^2}{2 \times 10} = h_2 + \frac{3^2}{2 \times 10} \]
\[ 2 + \frac{4}{20} = h_2 + \frac{9}{20} \]
\[ 2 + 0.2 = h_2 + 0.45 \]
\[ 2.2 = h_2 + 0.45 \]
Subtracting 0.45 from both sides:
\[ h_2 = 2.2 - 0.45 = 1.75\text{ m} \]
This result follows the principle that as fluid velocity increases (kinetic energy increases), the static pressure must decrease to conserve energy.
Step 3: Final Answer:
The pressure at the second point is \(1.75\text{ m}\) of water column.