Question

Water flows through a horizontal pipe. At one point, the velocity is \(2\ \text{m/s}\) and the pressure is \(2\ \text{m}\) of water column. What is the pressure at another point where the velocity is \(3\ \text{m/s}\)?
(Density of water \(=10^3\ \text{kg/m}^3\))

• A. \(1.75\ \text{m}\) of water
• B. \(1.2\ \text{m}\) of water
• C. \(1\ \text{m}\) of water
• D. \(1.5\ \text{m}\) of water

Hint:

For horizontal fluid flow problems, Bernoulli’s equation simplifies to: \[ P+\frac12\rho v^2=\text{constant} \] Thus, wherever the fluid velocity increases, the pressure decreases.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Bernoulli's Principle states that for an incompressible, non-viscous fluid in steady flow, the sum of pressure energy, kinetic energy, and potential energy per unit volume is constant along a streamline.
For a horizontal pipe, the potential energy (height) remains constant, so the equation simplifies to the relationship between static pressure and velocity (dynamic pressure).
As the velocity of the fluid increases, the static pressure must decrease to keep the total energy constant.
Step 2: Key Formula or Approach:
Bernoulli's Equation for horizontal flow:
\[ P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2 \]
In this problem, pressure is given in "meters of water column" (\(h\)).
Recall that \(P = \rho gh\). Substituting this into the Bernoulli equation:
\[ \rho gh_1 + \frac{1}{2}\rho v_1^2 = \rho gh_2 + \frac{1}{2}\rho v_2^2 \]
We can simplify this by dividing the entire equation by \(\rho g\):
\[ h_1 + \frac{v_1^2}{2g} = h_2 + \frac{v_2^2}{2g} \]
This is known as the "total head" equation, where \(h\) is the pressure head and \(v^2/2g\) is the velocity head.
Step 3: Detailed Explanation:
Given values:
\(h_1 = 2\) m
\(v_1 = 2\) m/s
\(v_2 = 3\) m/s
Assume \(g = 10\) m/s\(^2\) (common for CET exams unless 9.8 is specified).
Substitute these values into the 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:
\[ h_2 = 2.2 - 0.45 = 1.75 \text{ m} \]
Since the velocity increased from 2 m/s to 3 m/s, the pressure head decreased from 2 m to 1.75 m. This makes sense physically according to Bernoulli's principle.
Step 4: Final Answer:
The pressure at the second point is 1.75 m of water column.
Hence, the correct option is (A).