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

Concept: Since the pipe is horizontal, Bernoulli’s equation for incompressible fluid flow is: \[ P+\frac12 \rho v^2+\rho gh=\text{constant} \] For a horizontal pipe: \[ h_1=h_2 \] Hence the potential energy terms cancel, giving: \[ P_1+\frac12\rho v_1^2 = P_2+\frac12\rho v_2^2 \] Pressure is given in terms of height of water column, so: \[ P=\rho gh \]

Step 1: Writing Bernoulli’s equation between the two points.
At the first point: \[ v_1=2\ \text{m/s} \] \[ h_1=2\ \text{m of water column} \] At the second point: \[ v_2=3\ \text{m/s} \] Let the pressure head at the second point be \(h_2\). Using Bernoulli’s equation in pressure-head form: \[ h_1+\frac{v_1^2}{2g} = h_2+\frac{v_2^2}{2g} \] Substitute the values: \[ 2+\frac{2^2}{2g} = h_2+\frac{3^2}{2g} \]

Step 2: Simplifying the equation.
Taking: \[ g=10\ \text{m/s}^2 \] we get: \[ 2+\frac{4}{20} = h_2+\frac{9}{20} \] \[ 2+0.2 = h_2+0.45 \] \[ 2.2=h_2+0.45 \] \[ h_2=2.2-0.45 \] \[ h_2=1.75\ \text{m} \]

Step 3: Final conclusion.
Therefore, the pressure at the second point is: \[ \boxed{1.75\ \text{m of water column}} \] Hence, the correct answer is: \[ \boxed{(A)\ 1.75\ \text{m of water}} \]