Question

The reading of pressure metre attached with a closed pipe is \( 4.5 \times 10^4 \, N/m^2 \). On opening the valve, water starts flowing and the reading of pressure metre falls to \( 2.0 \times 10^4 \, N/m^2 \). The velocity of water is found to be \( \sqrt{V} \, m/s \). The value of \( V \) is _____.

Answer 1

Correct Answer: 50

Bernoulli's equation for fluid flow is applied to address this problem. The principle dictates that in a steady flow, the aggregate of all mechanical energy forms within a fluid along a streamline remains constant. The equation, expressed in terms of pressure \( P \), fluid density \( \rho \), velocity \( v \), and gravitational potential energy, is as follows:

\( P_1 + \frac{1}{2}\rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho gh_2 \)

For this specific scenario, with the valve closed, the water is stationary, meaning \( v_1 = 0 \). Furthermore, assuming that the heights \( h_1 \) and \( h_2 \) are identical, they cancel out. Consequently, the equation simplifies to:

\( P_1 = P_2 + \frac{1}{2}\rho v_2^2 \)

The given values are:

• \( P_1 = 4.5 \times 10^4 \, N/m^2 \)
• \( P_2 = 2.0 \times 10^4 \, N/m^2 \)

Substituting these values into the simplified equation yields:

\( 4.5 \times 10^4 = 2.0 \times 10^4 + \frac{1}{2}\rho v_2^2 \)

Rearranging to solve for \( v_2^2 \):

\( \frac{1}{2}\rho v_2^2 = 4.5 \times 10^4 - 2.0 \times 10^4 = 2.5 \times 10^4 \)

Utilizing the density of water, \( \rho = 1000 \, kg/m^3 \), the equation becomes:

\( \frac{1}{2} \times 1000 \times v_2^2 = 2.5 \times 10^4 \)

Simplification results in:

\( 500 v_2^2 = 2.5 \times 10^4 \)

\( v_2^2 = \frac{2.5 \times 10^4}{500} = 50 \)

Therefore, the velocity \( v_2 \) is \( \sqrt{50} \, m/s \). The value of \( V \) is 50, which falls within the expected range (50, 50).