Question

A pipe line has a diameter of \(5 \text{ cm}\) in which water flows at a pressure of \(10^5 \text{ N/m}^2\) with a mean velocity of \(2 \text{ m/s}\). If the head of the water at the cross section is \(5 \text{ m}\) above the datum line and consider \(g = 10 \text{ m/s}^2\), then the total head is

• A. \(15.2 \text{ m}\)
• B. \(5.2 \text{ m}\)
• C. \(18.2 \text{ m}\)
• D. \(21.2 \text{ m}\)

Hint:

Always ensure dimensional consistency in Bernoulli calculations: - Pressure unit must be in $\text{N/m}^2$ (Pascal). - Standard density of water $\rho_w = 1000\text{ kg/m}^3$. - Notice that the pipe diameter ($5\text{ cm}$) is extra information not needed to solve the problem, as the mean velocity is already explicitly provided.

Answer 1

The Correct Option is A

Concept: The total mechanical energy per unit weight of an incompressible, inviscid fluid flowing through a closed conduit is constant along a streamline according to Bernoulli's equation. This total energy per unit weight is expressed as the Total Head (\(H\)), which is the sum of three distinct energy heads: \[ H = \text{Pressure Head} + \text{Kinetic (Velocity) Head} + \text{Potential (Datum) Head} \] Mathematically, the equation is written as: \[ H = \frac{P}{\rho g} + \frac{v^2}{2g} + z \] Where:
• \(P\) = Static pressure of the fluid.
• \(\rho\) = Mass density of the fluid (for water, \(\rho = 1000 \text{ kg/m}^3\)).
• \(g\) = Acceleration due to gravity (given as \(10 \text{ m/s}^2\)).
• \(v\) = Mean flow velocity.
• \(z\) = Elevation height above a selected reference datum line.

Step 1: Calculating the Pressure Head (\(h_p = \frac{P}{\rho g}\)).
Given parameters:
• Pressure, \(P = 10^5 \text{ N/m}^2\)
• Density of water, \(\rho = 1000 \text{ kg/m}^3\)
• Gravity acceleration, \(g = 10 \text{ m/s}^2\) Substituting these values: \[ h_p = \frac{10^5}{1000 \times 10} = \frac{100,000}{10,000} = 10 \text{ m} \]

Step 2: Calculating the Velocity Head (\(h_v = \frac{v^2}{2g}\)).
Given parameter:
• Mean velocity, \(v = 2 \text{ m/s}\) Substituting these values: \[ h_v = \frac{2^2}{2 \times 10} = \frac{4}{20} = \frac{1}{5} = 0.2 \text{ m} \]

Step 3: Identifying the Datum Head (\(z\)).
The problem states that the cross-section is located at an elevation height of \(5 \text{ m}\) above the datum line: \[ z = 5 \text{ m} \]

Step 4: Summing the components to find the Total Head (\(H\)).
\[ H = h_p + h_v + z = 10 \text{ m} + 0.2 \text{ m} + 5 \text{ m} \] Adding the numbers step-by-step: \[ 10 + 0.2 = 10.2 \text{ m} \] \[ 10.2 + 5 = 15.2 \text{ m} \] The total mechanical head of the water at this cross-section is exactly \(15.2 \text{ m}\), which matches Option (1).