Water (density 1000 kg/m³) flows through an inclined pipe of uniform diameter. The velocity, pressure and elevation at section A are V_A = 3.2 m/s, p_A =186 kPa and z_A = 24.5 m respectively, and those at section B are V_B = 3.2 m/s, p_B = 260 kPa and z_B = 9.1 m, respectively. If acceleration due to gravity is 10 m/s² then the head lost due to friction is _________ m (round off to one decimal place).

The correct answer is 8.

Step-by-Step Explanation:

1. Understanding the System and Analyzing the Diagram:
Based on the provided schematic diagram, water flows through an inclined pipe of uniform diameter. The diagram labels the two sections of interest as section A (at a higher elevation) and section B (at a lower elevation). The vertical elevations from the reference datum to these sections are represented as z_A and z_B respectively. The density of water is represented as ρ_w.

To find the head lost due to friction between these two sections, we apply Bernoulli's equation, which states that the total head (total energy per unit weight) remains constant along a streamline in an ideal, frictionless flow. For a real fluid flow, the total head decreases in the direction of flow due to friction, and the loss of head is given by:

$H_{\text{loss}}=H_{\text{inlet}}-H_{\text{outlet}}$
where the total head at any section is the sum of the pressure head, velocity head, and datum (elevation) head:

$H=\frac{p}{\rho g}+\frac{V^2}{2g}+z$

2. Calculate Total Head at Section A:
Given parameters at section A:
- Pressure, p_A = 186 kPa = 186,000 Pa
- Velocity, V_A = 3.2 m/s
- Elevation, z_A = 24.5 m
- Density of water, ρ = 1000 kg/m³
- Acceleration due to gravity, g = 10 m/s²

Substituting these values into the head equation for section A:

$H_A=\frac{186\times {10}^3}{1000\times 10}+\frac{{3.2}^2}{2\times 10}+24.5$

$H_A=18.6+0.512+24.5=43.612\text{m}$

3. Calculate Total Head at Section B:
Given parameters at section B:
- Pressure, p_B = 260 kPa = 260,000 Pa
- Velocity, V_B = 3.2 m/s (note that velocity remains the same as the pipe has a uniform diameter)
- Elevation, z_B = 9.1 m

Substituting these values into the head equation for section B:

$H_B=\frac{260\times {10}^3}{1000\times 10}+\frac{{3.2}^2}{2\times 10}+9.1$

$H_B=26+0.512+9.1=35.612\text{m}$

4. Determine Flow Direction and Head Loss:
Since H_A (43.612 m) is greater than H_B (35.612 m), flow occurs from section A to section B.
The head lost due to friction (h_f) is the difference in total head between the inlet (section A) and the outlet (section B):

$h_f=H_A-H_B$

$h_f=43.612-35.612=8.0\text{m}$
Therefore, the head lost due to friction is exactly 8 m.