This question involves calculating the pressure at the lower end (A) of an inclined pipe, given the pressure at the upper end (B), the pipe's dimensions, and the flow rate of water. We can solve this using the principles of fluid mechanics, specifically Bernoulli's equation.

Understanding the Problem Setup

• Fluid: Water
• Flow Rate ($Q$): 0.5 m³/s (cumecs)
• Pipe Length ($L$): 12 m
• Pipe Orientation: Inclined, with end B above end A at an angle of 30° to the horizontal.
• Pressure at B ($P_B$): 20 kN/m²
• Goal: Find the pressure at A ($P_A$) in kN/m².

Applying Bernoulli's Equation

Bernoulli's equation relates pressure, velocity, and elevation between two points in a flowing fluid. For points A and B, it states:

$$P_A + \frac{1}{2}\rho v_A^2 + \rho g z_A = P_B + \frac{1}{2}\rho v_B^2 + \rho g z_B$$

Where:

• $P$ is the pressure
• $\rho$ is the fluid density
• $v$ is the fluid velocity
• $g$ is the acceleration due to gravity
• $z$ is the elevation

Simplifying the Equation

The problem states the pipe has a uniform cross-section. This means the cross-sectional area ($A$) is constant throughout the pipe. Since the flow rate ($Q = A \times v$) is also constant, the velocity ($v$) must be the same at both ends ($v_A = v_B = v$). Therefore, the kinetic energy terms ($\frac{1}{2}\rho v^2$) cancel out:

$$P_A + \rho g z_A = P_B + \rho g z_B$$

Rearranging to solve for $P_A$:

$$P_A = P_B + \rho g (z_B - z_A)$$

Calculating Elevation Difference

We need to find the difference in elevation ($z_B - z_A$) between points B and A. Let's set the datum at point A, so $z_A = 0$. Point B is above A, and the pipe length is 12 m at an angle of 30° to the horizontal. The vertical height difference ($h = z_B - z_A$) is:

$$h = L \sin(\theta)$$

Substituting the values:

$$h = 12 \text{ m} \times \sin(30^\circ)$$

Since $\sin(30^\circ) = 0.5$:

$$h = 12 \text{ m} \times 0.5 = 6 \text{ m}$$

So, the elevation difference $h$ is 6 meters.

Calculating Pressure at A

Now, we can plug the values into the simplified Bernoulli's equation. We'll use the standard density for water, $\rho = 1000$ kg/m³, and $g = 9.81$ m/s².

First, calculate the pressure difference due to the elevation head ($\rho g h$):

$$\rho g h = (1000 \text{ kg/m³}) \times (9.81 \text{ m/s²}) \times (6 \text{ m})$$ $$\rho g h = 58860 \text{ N/m²}$$

Convert this to kN/m² by dividing by 1000:

$$\rho g h = \frac{58860}{1000} \text{ kN/m²} = 58.86 \text{ kN/m²}$$

Now, calculate $P_A$:

$$P_A = P_B + \rho g h$$ $$P_A = 20 \text{ kN/m²} + 58.86 \text{ kN/m²}$$ $$P_A = 78.86 \text{ kN/m²}$$

Final Answer Check

The calculated pressure at end A is approximately 78.86 kN/m², which rounds to 78.9 kN/m². This matches one of the given options.