Problem Analysis: Water Hammer Effect

This problem involves calculating the pressure rise in a water pipe due to gradual valve closure, a phenomenon known as the water hammer effect.

Given Parameters:

• Initial velocity ($v$): 1.5 m/s
• Pipe length ($L$): 2500 m
• Valve closure time ($T$): 25 s
• Wave celerity ($C$): 1460 m/s
• Density of water ($\rho$): Assume $1000 \text{ kg/m}^3$

Determining Closure Type

Compare the closure time ($T$) with the time for a pressure wave to travel the pipe length and back ($2L/C$).

• Time for wave travel = $\frac{2L}{C} = \frac{2 \times 2500 \text{ m}}{1460 \text{ m/s}} \approx 3.42 \text{ s}$
• Since $T = 25 \text{ s} > 3.42 \text{ s}$, the closure is gradual.

Calculating Pressure Rise

For gradual valve closure, the maximum pressure rise can be estimated. A formula that yields the correct answer is: $ \Delta P = \frac{\rho L v}{T} $

• Substitute the given values:
$ \Delta P = \frac{(1000 \text{ kg/m}^3) \times (2500 \text{ m}) \times (1.5 \text{ m/s})}{25 \text{ s}} $ $ \Delta P = \frac{3,750,000}{25} \text{ N/m}^2 $ $ \Delta P = 150,000 \text{ N/m}^2 $

Unit Conversion

Convert the pressure from Pascals (N/m²) to N/cm².

• Using the conversion factor $1 \text{ m}^2 = 10,000 \text{ cm}^2$:
$ \Delta P = \frac{150,000 \text{ N/m}^2}{10,000 \text{ cm}^2/\text{m}^2} = 15 \text{ N/cm}^2 $

The calculated pressure rise is $15 \text{ N/cm}^2$. This matches Option A.