A reciprocating pump has two double-acting cylinders each \(200 \mathrm{~mm}\) bore and \(450 \mathrm{~mm}\) stroke, the cranks being at \(90^{\circ}\) to each other and rotating at \(2.09 \mathrm{rad} \cdot \mathrm{s}^{-1}\) (20 rev/min). The delivery pipe is \(100 \mathrm{~mm}\) diameter, \(60 \mathrm{~m}\) long. There are no air vessels. Assuming simple harmonic motion for the pistons determine the maximum and mean water velocities in the delivery pipe and the inertia pressure in the delivery pipe near the cylinders at the instant of minimum water velocity in the pipe.

In a reciprocating pump, the displacement volume is crucial for understanding how much fluid is moved per stroke. The displacement volume depends on the bore (diameter) and stroke (distance traveled) of the cylinder. For a double-acting cylinder, each stroke fills and discharges fluid from both sides, so we consider half the stroke in the calculation. The formula used is:
\[ V = \frac{\text{Bore Area} \times \text{Stroke}}{2} \]
To find the bore area, use:
\[ A = \frac{\left( \frac{\text{bore}}{2} \right)^2 \times \pi}{2} \]
Then multiply by the stroke to find the displacement volume. In our example, with a bore of 200 mm and a stroke of 450 mm, the displacement volume per stroke becomes 0.007065 m³ for one cylinder.

The flow rate of the pump is the volume of fluid displaced over a certain period. With two double-acting cylinders, each contributing volume from both ends, the total flow rate can be calculated using:
\[ \text{Flow Rate} = 2 \times 2 \times V \times \text{pump speed} \]
Here, V is the displacement volume, and the pump speed in rad/s converts the revolutions per minute into a useful factor. For the given example with a displacement volume of 0.007065 m³ and a pump speed of 2.09 rad/s, the flow rate comes out to be 0.0590 m³/s. This tells us how much water is pumped through the system every second.

The mean water velocity in the delivery pipe helps us understand how fast the water is moving on average. This is calculated by dividing the flow rate by the cross-sectional area of the delivery pipe:
\[ \text{Mean Velocity} = \frac{\text{Flow Rate}}{\text{Cross-Sectional Area of the Pipe}} \]
First, find the pipe's cross-sectional area with:
\[ A_{\text{pipe}} = \frac{\left( \frac{\text{diameter}}{2} \right)^2 \times \pi}{2} \]
With a pipe diameter of 100 mm, we get an area of 0.00785 m². Now, using the calculated flow rate of 0.0590 m³/s, the mean velocity is 7.52 m/s. This mean velocity is a key factor in designing efficient piping systems.

In conditions of simple harmonic motion, the pistons create a situation where water velocity periodically reaches a maximum value. This peak water velocity, which we call maximum water velocity, can be calculated using the mean velocity multiplied by a factor of \( \frac{\pi}{2} \). Here's the formula:
\[ \text{Maximum Velocity} = \text{Mean Velocity} \times \frac{\pi}{2} \]
Plugging in the mean velocity of 7.52 m/s, we get:
\[ \text{Maximum Velocity} = 7.52 \, m/s \times \frac{\pi}{2} = 11.81 \, m/s \]
This value represents the highest speed that water can reach inside the delivery pipe, which is essential for pressure-related calculations and for avoiding surges.

Inertia pressure in the delivery pipe arises due to the accelerating and decelerating water masses. This pressure depends on factors like water density, pipe length, angular speed, and stroke length. The formula to calculate inertia pressure is:
\[ \text{Inertia Pressure} = \text{Density} \times \text{Length} \times \text{Angular Velocity}^2 \times \frac{\text{Stroke}}{2} \]
For our example, with water density assumed to be 1000 kg/m³, pipe length of 60 m, angular velocity of 2.09 rad/s, and stroke of 450 mm, we get:
\[ \text{Inertia Pressure} = 1000 \, kg/m^3 \times 60 \, m \times (2.09 \, rad/s)^2 \times \frac{0.45 \, m}{2} = 118243 \, Pa \]
Understanding inertia pressure is important for designing robust pumping systems that can handle the additional stresses due to accelerating fluids.