A single-acting reciprocating pump has a cylinder of a diameter 150 mm and of stroke length 300 mm. The centre of the pump is 4 m above the water surface in the sump. The atmospheric pressure head is 10.3 m of water and pump is running at 40 r.p.m. If the length and diameter of the suction pipe are 5 m and 10 cm respectively, determine the pressure head due to acceleration in the cylinder : (i) At the beginning of the suction stroke, and (ii) In the middle of suction stroke.

Reciprocating Pump Acceleration Head Calculation

Problem Statement

Given Data & Constants

Cylinder diameter, $D = 150 \, \text{mm} = 0.15 \, \text{m}$
Stroke length, $L = 300 \, \text{mm} = 0.3 \, \text{m}$
Speed, $N = 40 \, \text{r.p.m.}$
Suction pipe length, $l_s = 5 \, \text{m}$
Suction pipe diameter, $d_s = 10 \, \text{cm} = 0.1 \, \text{m}$
Acceleration due to gravity, $g = 9.81 \, \text{m/s}^2$

Solution

1. Calculate Required Parameters

First, we calculate the crank radius, angular velocity, and the areas of the cylinder and suction pipe.

$$ \text{Crank radius, } r = \frac{L}{2} = \frac{0.3}{2} = 0.15 \, \text{m} $$ $$ \text{Angular velocity, } \omega = \frac{2 \pi N}{60} = \frac{2 \pi \times 40}{60} \approx 4.189 \, \text{rad/s} $$ $$ \text{Cylinder area, } A = \frac{\pi}{4} D^2 = \frac{\pi}{4} (0.15)^2 \approx 0.01767 \, \text{m}^2 $$ $$ \text{Suction pipe area, } a_s = \frac{\pi}{4} d_s^2 = \frac{\pi}{4} (0.1)^2 \approx 0.007854 \, \text{m}^2 $$

2. Formula for Acceleration Head

The pressure head due to acceleration ($h_a$) in the suction pipe is given by the formula:

$$ h_{as} = \frac{l_s}{g} \times \frac{A}{a_s} \times \omega^2 r \cos(\theta) $$ Where $\theta$ is the crank angle from the inner dead centre.

(i) Pressure Head at the Beginning of Suction Stroke

At the beginning of the suction stroke, the piston is at the inner dead centre, so the crank angle $\theta = 0^\circ$. At this point, acceleration is maximum.

$$ \cos(0^\circ) = 1 $$ $$ h_{as} = \frac{5}{9.81} \times \frac{0.01767}{0.007854} \times (4.189)^2 \times 0.15 \times 1 $$ $$ h_{as} = 0.5097 \times 2.25 \times 17.548 \times 0.15 $$ $$ h_{as} \approx 3.016 \, \text{m} $$

(ii) Pressure Head in the Middle of Suction Stroke

In the middle of the suction stroke, the piston is at its maximum velocity, and its acceleration is momentarily zero. The crank angle $\theta = 90^\circ$.

$$ \cos(90^\circ) = 0 $$ $$ h_{as} = \frac{l_s}{g} \times \frac{A}{a_s} \times \omega^2 r \cos(90^\circ) $$ $$ h_{as} = \text{Any value} \times 0 = 0 \, \text{m} $$

Final Results:

(i) Pressure head at the beginning of the stroke: $ \approx 3.02 \, \text{m} $

(ii) Pressure head in the middle of the stroke: $ 0 \, \text{m} $

Explanation of Acceleration Head

In a reciprocating pump, the piston doesn't move at a constant speed; it accelerates from rest, reaches a maximum velocity at mid-stroke, and decelerates to a stop. This acceleration and deceleration of the piston forces the entire column of water in the suction pipe to accelerate and decelerate with it.

According to Newton's second law ($F=ma$), a force is required to accelerate the mass of water in the pipe. This force translates to a pressure difference, which we express as the "acceleration head" ($h_a$).

At the beginning of the stroke ($\theta=0^\circ$): The piston's acceleration is at its maximum. This requires a large force, resulting in a significant drop in pressure at the cylinder inlet. This is the point where cavitation is most likely to occur.
In the middle of the stroke ($\theta=90^\circ$): The piston reaches its maximum velocity, and for an instant, its acceleration is zero. Therefore, no extra head is required to accelerate the water column, and the acceleration head is zero.