The cylinder of a single-acting reciprocating pump is 125 mm in diameter and 250 mm in stroke. The pump is running at 40 r.p.m. and discharge water to a height of 15 m. The diameter and length of the delivery pipe are 100 mm and 30 m respectively. If a large air vessel is fitted in the delivery pipe at a distance of 1.5 m from the centre of the pump, find the pressure head in the cylinder : (i) At the beginning of the delivery stroke, and (ii) In the middle of the delivery stroke. Take the co-efficient of friction = .01.

Reciprocating Pump with Air Vessel Analysis

Problem Statement

Given Data & Constants

Cylinder diameter, $D = 125 \, \text{mm} = 0.125 \, \text{m}$
Stroke length, $L = 250 \, \text{mm} = 0.25 \, \text{m}$
Speed, $N = 40 \, \text{r.p.m.}$
Static delivery head, $h_d = 15 \, \text{m}$
Delivery pipe: $d_d = 100 \, \text{mm} = 0.1 \, \text{m}$, Total length $l_{d,total} = 30 \, \text{m}$
Air vessel position, $l_{d1} = 1.5 \, \text{m}$ from cylinder
Friction factor, $f = 0.01$
Acceleration due to gravity, $g = 9.81 \, \text{m/s}^2$

Solution

1. Calculate Common Parameters

$$ \text{Crank radius, } r = L/2 = 0.25/2 = 0.125 \, \text{m} $$ $$ \text{Angular velocity, } \omega = \frac{2 \pi N}{60} = \frac{2 \pi \times 40}{60} \approx 4.189 \, \text{rad/s} $$ $$ \text{Cylinder area, } A = \frac{\pi}{4}D^2 = \frac{\pi}{4}(0.125)^2 \approx 0.01227 \, \text{m}^2 $$ $$ \text{Pipe area, } a = \frac{\pi}{4}d_d^2 = \frac{\pi}{4}(0.1)^2 \approx 0.007854 \, \text{m}^2 $$

2. Calculate Velocities and Head Losses

Mean velocity in pipe (due to air vessel):

$$ Q_{th} = \frac{A \cdot L \cdot N}{60} = \frac{0.01227 \times 0.25 \times 40}{60} \approx 0.002045 \, \text{m}^3/\text{s} $$ $$ v_{mean} = \frac{Q_{th}}{a} = \frac{0.002045}{0.007854} \approx 0.2604 \, \text{m/s} $$

Friction head in pipe after air vessel ($l_{d2} = 30 - 1.5 = 28.5$ m):

$$ h_{f2} = \frac{4 f l_{d2} v_{mean}^2}{2 g d_d} = \frac{4 \times 0.01 \times 28.5 \times (0.2604)^2}{2 \times 9.81 \times 0.1} \approx 0.039 \, \text{m} $$

Acceleration head (only for pipe before air vessel, $l_{d1} = 1.5$ m):

$$ h_{ad} = \frac{l_{d1}}{g} \frac{A}{a} \omega^2 r = \frac{1.5}{9.81} \times \frac{0.01227}{0.007854} \times (4.189)^2 \times 0.125 \approx 0.523 \, \text{m} $$

(i) Pressure Head at the Beginning of the Delivery Stroke

The head is the static head plus the friction from the steady flow after the vessel, plus the acceleration head from the short pipe before the vessel.

$$ H_{begin} = h_d + h_{f2} + h_{ad} $$ $$ H_{begin} = 15 + 0.039 + 0.523 \approx 15.562 \, \text{m} $$

(ii) Pressure Head in the Middle of the Delivery Stroke

At mid-stroke, acceleration is zero. However, friction occurs in the short pipe section ($l_{d1}$) due to the instantaneous maximum velocity, while the rest of the pipe ($l_{d2}$) experiences friction from the mean velocity.

$$ \text{Max velocity in pipe, } v_{max} = \frac{A}{a} \omega r = \frac{0.01227}{0.007854} \times 4.189 \times 0.125 \approx 0.818 \, \text{m/s} $$ $$ \text{Friction in first section, } h_{f1} = \frac{4 f l_{d1} v_{max}^2}{2 g d_d} = \frac{4 \times 0.01 \times 1.5 \times (0.818)^2}{2 \times 9.81 \times 0.1} \approx 0.020 \, \text{m} $$ $$ H_{middle} = h_d + h_{f1} + h_{f2} $$ $$ H_{middle} = 15 + 0.020 + 0.039 \approx 15.059 \, \text{m} $$

Final Results:

(i) Pressure head at the beginning of the delivery stroke: $ \approx 15.56 \, \text{m} $

(ii) Pressure head in the middle of the delivery stroke: $ \approx 15.06 \, \text{m} $

Explanation of the Air Vessel's Effect

An air vessel is a closed chamber containing compressed air, fitted to the pipe near the pump cylinder. Its purpose is to dampen the pressure fluctuations caused by the accelerating and decelerating flow.

Function: During the delivery stroke, the pump pushes out a pulsating flow. The excess water during the middle of the stroke (when piston velocity is high) enters the air vessel, compressing the air. Towards the beginning and end of the stroke (when piston velocity is low), the compressed air pushes this stored water out into the main pipe.
Constant Flow: The result is that the flow in the long section of the pipe *after* the air vessel becomes almost uniform and steady, with a velocity equal to the mean velocity ($v_{mean}$).
Reduced Head Loss: Because friction head is proportional to the square of velocity ($v^2$), the friction loss in the long pipe is significantly reduced by having a constant, lower mean velocity instead of a high, fluctuating peak velocity.
Reduced Acceleration Head: The acceleration head is only generated in the very short section of pipe between the pump cylinder and the air vessel. This dramatically reduces the pressure spikes at the beginning of the stroke, protecting the pump and pipework.