### Arrangement

Piston continuously makes forward stroke and backward stroke and these strokes happen between two extreme positions of the cylinder,

- Bottom dead center or BDC, which is on the crank side
- Top dead center or TDC, which is opposite to crank side

The piston moves from BDC to TDC and from TDC to BDC continuously. Now, let’s try to visualize & understand properly. The piston starts at BDC and increases its speed and reaches to the maximum speed in the middle. Then it has to stop at TDC, so it starts to reduce its speed from the middle of the cylinder and stops at the TDC. So, the piston speed shall be as follows:

### Diagram for Velocity

During suction stroke,

- Speed 0 at TDC ———-> Speed increases up to the middle towards BDC ———–> Maximum speed at the middle ———> Speed decreases towards the BDC ———-> Speed 0 at BDC

During the discharge stroke, it will be as follows:

- Speed 0 at BDC ———-> Speed increases up to the middle towards TDC ———–> Maximum speed at the middle ———-> Speed decreases towards the TDC ———-> Speed 0 at TDC

### Acceleration

We know that increase in velocity per unit time means acceleration. So, piston velocity increase means acceleration is coming into the picture. Now, as water is moved by the piston, the velocity, as well as acceleration of water inside the cylinder, will also be changed with the piston. This will also change the velocity and acceleration at the suction and discharge pipe. During suction stroke,

- Acceleration 0 at TDC ———> Acceleration increases up to the middle towards BDC ———–> Maximum acceleration at the middle ———> acceleration decreases towards the BDC ———-> Acceleration 0 at BDC

During the discharge stroke, it will be as follows:

- Acceleration 0 at BDC ———-> Acceleration increases up to the middle towards TDC ———–> Maximum acceleration at the middle ———-> acceleration decreases towards the TDC ———–> Acceleration 0 at TDC

### Derivation for Acceleration Head

Let’s try to calculate the acceleration. First, we will draw a simple diagram.

Let’s consider,

- ω = Angular speed of crank in Rad/sec
- r = Radius of the crank
- A = Cross sectional area of the pump cylinder
- a = Suction pipe or deliver pipe cross sectional area
- V = Velocity of water in the cylinder
- v = Velocity of water in the pipe
- l = Length of pipe
- x = Displacement of the piston in time t second

At the start of the suction stroke,

- Crank is at Point A
- Piston is at TDC

Now, the crank will rotate and the piston will move due to their arrangements. Consider, the crank will make an angle θ after t seconds. In this case,

- Crank creates angle θ with respect to horizontal line
- Time spent = t second
- Angular velocity ω means simply the rate of change of an angle, i.e., ω = θ/t or θ = ω.t

What is the velocity of the piston? Velocity = distance/time

V = dx/dt

Here, we have to find out the value of displacement of piston, x

- x = AO – PO
- x = r – r cosθ [cosθ =PO/QO, or PO = OQ cosθ = r cosθ]
- x = r – r cos (ωt) [θ = ω.t]

Velocity, V

- V = dx/dt
- V = d [r – r cos (ωt)]/dt
- V = rω d[1 – cos (ωt)]/dt
- V = rω [- (-sinωt)]
- V = rω sinωt

The fluid flow rate in the cylinder, Q1,

- Q1 = Velocity of water x cross sectional area of cylinder
- Q1 = V x A
- Q1= rω sinωt x A
- Q1= Arω sinωt

The fluid flow rate in the pipe, Q2,

- Q2 = Velocity of water x cross sectional area of pipe
- Q2 = v x a

As per the continuity equation, the fluid flow rate in the cylinder will be the same as the fluid flow rate within the pipe. So,

- Q1 = Q2
- Arω sinωt = v x a
- v x a = Arω sinωt
- v = A/a . rω sinωt

Acceleration of water in the pipe, fp

- fp = dv/dt
- fp = d(A/a . rω sinωt}/dt
- fp = A/a . rω ω cosωt
- fp = A/a . rω
^{2}cosωt

Now, we need to calculate the acceleration head, h_{a}.

h_{a} = P/W

Where,

- h
_{a}=Acceleration head, - P = Pressure intensity
- W =Weight density

Pressure intensity,

- P = Force / Area
- P = Mass of water x Acceleration / Area [Force = Mass of water x Acceleration]
- P = Volume x Density x Acceleration / Area [Mass of water = Volume x Density]
- P = Area x Length x Density x Acceleration / Area [Volume = Area x Length]
- P = a x l x ρ x fp /a [as, area =a]
- P = l x ρ x A/a rω
^{2}cosωt - P = l x ρ x (A/a) rω
^{2}cosωt

Weight density, W = ρg

Acceleration head in the pipe,

- h
_{a}= P/W - h
_{a}= l x ρ x (A/a) rω^{2}cosωt / ρg - h
_{a}= l /g x (A/a) rω^{2}cosωt - h
_{a}= (l /g) x (A/a) rω^{2}cosθ [θ = ωt]

Now, this acceleration head can be easily calculated, as follows:

The acceleration head at the suction pipe, h_{as}

- h
_{as}= (ls /g) x (A/as) rω^{2}cosθ

The acceleration head at the delivery pipe h_{ad}

- h
_{ad }= (ld /g) x (A/ad) rω^{2}cosθ

Maximum acceleration head, h_{max}

- h
_{max}= (l /g) x (A/a) rω^{2}[cosθ =1, where θ=90 deg]