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DERIVE AN EXPRESSION FOR ACCELERATION HEAD IN THE SUCTION PIPE OF A RECIPROCATING PUMP

We were discussing the basics of reciprocating pumpmain components of a reciprocating pumpworking principle of reciprocating pumpideal indicator diagram of reciprocating pump and effect of acceleration and friction on indicator diagram of reciprocating pump in our recent posts.  

Today we will derive an expression for acceleration head in the suction pipe of a reciprocating pump with the help of this post.  

Derive an expression for acceleration head in the suction pipe of a reciprocating pump 

Following figure indicates the mechanism of a reciprocating pump. When crank will start to rotate, piston will be started to reciprocate in forward and backward direction within the cylinder. At the two extreme positions i.e. inner dead center and outer dead center, the velocity of piston will be zero and velocity of piston will be maximum at the middle of stroke i.e. center of cylinder. 

We can say that at the start of stroke, suction or delivery, velocity of piston will be zero and it will be increasing and will be maximum at the middle of stroke and again it will be decreasing and finally velocity of piston will become zero at the end of the stroke. 

Therefore, at the start of suction stroke or delivery stroke, piston will have acceleration and this acceleration will be decreasing and will be zero at the middle of stroke. Further, retardation will be started and retardation will be max at the end of the stroke. In simple, piston will be having acceleration at the start of each stroke and piston will be having retardation at the end of each stroke. 

As water in the cylinder will be in contact with the piston and therefore velocity and acceleration of water in suction and delivery pipe will be dependent over the velocity and acceleration of the piston.
Therefore, water flowing through suction and delivery pipe will also have acceleration at the start of each stroke and will have retardation at the end of each stroke. Hence, velocity of liquid i.e. water in suction and delivery pipe will not be uniform. 

Therefore, acceleration head or retardation head will be acting on the water flowing through the suction or delivery pipe. Pressure inside the cylinder will be changing due to this acceleration head or retardation head. 

If length of connecting rod is much larger than the radius of crank or we can also say that if the ratio of length of connecting rod to the crank radius i.e. l/r is very large, then the motion of the piston inside the cylinder could be considered as simple harmonic motion. 

Let us consider the following terms, from above mechanism of reciprocating pump, as mentioned below. 

ω = Angular speed of the crank in Radian/sec, and let us consider that angular speed is constant
A = Area of the cylinder
a = Area of the suction pipe or delivery pipe
l = Length of the pipe  
r = Radius of the crank
V = Velocity of water in the cylinder
v = Velocity of water in the pipeline 

At the start, crank will be at position A i.e. at inner dead center and piston will be at extreme left position which is shown in above figure by dotted lines. 

Let us think that in time t seconds, crank is turned by an angle θ from its inner dead center i.e. A. Let us assume that piston is moved towards right i.e. towards outer dead center by x during time t seconds as displayed in above figure. 

Angle turned by crank in time t second, θ = ωt 

Distance travelled by the piston inside the cylinder will be given as mentioned here 

x = AO- FO
x = r – r cos θ
x = r – r cos ωt

Velocity of the piston, V = dx/dt
V = d(r – r cos ωt)/dt
V = rω Sin ωt 

Now we will apply the continuity equation, the volume of water flowing into the cylinder per second will be equal to the volume of water flowing from the pipe per second. 

Volume of water flowing into the cylinder per second = Volume of water flowing from the pipe per second 

Velocity of water in the cylinder x Area of the cylinder = Velocity of water in the pipeline x Area of the pipeline 

Vx A = v x a
v = V x A/a = A/a x rω Sin ωt 

Now we will find out the acceleration of water in the pipeline and it will be determined as mentioned below 

Acceleration of water in the pipeline = dv/dt 

Acceleration of water in the pipeline = d (A/a x rω Sin ωt)/dt   

Acceleration of water in the pipeline = (A/a) r ω2 Cos ωt 

Mass of water in the pipeline = Density of water x Volume of water in the pipeline 

Mass of water in the pipeline = ρ x (Area of pipeline x length of the pipeline) 

Mass of water in the pipeline = ρ x a x l = ρ a l 

Force required to accelerate the water in the pipeline = Mass of water in the pipeline x Acceleration of water in the pipeline  
Force required to accelerate the water in the pipeline = ρ a l x (A/a) r ω2 Cos ωt 

Intensity of pressure due to acceleration of water in the pipeline = Force required to accelerate the water in the pipeline / Area of pipeline
Intensity of pressure due to acceleration of water in the pipeline = ρ l x (A/a) r ω2 Cos ωt
Intensity of pressure due to acceleration of water in the pipeline = ρ l x (A/a) r ω2 Cos θ 

Acceleration head in the pipeline = Intensity of pressure due to acceleration / weight density of liquid
Acceleration head in the pipeline = ρ l x (A/a) r ω2 Cos θ / ρ g
Acceleration head in the pipeline = l/g x A/a x r ω2 Cos θ
Acceleration head in the pipeline = l/g x A/a x ω2 r Cos θ 


Acceleration head in the suction pipe and delivery pipe will be given by following equation 


Maximum pressure head due to acceleration or maximum acceleration head in the pipeline will be given by following equation as mentioned below. 

So, we have seen here the expression for the pressure head due to acceleration or acceleration head in the pipeline. We have also seen here the maximum value of this acceleration head or pressure head due to acceleration. 

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Reference: 

Fluid mechanics, By R. K. Bansal 
Fluid machines, By Prof. S. K. Som 
Image courtesy: Google  

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