Tafila Technical University

Faculty of Engineering

Mechanical Engineering Department

Name: Mohanad Yousef Al-Tahrawi.

Experiment: Double acting reciprocating pump.

Instructure: Dr. Khaled Rababaah.

Date: 28/10/2015

- 2 -

1- Objective

1- To calculate the volumetric efficiency.

2- To calculate the overall efficiency.

2- Introduction

A piston pump can be based on a single piston or, more likely, multiple

parallel pistons. The pistons are reciprocated using cams or crankshafts. The stroke

is generally adjustable. This type of pump can deliver heads of up to 1000 bar. The

largest sizes of piston pumps can deliver flows of 40m3 /hr. In practice these pumps

are more likely to be used for metering low flow rate fluids at more modest pressures

in laboratories and chemical process plants. Piston pumps are not generally suitable

for transferring toxic or explosive media.

There are two types of piston pump:

a) Single acting reciprocating pump

b) Double acting reciprocating pump

* Single Acting Reciprocating pump:

Figure 1 is a typical layout of single-acting piston pump. The power end,

which consists of the crankshaft, connecting rod(s), crosshead(s) and other

components, is to the right. The liquid end, including the suction and discharge

manifolds, piston(s) and cylinder(s), suction and discharge check valves and the fluid

chamber, is to the left.

FIGURE 1 Single Acting Reciprocating pump

- 3 -

* Double Acting Reciprocating Pump:

A more complex design, known as the double-acting pump, allows the piston

to discharge fluid during both its forward and back stroke, resulting in almost twice

the flow per cycle.

Figure 2 is a typical layout of a plunger pump. The power end is similar to the

one seen in Figure 1 but there are two distinct differences in the liquid end. The piston

is replaced with a plunger that is sealed by stationary sealing rings (A). Also, there is

no cylinder-just a fluid chamber between the check valves. Plunger pumps operate on

the principle that a solid will displace a volume of liquid equal to its own volume.

Instead of forcing the liquid out of a cylinder via a piston, the plunger enters the fluid

chamber and displaces an amount of liquid equal to the plunger volume entering the

chamber.

3- Theory

𝝁𝒗 = 𝑸𝒂𝒄𝒕

𝑸𝒕𝒉𝒆𝒐𝒓< 𝟏

𝑸𝒂𝒄𝒕 = 𝑽𝒕

V For all cylinder = number of cylinder *A * L

𝑸𝒕𝒉𝒆𝒐𝒓 = 𝑨 ∗ 𝑳 ∗ 𝑵𝒑𝒖𝒎𝒑

Shaft power (S.P) = 2 * π * N * T

𝑯𝒎 = 𝑷𝒅 − 𝑷𝒔

𝜸∗ 𝟏𝟎𝟓𝑷𝒂 = 𝒎𝒆𝒕𝒆𝒓

FIGURE 2 Double Acting Reciprocating pump

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Water Power (W.P) = 𝜸 ∗ 𝑸𝒕𝒉𝒆𝒐𝒓 ∗ 𝑯𝒎

𝝁𝒐 = 𝒘𝒂𝒕𝒆𝒓 𝒑𝒐𝒘𝒆𝒓

𝒔𝒉𝒂𝒇𝒕 𝒑𝒐𝒘𝒆𝒓

4- Data:

D = 0.04 m

L = 0.042 m

𝜸 = 9810𝑁/𝑚3 = 1000𝐾𝑔𝑓/𝑚3

N motor N Pump V (lit) t sec T (N.m) Ps (bar) Pd (bar)

4 1 3 38 0.2 -0.1 0.01

8 2 3 18 0.3 -0.17 0.05

12 3 5 18 0.5 -0.27 0.2

16 4 5 15 0.65 -0.45 0.4

5- Result:

N motor N Pump V (lit) T ( sec) Qact Qth Volumetric eff.

4 1 3 38 7.89 0.000105558 0.748

8 2 3 18 0.0167 0.000211115 0.79

12 3 5 18 0.0278 0.000316673 0.877

16 4 5 15 0.0333 0.00042223 0.7895

T (N.m) Ps (bar) Pd (bar) Hm W.P S.P Efficiency

0.2 -0.1 0.01 1.12 0.87 1.257 0.691

0.3 -0.17 0.05 2.243 3.67 3.77 0.9726

0.5 -0.27 0.2 4.791 13.1 9.42 1.3852

0.65 -0.45 0.4 8.66 28.33 16.33 1.7343

- 5 -

Slope = volumetric efficiency = 0.8282

0.00E+00

5.00E-05

1.00E-04

1.50E-04

2.00E-04

2.50E-04

3.00E-04

3.50E-04

4.00E-04

0.00E+00 5.00E-05 1.00E-04 1.50E-04 2.00E-04 2.50E-04 3.00E-04 3.50E-04 4.00E-04 4.50E-04

Q a

ct

Q th

0

1

2

3

4

5

0.00E+00 1.00E-04 2.00E-04 3.00E-04 4.00E-04 5.00E-04

rev

of

pu

mp

pe

r se

c

Q act

- 6 -

Slope = 1.8 = overall efficiency

y = 1.8387x - 2.6714

R² = 0.9883

0

5

10

15

20

25

30

35

40

0 2 4 6 8 10 12 14 16 18 20

W.P

S.P

0

1

2

3

4

5

6

0 0.5 1 1.5 2 2.5 3 3.5

rev

of

pu

mp

pe

r se

c

EfficencyHundreds

y = 11270x + 0.0862

R² = 0.9853

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

0.00E+00 5.00E-05 1.00E-04 1.50E-04 2.00E-04 2.50E-04 3.00E-04 3.50E-04

revo

f p

um

p p

er

sec

Q act

- 7 -

6- conclusion:

Volumetric efficiency in a hydraulic pump refers to the percentage of actual fluid flow

out of the pump compared to the flow out of the pump without leakage. In other words,

if the flow out of a 100cc pump is 92cc (per revolution), then the volumetric efficiency

is 92%. The volumetric efficiency will change with the pressure and speed a pump is

operated at, therefore when comparing volumetric efficiencies.

Hydraulic loss relates to the construction of the pump or fan, and is caused by the

friction between the fluid and the walls, acceleration and retardation of the fluid and

the change of the fluid flow direction.

The overall efficiency is the ratio of power actually gained by the fluid to the shaft

power supplied.

The losses in the pump or fan converts to heat transferred to the fluid and the

surroundings. As a rule of thumb the temperature increase in a fan transporting air is

approximately 1 oC.