volume

Question

5.2 The following data refers to a single -acting hydraulic press: Area of plunger = 0,09 m2 Stroke length of plunger = 0,13 m Force applied to the plunger = 300 N Area of the ram piston = 0,45 m2 Neglect ALL the losses and calculate the following: 5.2.1 The volume of liquid displaced by the plunger after 6 pumping strokes 5.2.2 The distance moved by the ram piston in mm after 1 pumping stroke of the plunger 5.2.3 The force exerted by the ram 5.2.4 The mechanical advantage of the press 5.2.5 The fluid pressure 5.3 A single -acting plunger pump with THREE cylinders has a plunger diameter of 150 mm and a stroke length of 250 mm and runs at a speed of 120 r/min. There is a slip of 3% and the static height is 110 metres. The density of water = 1 000 kg/m 3. Calculate the volume of water delivered in litre/minute.

Answer 1

5.2.1)

The  area of plunger is a = 0.09 m² .

Stroke length of plunger is 0.13 m .

Force applied to the plunger is 300 N .

Area of the ram piston is 0.45 m² .

The  volume of liquid displaced by the plunger for one stroke is

                                                                      = area of plunger * stroke length of plunger .

                                                                      = a * 0.13 

                                                                      = 0.09 * 0.13

                                                                      = 0.0117  m³ .

So the volume of liquid displaced by the plunger for one stroke is 0.0117  m³ .

The  volume of liquid displaced by the plunger for 6 stroke is =

                                         = ( volume of liquid displaced by the plunger for one stroke ) * 6

                                         = 6 * 0.0117

                                         = 0.0702 m³ .

So the  volume of liquid displaced by the plunger for 6 stroke is 0.0702 m³ .

Answer 2

 

5.2.2)

The  area of plunger is a = 0.09 m² .

Stroke length of plunger is 0.13 m .

Force applied to the plunger is 300 N .

Area of the ram piston is 0.45 m² .

Let distance move by the ram in one stroke is x .

The volume displace by the ram in one stroke 

                                = Area of ram * x 

So now calculate for volume displace by the ram in one stroke: 

Volume displace by the ram is equals to volume of liquid displaced by the plunger when no slip is present .

The  volume of liquid displaced by the plunger for one stroke is

                                                                      = area of plunger * stroke length of plunger .

                                                                      = a * 0.13 

                                                                      = 0.09 * 0.13

                                                                      = 0.0117  m³ .

The volume of liquid displaced by the plunger for one stroke is 0.0117  m³ .

So the volume displace by the ram is 0.0117  m³ .

0.0117  m³ =  Area of ram * x

0.0117 = 0.45 * x

x = 0.0117 / 0.45

x = 0.026 m

x = 26 mm .

So the distance moved by the ram piston in mm after 1 pumping stroke of the plunger  is 26 mm .

 

 

Answer 3

5.2.3)

The  area of plunger is a = 0.09 m² .

Stroke length of plunger is 0.13 m .

Force applied to the plunger is 300 N .

Area of the ram piston is 0.45 m² .

pressure on ram = pressure on plunger 

( force on ram / area on ram ) = ( force on plunger / area on plunger )  

(  force on ram / 0.45 ) = ( 300/0.09 )

 force on ram / 0.45  = 3333.33

 force on ram  = 3333.33 *  0.45 

 force on ram  = 1500 

So the  force on ram is 1500 N .

Answer 4

5.2.5)

The  area of plunger is a = 0.09 m² .

Stroke length of plunger is 0.13 m .

Force applied to the plunger is 300 N .

Area of the ram piston is 0.45 m² .

The fluid pressure is nothing but the pressure on ram .

            fluid pressure =  force on ram / area on ram 

we know that pressure on ram = pressure on plunger  .

pressure on ram = pressure on plunger 

pressure on ram = ( force on plunger / area on plunger )  

pressure on ram = ( 300/0.09 )

pressure on ram  = 3333.33

So the fluid pressure is 3333.33 pascals .

 

Answer 5

5.2.4)

The  area of plunger is a = 0.09 m² .

Stroke length of plunger is 0.13 m .

Force applied to the plunger is 300 N .

Area of the ram piston is 0.45 m² .

Pressure intensity on Ram = Pressure intensity on Plunger

Pressure intensity of the Ram = Weight of Ram/ Area of the Ram

(Weight of Ram / Area of Ram) = (Force on Plunger / Area of the Plunger)

Weight lift of Ram = (Force on Plunger / Area of the Plunger) * Area of Ram

Weight lift of Ram = ( 300/0.09 ) * 0.45

                           = 3333.33 *  0.45  

                          = 1500 kg

Weight  with respect to gravitational force g .

Weight lift of Ram = 1500*g 

                           = 1500*9.8

                          = 14700 N

Mechanical Advantage  = Weight Lift by Ram / Force applied on the Plunger

MA = 14700/ 300

MA = 49

Therefore Mechanical Advantage of the Hydraulic Press is 49. 

Answer 6

5.3)

Plunger diameter  (d) is 150 mm .

Stroke length (l) is 250 mm .

Speed of pump is n = 120 r/min .

There  is a slip of 3% .

Static height is 110 meters .

The density of water = 1 000 kg/m³ .

Volume per hour = (πd²/4)(l*n)60

Volume per minute = (πd²/4)(l*n) .

                                    = (π [150*10^-3]² /4) (250 *10^-3 [120] ) * slip 

                                    = (π [0.0225] /4) (30) * [97/100] 

                                    = (0.017671) (30) * [97/100] 

                                    = 0.5142 m³ / min .

 So the Volume per minute in liters = 0.5142*1000

                                                                 = 514.2 liter / min .

The Volume per minute in liters = 514.2 liter / min .