The document describes the design and working of a hydraulic jack. It consists of syringes that act as a pump and ram, tubes to connect them, a wooden block for support, and ball bearings as non-return valves. When the small syringe is pushed in, it transmits pressure undiminished to the large syringe due to Pascal's law, lifting a load. The document outlines the components, design process including determining forces and dimensions, and provides calculations showing it can lift a 0.523 kg load. It also discusses potential improvements such as increasing diameters or using pneumatics.

1. BY
14BME131D
14BME132D
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14BME 135D
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Hydraulic Jack

2. Definition Hydraulic Jack
 A hydraulic jack is a device used to lift heavy loads. The device
itself is light, compact and portable, but is capable of exerting great force. The
device pushes liquid against a piston; pressure is built in the jack's container. The
jack is based on Pascal's law that the pressure of a liquid in a container is the
same at all points.

3. Pascal’s Law
 Pressure on a confined fluid is transmitted undiminished and acts with
equal force on equal areas and at 90 degrees to the container wall.

4.  A fluid, such as oil, is displaced when either piston is pushed inward. The
small piston, for a given distance of movement, displaces a smaller
amount of volume than the large piston, which is proportional to the ratio
of areas of the heads of the pistons. Therefore, the small piston must be
moved a large distance to get the large piston to move significantly. The
distance the large piston will move is the distance that the small piston is
moved divided by the ratio of the areas of the heads of the pistons. This is
how energy, in the form of work in this case, is conserved and the Law of
Conservation of Energy is satisfied. Work is force times distance, and
since the force is increased on the larger piston, the distance the force is
applied over must be decreased.

5. Part List
 Syringes
 Tubes
 Wooden Block
 Two ball bearing
 CD Holder

6. Flow Chart
 Specify function of Element
 Determine forces Acting on Element
 Select Suitable Material for Element
 Determine Failure Mode
 Determine geometric dimension
 Modify dimension for Assembly
 Prepare working Drawing of element

7. Specify function of element
 Wooden Block: Provide support for structure and give a strength hole Assembly.
 Ball bearing: Behave like Non-return Valve.
 Syringes : Behave like Pump , By-pass and Hydraulic Ram.
 Tubes: provide Flow between two connection in original use hosepipes.
 CD Holder: Provide Proper Work Standing Area.
Non-return
Valve

8. Determine Forces
 Due to dead weight forces are acting up-word or-
Down-ward. But hydraulic pressure is there so-
weight is lifting.
 Actual time when pumping start then two forces
generate on syringes.
F
F
F

9. Selection Materials
 We are select wooden base so we can reduce weight.
 As a pump select syringes so cost reduce.
 For lifting any product use CD-holder. because here
application is small product.
 As Hose pipes use small tube. Carry pressure up to
ram position.
 Pipes use as hose pipes because here pressure
is not create to much.

10. Failure Criterion
 Under heavy load may be system is not lift proper manner.
 Suppose half of way it will be fail like bending .
 May be wooden block fail in sharp corner because stress concentration action.
 Tube is disengages with syringe because system pressure will be increase but it
not sufficient for lifting and tube is disengages then create leakage.
Heavy Load
Bend
Tube
dishsengag
es

11. Dimension
 Length of ram syringes = 118 mm
 Diameter “ = 30 mm
 Length of pump syringes=75 mm
 Diameter “ =18 mm
 Tube diameter = 3 mm

12. Actual Model Calculation
Suppose we apply .18828 kg weight on pump side syringe
P1= F1/A1
.18828
π/4(.018)^2
P1= 739.89Nm.
 Under .18828 kg load we create 739.89Nm pressure in system
Suppose we
apply Load

13.  P1=739.89Nm
P2=F2/A2
 According to Pascal's law P1=P2
so,
759.89 = F2/π/4(.03)^2
F2=.523kg
F1=.188kg
F2=.523kg
Lift
Input
Output

14.  Beyond the .523kg vibration and deflection
are come out the system means, not work
properly.
 If you want to increase the load carrying
capacity means increase diameter of syringe
so effective area will be increase .
 Second thing change the material select high
strength syringe so you can increase load carrying
capacity.
F1=.188kg
F2=.523kg
Lift

15. Costing
 Wooden block = 50rs.
 Small syringes(2)=100rs.
 Big syringe =150rs.
 Fevicol =50rs.
 Ball bearing = 5rs.
 Tube =10rs.
 CD-Holder =25rs.
Total =390rs.

16. Improvement
 Previous problem we can solve here increase diameter
 Here we use hydraulic system but we can also use pneumatic
system.
 Pneumatic is quick response system
 There is no require any reservoir because atmosphere it self
reservoir.
 In our model height is issue so ram water not go directly in reservoir
but we can do improve here, we can attach one push button bellow
the big syringe so flow direct go in reservoir.