Рави Додданнавар, Андрис Барнард. Practical Hydraulic Systems

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Appendix A

Practical problems

Exercise 1

1.1 Calculate the pressure due to a column of 0.5 m of

(a) Water

(b) Oil of specific gravity 0.8

1.2 The intensity of pressure at a point in a fluid is 4 N/cm^. Find the corresponding

height of fluid when the fluid is

(a) Water

(b) Oil of specific gravity 0.9.

1.3 A bullet of mass 8 g travels at a speed of 400 m/s. It penetrates a target, which

offers a constant resistance of 1000 N to the motion of the bullet. Calculate

(a) The kinetic energy of the bullet

(b) The distance through which the bullet has penetrated.

1.4 A hydraulic press has a ram of 25 cm diameter and a plunger of 4.0 cm diameter. Find

the weight lifted by the hydraulic press when the force applied at the plunger is 400 N.

1.5 Calculate the gage pressure and absolute pressure at a point 3 m below the free

surface of a liquid having a density of 1.53 x 10^ kg/m^, if the atmospheric pressure

is equal to 750 mm of mercury. The specific gravity of mercury is 13.6 and density

of water is 1000 Kg/ml

1.6 The diameters of a pipe are 15 and 20 cm at sections 1 and 2 respectively. The

velocity of water flowing through the pipe is 4 m/s at section 1. Determine the

discharge through the pipe and also the velocity of flow at section 2.

1.7 A pipe of 40 cm diameter carrying water, branches into two other pipes of diameters

25 and 20 cm respectively. If the average velocity in the 40 cm pipe is 2.0 m/s find

the flow rate in this pipe. Also determine the velocity in the 20 cm pipe if the

average velocity in the 25 cm pipe is 1.8 m/s.

Appendix A 209

Exercise 2

This exercise consists of problems involving designing and analyzing of simple hydraulic

circuits, along with their solutions.

Sample Problem 1

Deals with air to hydraulic pressure booster system: The following exercise will tell us

how to calculate the load-carrying capacity in a pressure booster system.

Problem:

The figure shows a pressure booster system used to drive a load T' via a hydraulic

cylinder. The following data is available:

Inlet air pressure (Pi) =100 psi

Air piston area (Ai) =20 sq.in.

Oil piston area (A2) = 1 sq.in.

Load piston area (A3) = 25 sq.in.

Find the load-carrying capacity of the system.

Problem 1a

Solve the following problem on the same lines as above (problem to be solved by

participants).

For the same system above, following data is given and you need to find the required load

piston area.

Inlet air pressure (Pi) =125 psi

Air piston area (Ai) = 20 sq.in.

Oil piston area (A2) = 1 sq.in.

Load-carrying capacity (F) = 75 000 lb

Sample Problem 2

Deals with the hydraulic horsepower analysis technique.

Problem:

A hydraulic cylinder is used to compress a car body down to a bale size in 10 s. The

operation requires a 10 ft stroke (S) and a force (Pbad) of 8000 lb. If a 1000 psi pressure

(P) pump is selected, find

(a) The required piston area

(b) The necessary pump flow rate

Rod

Load

210

Appendix

Problem 2a

Solve the following problem on the same lines as above (problem to be solved by

participants).

A hydraulic cylinder is to compress a car body down to bale size in 8 s. The operation

requires an 8 ft stroke and a 15 000 lb force. Find the following if the piston area is

12 sq. ins.

(a) The pressure to be developed by the pump

(b) The necessary flow rate of the pump

Sample Problem 3

In the following exercise, we shall study how a problem in a hydraulic system can be

solved using the metric-SI system units.

20 ft

elevation

Problem:

For the above hydraulic system,

The pump adds a power of

hp (3730 W) to the fluid.

Pump flow rate is 0.001896 mVs (6825.6 Iph)

The pipe diameter is 25.4 mm (0.0254 m)

The specific gravity of oil is 0.9

The elevation difference between station 1 and 2 is 20 ft (6.096 m)

6. The head loss between the two stations is 9.144 m of oil.

Find the pressure available at the inlet of the hydraulic motor located at station 2. The

pressure at station 1 in the hydraulic tank is atmospheric.

Problem 3a

Solve the following problem on the same lines as above (problem to be solved by

participants).

In the above hydraulic system, following data is available:

(a) The pump adds a power of 4 hp to the fluid

(b) Pump flow rate is 25 gpm

Appendix

A 211

(d) The specific gravity of oil is 0.9

(e) The head loss between the two stations is 40 ft of oil.

Find the pressure available at the inlet of the motor at station 2.

Sample Problem 4

Designing of a heat exchanger: the following example will show us how to calculate the

rise in temperature of the fluid as it flows through a restriction such as a pressure relief

valve, using the relation,

^ . Heat generated (Btu/min)

Temperature nse =

(Specific heat of oil x oil flow rate)

Problem:

Oil at 120

"^F

and 1000 psi is flowing through a pressure relief valve at a rate of 10 gpm.

What is the downstream oil temperature?

Specific heat of oil is 0.42 Btu/lb^F

Oil flow rate of

Ib/min equals 7.42 gpm.

Problem 4a

Solve the following problem on the same lines as above (problem to be solved by

participants).

Oil flows through a pressure relief valve at 15 gpm. Given,

(a) Oil temperature = 130 °F and

(b) Pressure = 2000 psi.

Calculate the rise in temperature of the oil as it flows through the relief valve.

Sample Problem 5

This example deals with the sizing of a heat exchanger in a hydraulic system.

Problems:

A hydraulic pump operates at 100 psi and delivers oil to a hydraulic actuator. Oil

discharges through the pressure relief valve (PRV) during 50% of the cycle time. The

pump has an overall efficiency of 85% and 10% of the power is lost due to frictional

pressure losses in the lines. What is the heat exchanger rating required to dissipate the

generated heat?

Problem 5a

Solve the following problem on the same lines as above (problem to be solved by

participants).

A hydrauUc pump operates at 2000 psi and delivers oil at 15 gpm to a hydraulic actuator.

Oil discharges through the PRV, 60% of the cycle time. The pump has an overall

efficiency of 82% and 10% of the power is lost due to frictional losses in the lines. What

rating heat exchanger is required to dissipate the heat generated?

Appendix B

Practical solutions

Exercise 1

1.1 Calculate the pressure due to a column of 0.5 m of

(a) Water

(b) Oil of specific gravity 0.8

Solution: Height of liquid column 'h'

0.5 m

5oii = 0.8

'^mercury = l-^.O

Density of water (p) = 1000 kg/m^

Gravitational force (g) = 9.8 m/s^

(a) We know according to hydrostatic law that pressure developed in a fluid due

to a height '/z' is given by

pgh

^ 1000

9.8

0.5 = 4900 N/m'

/. Pressure due to a column of 0,5 m of water = 4900 N/m^ (SI units)

[In metric units = 4900/9.8 x 100 x 100 = 0.05 kg/cm^ (where 9.81 is to

convert Newton into kgf and 100 x 100 is to convert m^ into cm^).]

(b) Pressure due to oil column: P(oii) =

p^on)

We know,

c r- .. ^ i/c ^ Density of oil

Specific gravity of oil

(S^-^)

Density of water

0.8 = pj 1000 =^

/?,,

= 0.8 X1000 = 800 kg/m'

P,,

= 800

9.8

0.5 = 3920 N/m'

Pressure due to oil column = 3920 N/m

Appendix B 213

Amercury)

A'^Cmercury) mercury r^(wa.ter)

13.6 X1000 =

600 kg/m^

•••^™e.u^)=13 600x9.8x0.5 = 66 64C

Pressure due to mercury column

66 640

N/m^

1.2 The intensity of pressure at a point in a fluid is 4 N/cm^. Find the corresponding

height of fluid when the fluid is

(a) Water

(b) Oil of specific gravity is 0.9.

Solution: P

4 N/cm^ = 40 000 NW

So.

=0.9

/?„,„= 1000 kg/m^

(a) P

pgh

^ 40 000 = 1000x9.81x/t

=> 40 000 = 9810/1

^ /i = 4.07 m.

This means that the height of water column required to produce a pressure

of4N/cm^is4.07m.

(b) S =

Poi\/PwateT

poll

= 0.9 x 1000 = 900 kg/m^

Since,

pgh

z:>40 000 = 900x9.81x;i

z:>

40 000 = 8829/1

=>h =

4.53 m.

This shows that the height of mercury column required to produce a pressure of 4 N/cm^

is 4.53, whereas using a water column of height 4.07 m can produce the same pressure.

1.3 A bullet of mass 8 g travels at a speed of 400 m/s. It penetrates a target, which

offers a constant resistance of 1000 N to the motion of the bullet. Calculate

(a) The kinetic energy of the bullet

(b) The distance through which the bullet has penetrated.

Solution: Mass of bullet = 8 g = 8/1000 kg

Velocity of bullet = 400 m/s

Force = 1000 N

(a) We know kinetic energy = y2mv^

(1x8x400^)/(2x1000)

= 640 J

.•.Kinetic energy of the bullet = 640 J

214

Appendix

(b) We know that work done = Force x Distance

640 = 1000

640/1000

= 0.64 m

.'.The distance the bullet has penetrated the fixed target = 0.64 m.

1.4 A hydraulic press has a ram of 25 cm diameter and a plunger of 4.0 cm diameter.

Find the weight lifted by the hydraulic press when the force applied at the plunger is

400 N.

Solution: Diameter of ram = 25 cm

^> Area of ram = 7rcf/4

^3.14(25)^/4

Area of ram (A) = 490.62 cm^

Diameter of plunger = 4 cm

=> Area of plunger (a) = 12.56 cm^

Force applied on plunger (F) = 400 N

Ram

M Plunger

Liquid

Now the force 'F applied to the plunger exerts a pressure on the liquid, which is given by

„ F

We know F

400 N,

= Area of the plunger (a) = 12.56 cm

400

P =

12.56

:31.9N/cm'

According to Pascal's law this pressure is transmitted to every point in the fluid and also

acts on the ram, as a result of which the ram will experience an upward force (W) given by

PxA

Where 'P' is the pressure exerted by the liquid on the ram and 'A' is the area of the ram.

W = 31.9x490.62 =

650.8 N = 15.65 kN

.•. By application of a force of 400 N on the plunger, the ram will be able to lift a weight

of 15.65 kN.

Appendix

B 215

1.5 Calculate the gage pressure and absolute pressure at a point 3 m below the free

surface of a Hquid having a density of 1.53 x 10^ kg /m^, if the atmospheric pressure

is equal to 750 mm of mercury. The specific gravity of mercury is 13.6 and density

of water is 1000 kgW.

Solution: /i = 3 m/7

(Hquid)

= 1.53 x 10^ kg/m^

Pressure head (atmospheric) = 750 mm of Hg

5'(mercury) = 13.6 kg/m

Awater) = 1000 kg/m^

We know, Absolute pressure = Gage pressure + Atmospheric pressure.

(a) Gage pressure =

Auquid)

=> 1.53

10^

9.81

3 = 45 028 N/m^

/. Gage pressure = 45 028 N/m^ = 4.5028 N/cm^

Atmospheric pressure =

^mercury)

We know,

A'^C mercury) (mercury) /^^(water)

An,ercu^. =13.6x1000

Amercury,=13 600kg/ni^

750 mm of mercury =

0.750

z=>

Atmospheric pressure =

600x9.81x0.750 = 100062 N/m'

/. Atmospheric pressure

10.0062 N/cm^

Hence (a) becomes

4.50+ 10.0 = 14.50 N/cm'

/. The absolute pressure is 14.50 N/cm and Gage pressure is 4.50 N/crn.

1.6 The diameters of a pipe are 15 cm and 20 cm at sections 1 and 2 respectively. The

velocity of water flowing through the pipe is 4 m/s at section 1. Determine the

discharge through the pipe and also the velocity of flow at section 2.

Solution: d\ = 15 cm

vi = 4 m/s = 400 cm/s (section 1.1)

^2 = 20 cm

216

Appendix

We know from continuity equation

_{3.14x(15')}

= 176.63 cm^

. _ (3.14x20x20)

= 314 cm'

/. (A.l) becomes

176.63x400=314xv

•v^

(176.63x400)

314

= 225 cm/s

= 22.5 m/s

We know

Discharge/flow rate (Q)

A,v,

A2V2

:^Q

= 176.63 X

400 = 70 652 cmVs

Thus it is seen that water in flowing from section 1 to 2, loses velocity, which is quite

obvious considering the increase in diameter. (However in doing so, it must be

understood that there will be a rise in the pressure energy.)

1.7 A pipe of 40 cm diameter carrying water branches into another 2 pipes of diameters

25 and 20 cm respectively. If the average velocity in the 40 cm pipe is 2.0 m/s find

the flow rate in this pipe. Also determine the velocity in the 20 cm pipe if the

average velocity in the 25 cm pipe is 1.8 m/s.

Solution:

= 1.8 m/s

1 di=40cm

= 2,0 m/s

0(3

= 20

Let Qi,

and

be the flow rates in pipes 1, 2 and 3 respectively.

(Note that the continuity equation is not applicable here, because the fluid in pipe 1 is

getting divided or removed.)

Appendix

B 217

We know,

Q^=Q2^Q3 (A.2)

Discharge

(Q^)

through pipe

A^v^

Where

A, =3.14x

looj

= 0.1256 m'

01 =0.1256x2 = 0.251 mV^

(3.14x0.25')

1.8 = 0.088 mVs

(A.2) becomes,

Now,

(22= 0.088 mVs

0.251 = 0.088 + 03

03 =0.251-0.088 = 0.163 mVs

11^03 =0.163 mVs

0.163

_ . (3.14x0.20') .^^ ,

03 =

A3V3

=>v^= =>

= 5.19 m/s

Exercise 2

A few problems and their solutions, which deal with, the designing and analyzing simple

circuits in a hydrauUc system.

Sample Problem 1

Deals with air to hydraulic pressure booster system: The following exercise will tell us

how to calculate the load-carrying capacity in a pressure booster system.

Problem:

The figure shows a pressure booster system used to drive a load 'F' via a hydraulic

cylinder. The following data is available.

Inlet air pressure (Pi) = 100 psi

Air piston Area (Ai) = 20 sq.in.

Oil piston area (A2) = 1 sq.in.

Load piston area (A3) = 25 sq.in.