HYDROSTATIC PRESSURESulham, Nurasia, Sarnah Ahmad, Nur Hidayah Tasrie.

ICP B PHYSICS 2014

Abstract

Has conducted lab hydrostatic pressure. Practicum aims to determine the effect of the depth and density of the liquid to the hydrostatic pressure, as well as understand the principle of hydrostatic pressure experiments. In this experiment, there are two activities that have been carried out. Activity 1 was to investigate the influence of the depth of the hydrostatic pressure. In this activity measurements were carried out repeatedly three times at each depth that is used is 2 cm, 3 cm, 4 cm, 5 cm, 6 cm, 7 cm and 8 cm. And the second activity is investigating the influence of the density of the liquid to the hydrostatic pressure. There are 5 types of liquids used in this experiment: water, glycerin, oil, salt solution 1 and salt solution 2. From these experiments, it can be concluded that the depth and density of liquid affect the hydrostatic pressure. The depth and density is directly proportional to the hydrostatic pressure. The higher the depth of a liquid substance, the higher the pressure hidrostatiknya, as well as the density of the liquid, the higher the density of the liquid, the higher the hydrostatic pressure.

Key words: hydrostatic pressure, fluid density, depth.

PURPOSE

1. To determine the influence of the depth of the hydrostatic pressure.2. To determine the influence of the density of the liquid to the hydrostatic pressure.3. To determine the principle of hydrostatic pressure experiments.

EXPERIMENT METHOD

Brief theory

Theory of hydrostatic pressure can also be explained by observing or glass

vessel containing water as an example. Consider the following picture:

So much pressure on the bottom of the vessel are:

P = FA

= wA

= m. g

A

= ( ρV ) . g

A

= ( ρ hA ) . g

A

P = ρgh

Thus, the magnitude of the hydrostatic pressure is generally formulated with:

P = ρgh

Explanation :

P = hydrostatic pressure (N/m2 atau Pascal)

ρ = density of fluid (kg/m3)

g = acceleration of gravity (m/s2)

h = depth of fluid (m)

If the atmospheric pressure at the surface of the liquid that is P0, the absolute pressure at

the place or point located at a depth h is:

P= P0 + ρgh

Hydrostatic force on a base force vessel is determined by the following formula:

F = P.A

F = ρgh.A

Explanation :

F = hydrostatic force (N)

A = area of the base vessel (m2)

As for the type of large pressure liquids in it depends on the depth. Each point that is at the same depth will undergo the same hydrostatic pressure anyway.

Tools and Material

1. U-shaped pipe 5. Funnel

2. Pycnometer 6. Ruler

3. Beaker 7. 311 gram ohaus balance

4. Plastic hose 8. Kinds of liquid (fluid)

Variables Identification

Activity 1 : The effect of depth to the hydrostatic pressure

1. Control variable : Kinds of liquid / density of liquid

2. Manipulation variable : Depth

3. Response variable : The difference in height of the liquid in the pipe U

Activity 2 : The effect of the density of liquid to the hydrostatic pressure

1. Control variable : Depth

2. Manipulation variable : Kinds of liquid / density of liquid

3. Response variable : The difference in height of the liquid in the pipe U

Operational Definition of Variables

Activity 1: The effect of depth to the hydrostatic pressure

1. Control variable

Kinds of liquid is a liquid that is used in this observation is water. The unit of

the density is gram /cm3

2. Manipulation variable

The depth is the height of liquid in the funnel when the funnel in placed in

distilled water. The unit is centimeter (cm)

3. Response variable

The difference in height of the U pipe is the differenced in height of the U

pipe measured with a ruler at the time of the funnel depth altered. The unit is

centimeter (cm)

Activity 2: The effect of the density of liquid to the hydrostatic pressure

1. Control variable

The depth is the height of the liquid in the funnel to the surface of liquid

when the funnel in place in distilled water. In this activity, the depth used

was 5,00 cm. The unit is centimeter (cm)

2. Manipulation variable

The kinds of liquid is the liquid by used in this activity. There were five

kinds of liquid was used. That are water, glycerin, oil, salt solutions 1 and

salt solutions 2. The density of liquid is gram /cm3

3. Response variable

The difference in height of the U pipe is the differenced in height of the

U pipe measured with a ruler upon the type of liquid used altered. The unit

is centimeter (cm)

Work procedure

Activity 1 : The effect of depth to the hydrostatic pressure

1. The density of liquid were determined and measuring the mass and volume.

2. The U pipe contain the liguid with the glass funnel by a plastic hose were

connected.

3. The funnel was put into the water, hitted with the certain depth. The depth

was measured by using a ruler (measure from the water surface to the surface

of the water in the funnel).

4. Inspecting the change in the surface height of liquid in the U-pipe. The

difference of liquid height was measured. The supervision result in the table

of supervision were recorded.

5. Repeated the experiment with the difference depth by three times

Activity 2 : The effect of the density of liquid to the hydrostatic pressure

1. Making a graph that showing the relationship between the surface height

with the hydrostatic pressure according to the supervision table.

2. The formula of hydrostatic pressure was determined, if tan α was obtained

from the graph is equel ρg, with ρ = the density of water and g =

acceleration of gravity.

EXSPERIMENT RESULT AND DATA ANALYSIS

Experiment Result

Table 1. Density of liquid

Nu Type of liquid Mass (gram) Volume (ml)

1 Air |97,000 ± 0,010| |100,00 ± 0,05|

2 Glycerin |103,000 ± 0,010| |100,00 ± 0,05|

3 Oil |87,000 ± 0,010| |100,00 ± 0,05|

4 Salt solution 1 |98,910 ± 0,010| |100,00 ± 0,05|

5 Salt solution 2 |121,800 ± 0,010| |100,00 ± 0,05|

Activity 1. The influence of the depth of the hydrostatic pressure

Type of liquid : water

Table 2. the relationship between the depth of liquid with hydrostatic pressure

Nu depth(cm) Height difference of liquid at U pipe (cm)

1 |2,00 ± 0,05|1 |2,10 ± 0,05|2|2,10 ± 0,05|3 |1,90 ± 0,05|

2 |3,00 ± 0,05|1 |3,00 ± 0,05|2|3,00 ± 0,05|3 |3,15 ± 0,05|

3 |4,00 ±0,05|1 |4,10 ±0,05|2|4,00 ± 0,05|3 |4,10 ±0,05|

4 |5,00 ± 0,05|1 |5,15 ± 0,05|2|5,25 ± 0,05|3 |5,10 ± 0,05|

5 |6,00 ± 0,05|1 |6,30 ± 0,05|2|6,25 ± 0,05|3 |6,20 ± 0,05|

6 |7,00 ± 0,05|1 |7,20 ± 0,05|2|7,15 ± 0,05|3 |7,20 ± 0,05|

7 |8,00 ± 0,05|1 |8,05 ± 0,05|2|8,10 ± 0,05|3 |8,15 ± 0,05|

Activity 2. The influence of the density of the liquid to the hydrostatic pressure

Depth =|5,00 ± 0,05|cm

Table 3. the relationship between the density of the liquid with the hydrostatic pressure

NuDensity of Liquid (

kg /cm3)Height difference of liquid at U pipe (cm)

1Water

| 970,0 ± 0,5849 |

1 |5,15 ± 0,05|2|5,10 ± 0,05|3 |5,10 ± 0,05|

2Glycerin

| 1030 ± 0,6150 |

1 |6,35 ± 0,05|2|6,40 ± 0,05|3 |6,35 ± 0,05|

3Oil

| 870,0 ± 0,5349 |

1 |4,75 ±0,05|2|4,70 ± 0,05|3 |4,65 ±0,05|

4Salt solution 1

| 989,1 ± 0,5945 |

1 |5,35 ± 0,05|2|5,30 ± 0,05|3 |5,25 ± 0,05|

5Salt solution 2

| 121,8 ± 0,7089 |

1 |5,20 ± 0,05|2|5,10 ± 0,05|3 |5,15 ± 0,05|

DATA ANALYSIS

Density of liquid.

1. Water

ρ = mV

= 97 x10−3

1x 10−4

= 970 kg/m3

dρ = |∂ ρ∂ m

|dm + |∂ ρ∂ v

|dv

dρ = |v¹∨dm + |mv²|dv

dρρ

= v ¹

mv ¹dm +

mv ²mv ¹

dv

=dmm

+ v¹ dv

∆ ρρ

= ∆ mm

+ v¹ ∆ v

∆ ρ = |∆ mm

+ ∆ vv

|ρ

= |0,0197

+ 0,05100

| 970

= 0,58491 kg/m3

RE = ∆ ρρ

×100%

= 0,58491

970 × 100%

= 0,06 % ( 4 SF )

DC = 100% - RE

= 100% - 0,06%

= 99,94 %

ρ = |970,0± 0,5849| kg/m3

2. Oil

ρ = mV

= 87 x10−3

1 x 10−4

= 870 kg/m3

∆ ρ = |∆ mm

+ ∆ vv

|ρ

= |0,0187

+ 0,05100

| 870

= 0,5349 kg/m3

RE = ∆ ρρ

×100%

= 0,5349

970 × 100%

= 0,06 % ( 4 SF )

DC = 100% - RE

= 100% - 0,06%

= 99,94 %

ρ = |870,0± 0,5349| kg/m3

3. Glycerin

ρ = mV

= 103 x 10−3

1 x10−4

= 1030 kg/m3

∆ ρ = |∆ mm

+ ∆ vv

|ρ

= |0,01103

+ 0,05100

| 1030

= 0,615 kg/m3

RE = ∆ ρρ

×100%

= 0,6151030

× 100%

= 0,06 % ( 4 SF )

DC = 100% - RE

= 100% - 0,06%

= 99,94 %

ρ = |1030± 0,6150| kg/m3

4. Salt Solutions 1

ρ = mV

= 98,91 x 10−3

1 x10−4

= 989,1 kg/m3

∆ ρ = |∆ mm

+ ∆ vv

|ρ

= |0,01

98,91 +

0,05100

| 989,1

= 0,5945 kg/m3

RE = ∆ ρρ

×100%

= 0,5945989,1

× 100%

= 0,06 % ( 4 SF )

DC = 100% - RE

= 100% - 0,06%

= 99,94 %

ρ = |989,1 ± 0,5945| kg/m3

5. Salt Solutions 2

ρ = mV

= 121,8 x 10−3

1 x10−4

= 1218 kg/m3

∆ ρ = |∆ mm

+ ∆ vv

|ρ

= |0,01

121,8 +

0,05100

| 1218

= 0,7089 kg/m3

RE = ∆ ρρ

×100%

= 0,70891218

× 100%

= 0,06 % ( 4 SF)

DC = 100% - RE

= 100% - 0,06%

= 99,94 %

ρ = |1218± 0,7089| kg/m3

Activity 1 Influence of depth to hydrostatic pressure

Type of liquid: Water

1. For depth 2 cm

h = h 1+h 2+h 3

3 =

2,1+2,1+1,93

= 2,03 cm = 0,0203 m

δ 1= |h1 - h| = |0,021 – 0,0203 | = 0,0007 m

δ 2= |h2 - h| = |0,021 - 0,0203 | = 0,0007 m

δ 3= |h3 - h| = |0,019 - 0,0203 | = -0,0013 m

∆ h=δ max ¿0,0007 m

h = | 0,0203 ± 0,0007 | m

2. For depth 3 cm

h = h 1+h 2+h3

3 =

3+3+3,153

= 3,05 cm = 0,0305 m

δ 1= |h1 - h| = |0,03 – 0,0305 | = -0,0005 m

δ 2= |h2 - h| = |0,03 - 0,0305 | = -0,0005 m

δ 3= |h3 - h| = |0,0315 - 0,0305 | = 0.001 m

∆ h=δ max ¿0,001 m

h = | 0,0305 ± 0,0010 | m

3. For depth 4 cm

h = h 1+h 2+h 3

3 =

4,1+4+4,13

= 4,07 cm = 0,0407 m

δ 1= |h1 - h| = |0,041 – 0,0407 | = 0,0003 m

δ 2= |h2 - h| = |0,040 - 0,0407 | = -0,0007 m

δ 3= |h3 - h| = |0,041 - 0,0407 | = 0.0003 m

∆ h=δ max ¿0,0003 m

h = | 0,0407 ± 0,0003 | m

4. For depth 5 cm

h = h 1+h 2+h 3

3 =

5,15+5,1+5,13

= 5,12 cm = 0,0512 m

δ 1= |h1 - h| = |0,0515 – 0,0512 | = 0,0003 m

δ 2= |h2 - h| = |0,051 - 0,0512 | = -0,0002 m

δ 3= |h3 - h| = |0,051 - 0,0512 | = -0,0002 m

∆ h=δ max ¿0,0003 m

h = | 0,0512 ± 0,0003 | m

5. For depth 6 cm

h = h 1+h 2+h 3

3 =

6,30+6,25+6,203

= 6,25 cm = 0,0625 m

δ 1= |h1 - h| = |0,063 – 0,0625 | = 0,0005 m

δ 2= |h2 - h| = |0,0625 - 0,0625 | = 0 m

δ 3= |h3 - h| = |0,062 - 0,0625 | = -0,0005 m

∆ h=δ max ¿0,0005 m

h = | 0,0625 ± 0,0005| m

6. For depth 7 cm

h = h 1+h 2+h 3

3 =

7,20+7,15+7,203

= 7,18 cm = 0,0718 m

δ 1= |h1 - h| = |0,072 – 0,0718 | = 0,0002 m

δ 2= |h2 - h| = |0,0715 - 0,0718 | = -0,0003 m

δ 3= |h3 - h| = |0,072 - 0,0718 | = 0,0002 m

∆ h=δ max ¿0,0002 m

h = | 0,0718 ± 0,0002| m

7. For depth 8 cm

h = h 1+h 2+h 3

3 =

8,05+8,10+8,153

= 8,1 cm = 0,081 m

δ 1= |h1 - h| = |0,0805 – 0,081 | = -0,0005 m

δ 2= |h2 - h| = |0,081 - 0,081 | = 0 m

δ 3= |h3 - h| = |0,0815 - 0,081 | = 0,0005 m

∆ h=δ max ¿0,0005 m

h = | 0,0810 ± 0,0005| m

Table Relationship between the depth and hydrostatic pressure

Nu

.

Depth (m) Difference in height of

liquid the U pipe (m)

1 | 0,02 ± 0,0005 | | 0,0203 ± 0,0007 |

2 | 0,03 ± 0,0005 | | 0,0305 ± 0,0010 |

3 | 0,04 ± 0,0005 | | 0,0407 ± 0,0003 |

4 | 0,05 ± 0,0005 | | 0,0512 ± 0,0003 |

5 | 0,06 ± 0,0005 | | 0,0625 ± 0,0005|

6 | 0,07 ± 0,0005 | | 0,0718 ± 0,0002|

7 | 0,08 ± 0,0005 | | 0,0810 ± 0,0005|

0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.090

0.010.020.030.040.050.060.070.080.09

f(x) = 0.976610055260887 x + 5.33714595146242E-05R² = 0.999281360115158

Chart relations between the depth of the hydrostatic pressure

the depth (m)the

diff

eren

ce in

hei

ght o

f th

e pi

pe

U (

m)

Activity 2 Influence the dencity of liquid to hydrostatic pressure

Depth = | 5,00 ± 0,05 | cm

1. Water

h = h 1+h 2+h3

3 =

5,15+5,10+5,103

= 5,12 cm = 0,0512 m

δ 1= |h1 - h| = |0,0515 - 0,0512 | = 0,0003 m

δ 2= |h2 - h| = |0,051 - 0,0512 | = - 0,0002 m

δ 3= |h3 - h| = |0,051 - 0,0512 | = - 0,0002 m

∆ h=δ max ¿0,0003 m

h = | 0,0512 ± 0,0003| m

2. Oil

h = h 1+h 2+h 3

3 =

4,75+4,70+4,653

= 4,7 cm = 0,047 m

δ 1= |h1 - h| = |0,0475 - 0,047 | = 0,0005 m

δ 2= |h2 - h| = |0,047 - 0,047 | = 0 m

δ 3= |h3 - h| = |0,0465 - 0,047 | = - 0,0005 m

∆ h=δ max ¿0,0005 m

h = | 0,0470 ± 0,0005| m

3. Glycerin

h = h 1+h 2+h 3

3 =

6,35+6,40+6,353

= 6,37 cm = 0,0637 m

δ 1= |h1 - h| = |0,0635 - 0,0637 | = -0,0002 m

δ 2= |h2 - h| = |0,0640 - 0,0637 | = 0,0003 m

δ 3= |h3 - h| = |0,0635 - 0,0637 | = - 0,0002 m

∆ h=δ max ¿0,0003 m

h = | 0,0637 ± 0,0003| m

4. Salt Solutions 1

h = h 1+h 2+h 3

3 =

5,35+5,30+5,253

= 5,3 cm = 0,053 m

δ 1= |h1 - h| = |0,0535 - 0,053 | = 0,0005 m

δ 2= |h2 - h| = |0,053 - 0,053 | = 0 m

δ 3= |h3 - h| = |0,0525 - 0,053 | = - 0,0005 m

∆ h=δ max ¿0,0005 m

h = | 0,0530 ± 0,0005| m

5. Salt Solutions 2

h = h 1+h 2+h 3

3 =

5,20+5,10+5,153

= 5,15 cm = 0,0515 m

δ 1= |h1 - h| = |0,052 - 0,0515 | = 0,0005 m

δ 2= |h2 - h| = |0,051 - 0,0515 | = - 0,0005 m

δ 3= |h3 - h| = |0,0515 - 0,0515| = 0 m

∆ h=δ max ¿0,0005 m

h = | 0,0515 ± 0,0005| m

Table Relationship between density and hydrostatic pressure

Nu. Liquid Density (kg/m3) Difference in height of liquid

the U pipe (m)

1 Water | 970,0 ± 0,5849 | | 0,0512 ± 0,0003|

2 Oil | 870,0 ± 0,5349 | | 0,0470 ± 0,0005|

3 Glycerin | 1030 ± 0,6150 | | 0,0637 ± 0,0003|

4 Salt Solutions 1 | 989,1 ± 0,5945 | | 0,0530 ± 0,0005|

5 Salt Solutions 2 | 1218 ± 0,7089 | | 0,0515 ± 0,0005|

0.046 0.048 0.05 0.052 0.054 0.056 0.058 0.06 0.062 0.064 0.0660

200

400

600

800

1000

1200

1400

f(x) = 5182.99611795253 x + 739.269966835489R² = 0.0641500887785357

Graph the relationship between the density of liquids with hydrostatic pressure

the difference in height of the pipe U(cm)

dens

ity (

kg/

m3)

From the graph, shows that the pressure is proportional to the density of the liquid and depth. Can be written:

Ph ~ ρ dan Ph ~ ρ

The factors that affect the pressure experienced by objects in liquids is the

depth and density of the liquid. Can be written:

Ph ~ ρ h

Ph= k ρ h

where: k = constant

Dimensional analysis to obtain the hydrostatic pressure equation:

Ph=k ρ h

kg

m. s2 =k ( kg

m3 ) (m )

M

LT 2 =k ( M

L3 ) ( L )

M L−1T−2=k M L−2

k=M L−1T −2

M L−2

k= L

T −2

a = L

T−2 = k

because a is not a constant, then that becomes a constant in the equation is the

acceleration of gravity (g) the value of 9,8 m/s2. Thus, the hydrostatic pressure

equation can be written as follows.

Ph=g ρ h

DISCUSSION

In this experiment we get the density value at any liquids from the biggest to

the smallest, respectively, are salt solution 2 with ρ = | 121.8 ± 0.7089 | kg / m3,

Glycerin with ρ = | 1030 ± 0.6150 | kg / m3, salt solution 1 with ρ = | 989.1 ±

0.5945 | kg / m3, with ρ = | 970.0 ± 0.5849 | kg / m3 and oil with ρ = | 870 , 0 ±

0.5349 | kg / m3

       The experiment consists of two activity is the activity of the first to

investigate the relationship between the depth of the hydrostatic pressure. The

second activity to investigate the relationship between the density of liquids with

hydrostatic pressure.

1. Relationship with the depth of hydrostatic pressure

In this activity, repeated measurements were performed three times at each

depth that is used is 2 cm, 3 cm, 4 cm, 5 cm, 6 cm, 7 cm and 8 cm but we only use

one type of liquid. Liquids used in this activity, namely water. In this experiment,

the difference in height of the liquid in the pipe U is assumed as the magnitude of

the hydrostatic pressure. From the analysis of data and graphs, the data obtained

showed that the depth is proportional to the hydrostatic pressure. Where the

funnel pressed deeper into the water, the greater the difference in height in the

pipe U. It can also be proved by looking at the graph where the linear-shaped

graph with R² = 0.9998, or in other words, the degree of credibility reaches

99.98%. This is in accordance with the theory that the depth is proportional to the

hydrostatic pressure or can be written p ~ h.

2. Relationship density of liquid with hydrostatic pressure

In this activity, we investigated the hydrostatic pressure using the mass of

different types at the same depth. There are 5 types of liquids used in this

experiment include water, glycerin, oil, salt and salt solution 1 solution 2. Prior to

the experiment, we first determine the depth is used. In this experiment, the depth

used is | 5.00 ± 0.05 | cm. As in the first activity, we observe the difference in

height of liquid in the pipe U is assumed as the magnitude of the hydrostatic

pressure at the time of a funnel inserted into the various types of fluids. This

activity is performed three times repeatedly for each liquid. From the analysis of

the data, the data obtained showed that the density of the liquid is directly

proportional to the hydrostatic pressure. Where the greater the density of the

liquid, the greater the difference in height of the liquid in the pipe U. This is

consistent with the theory that the density of the liquid is directly proportional to

the hydrostatic pressure, which can be written p ~ ρ. However, the analysis of

charts, graphs depicted are not too good with the degree of truth R² = 0.0642 or R²

= 6.42%. This is caused by an error in data collection in salt solution 2. However,

four other data indicate that the density of the liquid is directly proportional to the

hydrostatic pressure.

CONCLUSION

From these experiments, it can be concluded that the hydrostatic pressure is

affected by the depth and density of the liquid. The relationship between the depth

and density of liquid with a hydrostatic pressure that the pressure is directly

proportional to the depth and density of liquid, where the larger the depth and

density of the liquid, the greater its hydrostatic pressure.

REFERENCES

Herman . 2014. Penuntun Praktikum Fisika Dasar I. Makassar: Jurusan Fisika FMIPA

UNM.

Tipler, Paul A. 2001. Fisika untuk sains dan teknik edisi ketiga jilid 1(terjemahan).

Jakarta: Erlangga.