FACULTY:ENGI NEERI NG TECHNOLOGY LABORATORY:FLUI D MECHA HANI CS EXPERI MENT:HYDROS OSTATI C PRESSURE TEST
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FACULTY OF ENGI NEERI NG TECHN HNOLOGY DEPARTME MENTOF OFCH HE EMI MI CALENGI NEERI NG TECHNOLOGY
FLUI D MECHA HANI CS LABORATORY LABORATORY I NSTRUCTI ON SHEETS
COURSE CODE
BNQ 10304
EXPERI MENTNO.
EXPERI MENT6
EXPERI MENTTI TLE
HYDROSTATI C PRESSURE TEST
DATE GROUP UPNO. LECTURER/I NSTRUCTOR/ 1) TUTOR 2) DATEOF OFREPORT SUBMI SSI ON ATTENDANCE/PARTI CI PATI ON/DI SCI PLI NE: I NTRODUCTI ON:
/5%
PROC OCEDURE:
/5%
RESULTS& CALCULATI ONS S DI STRI BUTI ON OFMA MARKS ANALYSI DI SCUSSI ONS: FORLABOR ORATORY ADDI TI ONALQUESTI ONS REPORT: CONCLUSI ON
/15% /15% /20% /15% /10%
SUGGESTI ONS& RECOMENDATI ONS
/5%
REFERENCES:
/5%
TOTAL: EXAMI NERCOMM MMENTS:
/5%
RECEI VED DATE AND STAMP:
/100%
FACULTY:ENGI NEERI NG TECHNOLOGY LABORATORY:FLUI D MECHANI CS EXPERI MENT:HYDROSTATI C PRESSURE TEST
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KOD ETI KA PELAJAR ( KEP) JABATAN TEKNOLOGIKEJURUTERAAN KI MI A FAKULTITEKNOLOGIKEJURUTERAAN
Sayadengani nimengakubahawasay at el ahmenyedi akanl aporani nidenganday ausaha say ase ndi r i .Sayaj ugamengakut i dakmener i maat aumember ise bar angbant uandal am menye di akanl apor ani nidanmembuati kr ari nidengankeper cay aanbahawaapaapayang t er sebutdidal amnyaadal ah benar .
Ke t ua
Nama:
Kumpul an
No.Mat r i ks: ( Tandat angan)
Ahl i1
Nama: No.Mat r i ks: ( Tandat angan)
Ahl i2
Nama: No.Mat r i ks: ( Tandat angan)
Ahl i3
Nama: No.Mat r i ks: ( Tandat angan)
FACULTY:ENGI NEERI NG TECHNOLOGY LABORATORY:FLUI D MECHANI CS EXPERI MENT:HYDROSTATI C PRESSURE TEST
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1.0 OBJECTIVES 1. 2.
To determine the centre of pressure on both submerged and partially submerged plane surface. To compare the centre of pressure between experimental and theoretical.
2.0 LEARNING OUTCOMES
Demonstrate the ability to conduct experiments related to fluid flow by following standard operating procedure effectively in a group.
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3.0 INTRODUCTION / THEORY The hydrostatic pressure apparatus has been designed to study on the hydrostatic of an object immersed in fluid. It will enable students to measure the static thrust exerted by a fluid on a submerged surface while allowing them to compare the magnitude and direction of the force with theory. The calculation of the magnitude, and line of the hydrostatic forces exerted by a fluid on a submerged plane surface is 3.1 Hydros!"# $or#%s O&direction '(!&% S)r*!#% important in the design of structures such as drams, bulkheads, gates, tanks, submarines etc. The pressure which the force acting over a unit surface area and varies linearly with depth. This fact enables us to describe the pressure distribution over a &onsider a plane surface of arbitrary shape and orientation submerged in a static fluid as shown in 'igure 1. If ( represents submerged surface. hen the pressure is uniform over a surface, such as static pressure on a hori!ontal surface, the resultant the local pressure at any point on the surface and h represents the depth of fluid above any point on the surface, from the force is e"ual to area times the pressure and acts through the centroid of the area. #owever, in many cases, such as li"uids basic physics, we can easily show that the net hydrostatic force on a plane surface is given by) acting on a non$hori!ontal surface, the situation is more complex. % general approach has been developed to estimate resultant force and its line of action.
∫
F = PdA
A the reliability of this approach is examined by comparing analytical values for forces and moments In this experiment, acting *1+ on a plane surface with experimental measurements of these values on the same surface on a partially submerged and fully submerged plane surface, but for more clearly to get the result of the concept
$"+)r% 1, Hydros!"# 'r%ss)r% o& '(!&% S)r*!#%
The hydrostatic force on one side of a plane surface submerged in a static fluid e"uals to the product of the fluid pressure at the centroid of the surface times the surface area in contact with the fluid. Thus, basic physics says that the hydrostatic force is a distributed load e"ual to the integral of the local pressure force over the area. (ressure acts normal to a surface, therefore the direction of the resultant force will always be normal to the surface. In most cases, since it is the net hydrostatic force that is desired and the contribution of atmospheric pressure ( a will act on both sides of a surface, the result of atmospheric pressure ( a will cancel and the net force is obtained by)
F = ρ g hcg A F = Pcg A
*2+ *+
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(cg is now the gauge pressure at the centroid of the area in contact with the fluid. Therefore, to obtain the net hydrostatic force, ' on a plane surface, 1. 2. .
-etermine depth of centroid, hcg for the area in contact with the fluid -etermine the gauge pressure at the centroid ( cg &alculate ' (cg%
%ppendix & shows the centroid and other geometric properties of several areas. The effective point of application of force which is normally called the /&enter of (ressure, &(0 of the hydrostatic force and this is not necessary the same as the centroid. The location of the resultant force is determined by integrating the moment of the distributed fluid load on the surface about each axis and e"uating this to the moment of the resultant force about that axis. Therefore, for the moment about the y axis)
F y cp
∫ y P dA
=
A
*+
%pplying a procedure similar to that used previously to determine the resultant force, we obtain)
=−
Y cp
ρ g sin θ I xx
Pcg A
≤0 *+
here, Ixx is defined as the 3oment of Inertia, or
I xx
=
∫
2
nd
moment of the area
*4+
Therefore, the resultant force will always act at a distance y cp below the centroid of the surface *except for the special case of sin5 6+. (roceeding in a similar manner for the x location, and defining I xy product of inertia, we obtain)
X cp
=−
ρ g sin θ I xy
Pcg A
*7+
here 8cp can be either positive or negative since I xy can be either positive or negative. 'or areas with a vertical plane of symmetry through the centroid, the y$axis *e.g. s"uares, circles, isosceles triangles, etc.+, the center of pressure is located directly below the centroid along the plane of symmetry, 8 cp 6. 'or most problems where we have a single, homogeneous fluid and the surface pressure is at atmospheric, the fluid specific weight γ cancels in the e"uation for 9 cp and 8cp and we have the following simplified expressions)
Y cp
X cp
=−
=−
I xx sin θ h cg A
*:+
I xy g sinθ h cg A
*;+
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3.2 Hydros!"# 'r%ss)r% O& A V%r"#!( '(!&% S)r*!#% The following figure illustrates a hydrostatic pressure demonstration setup where the center of pressure of an immersed and partial immersed vertical plane surface can be determined.
$"+)r% 2, Hydros!"# 'r%ss)r% D%-o&sr!"o&
uspended 3ass and 'ulcrum - #eight of ?nd >urface @ idth of ?nd >urface # Total -epth of Auadrant & &entroid of ?nd >urface ( &enter of (ressure of (lane >urface
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3.2.1 '!r"!( I--%r+%d V%r"#!( '(!&% S)r*!#%
$"+)r% 3, '!r"!( I--%rs%d V%r"#!( '(!&% S)r*!#%
-epth of Immersion in ater #ydrostatic 'orce exerted on Auadrant -epth of &entroid from the ater >urface -epth of &enter of (ressure from the ater >urface -istance between 'ulcrum and the &enter of (ressure
'or a partially immersed plane surface as shown in 'igure ,
A = Bd h =
*16+
d 2
*11+
Therefore, the hydrostatic force, ' *?"uation 2+ is simplified as)
F = ρ g
Bd 2 2
*12+
If the system as shown in 'igure is e"uilibrium with the moment of hori!ontal arm, then,
∑M Fh
C
fulcrum
= Fh C−WL
= WL = mg L
>ubstituting ?"uation 12 into ?"uation 1, we obtain,
hC=
mg L F
=
2mL
ρ Bd 2
*1+
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Theoretically, the center of pressure from the water surface, hB is expressed in ?"uation :)
h D = h − Y cp
= h − −
I xx sin θ h cg A
*1+
The orientation of the surface from the hori!ontal ;6E gives 1 for sin5, then
hD= h +
I xx h A
*14+
%nd
I xx
=
Bd 3 12
*'or a urface+
*17+
here, I xx is the moment of inertia of the submerged surface *please refer to %ppendix &+ The center of pressure bellow the fulcrum is finally determined by,
hC = h D+H − d
*1:+
Thus, substituting ?"uation 14 and 17 into ?"uation 1: will finally give,
hC = H −
d 3
*1;+
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3.2.2 $)((y I--%r+%d V%r"#!( '(!&% S)r*!#%
$"+)r% , $)((y I--%rs%d V%r"#!( '(!&% S)r*!#%
-epth of Immersion in ater #ydrostatic 'orce exerted on Auadrant -epth of &entroid from the ater >urface -epth of &enter of (ressure from the ater >urface -istance between 'ulcrum and the &enter of (ressure
The hydrostatic force of a fully immersed vertical plane surface, ' is determined as)
F = ρ g Ah = ρ gBD d −
D 2
*26+
If the system as shown in 'igure is e"uilibrium with the moment of hori!ontal arm, then,
∑M Fh
C
fulcrum
= Fh C−WL
= WL = mg L
Thus,
hC=
mg L F
=
mL
ρ BD d −
D 2
*21+
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Theoretically, the center of pressure from the water surface, hB is expressed in ?"uation :)
h D = h − Y cp
= h − −
I xx sinθ h cg A
The orientation of the surface from the hori!ontal ;6E gives 1 for sin 5, then
h D= h +
I xx h A D d − D + 2 12 = d − D 2 2
D BD 1 1 h ' = d − + × × D 2 12 BD d − 2 3
2
*22+
The center of pressure bellow the fulcrum is finally determined by,
hC = hD+H − d D d − D + 2 12 + H − d hC = d − D 2 2
2
*2+
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.0 EUI'MENTS MATERIALS .1 D%s#r""o& !&d Ass%-(y
$"+)r% , Ass%-(y D"!+r!- o* Hydros!"# 'r%ss)r% a+
Auadrant 3aterial Total -epth of Auadrant, # #eight of 'ulcrum above Auadrant #eight of ?nd >urface, idth, @ =ength of @alance, =
) (F& ) 266 mm ) 166 mm ) 166 mm ) 7 mm ) 66 mm
b+
c+
@alance %rm -istance between >uspended 3ass and 'ulcrum, =) 2;6 mm >et of eights) 6 gGeach
d+
Hverall -imension =ength ) 6 mm idth ) 16 mm #eight ) 66 mm
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.0 'ROCEDURE 1. 2. . . .
(lace the #ydrostatic (ressure %pparatus on top of a hydraulic @ench. 3ake sure that the apparatus is installed properly as in 'igure . =evel the apparatus using the adjustable feet facilitate by the spirit level attached. %djust the counter$weight to level the balance arm to hori!ontal position. 3ake sure that the drain valve is closed and slowly add water into the tank until the surface just touches the "uadrant base, thus establishing a datum level. 4.
4.0 RESULTS CALCULATION
Height of Quadrant, D
0.100
m
Width of Quadrant, B
0.075
m
Length of Balance, L
0.275
m
Quadrant to pivot, H
0.205
m
Denit!
1000.00
%ravit!
&.'1
"g#m$ m#2
M!ss
D%5 o*
I--%rs"o&
T5r)s
2&d Mo-%&
2&d Mo-%&
-
I--%rs"o&6 d
7$)((/'!r"!(8
$
E9%r"-%&!(6 5:
T5%ory6 5:
7;+8
7-8
7N8
7-8
7-8
6.6 6.16 6.1 6.26 6.2 6.6 6. 6.6 6.
Error <
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$or-)(! '!r"!( I--%rs%d F =
Bd 2 ρ g
h"=
?xperimental)
F =
2
2mL
$)((y I--%rs%d D ρ gBD d −
h"=
2
ρ Bd
mg L F
2
mL
=
ρ BD d −
D d − D + 2 2 h" = d − D 2 2
h" = H −
Theoretically)
d 3
2
D 2
+ H − d
'ro%r"%s o* '(!&% S%#"o&s G%o-%ry
C%&ro"d
Mo-%& o* I&%r"! 'rod)# o* I&%r"! I 99 I 9 y
Ar%!
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=.0 DATA ANALYSIS
>.0 DISCUSSIONS CONCLUSION
Pre par edby/Di sahkanol eh:
Si gnat ur e/Tandat angan: Name/Nama:DR.NORFAI ZAH BI NTIRAZALI
Appr ov e dby/Di s ahkanol eh:
Si gnat ur e/Tandat angan: Name/Nama:PM.DR.ANGZZASSARI BI NTIMOHD KASSI M
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