Modul XI
Orifice.
Prinsip dasar orifice adalah dengan diketahuinya tinggi H dan luas lubang aliran keluar fluida
:a,maka dapat dihitung atau ditentukan jumlah aliran keluar fluida Q.Alat ukur dengan prinsip dasar
diatas dikatakan orifices.
Skema prinsip dasar tersebut dapat dipahami dengan mudah seperti pada Gambar 11.1 dibawah :
Gambar.11.1. Prinsip Dasar Orifices.
Berdasarkan bentuknya (Gambar 11.1.) yang umum dipakai dihidraulics adalah:
 Orifice lingkaran
 Orifice pesegi,dan
 Orifice segitiga
Bila didefinisikan posisi dipermukaan bebas :kecepatan air atau fluida dipermukaan sama dengan
nol,tekanan udara=p0, head potential =H; dan diposisi fluida keluar kecepatan air keluar diposisi
orifice =v,head potential=0,maka dari azas Bernoulli yaitu:
v2
p
0
H 
 0  0.....(11.1)

2g 
p0
Didapat:
v  2gh........(11.2)
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Persamaan diatas disebut Teori Toricelli.
(i). Hydraulic Coefficients:
Koefisien penyusutan,cc
cc 
a0
...(11.3)
a
Bila kecepatan di Vena Contract=v0,maka konstanta kecepatan cv didefinisikan :
cv 
v0
2 gH
.....(11.4)
Maka konstanta aliran c didefinisikan :
c  cc  cv ....(11.5)
Dan didapat debit aliran fluida Q sbb :
Q  a0v0  cc cv a 2 gH
atau
Q  c  a 2 gH ......(11.6)
Berbagai macam percobaan dilakukan dengan merubah nilai a dan H;untuk orifice lingkaran
dengan diameter d,Mawson menurunkan persamaan :
1


c  0.592  0.000677

d H 
0.7636
....(11.8)
dan JIS menurunkan formula:
 1 
c  0.592  0.00069

d H 
0.75
.....(11.9)
d  0.01,..dan
dengan ketentuan :
d
H  0.0025
(d dan H dalam meter ).
Koefisien discharge(cd).
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cd 
Q
disch arg e..actual
 a
disch arg e..theory
Q
v0  a0
va

v0
a
 0  c c  c v ....(11.10)
v
a
Koefisien hambatan,cr.
Kerugian..K .E...di..Orifice
K .E. Actual
Theoretical..K .E.  K .E. Actual

K .E. Actual
v 02
v2

v 2  v 02
2g
2g

v 02
v 02
2g
cr 
2
2
 v 
 1 




1

v 
c 

 0 
 v 
c
1
 2 1  ( c )2 1
cd
cv
c c2  c d2
.....(11.11).
c d2
Contoh-soal
1.Hitunglah debit fluida melalui orifice lingkaran dengan diameter 3 cm pada head potential 5
cm,bila koeficien discharge cd=0.62.Juga hitunglah jet fluida bila cv=0.94.
Jawab.
Debit aktual=cd x debit teori
Qa  c d  Q  c d  a  v

3 2
) 
4 100
 0.00434m 3 / det .
 0.62 
(
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2  9.81  5
Dr. Ir. Abdul Hamid M.Eng. MEKANIKA FLUIDA
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Dari pers.(11.4):
v0  cv 
2 gH
 0.94  98.1
 9.32m / det .
(ii ). Menentukan Koefisien Hydraulik di Laboratorium.
 Koefisien kecepatan cv (untuk vertikal orifice )
x  v0 t
y  1 / 2 gt 2
Maka.. y 
v0 
cv 
1
x 2
g(
)
2
v0
gx 2
2y
vo
v
1/ 2
 gx 2 


2y 


2 gH 1 / 2
 x2 
 

 4 yH 
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2.Bila dalam soal 1 diatas ,diameter jet fluida di vena contract terukur=2.4 cm,hitunglah koefisien
kecepatan,cv.
Jawab:
cv 
cc
cd
cc
1
 ( 2.4  10  2 ) 2
a0

 4
 0.64
1
a
2 2
 (3  10 )
4
untuk cd=0.62,maka:
cv 
0.62
 0.969
0.64
(iii). Discharge melalui Ukuran Besar Orifice Pesegi Vertikal.
Ambil suatu strip elemen kecil pada kedalaman h dengan tebal dh sehingga kecepatan melalui
strip tersebut=(2gh)
Discharge melalui strip :dQ=cd x luas x velocity
= cd x b x dh x (2gh).
Jumlah discharge:
Q
H2
c
d
 b  dh 
2 gh  c d .b. 2 g
H1

h
1/ 2
dh
H1

2
c d .b. 2 g H 23 / 2  H 13 / 2
3
3.Orifice pesegi
H2

dengan ukuran lebar =1.5m dan tinggi =0.75 m,dipasang disisi tanki.Tinggi
permukaan air=75cm,diukur dari pinggir atas orifice.Hitunglah dischargenya,bila cd=0.6.
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Jawab:


2
c d .b. 2 g H 23 / 2  H 13 / 2
3
3
3


2
2
  0.6  1.5  2  9.81 1.5  0.75 2 
3


3
 3.15m / sec .
Q
(iv). Totally Submerged Orifice.
Q  c d  luas..orifice  velocity
 cd  a 
gH
di sin i : H  H 2  H 1
(v). Partially Drowned Orifice.
Q1  c d  a 
2 gH

2
c d  b  2 gH H 3 / 2  H 13 / 2
3
Total..Disch arg e :
Q2 

Q  Q1  Q2
4.Submerged orifice keseluruhan dengan lebar 1.5m mempunyai tinggi air dari pinggir atas dan
bawah orifice masing2 setinggi 3m dan 3.75m.Hitunglah jumlah debit fluida bila perbedaan
permukaan fluida pada kedua sisi orifice adalah 0.5 m, dan diketahui cd=0.6.
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Jawab:
Debit fluida,Q=cd x a x (2gH)
a  bH 2  H 1 
Q  0.61.5(3.75  3)
2  9.81  5
 0.6  1.125 2  9.81  5  2.115m 3 / sec .
5.Suatu orifice berbentuk pesegi dengan ukuran lebar 1.2 m dan tinggi 0.6 m dipasang pada suatu
tangki besar.Bila tinggi permukaan air terukur 2.4 m diukur dari pinggir atas orifice dan tinggi
permukaan air pada sisi lain terukur 0.3 m dibawah pinggir atas orifice,hitunglah discharge per
second,bila c=0.62.
Jawab:
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Total Discharge=Discharge melalui bagian orifice yang tenggelam + discharge bagian yang bebas
atau:
Q  Q1  Q2
Q1  c d  a 
2 gH
di sin i : a  1.2  0.3
c d  0.62, dan..H  2.7

2
c d  b  2 g H 3 / 2  H 13 / 2
3
0.62  1.2  0.3 2  9.81  2.7 
Q2 


2
 0.6  1.2 2  9.81 2.7 3 / 2  2.4 3 / 2
3
 1.627 m 3 / sec .  1.605m 3 / sec .

 3.232m 3 / sec .

The co-efficient of discharge (Cd). This may be found by a. simple method. The quantity of
liquid coming out of an orifice under a known head can be collected for a known time and
its volume measured. Then the quantity converted to per unit time divided by the
theoretical discharge will give the co-efficient of discharge.
(c) The co-efficient of contraction (CC). This co-efficient can be found by actually measuring the
area of jet at vena contracta and then divided by the area of the orifice. The measuring of dia of the
jet at vena contracta is done by micro-screw equipment as shown in Fig. 5 But this is very difficult
job and needs time and patience. So this is generally determined from the following relation.
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Fig. 5
Example 5-2. If the dia of the jet at vena contract is measured by screw gauge and is found 2.4
cm. Find the coefficient of velocity.
Solution. Area of the orifice
a

3 2
)
4 100
(
Area of the jet at vena contracta
ac 
 2 .4
(
4 100
)2
Co-efficient of contraction
Cc 
ac  2.4 2
 ( )
a
4 3
=0-64
But co-efficient of discharge
Cd=0.62
Therefore co-efficient of velocity
Cv 
C d 0.62

CC 0.64
=0-969.
Ans.
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Example 5-3.
cm.
A jet of water is coming out of a sharp edged vertical under a constant head of 10
At a certain point of the jet, the horizontal and vertical co-ordinates measured from
vena contracta are 40 cm and 42 cm respectively.
the
Determine Cv.If Cd = 0.62 then find Cc.
[A.M.I.E. May 1965, converted to metric- units].
Solution. Given : Horizontal co-ordinate
x=40 cm
and Vertical co-ordinate, y=42 cm
Head of water, H= 10 cm
.*. Co-efficient of velocity,
Cv 

x2

4 yH
( 40) 2

4  42  10
=0.976.
Ans.
and also co-efficient of contraction
Cv 
Cd
0.62

CC 0.976
=0-635.
Ans.
Example 5 4. Two jets of water are issuing from two similar orifices fitted in a vertical side of a
tank. The head of water at upper and lower orifice is 3 m. and 6 m. respectively. Find the coordinates of the point where the two jets intersect. Take Cv for both the orifices 0.94.
Solution. Let x and y be the horizontal and vertical co-ordinates of the point of intersection for the
lower orifice.
Then the corresponding co-ordinates for the upper orifice be x and (y+3) as shown in Fig.
5-4.
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Fig.5-4
Then for lower orifice
x2

4 yH
Cv 

…….(i)
( x) 2

4 y 6
.and for upper orifice
Cv 

x2

4( y  3) H
…(ii)
2
( x)

4  ( y  3)  H
Since Cv for both the orifices is same, then from equations (i) and (ii), we have
Cv 
x2
x2

24 y
12( y  3)
whence y=3 metres.
Ans.
and putting the value of y in Eq. (i), we get
x2
 (0.94) 2 whence x = 7.97 m.
24  3
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Ans.
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
To Find out the Time of Emptying Vessels of Various Types
(i) Vessel of uniform section. In Fig. 5-9 is shown a vessel of uniform cross-sectional area
The vessel is
having
"A".
liquid originally up to the height of H1 and the .liquid is discharged
Fig. 5-9
through an orifice at the bottom of the vessel. Let its level falls to H2 in T seconds. Naturally, the
rate of discharge will decrease as the level of the liquid falls.
Let A=Level of the liquid in the vessel at a given instant.
dh=Fall of level of liquid
dT=Time taken in sec. for the fall of level dh.
v= Velocity of liquid passing through the orifice under head h
2 gh
dQ=Discharge of the liquid for the period dT sec.
Then
dQ   A.d h  Cd .av.d T
where a=Area of the orifice.
and minus sign indicates that the level of the liquid is decreasing as the discharge is taking place
through the orifice.
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Total time taken
If the vessel is to be emptied completely, then H2=0 and taking H1=H
'
Problem 1.A rectangular tank 5 metres long and 3 metres wide contains water to a depth of 1.2
metres. The water is discharged through an orifice of an area of 0.2 sq. metres, made at the bottom
of the tank. What will be the time taken in emptying the tank completely ? Take Cd=0-64.
Problem 2. A tank of cross-section 2m X m contains water upto the height of 4 m.
An orifice of
5 cms. dia is made at the bottom of the tank.Find the head of the water level after 3 min.if the coefficient of discharge is taken as 0.62.
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