Mata Kuliah FT2012A:Mekanika Fluida, Senin 10:30-13:00
Mekanika Fluida Kamaruddin Abdullah
Silabus Kode
Kredit:
Semester:
Bidang Pengutamaan:
Sifat:
2180
3
04
Teknik Mesin
Kuliah Wajib
Sifat Kuliah
Kuliah Mekanika Fluida
Nama Mata Kuliah
Dalam Bahasa Inggris Fluid Mechanics
Silabus Ringkas
Memberikan pemahaman yang komprehensif mengenai mekanika fluida, yang menyangkut sifat-sifat fluida, aliran fluida dalam pipa dan sekeliling benda padat, mesin-mesin hidraulika Comprehensive introduction concerning fluid mechanics, related to fluid properties, flow in pipes and on immersed bodies, and fluid machinery
Silabus
Silabus Lengkap
Kuliah ini memberikan pemahaman komprehensif mengenai teori dan aspek praktis mekanika fluida, dimulai dengan pembahasan sifat-sifat fluida, aliran laminer dan turbulen dalam pipa dan disekeliling benda padat, serta mesin-mesin hidraulika. Isi kuliah mencakup konsep perpindahan momentum, berupa sifat kekentalan fluida yang akan mempengaruhi distribusi aliran dalam pipa dan disekeliling benda padat, metoda pengukuran aliran fluida, perhitungan yang menyangkut penurunan tekanan yang terkait dengan daya pompa, atau kipas serta daya yang dapat dibangkitkan dari aliran fluida melalui turbin. This course introduces students to comprehensive understanding concerning fluid mechanics, and its practical aspect in industries, fluid properties, laminar and turbulent flow in pipes and surrounding The content of the course includes the concept of momentum transfer as related to viscosity, flow characteristics, flow distribution in pipes and surrounding solid body, method of fluid measurement, calculation of pressure drop and power required by pumps and blowers and power generated by turbine.
Silabus Tujuan Instruksional Umum (TIU) Luaran (outcomes) Mata Kuliah Terkait
Mahasiswa mempunyai dasar pengetahuan dan wawasan yang kuat mengenai mekanika fluida, serta penerapannya di dunia kerja. Mahasiswa mampu menerapkan dan memanfaatkan pengetahuannya untuk menyelesaikan persoalan mekanika fluida pada kuliah terkait atau pada dunia kerja sesuai profesinya. - Perpindahan panas dan massa 1.
Pustaka
Reymon C. Binder.:” Fluid Mechnics”, 4th edition, Prentice-Hall Asian Edition, 1962. 2. J.M. Gasiorek and W.G. Caeter, Mechnics of Fluids for Mechnical Engineers. London, Blackie &Sons-Glasgow. 3. Handout kuliah
SAP(Satuan Acara Perkuliahan) Minggu ke Isi kuliah
Dosen
1-2
Konsep dasar mekanika fluida, tekanan fluida, alat ukur tekanan, sifat transport terutama mekanisme perpindahan momentum,viskositas
KA
3-4
Aliran fluida, aliran laminer,turbulen, HagenPoiseulle, Stokes Pengukuran fluida
5-6
Penurunan tekanan dalam pipa, D’Arcy Weisbach, persamaan Colbrook, metoda online, tahanan dalam pipa, fitting
7-8
Pengukuran fluida dengan Re rendah
9
UTS
10-11
Kipas dan pompa, karakteristik pompa, Daya, efisiensi, Hukum Euler
Tugas
Praktikum Tugas
SAP Minggu ke
Isi kuliah
Dosen
Tugas
12
Aliran sekeliling benda padat, koefisien gesek (drag)
KA
PR
13-14
Mesin-mesin hidro, turbine, kincir air, kincir anginl
PR
15
Pembahasan umum, peranan mekanika di industri
PR
16
UAS
Konsep dasar dan sifat-sifat fluida • Fluida secara umum menyangkut baik gas maupun cairan. • Gas merupakan fluida yang dengan mudah dimampatkan dan dapat mengisi seluruh volume ruang/tangki pengisinya. • Cairan tidak dapat dimampatkan dan hanya dapat mengisi sebagian dari ruang atau tangki yang mengisinya. • Cairan mempunyai permukaan, seperti permukaan danau sungai sedangkan gas tidak mempunyai permukaan. • Dalam studi fluida cair baik dalam kondisi diam atau mengalir fluida dianggap sebagai fluida sempurna atau fluida ideal, yang mempunyai sifat tak termampatkan dan tidak memberikan hambatan terhadap aliran atau dengan kata lain tidak mempunyai tahanan gesek.
Tekanan Fluida • Hukum Pascal: • Bila sebuah benda piramida kecil ditenggelamkan kedalam sebuah bencana sampai ke kedalam h, maka tekanan luar dari arah vertikal, samping dan horizontal akan sama. • P1=p2=p3
p1
P2
P3
h
Hukum Hidrostatika • Tekanan oleh suatu kolom fluida dinyatakan oleh persamaan berikut: • • • •
P= h (1.1) Dimana , bobot spesifik (N/m3) P, Tekanan dan N/m2 h, ketinggian kolom (m)
Tekanan ukur (gage pressure) dan tekanan mutlak (absolut pressure)
Skala tekanan
Tekanan ukur positip
Tekanan atmosfir Tekanan ukur negatip Tekanan absolut
Tekanan mutlak
Datum line
Sistem hidraulika F1 W A1
A2 Pipa (bejana berhubungan)
F1 x A1= WxA2 F1=W A2/A1 atau W=F1*A1/A2
Pressure gage • Bordon gage – Gage pressure
Aneroid gage (absolute pressure)
Barometer air raksa Vacuum
Air raksa
Udara
Sensitive manometer • Sin = x/y • DP=P2-P1=x=ysin
Contoh soal manometer sensitip ketinggian kolom-h (5 mm) h (mm)
radian
250
y (mmm)
200
derajat
5
0.02618
1.5
191.008
y(m)
150
y (mmm)
100
0.031416
1.8
159.181 50
0.05236
3
95.5361 0
0.10472
6 47.83386
1.5
1.8
3
6
sudut kemiringan (derajat)
Contoh soal • Dari Gbr disebelah tentukan : • 1). tekanan ukur (gage pressure) udara dalam pipa dalam mm Aq, bila diketahui manometer berisi oli dengan BD=0.8 • Jawab: • Karena pada bidang yang sama dalam manometer tekanan ukur sebelah kiri kaki manometer sama dengan tekanan sebelah kanan: • (dengan mengabaikan tekanan udara pada manometer) • Hx+30 (mm)* 0.8=0 • Hx=-24 mm Aq
x
30 mm
Contoh soal •
Perbedaan tekanan antara titik A dan B pada pipa horizontal yang mengalirkan minyak dengan BD 0.89 diukur dengan menggunakan manometer U terbalik.
• •
.
Masing masing lengan manometer U dihubungkan dengan pipa pada A dan B.. Tekanan udara perada pada 100 lbf/in.2 dipompa kedalam bagian atas manometer untuk menekan minyak sehingga meniskus manometer berada pada kondisi yang seimbang. Bila bobot spesifik udara. Bila titik A berada 10 ft diatas meniskus yang terkoneksi dengan kaki manometer tentukan. A) Perbedaan tekanan antara A dan B dalm lbf/in.2 untuk perbedaan 10 ft diatas meniskus. B). Tekanan pada titik A dalam Lbf/in.2
A
B x
PA Z=10 ft m
Udara pada tekanan Pa n h=10 in.
PB
Jawab • A). Dengan menyamakan tekanan pada titik m dan n :
pm PA hwa zwo dan pn PB hwo zwo dimana wa bobot spesifik dari udara dan wo bobot spesifik dari minyak karena pm pn pa maka p A-hwa pB hwo
pB p A h( wo wa ) b).Karena
10 62.4 x0.89 0.52 ( ) 0.318lbf / in.2. 144 144
pm PA hwo zwo pa
Jadi PA pa hwa zwo pa 100
10 0.52 62.4 x0.89 x 10 x 96.14lbf / in.2 144 144
Perpindahan momentum dan kekentalan fluida • Hukum Newton mengenai kekentalan – Perhatikan suatu aliran fluida (baik gas atau cairan) yang mengisi dua plat sejajar berukuran besar dengan jarak antara keduanya adalah Y. Pada saat kondisi diam (t=0) plat bagian bawak kemudian ditarik ke-arah sumbu x dengan kecepatan konstan V. Sejalan dengan waktu maka fluida akan mendapatkan momentum dan akhirnya berada pada kondisi mantap (steady –state). Untuk mempertahankan kondisi ini diperlukan gaya F untuk mempertahankan aliran plat bagian bawah. Besarnya gaya tsb. dapat dinyatakan oleh hubungan
Atau
𝐹 𝐴
𝑉 𝑌
(1)
𝑑𝑉𝑥 𝑑𝑦
(2)
=
yx=
yx
dVx/dy
Teori Kinetika gas untuk kekentalan • = • • • •
2
3
2 (mT)/ d 32
Dimana m, massa gas , konstanta Boltzaman d, diamter tabrakan gas
(g/m det)
Hukum Kekentalan Newton
v=/r
=kg/m sKg/m3
Sifat Udara
Pr=Cp /l =J/kg C Kg/m.s/W/m C
Suhu -t(oC)=
Kerapatan -ρ(kg/m3)
Panas jenis - cp (kJ/kg K)
Konduktivitas termal -l(W/m K)
Kekentalan kinematik -ν(m2/s) x 10-6
Koeffisien pemuaian -b(1/K) x 10-3
Bilangan Prandtl's - Pr -
-150 -100 -50 0 20 40 60 80 100 120 140 160 180 200 250 300 350 400
2.793 1.980 1.534 1.293 1.205 1.127 1.067 1.000 0.946 0.898 0.854 0.815 0.779 0.746 0.675 0.616 0.566 0.524
1.026 1.009 1.005 1.005 1.005 1.005 1.009 1.009 1.009 1.013 1.013 1.017 1.022 1.026 1.034 1.047 1.055 1.068
0.0116 0.0160 0.0204 0.0243 0.0257 0.0271 0.0285 0.0299 0.0314 0.0328 0.0343 0.0358 0.0372 0.0386 0.0421 0.0454 0.0485 0.0515
3.08 5.95 9.55 13.30 15.11 16.97 18.90 20.94 23.06 25.23 27.55 29.85 32.29 34.63 41.17 47.85 55.05 62.53
8.21 5.82 4.51 3.67 3.43 3.20 3.00 2.83 2.68 2.55 2.43 2.32 2.21 2.11 1.91 1.75 1.61 1.49
0.76 0.74 0.725 0.715 0.713 0.711 0.709 0.708 0.703 0.70 0.695 0.69 0.69 0.685 0.68 0.68 0.68 0.68
Sifat air Temperature -t(oC)
Dynamic Viscosity -µ(N s/m2) x 10-3
Kinematic Viscosity -ν(m2/s) x 10-6
0
1.787
1.787
5
1.519
1.519
10
1.307
1.307
20
1.002
1.004
30
0.798
0.801
40
0.653
0.658
50
0.547
0.553
60
0.467
0.475
70
0.404
0.413
80
0.355
0.365
90
0.315
0.326
100
0.282
0.294
Metoda pengukuran kekentalan fluida •
Pers. Stokes
• •
Vs adalah kecepatan jatuh bola (m/s) (jatuh dari atas ke bawah bila ρp > ρf, mengambang keatas ρp < ρf), r ,adalah jari-jari bola yang mengikuti hukum Stoke (m), g, percepatan gravitasi bumi (m/s2), ρp ,kerapatan bola (kg/m3), ρf ,kerapatan fluida (kg/m3), dan μ , kekentalan dinamik (Pa s).
• • • • •
•
percobaan
Alat Ukur Oswald • Pers. Hagen-Poiseulle
( Po PL ) R 4 Q . 8L • Radius kapilari dihitung dari volume minyak yang diukur kekentalannya. V R
R 2 L 4 4V L
Alat ukur aliran fluida • • • •
Manometer Orifice Venturi Pitot Tube
Pipa pitot (pitot tube) Pitot tube ialah pipa terbuka kecil dimana permukaannya bersentuhan langsung dengan aliran. Terdiri dari 2 pipa, yaitu : • Static tube (untuk mengukur tekanan statis) Pipa ini membuka secara tegak lurus sampai ke aliran sehingga dapat diketahui tekanan statisnya.
• Impact/stagnation tube (untuk mengukur tekanan stagnasi = velocity head) Impact pressure selalu lebih besar daripada static pressure dan perbedaan antara kedua tekanan ini sebanding dengan kecepatan.
SKEMA PIPA PITOT (1)
SKEMA PIPA PITOT (2)
Cara kerja pitot tube : 1 rV 2 , ( Bernoulli ) Pipa yang mengukur tekanan statis 2 terletak secara radial pada batang yang dihubungkan ke manometer (pstat) V 2( p0 pstat ) / r p0 pstat
V C 2( p0 pstat ) / r P0 = stagnation pressure Pstat = static pressure
Prinsip dari pitot tube :
Energi kinetik dikonversikan menjadi static pressure head.
Tekanan pada ujung pipa di mana fluida masuk merupakan tekanan stagnasi(p0) Kedua pengukuran tekanan tersebut dimasukkan dalam persamaan Bernoulli untuk mengetahui kecepatan alirannya Sulit untuk mendapat hasil pengukuran tekanan stagnasi secara nyata karena adanya friksi pada pipa. Hasil pengukuran selalu lebih kecil dari kenyataan akibat faktor C (friksi empirik)
Kegunaan pipa pitot - mengukur tekanan fluida pada wind tunnel - menghitung profil kecepatan aliran pada pipa
Kegunaan Pipa pitot (2) V
1
z1-z2
2
l
y
Aplikasi pipa pitot • Mengukur kecepatan pada pesawat (airspeed) • Altimeter pesawat • Mengukur tekanan fluida pada wind tunnel
(terowongan angin)
Kelebihan dan Kekurangan Kelebihan : Susunan sederhana Relatif mudah dan murah
Tidak perlu adanya kalibrasi Pressure drop aliran kecil Kekurangan : Keakuratan rendah untuk beberapa aplikasi Pipa harus lurus dengan kecepatan aliran untuk mendapatkan hasil yang baik
Kesimpulan • Sensor aliran berfungsi untuk mengukur jumlah dari material fluida yang melewati sebuah titik pada suatu waktu dalam pipa • Jenis sensor aliran ada 3 yaitu : pressure-based flow sensors, turbine flow sensors, dan magnetic flowmeters
• Pitot tube merupakan sensor aliran jenis pressure-based flow sensors. Sensor ini menggunakan prinsip perbedaan tekanan
untuk mengetahui kecepatan aliran dengan persamaan Bernoulli.
Integral numerik-Metoda Simpson • Integral numerik untuk mencari laju rata-rata aliran setelah diukur dengan Pitot Tube
Untuk N=3 = (b-a)/3(fa)+2f(m)+ f(b))
Orifice meter • Persamaan untuk mengukur debit aliran
Bernoulli P1/ +z1+V12/2g= P2/ +z2+V22/2g M=r1V1A1= r2V2A2
Orifice meter vortices
D
d
Pressure tappings o
d1
1
Cd= area at vena contarcta/orea of orifice=a1/a=(d1/d)2
• Bernoulli’s equation 2
2
p0 v0 p v 1 1 where z 0 z1 0 w 2g w 2g p0 p1 ) 2 2 v1 v0 2 g ( w and from continuity Av0 a1v1 or v1 Av0 / a1 but the area at vena contracta a1 is usually not known and therefore it is easier to replace by area of the orifice a, and the coefficient of contraction which can be determined experimentally
• Since Cd a1 / a
or
a1 Cd a
we have v1
Avo v 1 2 p p1 2 o therefore, vo [( ) 1] 2 g 0 Cd a Cd m Cd m w
or the velocity in the pipe v0
Cd m 2 g [1 (Cd m) 2 ]
p0 p1 w
which is a theoreticl velocity and must be corrected thus actual velocity theoretical velocity and therefore the actual velocity in the pipe is given by Cv
(v0 )act
Cv C d m 2 g [1 (Cd m) 2 ]
p0 p1 w
Online calculation of Orifice Inputs Pipe (inlet) diameter upstream of orifice, Di: Orifice diameter (less than the inlet diameter), Do: Pressure difference across the orifice, Dp: Fluid density, r: Flow Coefficient, Cf:
Velocity at the inlet, Vi:
1.76 m/s
Volumetric Flowrate, Q:
13.9 l/s
Mass Flowrate:
0.0179 kg/s
10 8 10 1.29 0.7
cm cm Pa Kg/m^3
Tabung venturi • Pengukuran laju aliran
M1=rV1M=rV2M
a2 v2 p2 a1
v1 z1
p1
A B
D C
w1
h
z
z2
Hukum Bernoulli 2
2
p1 v1 p2 v2 ( z1 ) ( ) z2 w 2g w 2g p1 p2 2 2 v2 v1 2 g ( z1 z2 ) w Dari persamaan kontinyuitas a1v1 a2v2 2
v1 [(
atau
v2 a1v1 / a2
a1 2 p p2 ) 1] 2 g ( 1 z1 z2 ) a2 w
atau v1
2g ( a1 / a2 ) 2 1)
(
p1 p2 z1 z2 ) w
• Dan besarnya debit Q Q a1v1
a1 2 g p1 p2 ( z1 z2 ) w ( a1 / a2 ) 2 1
untuk fluida nyata Q
Cv a1 2 g p1 p2 ( z1 z2 ) 2 w ( a1 / a2 ) 1
• Dengan mempertimbangkan manometer p A p1 w( z1 z h) pB p2 wh p1 wz1 wz wh wh p1 wz1 wz PD p2 w( z2 z h) pC pD w1h p2 wz2 wz wh w1h
(1)
• Tetapi
pC pB
p2 wz2 wz wh w1h p1 wz1 wz p1 p2 wz2 wz1 h( w1 w) •
(2)
Dengan mesubstitusikan pers.(2) ke pers. (1) Q
Cv a1 2 g wz2 wz1 h( w1 w) [ z1 z2 ] 2 w (a1 / a2 ) 1 Cv a1 2 g h( w1 w) [ ] 2 w (a1 / a2 ) 1
• Yang menyatakan bahwa bacaan manometer tidak tergantung dari kemiringan dari manometer.
Contoh soal • Sebuah venturi meter vertikal digunakan untuk mengukur aliran minyak dengan bobot spesifik 0.82 dan mempunyai pimtu masuk 5 in. Dan leher 2 in. Terdapat pengukur tekanan pada pint masuk dan padaleher venturi yang berada 12 in. Diatas pintu masuk. Bila koefisien debit Cd =0.97 tentukan nilai debit aliran bila perbedaan tekanan adalah 4 lbf/in.^2. • Jawab: • Area pintu masuk 5 2 a1 1 / 4 (5) 2 a1 1 / 4 ( ) dan 6.25 12 a2 1 / 4 ( 2) 2 p1 p2 4 x144lbf / ft.2
w 0.82 x 62.4lbf / ft.3
z1 z 2 1 ft......................Cd 0.97 Q Q
25 144 [ 2 g ( 4 x144 1)] 0.82 x 62.4 [(6.25) 2 1
0.97 x1 / 4x
0.132 x 25.7 0.549 ft 3 / s 6.17
Callibration of venturi meter Bleed valve
Flow 2 in. Diam.pipe
Discharge valve
Mercury manometer
Collecting tank
Drain valve Pump
Sump
Contrilvalve
The apparatus • The venturi meter was situated in a 2 in.diametr pipe line and arranged in closed circuit with centrifugal pump supplying the flow of water from the sump. The rate of flow was controlled by the control valve at the discharge end of the 2 in. Diameter pipeline. It was measured by using a collecting tank which was situated above the sump so that the collected water could be drained from the colecting tank directly into the sump. • The venturi meter tested was a conical type manufactured by Messrs. Guest& Chimes Ltd, having a 2 in. Dimeter inlet and i in, diametr throat. The manometer used was a drowned U–tube type using mercury as the manometer liquid
The procedure • •
• •
•
With the control valve and the pump discharge valve shut, the pump motor was started and then the pump dischrage valve was opened gradually to the fully open positions. The control valve wa opened slightly so as to permit a small flow of water through the system. Any air present in the system was allowed to escape by bleeding off some water through bleed valve situtated at the highest point of the 2 in. Diameter pipe and at the top of the manometer. The control valve was fully shut and the zero reading of the manometer checked. The contril valve was again opened slightly and when steady conditions were obtained the reading of mercury manometer was taken and the discharge was determined using the collecting tank. This was done as follows:the drain valve was shut and when the water level in the tank had reached a 3 in. Mark on the level gage the stop clock was started. The time taken for the level to rise a specific height was measured and recorded. The collecting tank was then drained through the drain valve. The procedure was repeated for larger value opening and hence greater flows. In all 20 test were carried out
Results .
Tets No D h (in Hg) Level rise (in.) Time (s) Q(ft3/s) h(in.of water) 1 6.4 12 68 0.11 6.71 2 5.38 12 73 0.102 5.64 3 5.16 12 76 0.098 5.38 4 4.76 12 79 0.094 5.01 5 4.57 6 40.5 0.092 4.8 6 3.83 6 44 0.0846 4.01 7 3.65 6 45.5 0.082 3.83 8 3.38 6 46 0.081 3.54 9 2.88 6 50.5 0.0737 3.02 10 2.7 6 52.5 0.071 2.83 11 2.2 6 58 0.0643 2.31 12 2.13 6 59 0.0631 2.23 13 1.9 6 61 0.0611 1.99 14 1.51 6 69 0.054 1.58 15 1.3 6 77 0.0484 1.36 16 0.9 6 87.5 0.0426 0.945 17 0.85 6 97 0.0384 0.891 18 0.55 6 109 0.0342 0.576 19 0.35 6 145 0.0257 0.367 20 0.15 3 122 0.0153 0.157
Discharge Q (ft3/s)
0.11
0.102
0.098
0.094
0.092
0.0846
0.082
0.081
0.0737
0.071
0.0643
0.0631
0.0611
0.054
0.0484
0.0426
0.0384
0.0342
0.0257
Water head h in ft of water 8
7
6
5
4
3
2
1
0
Q(ft3/s) h(in.of water) log Q Log h 0.01 0.08 2 -1.09691 0.02 0.23 -1.69897 -0.63827 0.03 0.5 -1.52288 -0.30103 0.04 0.87 -1.39794 0.136721 0.05 1.37 -1.30103 0.296665 0.06 1.98 -1.22185 0.432969 0.07 2.71 -1.1549 0.553883 0.08 3.58 -1.09691 0.655138 0.09 4.52 -1.04576 0.745855 0.1 5.57 -1 0.822822 0.11 6.65 -0.95861 0.822822
• • • • • • •
Internal dimension of collecting tank : 29.9 in. X 35.9 in. Surface area of collecting tank: 7.45 ft^2 Venturi head h=[(13.61-1)/12] D h=1.05 D h ft of water Sample calculation: Test 1. H=1.05 D h=1.05 x 6.4 =6.71 ft. of water Q= volume collected/time= 7.45x level rise/time=7.45x12/(12x68)=0.11 ft3/s • The relation betwen Q and h can be represented by
Q Kh n Then LogQ log K n log h in which K CdxEA2 2 g
(2.10.1)
• N from theory is expected to be equal to 2. Eq. (2.10.1) is linear and thus if log h is plotted against log Q the slope of the line will give the value of n and the intercept will be equal to log K. • From the graphn=slope=97/48.7=1.993=2 and the intercept log K= 2.63 • And K=0.04266 (meter coefficient) • From the above value the means coefficient of discharge can be determined, since K CdxEA2 2 g
• So that Cd
K EA2 2 g
but 1 E and 2) (1 m
A2 ( D2 ) 2 m (1 / 2) 2 0.25 2 A1 ( D1 )
E
1 1 2 2 1 . 0327 and A 2 1 / 4 ( ) 0 . 00545 ft 12 (1 (0.25) 2
Cd
0.04266 0.951 1.0327 x0.00545 x 64.4
Pengukuran aliran fluida dengan Re kecil-kalibrasi dengan cup anemometer
5 10 15 20
L
V(m/s) mb
V=f()
L
Db
mb
Analisis Dimensi • Hasil percobaan dapat di sajikan dalam bentuk korelasi dari persamaan tak berdimensi, menggunakan teori Buckingham,. • Andakan: V=f(m,L,g,D,r ,) G
• V=[M0L1T-1]=[M1L0T0]a[M0L1T0]b[M0L1T-2]c[M0L1T0]d[M1L-3T0]e[M0L0T0]f
• M: 0=a+e • L: 1=b+c+d-3e • T: -1=-2c
Analisis dimensi • Dari hubungan diatas didapat, • c=1/2 3 V mL a D d • a=-e ( ) k( ) ( ) ( gL)1/ 2 sin( ) rG L • b+1/2+d-3 e=1 gL • b+d=0.5-3a • b=0.5+3a-d • d=d • sehingga kita akan memndapatkan 5 bilangan tidak berdimensi
BILANGAN REYNOLDS Re=rVD/ .............................(3.1.) Atau Re= VD/................................(3.2) Aliran sekeliling pipa silinder 49
•Vs,kecepatan fluida, (m/det) •L - panjang karakteristik, (m), diamater pipa atau diameter hidraulik, jarak tempuh aliran •μ – kekentalan mutlak, (Kg/m det) •ν - viskositas kinematik fluida: ν = μ / ρ, (m2/det) •ρ - kerapatan (densitas) fluida. (kg/m3)
Wetted perimeter • Untuk aliran dengan penampang bukan silinder berlaku persamaan untuk menentukan diameter hidraulik (kesetaraan diamater untuk silinder) 4 Ax De P
• Dimana Ax,luas penampang aliran dalam pipa/saluran (m2),P, keliling pipa/saluran terbasahi (m), De, diameter setara/wetted perimeter (m)
ALIRAN DALAM PIPA • Jarak aliran masuk pipa untuk mencapai aliran mantap (Entrance Length) • Jarak dari pintu masuk pipa sampai aliran membentuk ditribusi aliran mantap dapat dinyatakan dengan bilangan tak berdimensi yang disebut Bilangan Jarak Masuk (Entrance Length Number ) , EL dan dinyatakan oleh rumus berikut • E L = Le / d (4.1) • dimana • EL = Bilangan Jarak Masuk (-) • Le = Jarak mencapai profile kecepatan yang mantap (m) • d = diameter pipa (m)
Bilangan jarak masuk kedalam pipa • Bilangan Jarak Masuk Untuk Aliran Laminer • Nilai EL dapat dibuat persamaan korelasinya dengan bilangan Re untuk aliran laminer sbb. • ELaminar = 0.06 Re (4.2) • Bilangan Jarak Masuk Untuk Aliran Turbulen • Nilai EL dapat dibuat persamaan korelasinya dengan bilangan Re untuk aliran turbulen adalah sbb. • ELturbulent = 4.4 Re1/6 (4.3)
Penurunan Tekanan Dalam Pipa Saat Pemompaan/kipas • Penurunan tekanan dalam pipa /tabung atau saluran berbagai penampang pada saat aliran dipompa maka karena adanya gesekan oleh dinding pipa akan terjadi kehilangan energi sehingga daya dorong aliran menjadi berkurang. Perhitungan semacam ini sangat penting dalam perhitungan daya pompa atau blower. Besarnya penurunan sepanjang pipa dapat dihitung dengan menggunakan persamaan D'Arcy-Weisbach secara umum sebagai berikut. Pw=QH=QDP • Δp = λ (l / dh) (ρ v2 / 2) (4.4) • dimana • Δp = penurunan/kehilangan tekanan aliran (Pa, N/m2) • λ =koefisien gesekan D'Arcy-Weisbach (-) • l = panjang saluran atau pipa (m) • dh = diameter hydraulik (m)-luas penampang/panjang keliling terbasahi • ρ = kerapatan fluida (kg/m3)
Diagram Moody
Diagram Moody
Head Total K V= Re= e f= Z1-4= z1=
Z1-4 2.2 3.055775 1.17E+05 1.31E-06 0.0052 0.032 28.18479 32.18479
m/s
m2/s
Koefisien gesekan, l atau Cf • Aliran laminer Re<2100 • Cf= 64/Re • Turbulent Re>10000
Keterangan • f, is the Darcy friction factor • e, Roughness height, (m, ft) • Hydraulic diameter, DH(m, ft) — For fluidfilled, circular conduits, = D = inside diameter • Hydraulic radius, Rh (m, ft) — For fluid-filled, circular conduits, = D/4 = (inside diameter)/4 • is the Reynolds number
Perhitungan koefisien gesekan dalam pipa dan saluran. • Koerfisien gesekan yang digunakan untuk menghitung kehilangan tekanan dalam pipa atau salauran dapat dihitung dengan menggunakan persamaan Colebrook berikut. • • • • • •
1 / λ1/2 = -2 log [ 2.51 / (Re λ1/2) + (k / dh) / 3.72 ]
(4.5)
dimana λ = koefisien gesekan D'Arcy-Weisbach (-) Re = bilangan Reynolds k = kakasaran permukaan pipa, saluran (m) dh = diameter hydraulik (m)
• Persamaan Colebrook ini hanya berlaku untuk aliran turbulen.
Wlall drag and changes in height lead to pressure drops in pipe fluid flow. To calculate the pressure drop and flowrates in a section of uniform pipe running from Point A to Point B, enter the parameters below. The pipe is assumed to be relatively straight (no sharp bends), such that changes in pressure are due mostly to elevation changes and wall friction. (The default calculation is for a smooth horizontal pipe carrying water, with answers rounded to 3 significant figures.) Note that a positive Dz means that B is higher than A, whereas a negative Dz means that B is lower than A.
100 5 10 1 0 kPa m/s 0
Inputs
Pressure at A (absolute): 100 Average fluid velocity in pipe, V: 1 Pipe diameter, D: 10 Pipe relative roughness, e/D: 0 Pipe length from A to B, L: 50 Elevation gain from A to B, Dz: 0
kPa m/s cm m/m m m
Fluid density, r: 1
kg/l
Fluid viscosity (dynamic), : 1
Cp
Answers
Reynolds Number, R:
Friction Factor, f:
1.00 × 105
0.0180
Pressure at B:
95.5 kPa
Pressure Drop:
4.50 kPa
Volume Flowrate:
7.85 l/s
Mass Flowrate:
7.85 kg/s
Equations used in the Calculation Changes to inviscid, incompressible flow moving from Point A to Point B along a pipe are described by Bernoulli's equation,
where p is the pressure, V is the average fluid velocity, r is the fluid density, z is the pipe elevation above some datum, and g is the gravity acceleration constant. Bernoulli's equation states that the total head h along a streamline (parameterized by x) remains constant. This means that velocity head can be converted into gravity head and/or pressure head (or vice-versa), such that the total head h stays constant. No energy is lost in such a flow.
For real viscous fluids, mechanical energy is converted into heat (in the viscous boundary layer along the pipe walls) and is lost from the flow. Therefore one cannot use Bernoulli's principle of conserved head (or energy) to calculate flow parameters. Still, one can keep track of this lost head by introducing another term (called viscous head) into Bernoulli's equation to get,
where D is the pipe diameter. As the flow moves down the pipe, viscous head slowly accumulates taking available head away from the pressure, gravity, and velocity heads. Still, the total head h (or energy) remains constant.
For pipe flow, we assume that the pipe diameter D stays constant. By continuity, we then know that the fluid velocity V stays constant along the pipe. With D and V constant we can integrate the viscous head equation and solve for the pressure at Point B,
where L is the pipe length between points A and B, and Dz is the change in pipe elevation (zB - zA). Note that Dz will be negative if the pipe at B is lower than at A. The viscous head term is scaled by the pipe friction factor f. In general, f depends on the Reynolds Number R of the pipe flow, and the relative roughness e/D of the pipe wall,
The roughness measure e is the average size of the bumps on the pipe wall. The relative roughness e/D is therefore the size of the bumps compared to the diameter of the pipe. For commercial pipes this is usually a very small number. Note that perfectly smooth pipes would have a roughness of zero. For laminar flow (R < 2000 in pipes), f can be deduced analytically. The answer is,
For turbulent flow (R > 3000 in pipes), f is determined from experimental curve fits. One such fit is provided by Colebrook,
The solutions to this equation plotted versus R make up the popular Moody Chart for pipe flow,
Diagram Moody
The calculator above first computes the Reynolds Number for the flow. It then computes the friction factor f by direct substitution (if laminar; the calculator uses the condition that R < 3000 for this determination) or by iteration using Newton-Raphson (if turbulent). The pressure drop is then calculated using the viscous head equation above. Note that the uncertainties behind the experimental curve fits place at least a 10% uncertainty on the deduced pressure drops. The engineer should be aware of this when making calculations.
Pressure drop in fittings.... Head Loss in Fittings is frequently expressed as the equivalent length of pipe that is added to the straight run of pipe as shown below. This approach is used most often with the Hazen-Williams or Manning’s equations. The approach does not consider turbulence and subsequent losses created by different velocities. EQUIVALENT LENGTH IN METERS 150
200
250
300
350
400
90oElbow 8.5
6.4
7.9
9.4
10.7 12.2 14.0 17.0
23.0 28.0 32.5 37.1
45o Elbow
3.5
3.4
4.2
5.0
5.7
10.9
13.6 16.2 20.1 23.5
Tees
11.0 14.4 17.8 21.1 24.0 27.5 32.8 38.3
49.5 61.5 72.9 84.6
6.5
450
8.2
500
600
700
800
900
When more accuracy is required, head loss in fittings can be determined using loss coefficients (K factors) for each type of fittings. In this approach K-factor is multiplied by the velocity head of the fluid flow.
H = K (v²/2g) where, H = Head loss, m V = Velocity of flow, m/s
K-FACTOR FOR GRP FITTINGS Type of Fitting K-Factor 90° Elbow, standard 0.5 90° Elbow, single miter 1.4 90° Elbow, double miter 0.8 90° Elbow, triple miter 0.6 45° Elbow, standard 0.3 45° Elbow, single miter 0.5 Tee, straight flow 0.4 Tee, flow to branch 1.4 Tee, flow from branch 1.7 Reducer, single reduction 0.7
Bleed valave
Percobaan 2 inch. Dimater return pipe
Control valve 2way valve
Balance arm
Weighing tank Weighing machine 1 icn.diam. Test pipe 5 ft long Motor
4.0 ft. sump Pump
Drain valve
Procedure • The tappings on the test pipe were connected to the mercury manometer and with the control valve and the pump dischare valve shut, the pump motor was started and then the pump discharge valve was opened gradually to thefully open position • The control valve was opened slightly so as to permit a small flow of water through the system. Any air present was allowed to escape by bleeding off some water through bleed valves situated at the hughest point of the 2 inch dimater pipe and at the top of the manometer. • The control valve was fully shut and zero reading of the manometer checked. • The control valve was opened fully and when steady conditions were obtained the reading of mercury manometer (h’f) was taken and the discharge was determined using the weighing tank. This was done as follows; the drain valve was hut, the weight on the weighing machine was preset to 1 cwt and the water wasdirected to the weighing tank by means of the two-way valve.
Procedure • The balance arm of the weighing machine was watched and then re-set to 4 cwt and the stop watch was stopped when the balance arm rose again. Thus 3 cwt of water were collected during the time indicated by the stop watch. The water was directed to the sump and the weighing tank drained. • The procedure was repeated for smaller rate of flow. After Test No. 10, the tappings on the test pipe were disconnected from the mercury manometer and connected to the invertsed u\U-tube water manometer. In all 15 test were carried out/
Results. Table 1. Readings h'f(inch.H Test No g) W (lbf) t (s) 1 16.9 336 2 15.4 336 3 14 336 4 12.5 336 5 10.6 336 6 9.2 336 7 7.7 336 8 5.9 336 9 4.35 336 10 2.8 336 Inch.water 11 33.6 336 12 26.9 224 13 18.6 224 14 13.2 224 15 7.6 112
Calculated values hf (ft water) Q (ft3/s) v (ft/s) Log hf lg v 29.7 17.75 0.181 33.3 1.2492 1.5224 31.3 16.19 0.172 31.6 1.2092 1.4997 33 14.7 0.163 30 1.1673 1.4771 35 13.12 0.154 28.4 1.1179 1.4533 38 11.12 0.142 26.2 1.0461 1.4183 41.1 9.66 0.131 24.1 0.9850 1.3820 45.4 8.08 0.119 21.9 0.9074 1.3404 52.4 6.19 0.103 19 0.7917 1.2788 61.1 4.56 0.088 16.2 0.6590 1.2095 79.2 2.94 0.068 12.5 0.4683 1.0969 29.1 60.9 74.6 89.5 59.9
2.8 2.24 1.55 1.1 0.634
0.068 0.059 0.048 0.04 0.03
12.5 10.9 8.8 7.4 5.5
0.4472 0.3502 0.1903 0.0414 -0.1979
1.0969 1.0374 0.9445 0.8692 0.7404
Sample of calculations • Test No.1 • For the drwoned mercury manometer showing a reading of h’f in Hg, the friction head hf in feet of water is given by
13,6 1 12.6 h f h' f 16.9 17.75 ft of water 12 12 • This applies to test 1 to 10 • For test 11 to 15
h f h' f
1 12
• Volume flow is given by Q
W 336 0.181 ft 3 / s wt 62.4 x 29.7
• Mean velocity in pipe v=Q/a but cross sectional area of the pipe is
1 1 2 a ( ) 0.00545 ft 2 4 12 Q 0.181 v 33.3. ft / s 0.00545 0.00545 • Since hf kv n by taking logarithms of both sides of the equations we have
log10 hf log10 k n log10 v
• Which represented by the straight line drawn jn Fig. 4.10.2 so that n is the slope of the line and log10 k the intercept. • From fig 4.10.2
n
68 1.83 37.2
Log hf vs log v 1.4000 1.2000 1.0000
Log hf
0.8000
0.6000 Series1
0.4000 0.2000
0.0000 -0.2000 -0.4000
Log v
• Taking log hf=1.2 corresponding to log v=1.5 • Log k=log hf –n log v= 1.2 -1.83 x 1.5=1.2 2.745= -1.545 • K= 0.02851 • Therefore the law connecting the frictional loss of head and the mean velocity for this pipe and over the range tested is hf 0.02851v1.83
Table 4.10.2. Variation of f with Re V (ft/s) 33.3 31.6 30 28.4 26.2 24.1 21.9 19 16.2 12.5 10.9 8.8 7.4 5.5
hf (ft of water) V^2 17.75 16.19 14.7 13.12 11.12 9.66 8.08 6.19 4.56 2.87 2.24 1.55 1.1 0.634
1108.9 998.56 900 806.56 686.44 580.81 479.61 361 262.44 156.25 118.81 77.44 54.76 30.25
f(=0.268 hf/v^2) 0.00429 0.00435 0.00438 0.00436 0.00434 0.00446 0.00452 0.00460 0.00466 0.00492 0.00505 0.00536 0.00538 0.00562
Re (6.95x10^3v) 231435 219620 208500 197380 182090 167495 152205 132050 112590 86875 75755 61160 51430 38225
• The calculation for Table 4.10.2 are as follows: 4 flv 2 hf d 2g h f 2 gd 64.4 x1xh f hf f 0.268 2 2 2 4lv 48 x5v v
• Re=Vd/ and since water at the test temperature of 16 C the kinematic viscosity • =0.12x10^-4 ft2/s then • 104 3 Re
12 x0.12
v 6.95 x10 v
Figure 4.10.3 0.006
e/d=0.0004
0.005
f
0.004
e/d=0.0001
0.003 0.002 0.001 0
Re
Conclusions • The relationship between the head loss due to friction and the mean velocity in the pipe was determined with the aid of a logartihmic graphs shown in Fig. 4.10.2 and was found to be hf 0.02851v1.83
• In which hf is in ft of water and v in ft/s. This results applies only to the pipe tested and over the test range, that is for • 0.03
Experimental error 1) Discharge: weight ¼ lbf in 112 lbf error 0.22% • Timing ½ s in 29.7 s error 1.7% • Error in discharge =(0.22+1.7)=1.92% say 2% 2) Velocity :since the pipe diameter was not actually measured and the manufacturing tolerance is not known, assume 1% tolerance in diameter. Therefore the error in cross sectional area is 2% and hence the error in velocity =(2.0+2.0)=4%. 3) Manometric readings-mercury manometer 1/20 in. In 4.35 in., therefore error 1.1%: water manometer 1/20 in. In 7.6 in. Which being less then the error for the mercury manometer need not be taken into consideration. 4) Error in f: since f hf/v^2 , the maximum error may be (1.1+2x4)=9.1%. 5) Error in Re: error in kinematic viscocity of water corresponding to a 1/2 C error in temperature measurement at 16 C is 1.2%.Therefore, since Re=Vd/v the error in Re is (4.0+1.0+1.2)=6.2%
• This results appears to be rather large and is not supported by the much smaller scatter of experimental points, plotted in Fig. 4.10.3. However the calculated maximum error allows a 1% error in pipe diameter which contributes substantially to the final results. It is reasonable to state that the error in diameter is a biassed or systematic error since it probably affects the whole length of the test pipe
• Also in the calculations one value forthe pipe area was used for all tests and therefore the error the maximum probable error in f would be 5% and in Re3.2% • The areaerror however becomes relevant when the experimental points are considered in relation to the e/d curves drown in Fig. 4.10.3. Therefore the estimated relative roughness values for the pipe are subject to amuch greater error due to the possible 4% vertical draft shift and 3% horizontal shift of all the experimental points.
Figure 4.11.1.Pipe in series
h
H1
H2
Z1 Separation and friction loss 1
Datum
z2 2
• Consider water flowing from higher reservoir to a lower reservoir through a system of pipe in series. Let the water levels in the two reservoir be kept constant at H1 and H2 above the center lines of the pipes. • Applying Bernoulli’s equation to section 1 and 2 2
p1 v1 p2 z1 z 2 all friction and separation losses w 2g w
• But p1/w=H1, and p2/w=H2 also v1=0 and v2=0 so that the above equations become • H1+z1=H2+z2+ all friction and separation losses
• • • • •
Rearranging (H1+z1)-(H2+z2)= losses But referring to fig. 4.11.1 it will be seen that (H1+z1)-(H2+z2)=h Where h is the difference between the water levels in the two reservoir. • Then for the above arrangement • H=losses 4 flv 2 H v d 2g
Figure 4.11.2 Sudden enlargement (v1-v2)^2/2g Friction loss 4f3(l3/d3)(v3^2/2g)
Entry loss 0.5 V^2/2g
Friction loss 4f1(l1/d1)(v1^2/2g) Friction loss 4f2(l2/d2)(v2^2/2g)
Sudden contraction (1/Cc1)^2(v3^2/2g)
Exit loss V3^2/2g
Total head loss • Thus the following equation can be written for the above example of piping system and reservoir. v1 4 f1l1 v1 (v1 v2 ) 2 4 f 2l2 v2 h 0 .5 2g d1 2 g 2g d 2 2g 2
2
2
2
2
2
1 v 4f l v v ( 1) 2 2 3 3 3 3 Cc 2g d3 2 g 2 g
• Also by the continuity equation a1v1 a2v2 a3v3
(4.11.2)
• Thus two of the unknown velocities can be expressed in terms of the third velocity, for example
a3 a3 v1 v3 and v2 v3 a1 a2 • By substitution for v1 and v2 into eq. 4.11.2 • It can be soved for v3 for agiven level of h. The discharge can then be calculated from
Q a3v3 The alternative way of solving eq 4.11.2 is by expressing each velocity in terms of Q and then solving for it
• Thus
H
Q Q Q v1 , v2 , v3 a1 a2 a3
Q
• Eq.(4.11.2) then takes the general form h k1Q 2 k2Q 2 k3Q 2 k4Q 2 k5Q 2 k6Q 2 k7Q 2 (k1 k2 k3 k4 k5 k6 k7 )Q 2
1 4 f1l1 • In which k1 0.5 2 , k2 , dst. 2 a1 2 g d1 2 ga1
• If now • K=k1+k2+k3+k4+k5+k6+k7, then h KQ 2
(4.11.3)
• Since eq.(4.11.3) represents the relationship between the flow of a fluid through a system of pipes or ducts and the head required to maintain this flow, the equation is known as system resistance.
• When expressing frictional loss in terms of Q an approaximate formula derived from D’Arcy’s equation, may usefully employed Since 4 flv 2 1 hf and Q av d 2v d 2g 4 so v 4Q / d 2 and substituting into D' Arcy's equation 4 fl16Q 2 1 32 flQ 2 hf 2 2 4 d d 2 g g d 2 1 2 but g 9.95 10 with 0.5% error 32 therefore flQ 2 hf 10d 2
Total energy line
Inlet duct
fan
diffuser
Flow
Dischrage duct
Soal • An axial flow fan 0.9 m in diameter is used to move air weighing 1.2 kgf/m3 through a system of horizontal ductwork at a rate of 11.5 m3/s. The system consists of an inlet duct 0.9 m dimater and 15 m long and a discharge duct 1.0 m dimater and 60m long. A diffuer, in which the loss of head can be taken as 1/3 of the distance between the velocity heads at its two ends, is fitted between the fan and the discharge duct. The loss at entry to the inlet duct is 0.5 v^2/2g and the friction factor f for the inlet duct is 0.0041 and for the outlet duct is 0.0035. Draw the total energy and the hydraulic gradien lines and determine in milimeter of water the total loss head rise in the fan.
Solution • Inlet duct area
( 0 .9 ) 2 A1 0.635m 2 4 v1 Q / A1 11.5 / 0.635 18.1m / s • Outlet duct area • 2
(1) A2 0.785m 2 4 v2 Q / A2 11.5 / 0.785 14.7 m / s
• To convert meters of air to milimetres of water
hmmwater
1,2 hmair x10 3 1.2hmair 10 3
• (since specific weight of water=1g/cm^3=10^6/1000kgf=10^3 kgf/m^3) • Therefore 2
v1 (18.1) 2 (18.1) 2 mair / 1.2mm water 19.98mm water 2 g 2 x9.81 19.62 2
v2 (14.7) 2 / 1.2mm water 13.2mm water 2 g 19.62
• Losses and outlet kinetic energy • Inlet loss 0.5(v1 ) 2 2g
• Inlet friction loss
(
4 fl )(v12 / 2 g ) (4 x0.0041x15 / 0.9)19.98 5.327 d 2
• Diffuser loss
0.5 x19.98 9.994
2
1 v v ( )( 1 2 ) 1 / 3(19.98 13.2) 2.26 3 2g 2g
• Outlet friction loss • Outlet velocity head
( 4 x 0.0035 x 60 / 1.0) x13.2 11.1
=13.2 •
Total head lrise through fan
_________________________________________________________________
41.881 mair 34.9 mmwater
Parallel and branching pipe • Q=QB+QC
A B HA
ZB C ZC
ZA Datum line
• Apply Bernoulli’s theorem to a stremline passing through point A and B 2
2
v v H A Z A A H B Z B B losses in pipeAB 2g 2g for constant level in the tank v A 0 and since the pressure at point B is atmospheric then H B 0. Also let the losses in pipe AB be h1 , then 2
v H A z A z B B h1 2g Similarly applying Bernoulli's theorem to a streamline passing though points A and C we have 2
v H A z A zC C h ' 2g where h' some of all losses in pipe AC.
• A special case is of interest, in which zB=ZC, that is when the pipes discharge at the same level. It follows from the two equations abobe that in such a case since H A z A z B H A z A zC then 2
2
vB v h1 C h' 2g 2g In such a case then when the pipes are connected in parallel the sum of velocity head and the losses is the same for all pipes
Figure 4.13.2.Branching pipes
HA B
1
hC 2
C
3 zA
zB
hD D
zC zD
Datum line
• Consider now a system of branching pipes as shown in Fig.4.13.2. • Applying the Bernoulli’s equationto point A and C and considering a steramline passing through A,B and C. 2 2 • v v H A z A h11 h12
2
zC or H A z A zC h11 h12
2g but H A z A-zC hC and therefore
2
2g
2
v hC h11 h12 2 B 2g Similarly for a streamline passing through A.B and D
(4.13.2)
2
v H A z A h11 h13 3 D z D 2g or 2
v hD h11 h13 3 D 2g
(4.13.3)
• Since as was shown previously losses in pipe due to friction and separation are expressed in the form kv2/2g,then the sum of all such losses in one pipeline (which may be consist of a numberofpipes in series) can also be expressed as kv2/2g. Therefore, 2
2
2
v v v h11 k1 1 , h12 k2 2 , h13 k3 3 2g 2g 2g Substituting these values into eqs(4.13.2) and (4.13.3) 2
2
2
2
v v v v hC k1 1 (k 2 1) 2 hD k1 1 (k3 1) 3 2g 2g 2g 2g It is seen now that these equations are not independent and can only be solved simultaneously. Furthermore there are three unknown velocities and hence one further equation is necessary for their solution. This is the continuity equation
Q1 Q2 Q3 or a1v1 a2 v2 a3v3
(4.13.4)
Problems
30 ft v1
v2 v3
30 ft
Soal • Water flows from a reservoir through a 1 in.diameter pipe 50 ft long to a large junction box A. The differences in levelbetween the water surface in the reservoir and the box is 30 ft. Two ½ in.dimeter and 50 long pipes are connected in parallel to the junction box and discharge to atmosphere at a level 30 ft below the box.Calculate the flow arte from the reservoir when (a) one branch only is open (b) both branches are open. • Take f=0.01 for all pipes and take into account losses at entry to the pipes, but negelct the enlargement loss in the junction box.
Jawab • A).v3=0 2
2
2
2
2
v 4 fl1 v1 v 4 fl2 v2 v h 0 .5 1 0 .5 2 2 2g d1 2 g 2g d2 2 g 2 g 2
2
4 x0.01x50 v1 4 x0.01x50 v2 60 (0.5 ) (1.5 ) 1 / 12 2g 1 / 12 2g 3860 24.5v1 40.5v2 2
(1)
2
Tetapi dari persamaan kontinyuitas
v1d1 v2 d 2 2
2
sehingga v1 (1 / 2)v2 3860 24.5(1 / 4v2 ) 2 49.5v2 ) 51.03v2 2
Jadi
v2 3860 / 51.03 75.6, maka 2
Q 1 / 4 (1 / 24) 2 x8.7 0.012 ft 3 / s
2
v2 8.7 ft / s
• B). V2=v3 v1d1 v2 d 2 v3d 3 2v2 d 2 2
2
2
v1 1/2 v 2
2
substituting into eq.(1) 3860 24.5((1 / 2)v2 ) 2 49.5v2 (6.12 49.5)v2 55.62v2 2
v2 3860 / 55.62 69.4, maka 2
2
v2 8.35 ft / s
Q 2a2v2 2 x1 / 4 (1 / 24) 2 x8.35 0.0228 ft 3 / s
2
Soal • In a horizontal duct system a 12 inch diameter main duct 200 ft long branches in to a 6 inch diameter duct 100 ft long and a 9 inch diameter duct 150 ft long. Air is supplied to the main C
v2
v1
v3 A
B D
• Duct at a total head of 1.5 inch of water. If the two • branches discharge to the atmpsphere, calculate the air velocity and sicharge in each duct. Assume specific weight of air 0.0756 lbf/ft3 and friction coefficient f=0.0044 for each duct. Neglect all losses other than friction.
Jawab • Pressure at A • Route ABC
hA
1.5 62.4 x 103 ft of water 12 0.0756 2
2
2
103 0.0546v1 0.0704v2 2
• Route ABD
2
2
(1)
2
2
4 fl3 v3 v 4 x0.044 x150 v 103 0.0546v1 3 0.0546v12 ( 1) 3 d3 2 g 2 g 0.75 2g 2
103 0.0546v1 0.0704v2 2
• Equating (1) and (2) then • By continuity
2
4 fl1 v1 4 fl2 v2 v2 4 x0.88 v12 4 x0.044 v 103 ( 1) 2 d1 2 g d 2 2 g 2 g 1 2g 0 .5 2g
2
v2 v3
v1 0.25v2 0.56v3 0.81v2
(2)
Jawab • Substitute into (1)
103 0.0546(0.81v2 ) 2 0.0704v2 0.1062v2 2
v2
• Therefore
103 31,2 ft / s 0.1062
v1 0.81x31.2 25.21 ft / s
and
v3 31.2 ft / s
a1 (1 / 4) (1) 2 0.785 ft 2
and
Q1 19.8 ft 3 / s
a2 (1 / 4) (0.5) 2 0.196 ft 2
and
Q2 6.1 ft 3 / s
a3 (1 / 4) (0.75) 2 0.44 ft 2
and
Q3 13.7 ft 3 / s
2
Flow under varying head • Let us consider the simplest case of flow under varying head that of the flow through an orifice in a side of a tank having a constant cross sectional area shown in fig.4.15.1. dh H1 h H2 Q
• Let the cross sectional area of the tank to be A and that of the orifice be a . Also let a be small incomparison to A and the head of liquid above the orifice at any time be h
• By the continuity consideration it is clear that the decrease in the volume of liquid in the tank in a given time is equal to the quantity of liquid discharged through the orifice in that time. Let us assume that the level of the liquid in the reservoir falls by an elemental distance dh in an elemental time dt resulting in a dicrese in volume of A dh.Let the rate of flow through the orifice be Q, and therefore the quantity of liquid discharged through the orifice in time dt will be Q dt. • Thus -Adh=Q dt in which the negative sign indicate the decrease in head. By this then, it is possible to say the element of time taken for the liquid to decrease by dh is given by (4.15.1) • dt=-A dh/Q
• The total timetaken for the liquid to fall between given limits is obtaned by integrating this expression between these limits. • However before integration is possible it is essential to establish whether or not the area A and the discharge Q are influenced by the head h, and if they are so influenced A and Q must be expressed in terms of h. • For this example the area A is constant , and eq. (2.13.3) the instantaneous discharge through a small orifice is given by • Q= Cd a2gh • Where Cd is the coefficient of discharge of the orifice.
• Therefore
dt
A A dh h 1 / 2 dh Cd a 2 gh Cd a 2 g
• By integration, the total time required for the head to fall from H1 to H2 is T dt
A 2A H 2 1 / 2 h dh h1 / 2 H 1 Cd a 2 g Cd a 2 g
2A 1/ 2 1/ 2 ( H1 H 2 ) Cd a 2 g
H1 H2
Soal • Tentukan waktu penurunan permukaan air dalam tangki dengan penampang konstan dengan radius 15 cm. Dari H1=10 cm menjadi 5 cm. Air keluar dari tangki melalui orifice yang mempunyai penampang 2 cm2 dan koefisien Cd=0.8
Jawab A 2A H 2 1 / 2 1/ 2 T dt h dh h H 1 Cd a 2 g Cd a 2 g 2A 1/ 2 1/ 2 ( H1 H 2 ) Cd a 2 g 2 / 4(152 ) 1 / 2 1 / 2 (10 5 ) 0 .8 x 2 204.6detik
H1 H2
• Consider now the case in which the dischargeis through a pipe rather than an orifice. Let the pipe diameter be d and the length l. Area a
dh
h
l
d
Q
• It was shown previously that the discharge through a pipe depends upon the head h and the losses incurred in the system
• Thus for the above case v 2 4 fl v 2 v2 h 0.5 2g d 2g 2g
or
v 2 4 fl h ( 1.5) 2g d
• From which the velocity of the liquid in the pipe v [
2 gh 2g ] [ ] h1 / 2 ( 4 fl / d 1.5 ( 4 fl / d 1.5
• Therefore the discharge Q (1 / 4)d 2 [
2g ] h1 / 2 ( 4 fl / d 1.5
• Just as in the first example the element of time dt taken for theliquid level in the reservoir to fall by dh is given by dt
A dh Q
• Substituting the value of Q expressed in terms of h for this case and integrating H2
T dt
H1
(1 / 4)a 2
A h 1 / 2 dh [ 2 g /( 4 fl / d 1.5)]
• For a given pipeand a reservoir of constant cross sectional area, all values under the integral sign except h are constant T
(1 / 4)a 2
2A h1 / 2 [ 2 g /( 4 fl / d 1.5)]
2 A( H1 H 2 ) (1 / 4)a 2 [ 2 g /( 4 fl / d 1.5)] 1/ 2
H1 H2
1/ 2
soal • The square tank have a common wall to which an orifice with area of 0.25 ft2 and a coefficient of discharge of 0.8. Tank A measures 8ft on a side and the initial depth above the orifice is 10 ft, tank B measures 4.0 ft on theside and the initial depth above the orifice is 3 ft. How long it take for the water surface to be at the same level? 8ft
4 ft
A dy
10 ft
h 3ft
Orifice
B
• At any instant the differencein level of the surface is h. • Discharge through orifice : Q Cda 2 gh 0.8 x0.25 64.4h 1.605 h
• In time dt the change in head (that is the change in the distance ) is dh. • Let the change in level in tank A be dy and in tank B be kdy. Since the decrease in volume in tank A equals the increase in volume in tank B. • 8x8x dy=4x4xkdy k=4 • So that the total change in head is • dh=dy + kdy=(1+4)dy=5 dy or dy=1/5 dh
• Now change in volume in tank A equals the discharge through the orifice in time dt, therefore, • -8x8xdy= Qdt or -8x8x1/5dh=1.605hdt dt 7.98h1 / 2 dh
• Integrating between 7 ft and 0 the total time is 7
T 7.98h1 / 2 dh 7.98 2h1 / 2 0
7 0
7.98 x 2 7 42.1s
Soal • A sample of oil, specific gravity 0.86, is discharged from a cylinder 5 cm diameter through a tube 1.5 mm internal diameter and 1 cm long. If the time taken for the free surface of oil in the cylinder to fall from 10 cm to 7.5 cm is 292 s calculate, working from principles , the coefficient of absolut in viscocity of the oil in poises and that coefficient of kinematic viscocity in centistokes.
Jawab • Qdt=-Adh jadi dt=-A dh/Q • Untuk aliran kental (viscous flow) 32 vl d2 tetapi h p / w and p
Jadi
v Q / a 4Q /(d 2 )
h 128Q /( wd 4 )
dan Q d 4 wh / 128d
• Dengan memasukkan ke persamaan diferensial dt
A dh Q
128lA H 2 1 128lA 32 D 2 10 t h dh (log H 1 log H 2 ) l log ( ) e e e 4 4 4 d w H 1 d w d w 7 .5
• Also
/w
/ g rg
32 D 2 10 32 x 25 x1 10 t 4 l log e ( ) 2 . 303 log ( ) 460 10 4 d g 7.5 (0.15) 981 7 .5 292 / 460 0.635S 63.5cS
• Sekarang karena r dan kerapatan air r water 1g / cm3 then r oil 0.86 g / cm3 0.86 x0.635 0.545 Poise
Soal • A water container in the form of a frustrum of a cone as shown in Fig. 4.16.2 is fitted in the bottom with a 3 inch diamter orifice. Determine the average coefficient of the discharge for the orifice if the tank empties in 208 s from the initial head of 9ft. 3ft 6 ft dia
9 ft
x
x
9ft h
0.75 ft
1.5 ft dia
3ft
Jawab • • • • •
Qdt= -A dh But Q=Cda(2gh) A x 2 And 1 1 ( 1 / 3 ) x ( h 3 ) x (h 3) Also from similar triangles 12 4 Substituting into the differential equation 1 1 Cd d 2 2 ghdt (h 3) 2 dh 4 12 9 1 2 1 / 2 Cd dt 2 (h 3) h dh 4d 2 g 0 where dt 208s Therefore
9 1 3/ 2 1/ 2 1 / 2 Cd ( h 6 h 9 h )dh 2 208 x 4(0.25) 64.4 0
1 2 5/ 2 2 3/ 2 h 6h 18h1.2 417 5 3 Cd 0.62
9 0
97.1 108 54 0.62 417
Forces resulting from fluid flow • Newton’n second law of linear motion:The incerement of total linear momentum per unit time is equal to the resultant external forces. • Or mathematically (6.1.1.) • F=d/dt (mv) • Where m is mass of the body and v=its velocity. The product (mv) is known as the momentum. V2=0
v=v1
Figure 6.1.1. R
F
• If the mass of the body remains constant and only its velocity changes uniformely during a small time t then the expression above can be written as m F (v1 v2 ) t
(6.1.2)
• In which v1 and v2 are the velocities of the body at the beginning and the end of time t. The force F acts in the direction of the change of velocity but it is also important to know its sense. • Consider a body of mass m having an initial velocity v being brought to rest by impact on a stationary plate (Fig.6.1.1)
• Assuming that the velocity of the body after impact is equal to zero the change in momentum is
m(v1 v2 ) m(v 0) mv • Which is a positive quantity, and thus the force due to the rate of change of this momentum has the same sense as the body velocity and is the force with which the body acts on the plate. • Alternatively the direction R is the force with which the plate acts on the body in destroying the momentum of the body by bringing it to rest.
• Thus to calculate the reaction of the plate the order of velocities is changed.
l
Figure 6.1.2
• The principle of momentum is then of ten applied to a moving fluid. One such application is the determination of the force due to the inertia of the fluid when it is suddenly brought to rest. Imagine a pipe of length l and the cross sectional area a to be fitted with a valve at its end (Fig.6.1.2). Let the velocity of the fluidin the pipa be v initially and then imagine the valve
• To be closed completely during time t. The whole body of the fluid in the pipe will then be brought to rest in this time. For simplicity assume that the fluid is not compressible and that the pipe material is non-elastic. Then the force with which the fluid will act on the valve will be equal to the rate of change of momentum of the fluid. Then F
m (v1 v2 ) t
• But m=rla, v1=v and v2=0 and hence F
rla (v1 0) t
or since r w / g then F
wla v g t
• And the pressure rise in the fluid at the valve is F wlv P a gt
• This can be expressed in terms of pressure head which is then known as inertia head hl
p lv w gt
(6.1.3)
• This expression although useful in that it gives some idea of the order of magnitude of the inertia head is not applicable in certain cases, or when greater accuracy is required. Its limitations are the result of simplifying assumptions made in the derivation, namely that
a) The fluid is incompressible b) The pipe material is non-elastic, and c) The velocity change uniformely 2
B’ B a2 B
B’
A’
A a1 v1
1 A’
A
Figure 6.1.3
• The momentum principle is used in cases where there is a change in magnitude of fluid velocity or its direction or both. It is useful therefore to derive a set of general momentum equations for moving fluids. These can then be simplified in some special cases. • Consider a stream as shown in Fig. 6.1.3. Suppose that in a small time interval t the fluid contained between A-A and B-B moves toA’-A’ and B’-B’, the flow being steady tube (no change with time).Since the streamtube is curved, which means that the flow changes in direction, it will be more convenient to resolve the resultant forces into X and Z components.Thus the rate of change of momentum will be considered along these two direction separately.
• Let the volume flowing in unit time be Q and the crosssectional area at A-A and B-B be a1 and a2 respectively. The velocities at these sections will be v1 and v2. These angles increase positively in the anti-clockwise direction. • The momentum of the fluid contained between A-A and B-B before time t is MAB and the momentum of the same fluid after time t, now contained between A’-A’ and B’-B’ is MA’B’ . • Thus the change of momentum of the direction of flow is M M AB M A' B ' But
M A' B ' M AB M BB ' M AA'
Therefore
M M AB M AB M BB ' M AA' But or M ' M AA' M BB '
Z Fmz
(6.1.4)
Fm
Fmx
Figure 6.1.4.
X
• Thus if the time interval t is small so that section A’-A’ is very near A-A and B’-B’ is very near B-B, then eq.(6.1.4) means that the change of momentum of the fluid, isolated from the rest of the system by a control volume within the chain-dotted line, is equal to the difference between the momentum entering the control volume and the momentum leaving it. • The concept of the control volume will be found useful in later applications. It uses an arbitrary chosen space defined by fictitious boundaries which enables the motion of the fluid to be considered in relation to this space. • The above general statement can now be applied in particular to our stremtube in Fig.6.1.3. Momentum is the product of mass and velocity, M=mv, and the mass of fluid contained between A-A and A’-A’ is equal to the volume of this element multiplied by the fluid density.
• We have then Distance A-A' v1t Volume A-A' a1v1t mass in A-A' ρ1a1v1t but a1v1 Q, and therefore mA-A' ρ1Qt
• The velocity of fluid entering the control volume in the Xdirection is v1cos1 ,and therefore the momentum entering the control volume in the X-direction is ( M A-A' ) x ρ1Qtv1 cos1
• Similarly the momentum leaving the control volume in the Xdirection is ( M B-B' ) x ρ2Qtv2 cos 2 • Therefore the change of momentum in the X-direction is
M x ρ1Qtv1 cos1 ρ2Qtv2 cos 2 or M x Qt ( ρ1v1 cos1 ρ2v2 cos 2 ) • And the rate of change of momentum in the X-direction is M x Qt ( ρ1v1 cos1 ρ2v2 cos 2 ) t
• Similarly the rate of chnge of momentum in the Z-direction is
M z Qt ( ρ1v1 sin 1 ρ2v2 sin 2 ) t • If the flow is compressible then r1= r2= r and since r=w/g we have M x Qw (v1 cos1 v2 cos 2 ) t g M z Qw (v1 sin 1 v2 sin 2 ) t g • Finally since the rate of change of momentum is equal to the force in that direction
Qw (v1 cos1 v2 cos 2 ) g Qw (v1 sin 1 v2 sin 2 ) g
FMX
(6.1.5)
FMZ
(6.1.6)
• And the resultant force is given by FM
FMX FMZ 2
2
and its direction by
(6.1.7)
FMZ (6.1.8) tan FMX In the above expressions the quantity wQ/g is the mass flow of the fluid and v cos or v sin are the velocity components in the direction of force.
• It follows then that in the moving fluid the force due to its rate of change of momentum is equal to the product of the mass flow and the change in velocity in the direction of the force.
Flow on conduit • Reducers, expanders, bends, etc. produce a change in the fluid’s momentumand hence are subjected to a force with which the fluid acts on them. pa v v pa
Figure 6.2.1.
• Fig. 6.2.1. shows a pipe bend in which the cross-sectional area is constant but the direction of flow is changed. Using eq.(6.1.5) and (6.1.6) the X and Z component of the force due to the rate of change of momentum are obtained.v1=v2 Qw (v1 cos1 v2 cos 2 ) g Qw (v1 sin 1 v2 sin 2 ) g
FMX FMZ
• But in this case v1=v2=v and 1=0 and 2=, therefore Qwv (1 cos ) g Qwv ( sin ) g
FMX FMZ
• However these componenets are not the only forces acting on the bend.It is important to bear in mind that the fluid is conveyed under pressure and therefore this pressure acting on the cross-sectional area of flow gives rise to a force Fp. Let the pressure in the bend be p and neglecting the internal losses let it be the same at inlet and outlet. Thus at each end of the bend there will be a force numerically equal to pa but each acting along the center-line of the pipe. Considering again the X and Z components. FpX pa pa cos pa (1 cos ) and FpZ 0 pa sin pa sin
Z
Fx Fz
X
• Thus combining the pressure and momentum force FX FMX Fpx
wQv (1 cos ) pa (1 cos ) g
or FX (
Qwv pa )(1 cos ) g
FZ FMZ FpZ
wQv sin pa sin ) g
or FZ (
Qwv pa ) sin g
• And the resultant force is given by the magnitude F=(FX2+FZ2 ) and direction tan =FZ/FX
Z
P2 a2 v2
P1 a1 v1 Z1
Z2
Figure 6.2.3
X
• Consider now a case of a reducer bend as shown in Fig. 6.2.3. There is a change of velocity, both in magnitude and direction, and also a change in pressure. If the velocity and pressure are known at either section 1 or section 2 and if the geometry of the conduit is known (a1,a2 and ) then the velocity and pressure at the section can be determined using the continuity and Bernoully’s equation. Thus a1 v1=a2 v2 and if the loss of head between section 1 and 2 is hl then 2
2
p1 v1 p2 v2 z1 z2 hl w 2g w 2g • Having determined the velocity and pressure at the other section the forces in the X and Z direction can now be calculated
• Thus FX FMX
wQ FpX p1a1 - p 2a 2cos (v1 v2 cos ) g
and FZ FMZ FpZ p2 a2 sin
• The resultant is F
FX FZ 2
2
and its direction by tan
FZ FX
wQ wQ v2 sin ( p 2 a2 v2 ) sin g g
• Thus the force acting on the conduits due to flow of a fluid through them can be determined using the momentum principle, taking into account the effect of pressure in the fluid. Such forces are considered when designing pipe support or allowing for pipe tension or compression by the use of expansion joints. However, it is important to realize that sometimes other effects may give rise to additional forces which will then have to be taken into account. • These may include 1) The weight of the conduit itself 2) The weight of the fluid in the conduit,and 3) Inertia forces such as due to valve closure or pulsating flow
Contoh soal • At the end of a 2.0 ft diameter pipa there is a reducer connection to a 1.5 ft diameter pipa. The water velocity at the entrance to the reducer is 10 ft/s and the pressure there is 75 lbf/in.2. If the loss in the reducer is 10 ft of water find the resultant thrust on the reducer. a 2 • Jawab v1a1 v 2 a2 jadi v2 1 10 x( ) 2 17.8 ft / s a2
By Bernoulli 2
2
p1 v1 p v 2 2 hL w 2g w 2g or p1 p2 v1 v2 hL w 2g 2
2
1 .5
2ft
a1 p1
v1
• Gambar 6.3.1.
v2
a2 p2
1. 5 ft
p1 p2 v1 v2 (17.82 102 ) hL 10 w 2g 64.4 3.37 10 13.37ft of water 2
• Or
2
p1-p2 62.4 x13.37 833 lbf/ft 2
• But p1 75 x144 10800 lbf / ft 2 a1 (1 / 4) (2) 2 3.14 ft 2
and
and
p2 10800 833 9967 lbf / ft 2
a2 (1 / 4) (1.5) 2 1.763 ft.2
so that wQ (v1 v2 ) g 62.4 x3.14 x10 10800 x3.14 9967 x1.763 (10 17.8) 32.2 33900 17600 475 15825 lbf / ft 2 FX p1a1 p2 a2
Contoh soal • Water discharge from a tank through a 12 in. Diamter pipe 200 ft long, the entrance of which is located 50ft below the water surface in the tank. The first 100ft of pipe is horizontal and at the end of this length it is bent downwards through an angle of 10o in the vertical plane. If the friction factor for the pipe is 0.01, determine the resultant force on the bend.Assume the entry loss to be 0.5 v 2/2g and neglect the loss in the bend.
50 ft
100 ft
100ft 10o
• Gambar 6.3.2.
Jawab • Applying Bernulli’s equation to the water surface in the tank and the pipe outlet 2
2
2
2
p1 v1 p v v v z1 2 2 z2 0.5 1 4 fl 2 w 2g w 2g 2g d 2g where p1 0, v1 0, z2 0, p2 0 and z1 50 100 sin 10o 50 17.4 67.4 ft 2
200 v2 67.4 (1 0.5 4 x0.01x ) 12 / 12 64.4 67.4 x64.4 2 v2 456 or v2 456 21.4 ft / s (1.5 8)
• Apply Bernoulli to the water surface and the bend in the pipe 2
p 4 x0.01x100 v2 z1 2 (1 0.5 ) w 1 2g p 456 or 50 2 (1.5 4) 64.4 64.4 p2 (50 39)62.4 6871lbf / ft 2
• Assume p2 to be the mean pressure acting on the bend. Pressure force in X-direction PX (687 x(1 / 4) (1) 2 (1 cos10o ) 540(1 0.985) 8.1 lbf Pressure force in the Z - direction PZ 0-( 687 )( 1/ 4 )π (sin 10o ) 540 x0.1736 84 lbf
• Forcedue to the rate of change of momentum in the Xdirection 2
wav2 62.4 FX (1 cos10o ) x(1 / 4) (1) 2 (21.4) 2 (10.985) 10.5lbf g 32.2 Force due to rate of change of momentum in the Z - direction 62.4 ( 1/ 4 )π (21.4) 2 [(0 ( sin 10o )] 121 lbf 32.2 Total force in the x - direction PX FX 8.1 10.5 18.6 lbf FZ
Total force in the Z - direction PZ FZ 94 121 215lbf Resultant force on the bend [(18.6) 2 (215) 2 216lbf
Jet reaction and propulsion • The reaction which a free jet exerts on a nozzle from which it is issuing can be dtermined by the application of the basic momentum principles.
vo
Po A
a
v
Figure 6.4.1
• Let the nozzle be stationary and attached to the end od a pipe
• Of cross-sectional area A in which the fluid velocity is vo. Let the cross-sectional area of the jet be a and its velocity be v (Fig. 6.4.1). • The momentum entyering the chain dotted control volume mass entering w wQvo x velocity AVoxvo s g g The momentum leaving the control volume per second Mo
M
w wQv avxv g g
• Hence the force due to the rate of chnage of momentum which which the fluid acts on the nozzale is
F Mo M
wQ (vo v) g
• Since v>vo, the force F is negative indicating that it acts in the direction opposite to that of the jet velocity. The relationship between vo and v is given by the continuity equation Avo av • Where a is thearea of the jet and that of the nozzle. T his would be thecaseonly if the coefficient of contraction for the nozzle Cv was one. • It is important to realize that the velocity of the jet is produced by by conversion of the total energy (or total head)
• Available upstream and the nozzle into the kinetic energy of the jet. • By application of Bernoulli equation to the pipe just upstream of the nozzle and to the jet, assuming that there is no loss of energy, we obtain 2 2 Po vo v p th w 2g 2g w • Vth is the theoretical velocity of the jet obtainable only under the ‘no loss’ assumption. Let the total head upstream of the nozzle be denoted by H so that 2
H
Po vo w 2g
• And 2
v H th and vth 2 gH 2g However, since in practice there is a loss of energy in the nozzle the actual velocity of the jet is v Cv vth Cv 2 gH
(6.4.2)
in which Cv is the coefficient of velocity for the nozzle
• The loss of energy EL can nowbe determined by using Bernoulli’s equation again but allowing the loss Po vo 2 v 2 v2 EL and H EL w 2g 2g 2g
• Therefore, [(Cv 2 gH ) ]2 v2 EL H H H HCv 2 2g 2g or EL H (1 Cv 2 )
(6.4.3)
• In practice if the nozzle is well designed the value of Cv approaches unity (0.96 to 0.98) and thus the loss of energy is quite small. Referring now to eq.(6.4.1), if the upstream velocity v is negligible or zero, suchas may be the case when the discharge is from a large tank or in arocket, the jet reaction is given by F
wQv Wv g g
(6.4.4)
• Where W is the mass flow of the fluid. 2
1
U U
Engine
vg
Figure 6.4.2
• The jet reaction is hydrostatically unbalanced and therefore the nozzle will move unless it is held in position by external application of the required balancing force.This fact is made use of in jet propulsion and eq.(6.4.4) is directly applicable to rockets. • Howeverfor propulsion of ships or for a ram-jet engine the fluid
• Intake may be at the bow or in the nose so that there is a fluid momentum entering the control volume, the velocity being that of theship or theaircraft. • Consider the ram-jet engine represented diagrammatically in Fig. (6.4.2). Let the flight be U so that that air enters the engine with that velocity. In the engine energy E is added to the air by compression and addition of heat so that it leaves the engine at section 2 with high velocity Vg. Also because of the addition of fuel in the engine the weight of the air leaving (W2) is greater than that entering (W1). • Thus momentum entering momentum leaving W1u W2u (6.4.5) F
s
s
or F
g
g
• Here again the second term is greater than the first so that the direction of F is opposite to that of the air velocity. It is seen that in fact the jet reaction is reduced by the intake momentum and hence the term W1u/g is sometimes referred to as drag.
Contoh soal • A 2 in, diamter nozzle is fitted at the end of a 4 in. diameter pipe in which the mean velocity of water is 50 ft/s and the pressure upstream of the nozzle is 300 lbf/in.2 . If the coefficient of velocity for the nozzle is 0.86 clculate the jet reaction on the nozzle. What is the coefficient of contraction for the nozzle?
Jawab aj v
Po A
Vo
• ho=po/w=300 x 144/62.4=692 ft of water 2
vo 2500 / 64.4 38.8 ft of water 2g 2
v H ho o 692 38.8 730.8 ft of water 2g but v Cv 2 gH 0.96 x 64.4 x730.8 206.1 ft / s therefore the jet reaction F
wQ wAvo 62.4 x(1 / 4) (4 / 12)(50 206.1) (vo v) (vo v) 1320lbf g g 32.2
• Now from Avo= ajv the cross-sectional area of the jet Avo 50 (1 / 4) (4 / 12) 2 x 0.0212 ft 2 v 206.1 2 also the area of the nozzle a 1/ 4 π ( ) 2 0.0218 ft 2 12 Therefore Cc a j / a 0.0212 / 0.0218 0.97 aj
Contoh soal • A ram-jet engine consumes 100 lbf of aor per second and 1.5 lbf of fuel per second.The exit velocity of the gas is 1600 ft/s relative to the engine and the flight velocity is 600 ft/s. What is the horse power developped? • Jawab: W1u W2v2 100 x600 101.5 x1600 1870 5050 3180lbf g g 32.2 32.2 Therefore the engine thrust T 3180 lbf F
and
Horsepower
Txu 3180 x600 3470 HP 550 550
Forces on an aerofoil in a flowing fluid • Chapter 5 cosider the ideal flow around bodies of various shapes immersed in a flowing fluid; the most useful of those shapes are the aerofoil and the hydrofoil. One of the many application of the aerofoil is in the design of blades for axial flow pumps and fans the main function of which is to supply energy to the fluid. In order to understand this transfer of energy we must have some knowldge of the forces arising from the flow of the fluid past an aerofoil or a hydrofoil. • Ig we wre to look at the pattern of the flow past a model aerofoil set at a small angle relative to the flow in a smoke tunnel, it would be similar to that shown in Fig. 6.6.1. In this • two-dimensional flow pattern we would observe that the streamlines, made visible by the smoke, have been forced
• To change their path both above and below the aerofoil and we would conclude that forces must exist on the top and bottom of the aerofoil in order to cause these changes in the shape of the flow. 2
A
B
2
B 1
1
A Figure 6.6.1.
• Whenever possible it is much easier to break a problem down into parts and consider such part separately. Therefore let us first consider the flow over the top surface such as B-B than diverge again as they move downstream and we could consider the motion of the fluid in the streamtube between
• Streamlines 1-1 and 2-2 to be similar to that through an unsymmetrical ventury meter. Applying the equation of continuity to sections A-A and B-B we can easily show that the velocity v at section B-B is greater than the velocity vo at section A-A. Now if we apply Bernoulli’s equation for ideal flow to one point on a streamline at section A-A where the static pressure is po, and to another point on the same stremline at section B-B where the pressure p, then 2
po vo p v2 w 2g w 2g
• Assuming that the change in potential head z-zo is negligible. • Multiplying by w, and since p=w/g, we have
po 1 / 2 rvo p 1 / 2 rv 2 2
or
po p 1 / 2 r (v 2 vo ) 2
• Theoretically by the principle of continuity v could be expressed in terms of vo and the cross-sectional area at section A-A and B-B, suggesting that the pressure difference could be given by 1/2rv2 multiplied by a constant.
Compressed Air - Pressure Drop in Pipe Lines - Online Pressure Drop Calculator Calculate pressure drops in compressed air pipe lines - Online Pressure Drop Calculator and Spreadsheets - metric and imperial units The pressure drop in compressed air lines can be calculated by using the formula dp = 7.57 q1.85 L 104 / (d5 p) (1) where dp = pressure drop (kg/cm2) q = air volume flow at atmospheric conditions (FAD) (m3/min) L = length of pipe (m) d = inside diameter of pipe (mm) p = initial abs. pressure (kg/cm2) •1 kg/cm2 = 98068 Pa = 0.98 bar = 0.97 atmosphere = 736 mm Hg = 10000 mm H2O = 10 m H2O = 2050 psf = 14.2 psi = 29 in Hg = 394 in H2O = 32.8 ft H2O
Online Compressed Air Pipeline Pressure Drop Calculator - Metric Units The calculator below can used to calculate the pressure drop in compressed air pipelines.
10 100 50.501
q - air volume flow at atmospheric conditions (FAD) (m3/min) L - pipe length (m) d - inside diameter pipe (mm) p - initial abs. pressure (kg/cm2)
7 Jawab : pressure drop : 0.19 kg/cm2
Compressed Air Pipeline Pressure Drop Spreadsheet Calculations can be done for other pressures and/or pipe lengths by using this excel spreadsheet (metric units). The same spreadsheet including different types of pipes (imperial units). Compressed Air Pipeline Pressure Drop Diagrams Or, alternatively - Compressed Air - Pressure Drop in Pipe Lines - in Google Docs. You can open, save and modify your own copy of the Google spreadsheet if you are signed into your Google Account. Compressed Air Pipeline Pressure Drop Table - Initial abs. Pressure 7 kg/cm2 (100 psig) Pressure drops in 100 m (330 ft) compressed air schedule 40 steel pipe lines are indicated in the tables below:
Flow through Multiple Changes of Elevation As described above, the pressure on a fluid at a point in a piping run changes with the elevation of the fluid. As the fluid rises there is a pressure loss and as it falls there is an equivalent pressure gain (for the same change in elevation). Therefore we only need to consider the net change in fluid elevation between the start and end point of flow, to calculate the pressure loss/gain due to the elevation change.
If fluid enters a run of pipe which has starting elevation of say 2m (relative to some zero point) and it then flows through the pipe system, rising and falling many times, before it finally exits at an elevation of say 5m, then the net change in elevation is 3m (5m - 2m), and the result will be a pressure loss due to change in elevation of of 3m fluid head (which could be converted to units of bar or psi, if required). Of course there would also be pressure losses due to pipe friction, and in the diagram above, the pump would need to produce enough additional fluid head (pressure) to overcome both the pressure loss due to the change in elevation and the pressure loss due to pipe friction.
Energy Grade Line The Energy Grade Line, also called the Energy Line (EL), is a plot of the Bernoulli equation or the sum of three terms in the work-energy equation. The EL is equal to the sum of the fluid's velocity head, the pressure, and the elevation head. EL = (V²/2g) + (p/γ) + h where V = velocity g = acceleration due to gravity p =static pressure (relative to the moving fluid) γ = specific weight h = elevation height A pitot tube can be inserted into a pipe such that the fluid initially flows into the tip of the tube, until the height of fluid in the tube balances the energy coming in, at which point the flow in to the tube stops and the fluid velocity at the very tip of the pitot tube becomes zero. The pressure and the velocity head of the fluid are in effect converted to the equivalent head in height of fluid (i.e. the fluid will rise to the elevation of the EL for that specific point in the flow).
Hydraulic Grade Line The Hydraulic Gradeline (HGL), is the sum of the pressure and elevation heads. This sum is known as the piezometric head and can be measured by inserting a piezometer tube into the side of a pipe so that it is flush with the edge of the pipe. HGL = (p/γ) + h where p =static pressure (relative to the moving fluid) γ = specific weight h = elevation height
Diagram of Hydraulic Grade Line and Energy Grade Line The HGL and EL of flow in a pipe can be illustrated on a diagram, with the pipe drawn at an angle and the associated elevation, pressure, and velocity heads marked up on the vertical axis at each end of the pipe. The resulting HGL and EL lines can then be drawn between specific points on each vertical axis to give a good visualisation of how these change for fluid flow at different points along the pipe.
Perhitungan Kehilangan Tekanan Secara Online • Perhitungan online berikut didasarkan atas pers.(4.6) untuk menentukan besarnya kehilangan tekanan dalam tinggi kolom manometer dalam pipa atau saluran. Aliran yang dipakai adalah aliran udara dengan kondisi suhu sama dengan soal sebelumnya yaitu 20oC, kerapatan, 1.2 kg/m3 dan laju aliran 6 m/s. Nilai koefisien gesekan dihitung menggunakan pers. (4.5) dari Colebrook . • Dengan memasukkan nilai yang sudah diketahui sebelumnya akan didapatkan hasil perhitungan online sbb.. • Top of Form koefisien gesekan - λ • panjang pipa atau saluran - l - (m) • diameter hidraulik - dh - (m) • kecepatan aliran - v - (m/s) • SI-units • Bottom of Form • =0,11 mm H2O
Resistance In Fluid Systems 4.2
Define Drag • For a solid object moving through a fluid or gas, drag is the sum of all the aerodynamic or hydrodynamic forces in the direction of the external fluid flow. It therefore acts to oppose the motion of the object, and in a powered vehicle it is overcome by thrust. FD=1/2 rV2CD A
The drag exerted on an object by a fluid depends on many factors
– The speed of the object
(or fluid) – The size and shape of the object – The physical properties of the fluid
Laminar Flow • A slow, smooth flow over a surface, in which the paths of individual particles do not cross. – Each path is called a streamline
Turbulent Flow • Irregular flow with eddies and whorls causing fluid to move in different directions. – Turbulence is produced by high speeds, shapes that aren’t streamlined and sharp bends in the path of a fluid.
Pressure Drag • Changing direction of fluid into eddies and whorls requires work. When fluid does work – pressure drops.
W = -V DP
Viscosity • The property of a fluid that describes an internal friction between atoms and molecules of a fluid. – These forces create internal friction in the fluid, causing resistance to movement.
= viscosity
The viscosity of a fluid can be measured by pulling a plate at constant speed across a layer of the fluid.
= F Dy
Av
Stokes’ Law • Applies to objects moving at low enough speeds that the flow of fluid around the objects is streamlined. – There is no turbulence and the only drag force on the objects is due to frictional drag.
F drag = 6 r
Terminal Speed • The terminal speed of a falling object is the constant speed that occurs when the drag force equals the gravitational force.
Poiseuille’s Law • Gives the volume flow rate of a fluid flowing through a tube or pipe. – Applies to laminar flow
Poiseuille’s Law (cont.) • The rate at which fluid flows through a tube increases proportionately to the pressure applied and to the fourth power of the radius of the tube.
4
V =- r DP 8 L
Factors Affecting Flow • Poiseuille’s Law shows how resistance of flow through a pipe depends on three factors: – Radius of the pipe – Length of the pipe – Viscosity of the fluid
R=
Pressure drop = - DP Volume flow rate V
Summary
• Drag is the force that opposes the motion of an object moving through a fluid or the force a moving fluid exerts on a stationary object. • Laminar flow is slow, smooth flow over a surface, where particles follow streamlines.
Summary (cont.) • Turbulent flow is irregular flow with eddies and whorls that mix the fluid. • Drag increases with speed. • Viscosity is the property of a fluid that describes its internal friction. • Stokes’ Law – used to calculate drag force on a sphere moving at constant speed in a viscous fluid.
Summary (cont.) • Terminal speed – when drag equals the gravitational force acting upon a falling body. • Poiseuille’s Law – used to calculate the volume flow rate or pressure drop of a viscous fluid flowing through a tube or pipe.
• Induced drag vs.lift
Flow around Immersed bodies •
•
•
•
No regular pattern of motion like the vortex street forms in the wake of a sphere (or of any threedimensional body), although there is a general impression that vorticity is shed from the standing ringvortex like a succession of distorted vortex loops not symmetrical around the central axis (see Vortex Shedding). This flow process does not give rise to vibrations of the sphere. At large Reynolds number, flow over the front half of a sphere may be divided into a thin Boundary Layer region where viscous forces are dominant, and an outer region, where the flow corresponds to that of an inviscid fluid. Pressure decreases over the front half of the sphere from the stagnation point onwards, thus having a stabilizing effect on the boundary layer, which remains laminar up to about Red = 5·105. Beyond the minimum pressure point on the sphere surface, the boundary layer is subjected to an adverse pressure gradient and the fluid decelerates. Later, flow separation occurs. At low ReD, the separation point is located at the rear stagnation point and with increasing ReD moves forward and reaches φ ≈ 80° from the front stagnation point at ReD ≈ 1000. Pressure drag begins to dominate and the Drag Coefficient becomes almost independent of the Reynolds number until about ReD = 5·105, when the transition from Laminar to Turbulent Flow occurs before separation. As a result, the point of separation moves to the rear, making the wake smaller and abruptly reducing drag coefficient.
Flow around immersed bodies •
• •
At very low ReD, during the so-called creeping flow, inertia forces are assumed to be negligible; hence, the governing equations for the flow (Navier-Stokes equations and the continuity equation) are greatly simplified. Stokes succeeded in obtaining a solution for these equations and the drag force has been found to be (1) where R is the radius of the sphere, U∞ the freestream velocity and η the dynamic viscosity of the fluid. See also Stokes' Law for Solid Spheres and Spherical Bubbles. This relationship may be used up to ReD = 0.5 with negligible error. This flow range is often called the Stokes flow. The drag coefficient is defined as:
Drag forces/Drag coefficients • •
(2)
• • •
where A is the projected area (equal to πR2), which can be expressed as:
• • •
where ReD is
• •
Equation (1) is known as Stokes' law. Sometimes Eq. (3) is also referred to as Stokes' Law. An extension of Eq. (3) has been formulated using a method of successive approximations to the governing equations (still at low ReD). This formula reads
(3)
(4)
•
•
(5)
CD as a function of ReD for a sphere. •
•
Equation (5) may be used up to ReD ≈ 100. At higher values of ReD, it is necessary to rely on empirical expressions based on experiments. Figure 2 provides a graph of CD versus ReD for a sphere. When applied to particle mechanics, the following regimes are introduced:
• •
•
•
Several other expressions for drag coefficient may be found in other literature. For blunt bodies like a sphere, an increase in surface roughness may cause a decrease in drag. The transition to turbulent boundary layer flow occurs at a lower Reynolds number than for a smooth sphere. One effect is the wake region behind the sphere becomes considerably narrower and overall drag is reduced. Similar to a circular cylinder, the inviscid flow field around a sphere can also be determined analytically. The velocity components are:
where θ is measured from the forward stagnation point, r is the radial coordinate, ro is sphere radius and U∞ is the free-stream velocity. The maximum velocity occurs at θ = π/2 and is
Aliran sekitar bola • •
• • • • •
Experimentally, however, the maximum value is about 1.3 U∞ and occurs upstream of θ = π/2. Velocity distribution in the laminar boundary layer over the front part of a sphere can be calculated using a series expansion technique. It requires, however, an experimentally determined pressure distribution if accurate values of skin friction and boundary layer thickness are to be achieved. References: Batchelor, G. K. (1970) An Introduction to Fluid Dynamics. Cambridge University Press. Munson, B. R., Young, D. F. and Okiishi, T. H. (1990) Fundamentals of Fluid Mechanics. J. Wiley & Sons. Taneda S.. (1956) Rep. Res. Inst. Appl. Mech., Kyushu Univ, 4, 99. White, F. M. (1994) Fluid Mechanics. 3rd Edn. McGraw-Hill.
Type of Object Laminar flat plate (Re=106) Dolphin Turbulent flat plate (Re=106)
Drag Coefficient - cd 0.001 0.0036 0.005
Subsonic Transport Aircraft Supersonic Fighter,M=2.5 Streamline body Airplane wing, normal position Long stream-lined body Airplane wing, stalled Modern Car like Toyota Prius Sports Car, sloping rear Common Car like Opel Vectra (class C) Hollow semi-sphere facing stream Bird Solid Hemisphere Sphere Saloon Car, stepped rear Convertible, open top Bus Old Car like a T-ford Cube Bike racing Bicycle
0.012 0.016 0.04 0.05 0.1 0.15 0.26 0.2 - 0.3
frontal area frontal area
0.29
frontal area
0.38 0.4 0.42 0.5 0.4 - 0.5 0.6 - 0.7 0.6 - 0.8 0.7 - 0.9 0.8 0.88 0.9
Frontal Area - A - (ft2) wetted area
π / 4 d2
frontal area π / 4 d2 frontal area frontal area frontal area frontal area s2 3.9
Tractor Trailed Truck Truck Person standing Bicycle Upright Commuter Thin Disk Solid Hemisphere flow normal to flat side Squared flat plate at 90 deg Wires and cables Person (upright position) Hollow semi-cylinder opposite stream Ski jumper Hollow semi-sphere opposite stream Passenger Train Motorcyvcle and rider Long flat plate at 90 deg Rectangular box
0.96 0.8 - 1.0 1.0 – 1.3 1.1 1.1
frontal area frontal area
1.17
π / 4 d2
5.5 π / 4 d2
1.17 1.0 - 1.3 1.0 - 1.3 1.2 1.2 - 1.3 1.42 1.8 1.8 1.98 2.1
frontal area frontal area
• Example - Air Resistance on a Normal Car • The force required to overcome air resistance for a normal family car with drag coefficient 0.29 and frontal area 2 m2 in 90 km/h can be calculated as: • Fd = 0.29 1/2 (1.2 kg/m3) ((90 km/h) (1000 m/km) / (3600 s/h))2 (2 m2) • = 181 N • compare car air resistance with car rolling resistance
Shape and flow
Form Drag
Skin friction
0%
100%
~10%
~90%
~90%
~10%
100%
0%
Figure 1. Drag coefficients of blunt and streamlined bodies
Figure 2. Drag coefficient as a function of Reynolds number for smooth circular cylinders and smooth spheres.
Figure 3. Flow patterns for flow over a cylinder: (A) Reynolds number = 0.2; (B) 12; (C) 120; (D) 30,000; (E) 500,000. Patterns correspond to the points marked on figure 2.
Immersed Bodies, Flow Around and Drag- Sunden, Bengt •
•
•
External flows past objects encompass a variety of fluid mechanics phenomena. The character of the flow field depends on the shape of the object. Even the simplest shaped objects, like a sphere, produce rather complex flows. For a given shaped object, the flow pattern and related forces depend strongly on various parameters such as size, orientation, speed and fluid properties. By the concepts of dimensional analysis, the important dimensionless parameters are the Reynolds Number (ratio of inertia forces and viscous forces), the Mach Number (ratio of flow velocity and speed of sound) and sometimes the Froude Number (ratio of inertia forces and gravity forces). The study of flow around immersed bodies has a wide variety of engineering applications but in terms of heat and mass transfer, spheres are the most important. Thus most of this article deals with spheres. There is a great similarity in the development of flow pattern at increasing Reynolds number between a sphere and a circular cylinder (or tube), except for the vortex street associated with the latter and other two-dimensional bodies, which is not formed for three-dimensional bodies, instead a vortex ring occurs, which for a sphere is formed at about ReD = 24 (see Figure 1) and becomes unstable at about ReD = 200 when it tends to move downstream of the body and is immediately replaced by a new vortex ring.
Figure 1. Observed lengths of the region of closed streamlines behind a sphere. From Taneda S. (1956).
Figure 4. Boundary layer profiles for laminar and turbulent flow
Examples of drag • Examples of drag include the component of the net aerodynamic or hydrodynamic force actin g opposite to the direction of movement of the solid object relative to the Earth as for cars, aircraft and boat hulls; or acting in the same geographical direction of motion as the solid, as for sails attached to a down wind sail boat, or in intermediate directions on a sail depending on points of sail. In the case of viscous drag of fluid in a pipe, drag force on the immobile pipe decreases fluid velocity relative to the pipe.[7][8]
Types of drag • Types of drag are generally divided into the following categories: • parasitic drag, consisting of – form drag, – skin friction, – interference drag,
• lift-induced drag, and • wave drag (aerodynamics) or wave resistance (ship hydrodynamics).
• The phrase parasitic drag is mainly used in aerodynamics, since for lifting wings drag is in general small compared to lift. For flow around bluff bodies, drag is most often dominating, and then the qualifier "parasitic" is meaningless. Form drag, skin friction and interference drag on bluff bodies are not coined as being elements of "parasitic drag", but directly as elements of drag.
•
• •
Further, lift-induced drag is only relevant when wings or a lifting body are present, and is therefore usually discussed either in the aviation perspective of drag, or in the design of either semi-planing or planing hulls. Wave drag occurs when a solid object is moving through a fluid at or near the speed of sound in that fluid—or in case there is a freely-moving fluid surface with surface waves radiating from the object, e.g. from a ship. Drag coefficient Cd for a sphere as a function of Reynolds number Re, as obtained from laboratory experiments. The solid line is for a sphere with a smooth surface, while the dashed line is for the case of a rough surface. Drag depends on the properties of the fluid and on the size, shape, and speed of the object. One way to express this is by means of thedrag equation:
•
FD=1/2 rV2CD A
• • • • •
where FD is the drag force,ρ is the density of the fluid,[ v is the speed of the object relative to the fluid, A is the cross-sectional area, and CD is the drag coefficient – a dimensionless number.
• Drag coefficient Cd for a sphere as a function of Reynolds number Re, as obtained from laboratory experiments. The solid line is for a sphere with a smooth surface, while the dashed line is for the case of a rough surface.
Lift-induced drag • Induced drag vs.lift • Lift-induced drag (also called induced drag) is drag which occurs as the result of the creation of lift on a three-dimensional lifting body, such as thewing or fuselage of an airplane. Induced drag consists of two primary components, including drag due to the creation of vortices (vortex drag) and the presence of additional viscous drag (lift-induced viscous drag). The vortices in the flow-field, present in the wake of a lifting body, derive from the turbulent mixing of air of varying pressure on the upper and lower surfaces of the body, which is a necessary condition for the creation of lift. • With other parameters remaining the same, as the lift generated by a body increases, so does the lift-induced drag. For an aircraft in flight, this means that as the angle of attack, and therefore the lift coefficient, increases to the point of stall, so does the lift-induced drag. At the onset of stall, lift is abruptly decreased, as is lift-induced drag, but viscous pressure drag, a component of parasite drag, increases due to the formation of turbulent unattached flow on the surface of the body.
Parasitic drag • ]Parasitic drag (also called parasite drag) is drag caused by moving a solid object through a fluid. Parasitic drag is made up of multiple components including viscous pressure drag (form drag), and drag due to surface roughness (skin friction drag). Additionally, the presence of multiple bodies in relative proximity may incur so called interference drag, which is sometimes described as a component of parasitic drag. • In aviation, induced drag tends to be greater at lower speeds because a high angle of attack is required to maintain lift, creating more drag. However, as speed increases the induced drag becomes much less, but parasitic drag increases because the fluid is flowing more quickly around protruding objects increasing friction or drag. At even higher speeds in the transonic, wave drag enters the picture. Each of these forms of drag changes in proportion to the others based on speed. The combined overall drag curve therefore shows a minimum at some airspeed - an aircraft flying at this speed will be at or close to its optimal efficiency. Pilots will use this speed to maximize endurance (minimum fuel consumption), or maximize gliding range in the event of an engine failure.
Power curve in aviation • ]The power curve: form and induced drag vs. airspeed • The interaction of parasitic and induced drag vs. airspeed can be plotted as a characteristic curve, illustrated here. In aviation, this is often referred to as the power curve, and is important to pilots because it shows that, below a certain airspeed, maintaining airspeed counterintuitively requires more thrust as speed decreases, rather than less. The consequences of being "behind the curve" in flight are important and are taught as part of pilot training. At the subsonic airspeeds where the "U" shape of this curve is significant, wave drag has not yet become a factor, and so it is not shown in the curve.
The power curve: form and induced drag vs. airspeed
Wave drag in transonic and supersonic flow • Qualitative variation in Cd factor with Mach number for aircraft • Wave drag (also called compressibility drag) is drag which is created by the presence of a body moving at high speed through a compressible fluid. In aerodynamics, Wave drag consists of multiple components depending on the speed regime of the flight. • In transonic flight (Mach numbers greater than about 0.8 and less than about 1.4), wave drag is the result of the formation of shockwaves on the body, formed when areas of local supersonic (Mach number greater than 1.0) flow are created. In practice, supersonic flow occurs on bodies traveling well below the speed of sound, as the local speed of air on a body increases when it accelerates over the body, in this case above Mach 1.0. However, full supersonic flow over the vehicle will not develop until well past Mach 1.0. Aircraft flying at transonic speed often incur wave drag through the normal course of operation. In transonic flight, wave drag is commonly referred to as transonic compressibility drag. Transonic compressibility drag increases significantly as the speed of flight increases towards Mach 1.0, dominating other forms of drag at these speeds
Flow in Tube Banks • Figure 1 shows the two basic tube-bank patterns involving either a rectangular or a rhombic primitive unit. These are referred to as in-line tube banks and staggered tube banks respectively. These are characterized by crosswise and streamwise pitch-to-diameter ratio's, a and b, • •
a=ST /D
b= SL/D
(1)
(2)
• where D is the cylinder diameter, sT the crosswise (transverse) pitch, and sL the streamwise pitch. Most commonly encountered tube banks are the in-line square (a = b), rotated square (a = 2b) and equilateral triangle (a = 2b/√3). Tube banks with the product axb <1.252 are referred to as compact, while those with axb> 4 are considered widely-spaced.
• The flow of a real fluid in the body of a tube bank also resembles flow past a single cylinder, though with significant differences. (See Tubes, Crossflow over.) The flow Reynolds Number, Re, is defined by (3)
• where ρ is the fluid density, ūmax is the maximum bulk velocity i.e. the bulk velocity in the minimum cross section (seeFigure 1), D is the cylinder diameter, and η is the fluid viscosity. At low Re, the flow is Laminar with separation occurring at around ρ = 90°, resulting in stable Vortices forming behind each cylinder. For staggered banks, the upstream flow is typically a maximum between the preceding tubes, so the impinging flow bifurcates at the front-leading edge of each cylinder. For inline banks, cylinders are in a comparatively dead zone downstream of the preceding cylinder's wake, and there are two re-attachment points at around φ = ±45°.
The mean superficial velocities is useful when simulating large-scale flow in heat exchangers. It can readily be shown that
(4)
for all in-line banks, and staggered banks with a < 2b2 – 1/2. For some compact staggered banks, where a > 2b2 – 1/2, the minimum cross section occurs across the diagonal and so (5)
The hydrodynamic parameter of most interest to the heat exchanger designer is the overall pressure loss coefficient, often expressed in terms of an Euler Number Eu,
(6) where is the mean pressure drop across a single row. Numerous alternative definitions abound in the literature, for instance, Bergelin et al. (1950, 1958) define a Friction Factor f = 4Eu. Others simply use the symbol f to denote Eu. The friction factor of Kays and London (1984) is such that f = (a – 1)Eu/π for large banks. ESDU (1979) use a different definition for pressure loss coefficient, again, based on Ūmean, i.e. there has been little effort to-date to standardize parameters.
Various empirical correlations of pressure drop data have been devised over the years. The Eu vs. Re correlations of the Lithuanian group are commendable; they have been reconciled with numerous sources of externally-gathered experimental data, in addition to data gathered by the authors themselves. They are included here. Others such as those based on the Delaware groups work, could have been reproduced equally well, having formed an integral part of the thermal design of shell-and-tube exchangers, in the West for many years; see the articles by Taborek in the Heat Exchanger Design Handbook (1983) and Mueller in the Handbook of Heat Transfer [Rohsenhow and Hartnett (1972)]. Figures 2 and 3 show Eu vs. Re (based on ūmax) for in-line square and equilateral triangle tube banks. The Heat Exchanger Design Handbook (1983) contains analytical expressions approximating these curves. These take the form of a power series, (7) for all in-line and rotated square banks, except in-line banks with b = 2.5, for which
(8)
Values of the coefficients cj for in-line and staggered tube banks are given in Table 1 and Table 2, respectively. The reader is cautioned that these equations render poor continuity across certain ranges of application. Other mathematical correlations also exist, for example, ESDU (1979). Agreement between ESDU/Delaware, and the Lithuanian group's curves is not particularly good, especially in the low-intermediate Re range [Beale (1993)].
Figure 2. Pressure drop coefficient vs. Reynolds number for in-line tube banks. From Heat Exchanger Design Handbook(1983).
Figure 3. Pressure drop coefficient vs. Reynolds number for staggered tube banks. From Heat Exchanger Design Handbook (1983). Table 1. Coefficients, ci, for use in Equations
Table 1. Coefficients, ci, for use in Equations (7) and (8) to generate pressure drop coefficients for in-line square banks. From Heat Exchanger Design Handbook (1983)
Table 2. Coefficients, ci, for use in Eq. (7) to generate pressure drop coefficients for equilateral triangle banks. FromHeat Exchanger Design Handbook (1983)
In most practical tube banks it is necessary to modify Eu for several effects. This usually done as follows:
(9)
where Eu' is the value calculated from the correlation for an ideal bank, and the ki are correction factors. Corrections are typically required to account for the geometry, size, and location of the tube bank, deviations from normal incidence of the working fluid, and variations in the fluid properties due to temperature and pressure changes. These are detailed below.
Mesin-mesin hydraulika
Pompa sentrifugal Pw=QH/
Persamaan Euler • Torsi= laju perubahan momentum angular • Momentum angular =(massa)(kecepatan tangential)(radius) jadi • Mementum angular memasuki impeler /det =(W/g)Vw1 r1 • Momentum angular yang keluar dari impeler/det=(W/g)Vw2r2 • Disini W adalah bobot dari fluida yang mengalir tiap detik
• • • • • • • • • •
Karena it laju perubahan momentum angular adalh (W/g)Vw2r2-(W/g)Vw1r1 Sehingga torsi menjadi =(W/g)(Vw2r2-Vw1r1) Kaerna kerja per satuan waktu adalah torsi dikalikan dengan kecepatan angular, maka WD/det=(torsi)=(W/g)(Vw2r2-Vw1r1) Tetapi =u/r sehingga r2=u2 dan r1=u1 sehingga WD/det=Et=(W/g)(u2Vw2-u1Vw1) Dimana satauannya adalah Ft Lbf/det. (7.2.3) Karena kerja yang dilakukan per detik oleh impeler terhadap fluida (dalam hal ini laju perpindahan energi ) maka
• • • • • • • • • •
Laju perpindahan energi/lbf dari fluida =E=(1/g)(u2Vw2-u1Vw1) (7.2.4) Dimana satuannya adalah ft Lbf/Lbf Persamaan (7.2.4) ini dinamakan persamaan Euler Lebih lanjut karena u2=D2N/60 dan u1=D1N/60 maka E=H=(N/g60)(D2Vw2-D1Vw1) Vw2=u2-Vf2cot2 dan Vf2=Q/A2 Karena pada kondisi normal Vw1=0, maka untuk pompa sentrifugal • H=(ND2 u2)/g60(u2-Q/A2 cot 2)
• atau
D2 N D2 N
Q cot 2 H ( ) 60 g 60 A2 karena
A2 d 2b2
D2 N D2 N
Q cot 2 H ( ) 60 g 60 d 2b2
Hubungan Head (H) dan debit (Q) H 180 160 Head (m H2O)
140 120 100 80
H
60 40 20 0 0.01
0.02
0.03
0.04 0.05 Q(m3/s)
0.06
0.07
0.08
Hubungan Head dengan Q 60
Head (m H2O)
50 40 30
H 2500 N3000
20 10 0 0.01
0.02
0.03
0.04 0.05 Debit Q(m3/s)
0.06
0.07
0.08
Persamaan Euler untuk pompa/kipas sentrifugal • Hubungan Head,H dan debit, Q, tebal sudu b2, jari-jari luar cakram suhu R2, dan sudut singgung sudu dan cakram 2 H
QCot ( 2 .). u2 (u2 .. g 2R2b2
Persamaan Euler • Diturunkan berdasarkan Hukum Bernoulli yaitu keseimbangan momentum dalam pompa/kipas U2 Q H [U 2 (cot 2)] g A2
• Dimana U 2 r 2
Karakteristik pompa sentrifugal Pw=QH/
H=U2/g{U2Qcot 2/br2}
Laju udara, diameter pipa dan Re pada suhu 20 oC Re 1.40E+04 1.20E+04 1.00E+04 8.00E+03 6.00E+03 4.00E+03 2.00E+03 0.00E+00 0.02
0.03
0.04
0.05
Pipe diameter D (m)
V (m/s)
Udara 20oC D (m) v (m2/s) Re Cf 3 0.02 1.51E-05 3.97E+03 0.03 5.96E+03 0.04 7.95E+03 0.05 9.93E+03 0.06 1.19E+04
0.06
Laju air, V, Diameter Pipa dan Re, untuk air yang mengalir pada suhu 30 oC dan Re Re 1.40E+04 1.20E+04 1.00E+04 8.00E+03 6.00E+03 4.00E+03 2.00E+03 0.00E+00 0.02
0.03
0.04
0.05
Pipe diameter D (m)
Air 30 oC V (m/s) D (m) v (m2/s) Re Cf 3 0.02 8.01E-07 7.49E+04 0.03 1.12E+05 0.04 1.50E+05 0.05 1.87E+05 0.06 2.25E+05
0.06
Minor loss coefficients for common used components in pipe and tube systems • • • • • • • •
Sponsored Links Minor head loss in pipe and tube systems can be expressed as hminor_loss = ξ v2/ 2 g (1) where hminor_loss = minor head loss (m, ft) ξ = minor loss coefficient v = flow velocity (m/s, ft/s) g = acceleration of gravity (m/s2, ft/s2)
Relationship Between K, Equivalent Length and Friction Factor
• •
The total pressure drop in the pipe is typically calculated using these five steps. (1) Determine the total length of all horizontal and vertical straight pipe runs. (2) Determine the number of valves and fittings in the pipe. For example, there may be two gate valves, a 90o elbow and a flow thru tee. (3) Determine the means of incorporating the valves and fittings into the Darcy equation. To accomplish this, most engineers use a table of equivalent lengths. This table lists the valve and fitting and an associated length of straight pipe of the same diameter, which will incur the same pressure loss as that valve or fitting. For example, if a 2” 90o elbow were to produce a pressure drop of 1 psi, the equivalent length would be a length of 2” straight pipe that would also give a pressure drop of 1 psi. The engineer then multiplies the quantity of each type of valve and fitting by its respective equivalent length and adds them together. (4) The total equivalent length is usually added to the total straight pipe length obtained in step one to give a total pipe equivalent length. (5) This total pipe equivalent length is then substituted for L in Equation 2 to obtain the pressure drop in the pipe.
Pressure drop in pipe fittings
The following discussion is based on concepts found in reference 1, the CRANE Technical Paper No. 410. It is the author’s opinion that this manual is the closest thing the industry has to a standard on performing various piping calculations. If the reader currently does not own this manual, it is highly recommended that it be obtained. As in straight pipe, velocity increases through valves and fittings at the expense of head loss. This can be expressed by another form of the Darcy equation similar to Equation 1:
The K coefficient in fitting • When comparing Equations 1 and 3, it becomes apparent that:
• •
K is called the resistance coefficient and is defined as the number of velocity heads lost due to the valve or fitting. It is a measure of the following pressure losses in a valve or fitting: – – – –
Pipe friction in the inlet and outlet straight portions of the valve or fitting Changes in direction of flow path Obstructions in the flow path Sudden or gradual changes in the cross-section and shape of the flow path
Pipe fittings and pipe exit
Screwed Fittings - equivalent length in meter Equivalent length (in meters) of straight pipe for fittings like bends, returns, tees and valves.
Equivalent length in feet of straight pipe for copper pipe fittings Sponsored Links
Size (in)
90o Elbow
45o Elbow
Line Tee
Branch Tee
1/2
0.9
0.4
0.6
2.0
5/8
1.0
0.5
0.8
2.5
7/8
1.5
0.7
1.0
3.5
1 1/8
1.8
0.9
1.5
4.5
1 3/8
2.4
1.2
1.8
6.0
1 5/8
2.8
1.4
2.0
7.0
2 1/8
3.9
1.8
3.0
10.0
2 5/8
4.6
2.2
3.5
12.0
The equivalent length in feet for the fittings used is to be added to the actual length of the piping run to get the total length.
The k values of fittings Elbow, Flanged Long Radius 45o
0.2
Return Bend, Flanged 180o
0.2
Return Bend, Threaded 180o
1.5
Globe Valve, Fully Open
10
Angle Valve, Fully Open
2
Gate Valve, Fully Open
0.15
Gate Valve, 1/4 Closed
0.26
Gate Valve, 1/2 Closed
2.1
Gate Valve, 3/4 Closed
17
Swing Check Valve, Forward Flow
2
Ball Valve, Fully Open
0.05
Ball Valve, 1/3 Closed
5.5
Ball Valve, 2/3 Closed
200
Diaphragm Valve, Open
2.3
Diaphragm Valve, Half Open
4.3
Diaphragm Valve, 1/4 Open
21
Water meter
7
Type of Component or Fitting
Minor Loss Coefficient -ξ-
Tee, Flanged, Line Flow
0.2
Tee, Threaded, Line Flow
0.9
Tee, Flanged, Branched Flow
1.0
Tee, Threaded , Branch Flow
2.0
Union, Threaded
0.8
Elbow, Flanged Regular 90o
0.3
Elbow, Threaded Regular 90o
1.5
Elbow, Threaded Regular 45o
0.4
Elbow, Flanged Long Radius 90o
0.2
Elbow, Threaded Long Radius 90o
0.7
fittings Valve/Fitting
Elbows : # 90o Short radius # 90o Long radius # 45o Short radius # 45o Long radius
Tees* :Branch flow Run flow Laterals* : Branch-Run flow Run-Branch flow Run-Run Run-Run Branch-Run reverse Run-Branch reverse
Equivalent length L/D
Valve/Fitting
30 20 16 12
Valves: # Globe-conventional # Globe-60o Y Pattern # Globe-45o Y Pattern # Angle-Conventional # Cock -Straight through -Three way-Run -Three way-branch Check -Conventional Swing -Clearway Swing -Globe Lift or Stop -Angle Lift or Stop -Inline Ball Gate-Disk
60 20 30 40 20 25 130 130
Equivalent length L/D 340 175 145 145 18 44 140 135 50 345 145 150 13
Pipe fitting Valves: # Globe-conventional # Globe-60o Y Pattern # Globe-45o Y Pattern # Angle-Conventional # Cock -Straight through -Three way-Run -Three way-branch Check -Conventional Swing -Clearway Swing -Globe Lift or Stop -Angle Lift or Stop -Inline Ball Gate-Disk
340 175 145 145 18 44 140 135 50 345 145 150 13
Fitting coefficient Typical Fitting K Coefficients Fitting
K Value
Pipe Entrance
Fitting
K Value
90° Smooth Bend
Bellmouth
0.03-0.05
Bend Radius / D = 4
0.16-0.18
Rounded Sharp-Edged
0.12-0.25 0.50
Bend Radius / D = 2 Bend Radius / D = 1
0.19-0.25 0.35-0.40
Projecting
0.80
Mitered Bend
Contraction—Sudden
= 15°
0.05
D2/D1 = 0.80
0.18
= 30°
0.10
D2/D1 = 0.50
0.37
= 45°
0.20
D2/D1 = 0.20
0.49
= 60°
0.35
= 90°
0.80
Contraction—Conical D2/D1 = 0.80
0.05
Tee
D2/D1 = 0.50
0.07
Line Flow
0.30-0.40
D2/D1 = 0.20
0.08
Branch Flow
0.75-1.80
Expansion—Sudden
Cross
D2/D1 = 0.80 D2/D1 = 0.50
0.16 0.57
Line Flow Branch Flow
D2/D1 = 0.20
0.92
45° Wye
Expansion—Conical D2/D1 = 0.80
0.03
D2/D1 = 0.50
0.08
D2/D1 = 0.20
0.13
0.50 0.75
Line Flow
0.30
Branch Flow
0.50
Union/pipe fitting
Pipe fittings
BAB VII. KINCIR AIR • Tipe kincir air: – Kincir yang berputar berdasarkan bobot dari air yang dipasok – Kincir yang perputar sebagian oleh gaya berat air dan sebagian oleh gaya pancaran air – Kincir yang digerakkan seluruhnya oleh pancaran aliran alir
• Ada dua cara pemberian air kedalam ember kincir: – Melalui ember/sudu bagian atas (Overshoot) – Melalui ember/sudu bagian bawah (undershoot)
Tipe overshoot
Daya kincir air • Persamaan dasar output daya kincir air P= ( Q H)
(7.1)
• Contoh Spesifikasi 1) 2) 3) 4) 5) 6)
Tinggi air jatuh, H Diameter kincir, D Jumlah ember, Putaran kincir,N Panjang ember, Efisiensi daya kincir,
: 10 – 25 m : 3 – 20 m : 8 – 10 D : 4 – 8 RPM : 50 cm -100 cm : 60%-80%
Contoh soal • 7.1. Sebuah kincir air tipe overshoot terpasang pada parit dengan lebar 1,5 m. Laju air dalam parit adalah 1,5 m/det. dengan kedalam parit 15 cm. Hitung besarnya daya yang dapat dihasilkan oleh kincir air tsb.bila air jatuh dari ketinggian 20 m dari kincir dan efisiensi kincir 75%.
Jawab • Persamaan debit aliran Q untuk kincir air Q= k b d v atau Q=k b d D/2 ........(7.2) disini – d=panjang ember, m – b=Lebar ember, m – k=porsi ember yang terisi air, (%) – =kecepatan angular dari roda kincir (rad/det) – v=Kecepatan putar ember (m/det)
• • • • •
Jawab: Lebar parit, W : 1,5 m Laju aliran, v : 1,5 m/det. Kedalaman air, t :0,15 m Debit air, Q : W x v x t= 1,5 x 1,5 x 0,15=0,3375 m3/det • Dari data diatas dapat dihitung daya riel kincir yaitu • P= x xQ x H, atau • P=0,75 x 9800 x 0,3375 x 20 =49.6 kW
Praktikum • Judul: Pengukuran kecepatan aliran dalam pipa • Tujuan: 1) Menentukan laju dan debit udara dalam pipa 2) Menentukan total penurunan tekanan dalam pipa 3) Menentukan daya kipas
Teori • • • • • •
Persamaan D’Arcy Weisbach DP= Cf (L/D)V2/2g .........(1) Dimanan koefisien gesekan , Cf dihitung sbb. Aliran Laminer Cf= 64/Re .....................(2) Aliran Turbulen Re>4000 (Karman and Nikuradse) • 1/Cf=2 log 10 (ReCf)-0.8 .......(3)
Diagram Moody
Peralatan (Experimental Set up)
KIPAS
Hasil pengukuran Run
1 2 3 4
DP (mmH2O)/ m
DP (N/m2)
V (m/s)-hitung
V (m/s)-ukur
