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Poglavlje 1
Osnove
1.1 Osnovni pojmovi o strujanjufluida
1.1.1 Brzina
Slika 1.1: Kretanje fludnog deli�a kroz prostor.
U vremenskom trenutku t vektor polo�aja fluidnog deli�aje ~x (x1, x2, x3), a u vremenskom trenutku t+Dt fluidni deli�se pomerio za ~Dx (Dx1, Dx2, Dx3). Brzina fluidnog deli�a,
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~u (u1, u2, u3), je koliqnik pomeranja deli�a i odgovaraju�egvremena:
ui =Dxi
Dt(1.1)
Poxto izvod nije u bilo kom pravcu ~dxl, ve� se pratifluidni deli�, sa ~Dx, brzina je materijalni izvod vektorapolo�aja.
1.1.2 Trajektorija
Slika 1.2: Trajektorije na hidrauliqkom modelu reqne kri-vine na Tisi, kod Adorjana.
Trajektorija, ili putanja, je geometrijsko mesto taqakakroz koje deli� prolazi u svom kretanju kroz prostor. Radise o jednom deli�u i o vremenu koje je potrebno da deli�pro�e kroz deo prostora koji se izuqava.
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1.1.3 Materijalni izvodY je bilo koja veliqina koja je funkcija polo�aja u prostorui vremenu:
Y = Y (x1, x2, x3, t) (1.2)
Izvod po vremenu od Y , kao izvod slo�ene funkcije, je:
dY
dt=
∂Y
∂t+
∂Y
∂x1
dx1
dt+
∂Y
∂x2
dx2
dt+
∂Y
∂x3
dx3
dtdY
dt=
∂Y
∂t+
∂Y
∂xi
dxi
dt(1.3)
Jednaqina 1.3 je napisana primenom pravila da ako se u jed-nom monomu isti indeks udvostruqi to predstavlja trinom,prema [1].
Ako se priraxtaj Y ve�e za fluidni deli�, onda dxi idt nisu nezavisni, jer je njihov odnos odre�en brzinom ui.Priraxtaj po vremenu od Y , vezan za fluidni deli�, je ma-terijalni izvod od Y i mo�e se napisati kao:
DY
Dt=
∂Y
∂t+ ui
∂Y
∂xi. (1.4)
Prvi qlan zbira ∂Y/∂t je lokalna komponenta, a ui ∂Y/∂xi kon-vektivna komponent materijalnog izvoda.
1.1.4 UbrzanjeUbrzanje je materijalni izvod brzine:
Duj
Dt=
∂uj
∂t+ ui
∂uj
∂xi. (1.5)
Za pravac j = 2 komponenta ubrzanja je:
Du2
Dt=
∂u2
∂t+ u1
∂uj
∂x1+ u2
∂uj
∂x2+ u3
∂uj
∂x3. (1.6)
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1.1.5 Koliqina kretanjaKoliqina kretanja je proizvod mase ρdV i brzine uj. Mater-ijalni izvod koliqne kretanja jednak je:
D
Dt(ρdV uj) = ρdV
Duj
Dt. (1.7)
1.1.6 StrujnicaStrujnica je geometrijsko mesto taqaka, koje u svakoj taqkiima tangentu u pravcu brzine. Strujnica se odnosi na jedanvremenski trenutak.
Slika 1.3: Strujnica (za vreme t3) i trajektorija, koje pro-laze kroz taqku A.
Iz definicije brzine, jednaqina (1.1), sledi:
ui =Dxi
Dt⇒ Dx1
u1=
Dx2
u2=
Dx3
u3= Dt (1.8)
xto je diferencijalna jednaqina strunice.
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Slika 1.4: Strujnice na hidrauliqkom modelu evakuacionihobjekata na akumulaciji Grlixte, kod Zajeqara.
1.1.7 Emisiona linija
Slika 1.5: Emisiona linija iz taqke A i trajektorije.
Emisionu liniju saqinjavaju svi deli�i koji su proxlikroz istu taqku. Emisiona linija se odnosi na jedan trenu-tak vremena.
Ako je strujanje ustaljeno trajektorija, strujnica i emi-siona linija se preklapaju. Strujanja prikazana na slikama1.3 i 1.5 su neustaljena.
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1.1.8 Proticaj
Slika 1.6: Proticaj kroz povrxinu A.
Elementarni proticaj, dQ, jednak je skalarnom proizvoduvektora brzine, ~u, i orijentisane elementarne povrxine, ~dA,xto je dato sa:
dQ = ~u · ~dA,
dQ = ni ui dA, (1.9)
gde je ni ort spoljne normale.Prema pravilu o udvajanju indeksa sledi:
ni ui = n1 u1 + n2 u2 + n3 u3.
Ako se elementarni proticaj, dQ, definisan jednaqinom(1.9), integrixe po celoj povrxini kroz koju struju fluid,A, dobija se proticaj, Q, kao:
Q =∫
Ani ui dA. (1.10)
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Jednaqina (1.10) odre�uje proteklu zapreminu fluida. Pro-tok mase, Qρ, na osnovu jednaqine (1.10), je:
Qρ =∫
Aρni ui dA (1.11)
Na sliqan naqin mo�e se definisati protok energije, zaga-�enja, ili neke druge skalarne veliqine.
1.1.9 Srednja brzina
Slika 1.7: Raspored brzina po popreqnom preseku i srednjabrzina v.
Ako su strujnice me�usobno paralelne i normalne na povr-xinu, A, kao na slici 1.7, jednaqina (1.10) se svodi na:
Q =∫
Au dA (1.12)
Uvodi se srednja brzina u popreqnom preseku, v, kao:
v =
∫
Au dA
A=
Q
A(1.13)
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1.1.10 Gustina, stixljiv i nestixljiv fluidProseqna gustina, ρpros, je:
ρpros =m
V. (1.14)
Na sliqan naqin gustina u taqki, ρ, je:
ρ = lim∆V →0
∆m
∆V=
dm
dV. (1.15)
Ako je fluid nestixljiv gustina ne zavisi od pritiska iva�i:
ρ = const. (1.16)
Dimenzija za gustinu je:
[ρ] = M L−3.
Za vodu, pri temperaturi t = 4◦ C, gustina je ρ = 1000 kg/m3.Voda se smatra da je nestixljiva, osim pri razmatranju hi-drauliqkog udara. Vazduh se smatra da je nestixljiv svedok brzina kretanja vazduha nije ve�a od 1/4 brzine kretanjazvuka u vazduhu. Da li �e se fluid smatrati stixljivim,ili ne, zavisi od problema koji se rexava.
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1.1.11 Jednaqina nepromenljivosti mase– Jednaqina kontinuiteta
Stav o nepromenljivosti mase je: ,,Masa se ne mo�e ni un-ixtiti ni stvoriti”. Na osnovu ovog stava masa fluidnogdeli�a je:
dm = ρdV = const (1.17)
Materijalni izvod fluidnog deli�a daje:
D
Dt(ρdV ) = 0 (1.18)
xto pretstavlja jednaqinu nepromenjivosti mase deli�a. In-tegrisanjem jednaqine (1.18) po konaqnoj zapremini, V , do-bija se:
∫
V
D
Dt(ρdV ) =
∫
V
∂ρ
∂tdV +
∫
Aρni ui dA = 0 (1.19)
jednaqina nepromenljivosti mase u konaqnoj zapremini, V ,ograniqenoj povrxinom A.
Jednaqina (1.19) mo�e se napisati i u obliku:
∫
V
∂ρ
∂tdV = −
∫
Aρni ui dA (1.20)
odakle je jasno da do promene gustine fluida, ρ, unutar kon-trolne zapremine, V , mo�e do�i samo ako su ulaz i izlazfluida, kroz kontrolnu zapreminu, A, razlikuju. Na slici1.9, na delovima gde je ugao izme�u brzine i normale povr-xine ve� od pravog ugla, proizvod ima znak ,,-” i tu je ulazfluida. Obrnuto, na delovima gde je ugao manji od pravognalazi se izlaz i proizvod ima znak ,,+”.
Pri razmatrenju nestixljivog fluida za parcijalni izvodgustine, po vremenu, va�i:
∂ρ
∂t= 0 (1.21)
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Slika 1.8: Proticanje fluida kroz telo zapremine V .
zbog qega zapreminski integral u jednaqini (1.20) mora biti0, pa sledi:
∫
Aρni ui dA = 0, (1.22)
xto predstavlja jednaqinu odr�anja mase za nestixljiv fluid,za konaqnu zapreminu, V , ograniqenu povrxinom, A.
Slika 1.9: Proticanje fluida a) na proxirenju cevi i b) naraqvanju cevi.
Ako se jednaqina (1.22) primeni na strujanje na proxirenju,
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dato na slici 1.9, pod ,,a”, dobija se:
−v1 A1 + v2 A2 = 0
A1 v1 = A2 v2 (1.23)
ili za primer ,,b”, sa iste slike:
−v1 A1 + v2 A2 + v3 A3 = 0 (1.24)
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1.1.12 Dinamiqka jednaqina za konaqnu masu
Slika 1.10: Sile koje deluju na masu fluida izme�u preseka,,1” i ,,2”.
Pretpostavlja se da je:
• fluid nestixljiv
ρ = const
• i strujanje ustaljeno
∂v
∂t= 0 .
Priraxtaj koliqine kretanja za vreme dt je:
−ρQdt ~v1 + ρQdt ~v2 = ρQdt (~v2 − ~v1) (1.25)
gde je Qdt zapremina vode koja protekne kroz kontrolne pre-seke ,,1” i ,,2”, a ρQdt odgovaraju�a masa. Ako se leva stranajednaqine (1.25) podeli sa dt dobija se priraxtaj koliqinekretanja u jedinici vremena:
ρQ (~v2 − ~v1) . (1.26)
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Na osnovu zakona o odr�anju koliqine kretanja, da je pri-raxtaj koliqine kretanja u jedinici vremena jednak rezul-tanti deluju�ih sila, sledi:
ρQ (~v2 − ~v1) = ~G + ~K + ~P1 + ~P2 (1.27)
gde je te�ina, ~G, zapreminska sila, a ~K+ ~P1+ ~P2 su povrxinskesile. ~K je sila kojom qvrsta granica deluje na fluid, a ~Pi
je sila pritiska na odgovaraju�em kontrolnom preseku ,,i”.Leva strana jednaqine (1.27) sa promenjenim znakom je:
~I = −ρQ (~v2 − ~v1) = ~I1 + ~I2 (1.28)
gde je ~I ,,fiktivna inercijalna sila”.
~I1 = ρQ ~v1 (1.29)
~I2 = ρQ (−~v2) (1.30)
~I1 + ~I2 + ~G + ~K + ~P1 + ~P2 = 0 (1.31)
Zadaci koji se rexavaju su:
1. Poznati su: Q, P1, P2 I1 i I2, iz jednaqine (1.31) odre�ujese sila kojom fluid deluje na qvrstu granicu − ~K.
2. Poznati su: Q, I1, I2, K i jedna sila pritiska, a odre�u-je se druga sila pritiska.
Ako se po popreqnom preseku brzina strujanja menja uvodise:
β =
∫
Au2 dA
v2 A(1.32)
U tom sluqaju inercijalna sila se definixe kao:
~I = ρQβ ~v (1.33)
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