Lecture Note
University
Swansea UniversityCourse
EG-086 | Engineering SciencePages
6
Academic year
2023
Lloyd Reader
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0
Wednesday Reynolds Transport theorem () system in Fluid mechasies Lagragion approach. Eulevian Difference bhw (.) Control Volume and Eulenan Lograngies ( differential system Laglorgion (.) Integral equation cularian. The description of a greening equation can be transformed between a system approach Theorem. and a control volume approach using the Reynold's Transport Ct at t=to tat Derivation at system it totAt system I / II III + Costed Volune at t= to N represents the amount of mass or momentum or Angular momentum, Energy or Entropy of the system Corresponding to this extensive property, we need to use Intensive property n ( yeta) Nigister = n dm = Snedy M(syllem) & (system) selecting an artiblery piece of flowing fluid at is taken some time as to the , Control volume. which is fixed in space After , the initial shape of the fluid system am infitidecimal time At the System would have moved to a new location,
for any fluid mechanics equation were dNs lim Ns least - Nslto - I = dt at At Our aim is to change System to control volume Ns 16++ = NI + NIII 6+2+ 6++ Ns lb+4t b+A+ = Nc+ Cv - Ns/+ + NI - 2 to+++ to +At Ns16 = Nc ltnb - 3 Ct t=to. 2 and 3 in I dNs lim NCH- NI) +NE/ - Neviltato tato = to+At total H At-70 At = lim (New - Nit NII) totat - (NCV), to At At = lim New/bost - N total + total Nculto At 0 At lin Ns/4: = At lan Ncv 16+A+ - Norle) + At his NIII At trans toket - Afto At At
= ONew + lan Na - At-90 At 61 A lim NI ot At-70 At bist lim No - lim NI a nsdv + AL 41-10 = at st so st +st total Cv Nn lim NJ - it-30 At 17/02/17 dNs at = at a fn Co pdv to him At-30 At tot at total CH Now converting second torm to ie; Now consider a small promate III lim NIII At At totAt cigarettes stop for vega na dA Asv dv = dA cost sl. Al-V.st dv = dA. VAt k al Now NJ = friend dv lim SnPV.XX.AE lam N AK = At-Do Ct contribution At-> At to+At Cs Control lin np surface I of-> CSIII ! Similarly for the NJ. dv = dA cosk. sl. , dv = -4A-VAt Negative sign came beaz lim At-so Snp 3 dA dn are in opp d.A direct COSISP (+)(dat)
dNs dt of a freedo + Snov dR t Snowden CS or CCT This together go the Control sartai of CH dNs = at 2/05 a Sn pdv + Snow dA a Cs if N Mass of System Hims than N-1 for a System, dNs 7 dM -O dt at O = oF o S Sdv t s SV.dA Conservation of mass ct Cs (Integial relation) (Costionity equation) Now assuming stady How Then S and V are not function of time alto a = O I PV.dA = Cs A Now Assuming Incompressible flow ( P.Constant) I dA =0 (This is snothing but) (volume flow rate Cs equator) Q = A Vg)
Q for steady Viscous flow through circular tube the imp Glay to God" axial velicity profile is gree approimally by u- (1.(1-4)" So that u varies from 0@ the wall (r=R), upto a maximum of for highly Viscous laminal flow flow my 1/2 , while (u = (b) at the center line. less Viscous turbutent if my 1/7 the is for canpute the average velocity density constant From the problem our observations are h is (i) steady are (+ (ii) Circulate cls of in ) compressible To forg Vang (average velocity) crosssector Q = A. Vang Taking Q. R Vang 1/4 x r 4140 - P2 A.V i Q- St di = uit vj+nr. d Cars, as ds A v. 2 uit y+ari U= Uo [1-2]" (a)(qr2) or V dr = J.ridA= = 10B - a J.JA. u [2xr-di] T dA) u.(2ndn) dA = 2xurdr 0 = Jaxards Q=
P Q= 2xu. a dr substitutn r:0 R let you r-(y-y) a-(1-y)r y:1 O dy- dr P dr.- Rdy 6 Q-2KU. = 2x46 R2° ( gr(y-1)dy u = 2kuo R "g" dy mt2 = 2xup R e E); mH 1mtH = pt mt2 + 1 = 2AU Q = = 2TLto 22 (mti) [mia] 2xuo Veg Q A TR Vang = 240 (m+) (mt2) Laminas. m= Vang = 0.53 U. MON Tasbulent m==f, m= Yang = 0.8164.
Reynolds Transport System
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