Solutions Manual For Computational Fluid Dynamics For Mechanical Engineering 1st Edition By George Qin (All Chapters)

Solutions Manual for Computational Fluid D
H H H H H




ynamics for Mechanical Engineering, 1e byG
H H H H H H




eorge Qin (All Chapters) H H H




Chapter 1 H


1. ShowHthatHEquationH(1.14)HcanHalsoHbeHwrittenHas
𝜕𝑢 𝜕𝑢 𝜕𝑢 𝜕2 𝑢 𝜕2 𝑢 1H𝜕𝑝
+H 𝑢H +H 𝑣H =H 𝜈H( 2 +H 2)H −H
𝜕𝑡H 𝜕𝑥H 𝜕𝑦H 𝜕𝑥 H 𝜕𝑦 𝜌H𝜕𝑥
Solution
EquationH(1.14)His
𝜕𝑢 𝜕(𝑢2) 𝜕(𝑣𝑢) 𝜕2 𝑢 𝜕2 𝑢 1H𝜕𝑝
+ + =H 𝜈H( 2H +H 2)H −H (1.13)
𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑦 𝜌H𝜕𝑥
TheHleftHsideHi
s

𝜕𝑢 𝜕(𝑢2) 𝜕(𝑣𝑢) 𝜕𝑢 𝜕𝑢 𝜕𝑢 𝜕𝑣
+ + = +H2𝑢 + H𝑣 + H𝑢
𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑦
𝜕𝑢 𝜕𝑢 𝜕𝑢 𝜕𝑢 𝜕𝑣 𝜕𝑢 𝜕𝑢 𝜕𝑢
= +H𝑢 +H𝑣 +H𝑢H + + H𝑢 + H 𝑣
𝜕𝑡 𝜕𝑥 𝜕𝑦 ( 𝜕𝑥 )H 𝜕𝑡 𝜕𝑥 𝜕𝑦
=
since 𝜕𝑦
𝜕𝑢 𝜕𝑣
+ =H 0
𝜕𝑥 𝜕𝑦
dueHtoHtheHcontinuityHequation.
2. DeriveHEquationH(1.17).
Solution:
FromHEquationH(1.14)
𝜕𝑢 𝜕(𝑢2) 𝜕(𝑣𝑢) 𝜕2 𝑢 𝜕2 𝑢 1H𝜕𝑝
+ + =H 𝜈H( 2 +H 2)H−H
𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑥 H 𝜕𝑦 𝜌H𝜕𝑥
Define 𝑥𝑖H 𝑡𝑈H 𝑝
𝑢H̃ H= 𝑢 ,H𝑣H̃ = 𝑣 ,H𝑥̃ =H ,H𝑡̃H =H ,H𝑝̃H =H H
𝑈 𝑈 𝑖 𝐿 𝐿 𝜌𝑈2
EquationH(1.14)Hbecomes
𝑈𝜕𝑢̃ 𝑈2 𝜕(𝑢̃2 ) 𝑈2 𝜕(𝑣̃𝑢 𝜈𝑈HHH 𝜕 2 𝑢̃ 𝜕 2 𝑢̃ 𝜌𝑈2H𝜕𝑝̃
+ + = ( +
𝐿 𝐿𝜕𝑥̃ )H− 𝜌𝐿HH 𝜕𝑥̃
𝑈 𝜕𝑡̃ 𝐿𝜕𝑦̃ 𝐿2 𝜕𝑥̃2 𝜕𝑦̃2
DividingHbothHsidesHbyH𝑈2/𝐿,HEquationH(1.17)Hfollows.

3. DeriveHaHpressureHPoissonHequationHfromHEquationsH(1.13)HthroughH(1.15):

, 𝜕2𝑝 𝜕2 𝑝 𝜕𝑢H𝜕𝑣 𝜕𝑣H𝜕𝑢
+ H =H 2𝜌H( −H )
𝜕𝑥 H 𝜕𝑦 H
2 2 𝜕𝑥H 𝜕𝑦H 𝜕𝑥H 𝜕𝑦
Solution:
𝜕𝑢 𝜕𝑣
+ =H 0 (1.13)
𝜕𝑥 𝜕𝑦
𝜕𝑢 𝜕(𝑢2) 𝜕(𝑣𝑢) 𝜕2 𝑢 𝜕2 𝑢 1H𝜕𝑝
+ + =H 𝜈H( 2H +H 2)H −H (1.14)
𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑦 𝜌H𝜕𝑥
𝜕𝑣 𝜕(𝑢𝑣) 𝜕(𝑣2) 𝜕2 𝑣 𝜕2 𝑣 1H𝜕𝑝
+ + =H 𝜈H( 2H +H 2)H −H (1.15)
𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑦 𝜌H𝜕𝑦
TakingH𝑥-derivativeHofHeachHtermHofHEquationH(1.14)HandH𝑦-
derivativeHofHeachHtermHofHEquationH(1.15),HthenHaddingHthemHup,HweHhave

𝜕 𝜕𝑢 𝜕𝑣 𝜕2(𝑢2) 𝜕2(𝑣𝑢) 𝜕2(𝑣2)
( + 𝜕𝑥 2 +HH 2
𝜕𝑡 𝜕𝑥 )H + 𝜕𝑥𝜕𝑦 𝜕𝑦
+ H

𝜕𝑦 2
𝜕2 𝜕2 𝜕𝑢 𝜕𝑣 1H 𝜕2𝑝 𝜕2 𝑝
=H 𝜈H( 2H +H 2)H( +H )H 𝜌H +H )
𝜕𝑥 𝜕𝑦 𝜕𝑥H 𝜕𝑦 ( 𝜕𝑥 2HH 𝜕𝑦2
DueHtoHcontinuity,HweHhav −
e
𝜕2 𝑝 𝜕2 𝑝 𝜕2(𝑢2) 𝜕2(𝑣𝑢) 𝜕2(𝑣2)
+H =H −𝜌 +H2 + ]
𝜕𝑥2H 𝜕𝑦2H 𝜕𝑥2 𝜕𝑥𝜕𝑦 𝜕𝑦2HH
H[

=H −2𝜌(𝑢𝑥𝑢𝑥H +H𝑢𝑢𝑥𝑥H +H 𝑢𝑥𝑣𝑦H+H 𝑢𝑣𝑥𝑦H+H 𝑢𝑥𝑦𝑣H +H𝑢𝑦𝑣𝑥HH+H 𝑣𝑦𝑣𝑦H +H 𝑣𝑣𝑦𝑦)
𝜕 𝜕 𝜕𝑢 𝜕𝑣
=H −2𝜌H[(𝑢𝑥H +H𝑢H +H𝑣H )H( +H )H+H𝑢𝑦𝑣𝑥H +H𝑣𝑦𝑣𝑦]
𝜕𝑥H 𝜕𝑦 𝜕𝑥H 𝜕𝑦
𝜕𝑢H𝜕𝑣 𝜕𝑣H𝜕𝑢
=H −2𝜌(𝑢𝑦𝑣𝑥H +H 𝑣𝑦𝑣𝑦)H =H −2𝜌(𝑢𝑦𝑣𝑥H −H 𝑢𝑥𝑣𝑦)H =H 2𝜌H( −H )
𝜕𝑥H 𝜕𝑦H 𝜕𝑥H 𝜕𝑦
4. ForHaH2-DHincompressibleHflowHweHcanHdefineHtheHstreamHfunctionH𝜙H byHrequiring
𝜕𝜙 𝜕𝜙
𝑢H = ;H 𝑣H =H
𝜕𝑦 − 𝜕𝑥
WeHalsoHcanHdefineHaHflowHvariableHcalledHvorticity
𝜕𝑣 𝜕𝑢
𝜔H = −
𝜕𝑥 𝜕𝑦
ShowHthat
𝜕2𝜙 𝜕2𝜙
𝜔H =H −H( 2HH +H 2H)
𝜕𝑥 𝜕𝑦
Solution:
𝜕𝑣 𝜕𝑢 𝜕 𝜕𝜙 𝜕 𝜕𝜙 𝜕2𝜙 𝜕2𝜙
𝜔H = − = (− ( )H =H −H( +H )
)H− 𝜕𝑦 𝜕𝑦 𝜕𝑥2HH 𝜕𝑦2H
𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑥

, Chapter 2
H


1. DevelopHaHsecond- onHaHnon-uniform
𝑑𝜙
orderHaccurateHfiniteHdifferenceHapproximationHforH( )
𝑑𝑥HH 𝑖
meshHusingHinformationH(𝜙H andH𝑥H values)HfromHmeshHpointsH𝑥𝑖−1,H𝑥𝑖HHandH𝑥𝑖+1.HSupposeH𝛿𝑥𝑖H =
𝑥𝑖+1H −H𝑥𝑖H =H 𝛼𝛿𝑥𝑖−1H =H 𝛼(𝑥𝑖H −H𝑥𝑖−1).

Solution:

AssumeHcloseHtoHtheH𝑖𝑡ℎH point,H𝜙(𝑥)H =H 𝜙𝑖H +H 𝑏(𝑥H −H𝑥𝑖)H+H 𝑐(𝑥H −H 𝑥𝑖)2H +H 𝑑(𝑥H −H 𝑥𝑖)3H…
nd H𝑑𝜙H=H𝑏H+H2𝑐(𝑥H−H𝑥H )H+H⋯Ha 𝑑𝜙) =H 𝑏.
Then
𝑖 (
𝑑𝑥 𝑑𝑥HH 𝑖

NowH𝜙𝑖+1H =H 𝜙(𝑥𝑖+1)H =H 𝜙𝑖H +H 𝑏(𝑥𝑖+1H −H 𝑥𝑖)H +H 𝑐(𝑥𝑖+1H −H 𝑥𝑖)2H +H⋯H =H 𝜙𝑖H +H𝑏Δ𝑥𝑖H +H 𝑐Δ𝑥2H +H 𝑑Δ𝑥3H…
𝑖 𝑖
AndH𝜙𝑖−1H =H 𝜙(𝑥𝑖−1)H =H 𝜙𝑖H +H𝑏(𝑥𝑖−1H −H 𝑥𝑖)H+H 𝑐(𝑥𝑖−1H −H 𝑥𝑖)2H +H ⋯H =H 𝜙𝑖H −H𝑏Δ𝑥𝑖−1H +H 𝑐Δ𝑥2
−H 𝑑Δ𝑥3 …
𝑖−1 𝑖−1
SoHΔ𝑥2HH 𝜙𝑖+1H−HΔ𝑥2𝜙𝑖−1H =H(Δ𝑥2 −HΔ𝑥2)𝜙𝑖H +H𝑏Δ𝑥𝑖Δ𝑥𝑖−1(Δ𝑥𝑖H +HΔ𝑥𝑖−1)H+H𝑑Δ𝑥2Δ𝑥2HH (Δ𝑥𝑖H +
𝑖−1 𝑖 𝑖−1 𝑖 𝑖 𝑖−1
Δ𝑥𝑖−1)H+H⋯
AndH𝑏H = Δ𝑥2𝑖−1
HHHH
𝜙𝑖+1−Δ𝑥2
𝑖 𝜙𝑖−1−(Δ𝑥
2HHHHHH
𝑖−1
2
𝑖 −Δ𝑥 )𝜙𝑖
−H𝑑Δ𝑥𝑖Δ𝑥𝑖−1H +H ⋯
Δ𝑥𝑖Δ𝑥𝑖−1(Δ𝑥𝑖+Δ𝑥𝑖−1)
𝑑𝜙
AH2ndHorderHfiniteHdifferenceHforH( isHtherefore
)
𝑑𝑥HH 𝑖

𝑑𝜙 H Δ𝑥𝑖−1
2HHH 𝜙𝑖+1H−HΔ𝑥2𝜙𝑖−1H−H(Δ𝑥2 −H Δ𝑥2)𝜙𝑖
𝑖 𝑖−1 𝑖 𝜙𝑖+1H +H (α2H −H 1)𝜙𝑖H−H α2𝜙𝑖−1Hα
( ) =H 𝑏H ≈ =
𝑑𝑥H𝑖 Δ𝑥𝑖Δ𝑥𝑖−1(Δ𝑥𝑖H +H Δ𝑥𝑖−1) (αH+H1)Δ𝑥𝑖−1

2. UseHtheHschemeHyouHdevelopedHforHproblemH1HtoHevaluateHtheHderivativeHofH𝜙(𝑥)H
=Hsin(𝑥H−H𝑥𝑖H +H1)H atHpointH𝑖.HSupposeHΔ𝑥𝑖−1H =H 0.02H andHΔ𝑥𝑖H =H 0.01.HCompareH
your
numericalHresultHwithHtheHexactHsolution,HwhichHisHcos(1).HThenHhalveHbothHΔ𝑥𝑖−1H andHΔ
𝑥𝑖,HandHredoHtheHcalculation.HIsHtheHschemeHtrulyHsecond-orderHaccurate?
Solution:
clear;Hclc;
dxiH=H0.01;Hdxim1H=H0.02;HalphaH=Hdxi/dxim1;Hfor
HiterH=H1H:H2
xH=H[-
dxim1,0,dxi];HphiH=Hs
in(x+1);

, dphidxH=H(phi(3)+(alpha^2-1)*phi(2)-
alpha^2*phi(1))/(alpha*(alpha+1)*dxim1);Herr(iter)H=Hdphidx-cos(1);
dxiH=Hdxi/2;Hdxim1H=Hdxim1/2;Hen
d
err(1)/err(2)


ItHisHtrulyH2nd-orderHaccurate.

3. ReproduceHtheHcalculationHpresentedHinHSectionH2.1.2HExample:HLaminarHChannelHFlow.
Solution:
% %
%HFULLY-DEVELOPEDHCHANNELHFLOW %
%HByHGeorgeHQinHforH"AHCourseHofHComputationalHFluidHDynamics"H%
%
%H
tic
clear;Hclc;
%HPARAMETERSH
HH=H1;HNH=H5;
%HFACEHLOCATIONS
yfH=Hlinspace(0,H,N+1);
%HNODEHLOCATIONS
yH=H0.5*(yf(1:end-1)+yf(2:end));
%HDELTAHY
dyH=Hyf(2)-yf(1);
%HTHEHTHREEHDIAGONALHVECTORSHINHTHEHCOEFFICIENTHMATRIX
asH=H-
(1/dy^2)*ones(1,N);HapH=H(
2/dy^2)*ones(1,N);HanH=H-
(1/dy^2)*ones(1,N);HbH =
ones(1,N);
%HSPECIALHVALUESHATHTHEHBOUNDARIESH(BOUNDARYHCONDITIONS)
%ap(1) =H3/dy^2;Has(1) =H0;
%ap(1) =H4/dy^2;Han(1)H=H-H(4/3)/dy^2;Has(1)
=H0;Hap(1) =H3/dy^2;Has(1)
=H0;Hb(1)H=H3/4;
ap(end)H=H1/dy^2;Han(end)H=H0;
%HSOLVEHTHEHSYSTEMHOFHLINEARHEQUATIONSHWITHHTDMAHALGORITHM
uH=HTDMA(as,ap,an,b);H%HthisHisHinHtheHappendixHofHtheHtextbookHtoc
%HCOMPAREHWITHHEXACTHSOLUTION
u_exactH=Hy.*(1-0.5*y);
errH=H(u_exact-u)./u_exactH*H100;
%HRECALCULATEHEXACTHSOLUTIONHVECTORHFORHPLOTTING
y_exactH=H0:0.01:1;
u_exactH=Hy_exact.*(1-
0.5*y_exact);Hplot(y_exact,u_exact,y,u,'ro')Hxlabel('$y$','FontS
ize',20,'Interpreter','Latex')Hylabel('$u$','FontSize',20,'Inter
preter','Latex')Hh_legend=legend('ExactHSolution','NumericalHSolu
tion');
set(h_legend,'FontSize',14,'Interpreter','Latex','Location','NorthWest')


4. ShowHthatHtheHmethodHusedHinHSectionH2.1.2HExample:HLaminarHChannelHFlowHisHfir
st-HorderHaccurateHbyHusingHtheHglobalHerrorHestimateHtechnique.
Solution:

TheseHfiniteHdifferenceHequationsHandHtheirHtruncationHerrorsHareHreproducedHhere: