MATH529 Lab04:
Heat transfer in cylinder head fins

1Theory

Reciprocating, internal combustion engines dissipate heat into the environment through specific systems to eliminate thermal stress in the engine.

Figure 1. Aircraft engine schematic showing cooling flow [?]

2Methods

Figure 2. Model system

2.1Dirichlet boundary conditions

Mathematical problem

ut=kuxx+q(x,t) 0<x<L u(0,t)=u0(t)=U0+Ucos(Ωt), u(L,t)=0(ambienttemperature) u(x,0)=0, q(x,t)=xLV[1+Wcos(ωt)](convectivecooling) (1)

The homogeneous problem

wt=kwxx 0<x<L w(0,t)=0, w(L,t)=0(ambienttemperature)

has eigenfunctions

yn(x)=sin(nπx)

Expand u(x,t) and q(x,t) in terms of these eigenfunctions with time-dependent coefficients

u(t,x)=n=1un(t)sin(nπx),q(t,x)=n=1qn(t)sin(nπx). (2)

Replacing (2) into (3) obtain

(n=1un(t)sin(nπx))t=k(n=1un(t)sin(nπx))xx+n=1qn(t)sin(nπx)
n=1u˙n(t)sin(nπx)=-kπ2n=1n2un(t)sin(nπx)+n=1qn(t)sin(nπx)

Since {sin(πx),sin(2πx),} is linearly independent

n=1[u˙n(t)+kπ2n2un(t)-qn(t)]sin(nπx)=0u˙n+kπ2n2un-qn=0,

an inhomogeneous system of ODEs

Obtain qn(t) by Fourier coefficients of

n=1qn(t)sin(nπx)=q(x,t)=xLV[1+Wcos(ωt)]
qn(t)=2VL2[1+Wcos(ωt)]0Lxsin(nπx)dxAn=sin(πLn)-πLncos(πLn)π2n2.
qn(x,t)=An2VL2[1+Wcos(ωt)]
In[40]:= 
A[n_]=Integrate[x Sin[n Pi x],{x,0,L}]

sin(πLn)-πLncos(πLn)π2n2

In[41]:= 

The system to solve is

u˙n+kπ2n2un=An2VL2[1+Wcos(ωt)]=Bn

In[42]:= 
sol=DSolve[{D[un[t],t]+k Pi^2 n^2 un[t] == B[n] (1+ Cos[omega t]),un[0]==U0},un[t],t][[1,1]]

un(t)e-kn2π2t(k5L2π12U0n12-3ekn2π2tk4Lπ9Vcos(Lnπ)n9+8k4Lπ9Vcos(Lnπ)n9-4ekn2π2tk4Lπ9Vcos(Lnπ)cos(ωt)n9-ekn2π2tk4Lπ9Vcos(Lnπ)cos(2ωt)n9+5k3L2ω2π8U0n8+3ekn2π2tk4π8Vsin(Lnπ)n8-8k4π8Vsin(Lnπ)n8+4ekn2π2tk4π8Vcos(ωt)sin(Lnπ)n8+ekn2π2tk4π8Vcos(2ωt)sin(Lnπ)n8-4ekn2π2tk3Lωπ7Vcos(Lnπ)sin(ωt)n7-2ekn2π2tk3Lωπ7Vcos(Lnπ)sin(2ωt)n7+4ekn2π2tk3ωπ6Vsin(Lnπ)sin(ωt)n6+2ekn2π2tk3ωπ6Vsin(Lnπ)sin(2ωt)n6-15ekn2π2tk2Lω2π5Vcos(Lnπ)n5+32k2Lω2π5Vcos(Lnπ)n5-16ekn2π2tk2Lω2π5Vcos(Lnπ)cos(ωt)n5-ekn2π2tk2Lω2π5Vcos(Lnπ)cos(2ωt)n5+4kL2ω4π4U0n4+15ekn2π2tk2ω2π4Vsin(Lnπ)n4-32k2ω2π4Vsin(Lnπ)n4+16ekn2π2tk2ω2π4Vcos(ωt)sin(Lnπ)n4+ekn2π2tk2ω2π4Vcos(2ωt)sin(Lnπ)n4-16ekn2π2tkLω3π3Vcos(Lnπ)sin(ωt)n3-2ekn2π2tkLω3π3Vcos(Lnπ)sin(2ωt)n3+16ekn2π2tkω3π2Vsin(Lnπ)sin(ωt)n2+2ekn2π2tkω3π2Vsin(Lnπ)sin(2ωt)n2-12ekn2π2tLω4πVcos(Lnπ)n+12Lω4πVcos(Lnπ)n+12ekn2π2tω4Vsin(Lnπ)-12ω4Vsin(Lnπ))kL2n4π4(k2π4n4+ω2)(k2π4n4+4ω2)

In[43]:= 
u[x_,t_,nT_] := Sum[ un[t] Sin[n Pi x] /. sol, {n,1,nT}]

In[44]:= 
u[x,t,1]

eπ2(-k)tsin(πx)(π12k5L2U0-4π9k4LVeπ2ktcos(πL)cos(ωt)-π9k4LVeπ2ktcos(πL)cos(2ωt)+4π8k4Veπ2ktsin(πL)cos(ωt)+π8k4Veπ2ktsin(πL)cos(2ωt)+3π8k4Veπ2ktsin(πL)-3π9k4LVeπ2ktcos(πL)-8π8k4Vsin(πL)+8π9k4LVcos(πL)+5π8k3L2ω2U0+4π6k3ωVeπ2ktsin(πL)sin(ωt)+2π6k3ωVeπ2ktsin(πL)sin(2ωt)-4π7k3LωVeπ2ktcos(πL)sin(ωt)-2π7k3LωVeπ2ktcos(πL)sin(2ωt)+15π4k2ω2Veπ2ktsin(πL)-15π5k2Lω2Veπ2ktcos(πL)-16π5k2Lω2Veπ2ktcos(πL)cos(ωt)-π5k2Lω2Veπ2ktcos(πL)cos(2ωt)+16π4k2ω2Veπ2ktsin(πL)cos(ωt)+π4k2ω2Veπ2ktsin(πL)cos(2ωt)-32π4k2ω2Vsin(πL)+32π5k2Lω2Vcos(πL)+4π4kL2ω4U0+12ω4Veπ2ktsin(πL)-12πLω4Veπ2ktcos(πL)+16π2kω3Veπ2ktsin(πL)sin(ωt)+2π2kω3Veπ2ktsin(πL)sin(2ωt)-16π3kLω3Veπ2ktcos(πL)sin(ωt)-2π3kLω3Veπ2ktcos(πL)sin(2ωt)-12ω4Vsin(πL)+12πLω4Vcos(πL))π4kL2(π4k2+ω2)(π4k2+4ω2)

In[46]:= 
u[x,t,1] /. {k->1,L->1,omega->1,OMEGA->5,V->1,U0->2,U->1}

e-π2tsin(πx)(3π9eπ2t+15π5eπ2t+12πeπ2t+4π7eπ2tsin(t)+16π3eπ2tsin(t)+2π7eπ2tsin(2t)+2π3eπ2tsin(2t)+4π9eπ2tcos(t)+16π5eπ2tcos(t)+π9eπ2tcos(2t)+π5eπ2tcos(2t)+2π12-8π9+10π8-32π5+8π4-12π)π4(1+π4)(4+π4)

In[54]:= 
Plot[Evaluate[Table[u[x,t,3] /. {k->0.001,L->1,omega->180,OMEGA->30,V->10,U0->1,U->1},{t,0,1,0.1}]],{x,0,1},PlotRange->{All,{-1,15}}]

In[22]:= 
Evaluate[Table[u[x,t,5] /. {k->1,L->1,omega->1,Cn->1/n^2,c[1]->0},{t,0,0.01,0.05}]]

{(0.201613n2+1.c1)sin(πx)+(0.0506443n2+1.c1)sin(2πx)+(0.0225144n2+1.c1)sin(3πx)+(0.0126649n2+1.c1)sin(4πx)+(0.00810563n2+1.c1)sin(5πx)}

In[8]:= 
u[x,t,3] /. {k->1,L->1,omega->1,Cn->1/n^2,c[1]->0}

e-π2tsin(πx)(eπ2t+π4eπ2t+π2eπ2tsin(t)+π4eπ2tcos(t)-2π4-1)π2(1+π4)n2+e-4π2tsin(2πx)(e4π2t+16π4e4π2t+4π2e4π2tsin(t)+16π4e4π2tcos(t)-32π4-1)4π2(1+16π4)n2+e-9π2tsin(3πx)(e9π2t+81π4e9π2t+9π2e9π2tsin(t)+81π4e9π2tcos(t)-162π4-1)9π2(1+81π4)n2

In[9]:= 

Introduce

ψ(x,t)=u0(t)q(x,t),

that satisfies

ψxx=0

and v=u-ψ leads to the problem

vt=kvxx+q(x,t)-ψt 0<x<L v(0,t)=u0(t)=U0+Ucos(Ωt), u(L,t)=0(ambienttemperature) u(x,0)=0, q(x,t)=xLV[1+Wcos(ωt)](convectivecooling)

2.2Robin boundary condition

ut=kuxx+q(x,t) 0<x<L α[u(0,t)-u0]=kux(0,t), u(L,t)=0(ambienttemperature) u(x,0)=0, q(x,t)=xLV[1+Wcos(ωt)](convectivecooling) (3)

3Results

Figure 3.

In[58]:= 
fig=Plot[Cos[x],{x,0,2Pi}];

In[61]:= 
SetDirectory["/Users/mitran/courses/MATH529L"]

/Users/mitran/courses/MATH529L

In[62]:= 
Export["fig.png",fig]

fig.png

In[64]:= 
fig=Plot[Cos[x],{x,0,2Pi},GridLines->Automatic]; Export["fig.png",fig];

In[65]:= 

4Conclusion

Bibliography