Example 11.4-1 Heat Transfer in Forced Convection Laminar Flow along a heated Flat Plate.

We start with a few equations from example 4.4-1

>  restart;
>  emotx:=(x,y)->vx(x,y)*D[1](vx)(x,y)+vy(x,y)*D[2](vx)(x,y)-
nu*D[2](D[2](vx))(x,y); The equation of motion
>  vy:=(x,y)->-int(D[1](vx)(x,y1),y1=0...y); From integration 
of the continuity equation
>  vx:=(x,y)->vinf*phi(y/delta(x)); Assumed form for vx(x,y)
>  eq19:=int(emotx(x,y),y=0...delta(x)): Eq. 19 on page 144
>  eeng:=(x,y)->vx(x,y)*D[1](T)(x,y)+vy(x,y)*D[2](T)(x,y)-
alpha*D[2](D[2](T))(x,y); The energy equation
>  T:=(x,y)->T0+(Tinf-T0)*Theta(y/(Delta*delta(x))); The 
assumed form for the Temperature variations in the boundary layer.
>  eq11:=int(eeng(x,y),y=0...Delta*delta(x)); Eq. 11 on page 368
>  phi:=eta->2*eta-2*eta^3+eta^4; One polynomial that fits BC for 
vx
>  phi(0); phi(1); D(phi)(1); D(D(phi))(1); This vanishes at the 
wall, is equal to one at the outer edge of the bounday and has zero 
first and second derivatives there.
>  Theta:=etaT->2*etaT-2*etaT^3+etaT^4; Using the same polynomial 
for T
>  simplify(eq11/(Tinf-T0)); This gives a first order ODE in 
delta(x), but it involves Delta.
>  eq19; This gives a first order ODE in delta(x) alone.
>  s:=dsolve({eq19,delta(0)=0},delta(x));
>  assign(s[1]);delta:=unapply(delta(x),x);
>  simplify(eq11/(Tinf-T0)); Now we get a polynomial equation that 
determines Delta.
>  alpha:=nu/Pr;
>  Pr:=solve(eq11,Pr);
>  simplify(37/(630*Pr)-Delta^3/15+3*Delta^5/280-Delta^6/360); 
Compare to eq. 20 on page 369