& 9.9 F R E E C O N V E C T I O N

This maple session explains how the book found some of the results on section 9.9 that the book seems to get out of nowhere..

Transport Phenomena , Bird, Steward and Lightfoot. John Wiley & Sons: 1976.

> restart;

Start by doing a thermal energy balance over a shell of thikness Delta(y)

> BALANCE:=Thermal_energy_IN - Themal_energy_OUT + Thermal_energy_produced = 0;

>

Define each of the terms in the previous equation

> Thermal_energy_IN:=(W*L*y)*q(y,y); Themal_energy_OUT:= W*L*(y + Delta(y))*q(y,y+Delta(y)); Thermal_energy_produced:= 0;

> BALANCE;

Divide the previous equation by W*L*Delta(y)

> LIMIT:=BALANCE /(W*L*Delta(y));

By taking the limit as Delta(y) --> 0, the previous equation can easily be seen to be the derivative:

> delta(y*q)/delta(y) =0; (EQ***)

>

Now define Fourier's Law and insert it in the previous eqaution

> q[y]:=-k*(diff(T(y),y)); Fourier's Law

> Eq1:=(diff(y*q[y](y),y))=0; Substitute Fourirer's law into (EQ***)

> Eq1:=subs(({diff(k(y),y)=0},%)); The second term cancels bc "k" is assumed to be constant

>

> Eq9_9_1:= k*(diff(T(y),y$2))=0; This equation is the same as equation 9.9-1 on book. This differential equation gives the temperature distribution according to delta(y). We will proceed to solve it.

> BC:= T(-b)=T[2], T(b)=T[1]; Define the Boundary conditions to solve previous differential equation

> dsolve(Eq9_9_1,T(y)); Solve the differential eqn without the BC. (just to see what it looks like)

******To learn more about "dsolve" type: "?dsolve" at the maple prompt for help.******

> EQ9_9_4:=dsolve({Eq1,BC},T(y)); Now solve applying the BC to solve for _C1 and _C2 and get the final answer.

This equation is easily converted to the form of eqn. 9-9.4 and its then easily seen that both are equal. ======>

> EQ9_9_4:=T(y)= T[m] - 1/2*Delta(T)*(y/b); where: Tm= 0.5*(T1 + T2) or average temperatrure and, Delta (T) =T2-T1

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> restart;

NOW WE WILL DO A MOMENTUM BALANCE OVER THE SAME SLAB OF THICKNES DELTA (Y).

We start by using the general eqns. 3.2-9, and 3.2-11 through 16

> EQ3_2_9:=rho*D(Vz)/D(t)= - delta(p)/ delta(z) - (delta(tau[xz])/delta(x) + delta(tau[yz])/delta(y) + delta(tau[zz])/delta(z)) + rho*g[z];

>

Now define each of the tau terms using the definitions given in equations 3.2-11 through 16 on page 79 of book

> tau[xz]:=-mu*(delta(v[z])/delta(x) + delta(v[x])/delta(z)); tau[yz]:=-mu*(delta(v[y])/delta(z) + delta(v[z])/delta(y)); tau[zz]:=-2*mu*(delta(v[z])/delta(z)) + 2/3*mu*(gradient(v)); rho*D(Vz)/D(t)=0;

>

> EQ3_2_9; Lets see how eqn 3.2-9 looks now after inserting the values of tau

The left side is easily seen to be equal to zero from the conditions of the problem so lets fix that

> EQ3_2_9:=subs({rho*D(Vz)/D(t)=0},%);

from the conditions of the problem we are interested only of the variation in counjunction with delta (y) so, the second and fourth term on the previous equation become zero. Also the first term in the innermost parenthesis of the third term becomes zero.. Thus the eqn becomes --->

> EQ3_2_9:=subs({delta(tau[xz])=0,delta(tau[zz])=0, delta(v[y])/delta(z)=0},%);

By rearanging this equation we finally get equation 9,9-5 from the book: NOTE: There seems to be a sign error, but I cannot find it in my work. ASSUME: the book's equation is right, so I continue from there.

Lets rearrange EQ3_2_9 to make it look like equation 9.9-5. BOOK ASSUMPTION: mu is constant so it can be taken out of the derivative's parenthesis ===>

> EQ3_2_9:=mu*(delta^2*(v[z])/delta(y^2))= delta(p)/delta(z) + rho*g;

>

>

>

Now we need to expand rho in a Taylor series in T about the reference temperature T_ref.

NOTE: The coment "ref" next to a variable indicates that the variable value is calculated at a specific reference temperature

> rho:=rho[T_ref] + delta(rho)/delta(T)*(T-T[ref]);

Warning, recursive definition of name

>

This equation may be rewritten in terms of an expansion coefficient Beta

> rho:=rho_ref - rho_ref*beta[ref]*(T-T[ref]);

Now, redefine EQ3_2_9 using the previous rho to get equation (9.9-7)

> EQ9_9_7:= mu*(delta^2*(v[z])/delta(y^2))= delta(p)/delta(z) + rho*g;

Now assume that the pressure gradient in the system is due solely to the weight of the fluid in the slit, This makes: dp/dz=-rho_ref*g

Thus the previous equation becomes equation 9.9-8 ==>

> EQ9_9_8:=simplify(subs({delta(p)/delta(z)=-rho_ref*g},%));

What does the previous equation mean? The viscous forces are just balanced by the buoyancy forces.

>

Now take equation 9.9-4 (EQ9_9_4 on top of this maple session) and insert int into the previous equation to get eqn 9.9-9 ==>

> EQ9_9_9:=mu*(delta^2*(v[z])/delta(y^2))=-rho_ref*beta[ref]*g*((T[m]-T_ref)- 1/2*Delta (T)*(y/b));

>

Now, lets convert the previous equation into a diferential equation that maple can recognize:

> restart;

From now on v(y) will represent v[z](y)

> EQ9_9_9:= mu*(diff(v(y),y$2))=- rho[ref]*beta[ref]*g*((T[m]-T[ref])-1/2*Delta(T)*(y/b));;

>

> BC:=v(-b)=0, v(b)=0; Define the boundary conditions to use to solve the differential equation

> EQ9_9_12:=dsolve({EQ9_9_9,BC},v(y)); Solve for the previous differential equation using the specified boudary conditions (BC).

Compare this equation to the one given in the book to see if they are equal. This is done by inputing the equation from the book and subtracting it from the previous one on this maple session. If the answer is zero then they are equal

>

>

> EQ9_9_12_BOOK:=v(y)=rho[ref]*beta[ref]*g*b^2*Delta(T)/(12*mu)*(eta^3-A*eta^2-eta +A); Equation given in book

> A:=6*(T[m]-T[ref])/Delta(T); eta:=y/b; Input the equivalents of A and eta from the book:

> EQ9_9_12_BOOK; See the way equation 9.9-12 looks like when "A" and "eta"l values are inputted

> simplify(EQ9_9_12 - EQ9_9_12_BOOK); Check if my equation and the books' equation are equal

Beauuuuuutiful!!!!, they are equal.

>

It is required in the problem for the net volume flow in the "z-direction" to be zero; This is mathematically interpreted as the integral:

integral of: Vz d(eta) from -1 to 1

>

Click and paste the equation for Vz (9.9-9) from book inputed previously and integrate as already specified

> EQ_9_9_13:=int(rho[ref]*beta[ref]*g*b^2*Delta(T)/(12*mu)*(eta^3-A*eta^2-eta +A),y=-1...1)=0;

By inspection, the only way for the previous equation to be equal to zero if for Tm and Tref to be equal to each other. Therefore equation 9.9-12 from the book further reduces to its final form (9.9-15) =======>:

>

> EQ9_9_15:=subs({T[m]-T[ref]=0},EQ9_9_12_BOOK);

This equation can easily be seen to be equal to the one in the book if we simply change the terminology:

> EQ9_9_15_BOOK:=subs({(y/b)^3=n^3,y/b=n},%);

THE END .'. ALGDGADU

HECTOR R. PEREZ .'.