Project.

> restart;

Start with a shell of unspecified boundary temps and radii.

have surface area of each tube:

> Area := r -> 2*Pi*r*L;

Energy balance gives {E in} - {E out} + {E formed} = 0

> Q := (r) -> q(r) * Area(r);

> EnBal := Q(r) - Q(r+dr);

now consider division by 2*pi*L*dr, with limit dr -> 0

> Eq := limit ( EnBal / (2*Pi*L*dr) , dr = 0 );

> Deq := 0 = Eq;

this, of course, is equivalent to d(r*q)/dr.

define q(ro) = qo. This assumes the innermost radius (r = ro) has heat flow of qo.

> eq := dsolve( { Deq, q(ro) = qo } , q(r) );

note that q(r) is the heat flux through any region, as we defined earlire a generic shell only.

now we consider Fourier's law of heat conduction:

> q(r) := -k * D(T)(r);

substitute into the existing equation for q(r).

> eq;

Here we have seperated variables and integrate from the inner to outer interface of each region.

> LHS := int( -k , T = Tin ... Tout);

> RHS := int(rhs(eq),r = rin ... rout);

> region := LHS = RHS;

> simplify(region);

This expression holds for every region in which there is no fluid - solid interface.

In these cases (inner surface, outer surface), we will use Newton's law of cooling as representative of the physical phenomenon.

> q := h*(T - Tfliud);

The results can now be generalized, for N layers. Here we will solve a situation analogous to the book, with three insulation types.

The inner and outer layers will be modeled using Newton's law of cooling.

> T := array(0...3): r := array(0...3): k := array(0...3): expr := array(0...4): qo := array(0...0):

> for i from 0 to 2 by 1 do

> expr[i+1] := T[i] - T[i+1] = qo * r[0] / k[i+1] * ln( r[i+1] / r[i] )

> od;

Properties of the logarithm were used -> log(A) - log(B) = log(A/B)

Now define the first and last expressions for heat transfer - at the fluid - solid interfaces.

> expr[0] := qo = hin*(Tfluid - T[0]);

> expr[4] := qo*r[0]/r[3] = hout*(T[3] - Tair);

Now solve for Qo, the overall heat transfer.

> S := 0;

> S := sum(expr[j] , j = 0 ... 4);

We now want to solve for qo, from which Qo can be determined.

> q0 := solve (S,qo);

This value multiplied by the surface area gives the total heat flow, Qo.

> Qo := q0 * Area( r[0] );

Define over-all heat transfer coefficient based on inner surface :

> Uo := Qo / (Area(r[0])) / (Tfluid - Tair);

Now for the down, dirty and boring work.

FOR ALUMINUM:

Inner Pipe Diameter = 2.54 cm

Thickness = 0.635 cm

> r[0] := 2.54*cm: r[1] := r[0]+0.635*cm:

> hin := 0.27*cal/cm^2/sec/K: hout := 1.27*cal/cm^2/sec/K:

> Tfluid := (1000+273.15)*K: Tair := (50+273.15)*K:

> T[1] := (500+273)*K: T[2] := (300+273.15)*K:

> k[1] := .645*cal/sec/cm/K: k[2] := .540*cal/sec/cm/K: k[3] := .492*cal/sec/cm/K:

> expr[0]; expr[1];

> solve( {expr[0] , expr[1]} , {T[0] , qo} );

> T[0] := 868.91*K; qo := 109.14*cal/cm^2/sec;

> expr[2];

> r[2] := solve(expr[2] , r[2]);

> expr[3]; expr[4];

> solve({ expr[3] , expr[4]} , { r[3] , T[3] });

Thus we have accounted for all variables - clearly the second expressoin for T3, r3 does not hold.

>

> restart;

FOR COPPER

> r[0] := 2.54*cm: r[1] := r[0]+0.635*cm:

> hin := 0.27*cal/cm^2/sec/K: hout := 1.27*cal/cm^2/sec/K:

> Tfluid := (1000+273.15)*K: Tair := (50+273.15)*K:

> T[1] := (500+273)*K: T[2] := (300+273.15)*K:

> k[1] := .856*cal/sec/cm/K: k[2] := .876*cal/sec/cm/K: k[3] := .901*cal/sec/cm/K:

> solve( {expr[0] , expr[1]} , {T[0] , qo} );

> T[0] := 848.85*K; qo := 114.56*cal/cm^2/sec;

> r[2] := solve(expr[2] , r[2]);

> solve({ expr[3] , expr[4]} , { r[3] , T[3] });

>

>

>

> restart;

FOR CAST IRON

> r[0] := 2.54*cm: r[1] := r[0]+0.635*cm:

> Tfluid := (1000+273.15)*K: Tair := (50+273.15)*K:

> hin := 0.27*cal/cm^2/sec/K: hout := 1.27*cal/cm^2/sec/K:

> T[1] := (500+273)*K: T[2] := (300+273.15)*K:

> k[1] := .103*cal/sec/cm/K: k[2] := .114*cal/sec/cm/K: k[3] := .131*cal/sec/cm/K:

> solve( {expr[0] , expr[1]} , {T[0] , qo} );

> T[0] := 1071.94*K; qo := 54.33*cal/cm^2/sec;

> r[2] := solve(expr[2] , r[2]);

> solve({ expr[3] , expr[4]} , { r[3] , T[3] });

>