Example 8.4-1. Prediction of the Thermal Conductivity

of a Liquid

from Bird, Stewart and Lightfoot, pg. 261.

> restart;

Thermal conductivity in a liquid is given in equation 8.4-3:

> k:=2.80*(N/V)^(2/3)*kappa*vs;

where N is Avogadro's number, V is the molar volume, kappa is the Boltzman constant, and vs is the sonic velocity, defined as:

> vs:=sqrt(C[p]/C[v]*diff(p(rho),rho)[T]);

The problem in example 8.4-1 is concerned with finding the thermal conductivity of carbon tetrachloride at 20 C and 1 atm. First, define the compressibility as:

> c:=rho1->inv(rho1)*diff(rho1(p),p)[T];

a dummy variable rho1 is used here so, when we plug into the equation, maple does not also define the rho that appears in the expression for vs.

Then we have,

> eqn:=diff(p(rho),rho)[T]=1/(rho1*c);

The density of liquid CCl4 at 20 C and 1 atm is given as 1.595 g/cm^-3, and it's compressibility is given as 90.7 x 10^-6 atm

> c:=90.7e-6/atm;rho1:=1.595*g/cm^3;assume(cm > 0);assume(s>0);atm:=1.0133e6*g/cm/s^2;

and the equation becomes

> eqn;

This agrees with the result obtained in the book of 7.00 x 10^9 cm^2/s^2;

With this, we can now find vs

Recall:

> vs;

if we assume Cp/Cv = 1.0, and substitute in for diff(p,rho):

> vs:=algsubs(C[p]/C[v]=1.0,vs);

> vs:=simplify(algsubs(diff(p(rho),rho)[T]=7004385949.*cm^2/(s^2),vs));

Compare with the value obtained in the book of 8.37 x 10^4 cm/s

find the molar volume using the molecular weight and the density:

> M:=153.84*g/mole;V:=M/rho1;

Avogardo's number:

> N:=6.023e23/mole;

(note: N is define as per mole instead of molecule/mole since molecule will simply cancel out with the molecule term in the Boltzman constant)

The Boltzman Constant:

> kappa:=1.3805e-16*erg/K;

Plug everything into thermal conductivity equation:

> simplify(k);

(book: 1.10 x 10^4 (cm^-2)(erg/K)(cm/s)

redefine energy units:

> erg:=2.3901e-8*cal;k:=simplify(k);

Compare with book answer of k=2.62 x 10^-4 cal/s/cm/K