ME 353 Heat Transfer 1

M.M. Yovanovich

EX3P6.MWS

Example 3.6 of 4th edition of Incropera and DeWitt.

A composite plane wall consists of two materials, denoted here as 1 and 2 rather than A and B.

The first wall denoted as wall 1 has uniform heat generation , thermal conductivity , and thickness . The left boundary is adiabatic and the right boundary is in contact with a second plane wall denoted as wall 2. This wall has no heat generation, its thermal conductivity is , and its thickness is . There is negligible themal contact resistance at the common interface. The right boundary of plane wall is cooled by a water stream with temperature and . Let the temperature of the adiabatic boundary be , the temperature of the interface is , and the temperature of the boundary in contact with the water is denoted as .

1. Sketch the temperature distributions in the two walls under steady-state conditions.

2. Determine the temperatures and .

> restart:

Assumptions are

1. Steady-state

2. Constant properties

3. One dimensional conduction; therefore in both walls

4. Negligible contact resistance at common interface

5. Left boundary of wall is adiabatic

6. Unit heat transfer area

> syspar:= (L1 = 50/1000, k1 = 75, P = 1.5e6, L2 = 20/1000, k2 = 150, h = 1000, Tf = 30);

Temperature relations.

Begin with the right boundary of wall 2 and work to the left getting relations for and .

> T2:= Tf + Delta_Tfilm;
T1:= T2 + Delta_Twall2;
T0:= T1 + Delta_Twall1;

Temperature drops across the film, second wall and the first wall.

> Delta_Tfilm:= Qsys*Rf;

> Delta_Twall2:= Qsys*Rwall2;

> Delta_Twall1:= Qsys*Rwall1;

Component thermal resistances.

> Rf:= 1/(h*A);
Rwall2:= L2/(k2*A);
Rwall1:= L1/(k1*A)/2;

> Qsys:= P*A*L1;

Temperature relations again.

> T2:= T2;

> T1:= T1;

> T0:= T0;

Calculation of temperatures.

> T2:= evalf(subs(syspar, T2), 5)*C;

> T1:= evalf(subs(syspar, T1), 5)*C;

> T0:= evalf(subs(syspar, T0), 5)*C;

The calculated values are in agreement with the values given in the text.