ME 353 Heat Transfer 1

M.M. Yovanovich

EX1P6.MWS

Example 1.6 of 4th Edition of Incropera and DeWitt.

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The coating on a plate is cured by exposure to an infrared lamp providing

an irradiation of 2000 W/m^2. It absorbs 80 % of the irradiation and has an

emissivity of 0.50. It is also exposed to an air flow and large surroundings

for which the temperatures are 20 deg. C and 30 deg. C, respectively.

1. If the convection coefficient between the plate and the ambient air is

15 W/m^2 K, what is the cure temperature of the plate.

2. Final characteristics of the coating, including wear and durability, are

known to depend on temperature at which curing occurs. An air flow system

is able to control the air velocity, and hence the convection coefficient, on the

cured surface, but the process engineer needs to know how the temperature

depends on the convection coefficient. Provide the desired information by

computing and plotting the surface temperature as a function of h for

2 <= h <= 200 W/m^2 K. What value of h would provide a cure temperature

of 50 deg. C?

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> restart:

Assumptions:

1. Steady-state conditions.

2. Negligible heat loss from back surface of the plate (adiabatic back surface).

3. Plate is small object in large surroundings, and coating has an absorptivity

of alpha = 0.8 with repsect to irradiation from the surroundings.

Set the values of the input parameters for part 1.

> case1:=
(A = A*m^2, qrad = 2000*W/m^2, alpha = 0.8,
epsilon = 0.5, h = 15*W/m^2/K,
sigma = 5.67*10^(-8)*W/m^2/K^4,
Tinfty = 293*K, Tsur = 303*K);

Define the heat transfer rate into and from the exposed surfaces of the plate.

> Qabsorbed:= alpha*A*qrad;

> Qconv:= h*A*(Ts - Tinfty);

> Qrad:= epsilon*A*sigma*(Ts^4 - Tsur^4);

Steady-state heat balance at the exposed surface.

> HeatBalance:= Qabsorbed - Qconv - Qrad;

Define the equation for the unknown surface temperature Ts.

> eq1:= expand(simplify(subs(case1, HeatBalance = 0)));

To assist Maple set the values of the units W, K
and the area to 1. Maple can now find the four roots.

Take the positive, real root.

> eq2:= subs(A = 1, W = 1, K = 1, eq1);

> rts:= solve(eq1, Ts):

> rts2:= [solve(eq2, Ts)];

Extract the fourth element of the list called rts2.

> Ts:= rts2[4]*K;

Check the value of Ts by substitution into the heat balance.

> evalf(subs(case1, A = 1, Ts = Ts, HeatBalance));

>