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Examples for one- diamention heat transfer

الكلية كلية هندسة المواد     القسم قسم البوليمرات والصناعات البتروكيمياوية     المرحلة 3
أستاذ المادة صالح عباس حبيب الشبيب       02/12/2018 18:10:16
Examples for one- diamention heat transfer
A short cylindrical metal billet of radius R and height h is heated in an oven to a temperature of 600°F throughout and is then taken out of the oven and allowed to cool in ambient air at T = 65°F by convection and radiation. Assuming the billet is cooled uniformly from all outer surfaces and the variation of the thermal conductivity of the material with temperature is negligible, obtain the
differential equation that describes the variation of the temperature in the billet during this cooling process.
SOLUTION The billet shown in Figure blow is initially at a uniform temperature and is cooled uniformly from the top and bottom surfaces in the z-direction as well as the lateral surface in the radial r-direction. Also, the temperature at any point in the ball will change with time during cooling. Therefore, this is a two-dimensional transient heat conduction problem since the temperature within the billet will change with the radial and axial distances r and z and with time t.


That is, T = T (r, z, t ). The thermal conductivity is given to be constant, and there is no heat generation in the billet. Therefore, the differential equation that governs the variation of temperature in the billet in this case is obtained from Eq. 48 by setting the heat generation term and the derivatives with respect to equal to zero. We obtain

In the case of constant thermal conductivity, it reduces to
Which is the desired equation.

Examples for one –dimension heat transfer conduction
Consider a shielding wall for a nuclear reactor. The wall receives a gamma-ray flux
such that heat is generated within the wall according to the relation

Where .q0 is the heat generation at the inner face of the wall exposed to the gamma-ray
Flux and a is a constant. Using this relation for heat generation, derive an expression
for the temperature distribution in a wall of thickness L, where the inside and outside temperatures are maintained at Ti and T0, respectively. Also obtain an expression for the maximum temperature in the wall.
Solution








Example
The temperature distribution in a certain plane wall is

Where T1 and T2 are the temperatures on each side of the wall. If the thermal conductivity
of the wall is constant and the wall thickness is L, derive an expression for the heat generation per unit volume as a function of x, the distance from the plane where T =T1. Let the heat-generation rate be ?q0 at x = 0.
Solution
K = constant q = qo at x = 0 Assume one directional with no heat storage.

Example : Consider a large plane wall of thickness L = 0.2 m, thermal conductivity k = 1.2 W/m • °C, and surface area A = 15 m2. The two sides of the wall are maintained at constant temperatures of T1 = 120°C and T2 = 50°C, respectively, as shown in blow. Determine (a) the variation of temperature within the wall and the value of temperature at x = 0.1 m and (b) the rate of heat conduction through the wall under steady conditions.

SOLUTION A plane wall with specified surface temperatures is given. The variation of temperature and the rate of heat transfer are to be determined.
Assumptions 1 Heat conduction is steady. 2 Heat conduction is one dimensional since the wall is large relative to its thickness and the thermal conditions on both sides are uniform. 3 Thermal conductivity is constant. 4 There is no heat generation.
Properties The thermal conductivity is given to be k = 1.2 W/m • °C.
Analysis (a) Taking the direction normal to the surface of the wall to be the
x- direction, the differential equation for this problem can be expressed as
differential equation , , Integral
Integrated again
with boundary conditions
T (0) = T1 = 120°C
T (L) = T2 = 50°C
The first boundary condition can be interpreted as in the general solution, replace all the x’s by zero and T (x ) by T1.
T(0) = C1 * 0 + C2 ? C2 = T1
The second boundary condition can be interpreted as in the general solution, replace all the x’s by L and T (x ) by T2. That is,

Substituting the C1 and C2 expressions into the general solution, we obtain

Substituting the given information, the value of the temperature at x = 0.1 m is determined to be

(a) The rate of heat conduction anywhere in the wall is determined from Fourier’s law to be

The numerical value of the rate of heat conduction through the wall is determined
by substituting the given values to be

Example









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