Heal conduction steady

Various other physical processes lead in their mathematical description to equations of the same form as Flq (2). especially in its steady-state form, Such processes include the conduction of electricity in a conductor, or the shape of a thin membrane stretched over a curved boundary. This situation has led to the development of analogies (electric analogy, soap film analogy) to heal conduction processes, which are useful because they often offer the advantages of simpler experimentation. 

Consider a 2-m-hlgh and 0.7-m-wide bronze plate whose thickness is 0.1 m, One side of the plate is maintained at a constant temperature of 600 K vrhile the other side is maintained at 400 K, as shown in Fig. 2-65. The thermal conductivity of the bronze plate can be assumed to vary linearly in that temperature range as k T) = fed + pT) where fe = 38 W/m K and p = 9,21 x lQ-4 ) -i Disregarding the edge effects and assuming steady one-dimensional heat transfer, determine the rate of heal conduction through the plate. 

Starting with an energy balance on a cylindrical shell volume element, derive the steady one-dimensional heal conduction equation fora long cylinder with constant tliemial conductivity in which heat is generated at a rate of... 

Z-Xl Water flows through a pipe at an average temperature of VO C. The inner and outer radii of the pipe are rf = 6 cm and rj = 6.5 cm. respectively. The outer surface of the pipe is wrapped with a thin electric heater that consumes 300 W pet m length of the pipe. The exposed surface of the heater is heavily insulated so that the entire heal generated in the heater is transferred to the pipe. Heat is transferred from the inner surface of the pipe to the water by convection with a heat transfer coefficient of h = 85 W/m K. Assuming constant thermal conductivity and one-dimensionat heat transfer, express the mathematical formulation (the differential equation and the boundary conditions) of the heal conduction in the pipe during steady operation. Do not solve. 

W/m . Assuming steady one-dimensional heat transfer, (a) express the differential equation and the boundary conditions for heal conduction through the wall, (b) obtain a relation for the variation of temperature in the wall by solving the differential... 

C Consider steady oac-dimeosional heal conduction in a plane wall, long cylinder, and sphere with constant thermal conductivity and no heat generation. Will the temperature in any of these mediums vary linearly Explain. 

Consider a 20-cm-ihick large concrete plane wall k 0.77 V/in °C) subjected to convection on both sides with r, = 27"C and A, = 5 W/m °C on the inside, and = 8°C and A2 = 12 W/m °C on the outside. Assuming constant thermal conductivity with no heat generation and negligible radiation, [a) express the differential equations and the boundary conditions for steady one-dimensional heal conduction through the wall, (A) obtain a relation for the variation of temperature in the wall by solving the differential equation, and (c) evaluate the temperatures at the inner and outer surfaces of the wall. 

Steady one-dimensional heal conduction takes place in a long slab of width U (In the direction of heat flow, x) and thickness Z. The slab s thermal conductivity varies with temperature ask = k /(T -t- T), where Tis the temperature (in K), and k (in W/m) and T (in K) are two constants. The temperatures at X = 0 and x IV ate To and T,, re.spectively. Show that the heat flux in steady operation is given by... 

You are probably thinking that if heat is conducted into the element from both sides, as assumed in the formulation, the temperature of the medium will have to rise and thus heal conduction cannot be steady. Perhaps a more realistic approach would be to assume the heat conduction to be into the elemcnl on the left side and out of (he element on the right side. If you repeat the formulation using this assumption, you will again obtain the same result since the heat conduction term on the right side in this case involves T , - + instead... 

TliCf riiiite difl erence fomiulatioii of steady heal conduction problems usu ally results in a system of iV algebraic equations in /V unknown nodal temperatures that need to be solved siiiiullaneously. When Af is small (such as 2 or 3), we can use the elementary elimination method to eliminate ail unknowns except one and then solve for that unknown (sec Example 5-1). The other unknowns are then determined by back substitution. When W is large, which is usually Uie case, the elimination luelliod is not practical and we need to use a more systematic approach that can be adapted to computers. 

The finite difference fonnulalion of steady two-dimensional heal conduction in aniediuin with heat generation and constant thermal conductivity is given by... 

C In the eiieigy balance fonnulalion of the finite difference method, it is recommended that all heal transfer at the boundaries of the volume element be assumed to be into the volume element even for steady heal conduction. Is this a valid recommendation even though it seems to violate the conservation of energy principle ... 

Insulation and Heat-Flow Principles. Heat flows from places of higher temperature to those of lower temperature hy one or more of three modes 11) Conductance through solids (2 convection by induced motion of fluids carrying heat and (3) radiation by heal waves emitted from a surface. The rate of heal flow in solids depends upon temperature difference 7j - 7j and the resistances encountered. The heal flow, under steady stale, is expressed by ... 

Consider steady heat conduction through a large plane wall of thickness Alv = L and area A, as shown in Fig. 1 22. The temperature difference across the wall is AT =7 — 7V Experiments have shown that the rale of heat transfer Q through the wall is doubled when the temperature difference AT" across the wall or the area A normal to the direction of heat transfer is doubled, but is halved when the wall lliickness L is doubled. Thus we conclude that the rate of heat conduction through a plane layer is proportional to the temperature difference across the layer and the heal transfer area, but is inversely proportional to the thickness of the layer. That is. 

A layer of material of known thickness and area can be heated from one side by an electric resistance heater of known output. If the outer surfaces of the heater are well insulated, all the heat generated by the resistance heater will be transferred through the material whose conductivity is to be determined. Then measuring the two surface temperatures of the material when steady heal transfer is reached and substituting them into Bq. 1-21 togelher with other known quantities give tlie thermal conductivity (Fig. 1-26). 

B Understand mullidimensionality and time dependence of heat transfer, and the conditions under which a heal transfer problem can be approximated as being one-dirnensional, B Obtain the differential equation of heat conduction in various coordinate systems, and simplify it for steady one-dimensional case,... 

Consider a large plane wall of thickness L = 0.06 m and thermal conductivity k = 1.2 W/m C in space. The wall is covered with white porcelain tiles that have an emissivity of e = 0.85 and a solar absorptivity of a = 0.26, as shown in Fig. 2-48. The inner surface of the v/all is maintained at Ti = 300 K at all times, while the outer surface Is exposed to solar radiation that is incident at a rate of 800 W/m. The outer surface is also losing heal by radiation to deep space at 0 K. Determine the temperature of the outer surface of the wall and the rate of heat transfer through the wall when steady operating conditions are reached. What would your response be if no solar radiation was incident on the surface ... 

Consider a spherical container of inner radius r, outer radius rj, and thermal conductivity k. Express the boundary condition on the inner surface of the container for steady one-dimensional conduction for the following cases (a) specified temperature of 50°C, (b) specified heat flux of 30 W/m toward the center, (c) convection to a medium at 7. with a heal transfer coefficient of/i. 

Consider a spherical shell of inner radius r, and outer radius whose ihennal conductivity varies linearly in a specified temperature range as k(X) = (1 + fiT) where tfeo and /3 are two specified constants. The inner surface of the shell is maintained at a constant temperature of while the outer surface is mainlaincd al Tj. Assuming steady one-dimensional heat transfer, obtain a relation for (n) the heal transfer rate through the shell and (fe) the temprerature distribution 7(r) in the shell. 

Consider steady one-dimensional heat conduction through a plane wall, a cylindrical shell, and a spherical shell of unifonn thickness with constant thcimophysical propenies and no thermal energy generation. The geometry in which the variation of temperature in the direction of heal transfer will be linear is... 

SOLUTION A spherical container filled with iced water is subjected to convection and radiation heat transfer at its outer surface. The rate of heat transfer ans the amount of ice that melts per day arc to be determined. Assumptions 1 Heal transfer is steady since the specified thermal conditions at the lioundaries do not change with time. 2 Heat transfer is one-dimensional since there is thermal symmetry a out the midpoint. 3 Thermal conductivity is constant. 

To demonstrate the approach, again consider steady one-dimeiisional heal transfer in a plane wall of thickness L with heat generation. r) and constant conductivity k. The wall is now subdivided into M equal regions of thickness Ax = UM in the x-direction, and the divisions between the regions are selected as the nodes. Therefore, we have A/ + I nodes labeled 0, 1, 2,..., m, m,m + 1,... , A/, as shown in Figure 5-10. The. r-coordinate of any node m is simply x = nAx, and the temperature at that point is T x = Elements are formed by drawing vertical lines through (he midpoints between the nodes. Note that all interior elements represented by interior nodes are full-size elements (they have a thickness of A.t), whereas the two elements at the boundaries are half-sized. 

---