Wall surface temperatures, calculation

In the finite-difference appntach, the partial differential equation for the conduction of heat in solids is replaced by a set of algebraic equations of temperature differences between discrete points in the slab. Actually, the wall is divided into a number of individual layers, and for each, the energy conserva-tk>n equation is applied. This leads to a set of linear equations, which are explicitly or implicitly solved. This approach allows the calculation of the time evolution of temperatures in the wall, surface temperatures, and heat fluxes. The temporal and spatial resolution can be selected individually, although the computation time increa.ses linearly for high resolutions. The method easily can be expanded to the two- and three-dimensional cases by dividing the wall into individual elements rather than layers. 

For a person at a certain location in a room, direct radiation from internal heat sources may significantly affect the thermal comfort level. However, in the codes, room (or operative) temperatures are calculated on the basis of the room air and the wall surface temperatures only (both calculated considering the internal heat source, however). 

As explained above, one long side of the compartment wall was split into a large number of thin strips and the heat flux to the center of each strip calculated. For a constant heat flux, assuming the wall material to be semi—infinite, the wall surface temperature... 

No wall surface temperatures were measured in the full scale test series, scenario A and in the 1/3 scale test series, scenario B. Figure 6 a) shows the experimental and calculated wall surface temperatures, at a height of 0.45 m from the floor, for material no. 3, 1/3 scale test, scenario A. Figure 6 b) shows the same, but at a height of 1.2 m from the floor, for the full scale test, scenario B. 

Figure 6. Comparison of experimental and calculated data (wall surface temperatures and downward flame spread) in two different experimental setups. Continued on next page.

EXAMPLE 43-1. Heat Flow Through an Insulated fVall of a Cold Room A cold-storage room is constructed of an inner layer of 12.7 mm of pine, a middle layer of 101.6 mm of cork board, and an outer layer of 76.2 mm of concrete. The wall surface temperature is 255.4 K inside the cold room and 297.1 K at the outside surface of the concrete. Use conductivities from Appendix A.3 for pine, 0.151 for cork board, 0.0433 and for concrete, 0.762 W/m K. Calculate the heat loss in W for 1 and the temperature at the interface between the wood and cork board. 

Calculate the airflow rate along an external wall with a surface temperature 3°C above room remperature, at a height of 4 meters above the lower edge of the surface. 

The response factors are characteristic for the layer buildup of the selected wall and are calculated before (by a preprocessor program) or at the beginning ol the simulation. Numerical reasons limit the time step to approximately 10 to 60 min, depending on the thickness and material properties of the wall layers. The method allows the calculation of surface temperatures and heat fluxes bur not the determination of the temperature distribution within the wall. Due to the precalculation of these response factors, the computer time for the simulation might be significantly reduced. 

The wall temperature maps shown in Fig. 28 are intended to show the qualitative trends and patterns of wall temperature when conduction is or is not included in the tube wall. The temperatures on the tube wall could be calculated using the wall functions, since the wall heat flux was specified as a boundary condition and the accuracy of the values obtained will depend on their validity, which is related to the y+ values for the various solid surfaces. For the range of conditions in these simulations, we get y+ x 13-14. This is somewhat low for the k- model. The values of Tw are in line with industrially observed temperatures, but should not be taken as precise. 

These heat sources were derived from the reaction enthalpy and related to the wall surface area. This delivered a specific heat of 259.15 W m 2 developed at the walls. All other surfaces were set to be isothermal with a surface temperature of 500 °C. The justification for this boundary condition will be given by executing a temperature calculation inside the titer-plate. As fluidic boundary conditions, the flow velocity at the inlet to the well was fixed and the outlet pressure was set equal to the ambient pressure. 

Example Calculation of Heat Loss through a Composite Wall. A furnace wall is constructed of a 3-cm thick, flat steel plate, with a firebrick insulation 30-cm thick on the inside, and rock wool insulation 6-cm thick on the outside. The inside surface temperature of the firebrick insulation is 700°C. If the temperature of the outer surface of the rock wool insulation is 50°C, what is the heat flux through the wall ... 

Figure 34 The steps involved in determining the depth of container wall penetration under Canadian nuclear waste disposal conditions using data obtained in an electrochemical galvanic coupling experiment. (A) Crevice propagation rate (R cc Ic) as a function of temperature (T) (B) RCc as a function of 02 concentration [02] (C) calculated evolution of container surface temperatures and vault 02 concentrations with time in the vault (D) flux of 02 (Jo2) to the container surface as a function of time (E) predicted evolution of Rcc up to the time of repassivation (i.e., at [02]p) (F) total extent of crevice corrosion damage expressed as the total amount of 02 consumed (Q) up to the time of repassivation (G) experimentally determined maximum depth of wall penetration (Pw) as a function of 02 consumed (Q) (H) predicted maximum value of Pw up to the time of repassivation (fP)-...

A vertical cylinder 6 ft tall and I ft in diameter might be used to approximate a man for heat-transfer purposes. Suppose the surface temperature of the cylinder is 78°F, h = 2 Btu/h - ft2 °F, the surface emissivity is 0.9, and the cylinder is placed in a large room where the air temperature is 68°F and the wall temperature is 45°F. Calculate the heat lost from the cylinder. Repeat for a wall temperature of 80°F. What do you conclude from these calculations ... 

A wall 2 cm thick is to be constructed from material which has an average thermal conductivity of 1.3 W/m °C. The wall is to be insulated with material having an average thermal conductivity of 0.35 W/m °C, so that the heat loss per square meter will not exceed 1830 W. Assuming that the inner and outer surface temperatures of the insulated wall are 1300 and 30°C, calculate the thickness of insulation required. 

A horizontal pipe IS cm in diameter and 4 m long is buried in the earth at a depth of 20 cm. The pipe-wall temperature is 75°C, and the earth surface temperature is S°C. Assuming that the thermal conductivity of the earth is 0.8 W/m °C, calculate the heat lost by the pipe. 

A furnace of 1 by 2 by 3 ft inside dimensions is constructed of a material having a thermal conductivity of 0.5 Btu/h ft °F. The wall thickness is 6 in. The inner and outer surface temperatures are 1000 and 200°F, respectively. Calculate the heat loss through the furnace wall. 

A tube has diameters of 4 mm and S mm and a thermal conductivity 20 W/m2 °C. Heat is generated uniformly in the tube at a rate of 500 MW/m3 and the outside surface temperature is maintained at 100°C. The inside surface may be assumed to be insulated. Divide the tube wall into four nodes and calculate the temperature at each using the numerical method. Check with an analytical solution. 

In order to reduce the heat loss through the house wall in exercise 1.11, an insulating board with <53 = 6.5cm and A3 = 0.040W/Km along with a facing of 84 = 11.5cm and A4 = 0.79 W/K m will replace the outer plaster wall. Calculate the heat flux q and the surface temperature 5 vi of the inner wall. 

Example 2.3 A flat wall of thickness <5 has a constant temperature o- At time t = 0 the temperature of the surface x = 6 jumps to s, whilst the other surface x = 0 is adiabatic, Fig. 2.19. Heat flows from the right hand surface into the wall. The temperature rises with time, whereby the temperature of the left hand surface of the wall rises at the slowest rate. The temperature increase at this point, i.e. the temperature (x = 0, t) is to be calculated. 

Example 2.8 A wall of thickness <5 surrounds a room with a square base the length of the internal side of the square is 2.5 <5. The wall has constant surface temperatures i and i 0 < A respectively. Calculate the heat flow out of the room due to the temperature difference — 0. 

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