Convection losses

In buildings away from outside perimeter walls, air and surface temperatures are usually approximately equal. The heat losses from a person by radiation (q ) and convection (q ) are then flowing to the same temperature level. In such uniform spaces, the radiant and convective losses are about equal and together account for about 80-90% of the total heat loss of a sedentary comfortable individual. In the presence of hot or cold surfaces, as may occur in perimeter or other locations in a building, the average surface temperature of the surroundings (called mean radiant temperature) as seen by the person s body may be substantially different from air temperature. If the mean radiant temperature (MRT) is greater or less than air temperature (T,) the person will feel warmer or colder than in a thermally uniform space where MRT =. ... 

In calculating heat loss from surfaces freely exposed to air it is necessary to deal separately with both radiant and convective losses. 

Nutrient Losses Associated With Biomass Burning. Nutrient losses associated with slash fires occur through volatilization and convective losses of ash. Elements with low temperatures of volatilization (e.g. N, K, S, and some organic forms of P) will be lost in the highest quantities (Table III) (57). Conversely, Ca and Mg have volatilization temperatures higher than that recorded during most vegetation fires. Almost all fire-induced losses of these elements are due to particulate transfer by convective processes. 

The efficiency of this preprototype device was low due principally to the thermal mass of the copper buss bars, convection losses associated with the air cooled design and radiation losses at the high operating temperatures. Current work is directed at staged compressors (more than one alloy) operating over smaller temperature ranges supplied by a liquid heat transfer media. 

If the relative humidity of the air in the distiller enclosure is considerably less than 100% at the salt water surface and at the cover, more convective loss is actually occurring than estimated. Increasing these air moisture contents by better contact of air and cover, or by otherwise altering circulation patterns by fan, baffles, or enclosure shape could reduce air circulation loss and increase productivity. Whether the economics of such steps might be attractive remains to be determined. 

In view of the several possibilities for reducing energy losses, solar distiller yields might be substantially improved. Decrease of radiation and convection losses to half their present levels, along with the use of low-reflection cover surfaces, would result in about 50% increase in summer production and roughly double the winter per-... 

Distribution of Energy. During a 3-day period, October 7 to 9,1959, a continuous performance run was made on the deep-basin still for the purpose of computing an energy balance. Each item pertinent to the energy balance was measured, except convection loss to the atmosphere, which was obtained by calculations. The experimentally determined losses were then compared with the corresponding calculated losses. These showed remarkably close agreement. 

The internal convection loss is moderate, so that measures suggested to reduce this loss, such as replacing the air inside the still with a gas of lower specific heat or operating the still under vacuum, may not be economically feasible. 

Fluidized-bed Coating of an Article A rectangular metal article with dimensions of 0.5 x 5.0 x 10.0 cm is to be coated with PVC powder to a uniform coat thickness of 0.01 cm, using the fluidized-bed coating process. The fluidized-bed temperature is 20°C and the initial metal temperature is 150°C. (a) Assuming no convective losses to the fluidized bed, what would the metal temperature decrease need to be to form the desired coat thickness (b) Estimate the effect of convective heat losses on the temperature decrease of the metal. 

The diffusivity, a2, is subsequently determined under the condition that the intercepts of the linear fits for the thermally thick and thin conditions are equal. For the PA6 nanocomposite, when the diffusivity is equal to 0.9 x 10 7m2/s, the intercepts are almost the same at about 11.5kW/m2. These intercepts are equal to the 0.64 fraction of the critical heat flux (below which there is no ignition) for ignition [21], and thus the critical heat flux can be calculated equal to 11.5/0.64 = 17.9 kW/m2. The ignition temperature can then be calculated by considering the critical heat flux equal to surface reradiation and convection losses ... 

The total heat loss is the sum of convection and radiation. From Table 1-2 we see that an estimate for the heat transfer coefficient for free convection with this geometry and air is h = 6.5 W/m2 °C. The surface area is ir d L, so the convection loss per unit length is... 

The results of the model and calculations may be checked by calculating the convection heat lost by the top surface. Because all the energy generated in the small heater strip must eventually be lost by convection (the bottom surface of the glass is insulated and thus loses no heat) we know the numerical value that the convection should have. The convection loss at the top surface is given by... 

The heat flow out the top face is obtained by summing the convection loss from the nodes ... 

The heat loss for the numerical model is computed by summing the convection loss from the six nodes (including base node at 200°C). Using the temperatures for the first iteration corresponding to h = 22.11 W/m2-°C,... 

A spherical balloon gondola 2.4 m in diameter rises to an altitude where the ambient pressure is 1.4 kPa and the ambient temperature is — 50°C. The outside surface of.the sphere is at approximately 0°C. Estimate the free-convection heat loss from the outside of the sphere. How does this compare with the forced-convection loss from such a sphere with a low free-stream velocity of approximately 30 cm/s ... 

A large vat used in food processing contains a hot oil at 400°F. Surrounding the vat on the vertical sides is a shell which is cooled to 140°F. The air space separating the vat and the shell is 35 cm high and 3 cm thick. Estimate the free-convection loss per square meter of surface area. 

The cavity of Prob. 8-73 has a fused-quartz window placed over it, and the cavity is assumed to be perfectly insulated with respect to conduction and convection loss to the surroundings. The cavity is exposed to a solar irradiation flux of 900 W/m2. Assuming that the quartz is nonreflecting and r = 0.9, calculate the equilibrium temperature of the inside surface of the cavity. Recall that the transmission range for quartz is 0.2 to 4 /xm. Neglect convection loss from the window. The surroundings may be assumed to be at 20°C. 

A slab of white marble is exposed to a solar radiation flux of 1070 W/m2. Assuming the effective radiation temperature of the sky is -70°C, calculate the radiation equilibrium temperature of the slab, using the properties given in Table 8-3. For this calculation neglect all conduction and convection losses. 

It is often observed experimentally that if the relative concentrations of the reactants and the initial temperature of the mixture are kept fixed but the pressure p is decreased, then a limiting pressure is reached, below which flame propagation cannot be achieved. The flammable range of 0 usually narrows as the pressure is decreased, and below a critical pressure flame propagation does not occur for any value of 0. These observations are consistent with equation (24), in which the main pressure-dependent quantities are ml and L(Tf In terms of the overall order n (the pressure exponent) of the overall reaction rate, ml p . It will be seen in the following section that L(Tf) p", where m 0 for conductive or convective losses and 0 < m < 1 for radiative losses. Since n > 1 for practically all flames, the right-hand side of equation (24) decreases more rapidly than the left-hand side as p decreases, and the limiting equality may therefore be expected to be surpassed at a sufficiently low pressure. 

Unlike the radiant loss from an optically thin flame, conductive or convective losses never can be consistent exactly with the plane-flame assumption that has been employed in our development. Loss analyses must consider non-one-dimensional heat transfer and should also take flame shapes into account if high accuracy is to be achieved. This is difficult to accomplish by methods other than numerical integration of partial differential equations. Therefore, extinction formulas that in principle can be used with an accuracy as great as that of equation (21) for radiant loss are unavailable for convective or conductive loss. The most convenient approach in accounting for convective or conductive losses appears to be to employ equation (24) with L(7 ) estimated from an approximate analysis. The accuracy of the extinction prediction then depends mainly on the accuracy of the heat-loss estimate. Rough heat-loss estimates are readily obtained from overall balances. 

Material Balance and Heat Balance, Required heat was mainly supplied by incineration of char, and some amount of produced combustible gas was fed as auxiliary fuel to the regenerator, as the amount of char was not sufficient for continuous thermal cracking. The material balance around the reactors is shown in Table-Ill and heat balance in Table-IV. Radiation and convection loss in Table-IV is larger than that of usual incinerators because of the thin refractory. It can be decreased in case of commercial plants. Energy balance of the total plant is shown in Fig-3. 

Exclusive of radiation and convection losses. Mainly 2-ethyl-3-hydroxyhexan-l-al. 

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