Total heat loss

The combustor is designed to operate at a pressure of 600 kPa (6 atm) with 1867 K preheated air. First stage heat loss of the 250 MW combustor is about 4.3% and the total heat loss is about 6%. The relative pressure drop is 3%. More complete discussions of the design and scale-up of the combustor are available (75). 

Radiation differs from conduction and convection not only in mathematical structure but in its much higher sensitivity to temperature. It is of dominating importance in furnaces because of their temperature, and in ciyogenic insulation because of the vacuum existing between particles. The temperature at which it accounts for roughly half of the total heat loss from a surface in air depends on such factors as surface emissivity and the convection coefficient. For pipes in free convection, this is room temperature for fine wires of low emissivity it is above red heat. Gases at combustion-chamber temperatures lose more than 90 percent of their energy by radiation from the carbon dioxide, water vapor, and particulate matter. 

The second context is the process reac tor. There is a potential for a runaway if the net heat gain of the system exceeds its total heat loss capabihty. A self-heating rate of 3°C/day is not unusual for a monomer storage tank in the early stages of a runaway. This corresponds to 0.00208°C/min, 10 percent of the ARC s detection limit. ARC data for the stored chemical would not show an exotherm until the self-heating rate was 0.02°C/min. Therefore, onset temperature information from ARC testing must be used with considerable caution. 

The total heat loss over 30 minutes, tf, per unit area where k is the drermal conductivity of the container, is given by... 

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 =. ... 

Determine the air temperature that is necessary to maintain a volume of 1200 m- with six air changes per hour at 20 "C, if the total heat loss is 45 kW. 

Calculate the total heat loss by radiation and convection from an unlagged horizontal steam pipe of 50 mm outside diameter at 415 K to air at 290 K. 

The total heat losses in a fully developed fire can then be approximated as... 

Total heat gain = Total heat loss... 

Direct calorimetry (heat production) represents a reliable method for the estimation of total benthic community metabolism, because it is a direct measurement of the energy flow through the system. Although the heat release from the activity of extracellular enzymes and from chemical oxidations are also included, the latter two components are thought to be of minor importance for total heat loss from sediments (Pamatmat, 1982). 

A value of lossnorm = 1 indicates that the total heat loss via the exit streams of the exothermic and the endothermic stage is equal to the amount of heat required to heat up the feed of the endothermic stage about AT = Tequ-To. lossmrm can be expressed explicitly as a function of h and NTUenio taking into account the theory of ideal countercurrent heat exchangers [7]. 

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... 

A 5.0-cm-diameter cylinder is heated to a temperature of 200°C, and air at 30°C is forced across it at a velocity of 50 m/s. If Lhe surface emissivity is 0.7, calculate the total heat loss per unit length if the walls of the enclosing room are at 10°C. Comment on this calculation. 

A plane wall of thickness 2L has an internal heat generation which varies according to q = q0 cos ax, where q0 is the heat generated per unit volume at the center of the wall (x = 0) and a is a constant. If both sides of the wall are maintained at a constant temperature of Tw, derive an expression for the total heat loss from the wall per unit surface area. 

The total heat loss from the wire may be calculated as the conduction through R + at node 4. Then... 

The total heat loss during the 12-s time interval is calculated by summing the heat loss of each node relative to the initial temperature of 300°C. T/ius... 

Helium at 1 atm and 300 K. is used to cool a 1-m-square plate maintained at 500 K. The flow velocity is 50 m/s. Calculate the total heat loss from the plate. What is the boundary-layer thickness as the flow leaves the plate ... 

Air at 20°C flows across a 50-cm-diameter cylinder at a velocity of 25 m/s. The cylinder is maintained at a temperature of I50°C and has a surface emissivity of 0.7. Calculate the total heat loss from the cylinder per unit length if the effective radiation temperature of the surroundings is 20°C. 

The total heat loss in time t corresponding to Figs. 4-16 to 4-18 can also be obtained as follows. 

An average value of hr = 1.2 Btu/(hXft2X°F) may be used to determine heat losses by radiation. This is an adjusted value such that total heat loss per... 

The correct answer is (E). In this problem the enthalpy of reaction has already been determined. The only step, therefore, is to determine the effect of the additional heat loss of 44 kJ mol as the water condenses. For each mole of gaseous water vapor, an additional 44 kJ of heat will be lost. Because the balanced equation shows two moles of water vapor, the total released during condensation is 88 kJ. Therefore, the total heat loss in the reaction will be -802 kJ + (-88 kJ) = -890 kJ. 

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