11/21/2023 0 Comments Heat transfer ratebut "A" will reach that temperature quicker due to the high heat transfer coefficient. in this case, both "A" and "B" require the same amount of heat to get them to a specific temperature. A different material "B" can have low coefficient and high capacity. Returning to heat transfer a material "A" can have a high coefficient of heat transfer and a high heat capacity. you can change the resistor and the capacity of the capacitor stays the same. it increases the time required to charge the capacitor).Īs you can see here, there is no relationship between the capacity of the capacitor and the value of the resistor. the higher the value of the resistor, the less the value of the electric current (i.e. The resistor affects (restricts) the flow rate of electric charge (The opposite of coefficient of heat transfer). The higher the capacity (which depends on surface area and other factors) the more charge it can store, and, the more time it requires for it fully charge. The process of conduction in thermal dynamics is dependent on the factors such as heat transfer coefficient of the material, area, thickness through which the heat is transferred and the change in temperature.Īn analogy that could work is an electric circuit that has a battery, a capacitor and a resistor. Heat transfer takes place in 3 main ways: conduction, convection and radiation. This is slightly different than specific heat $$Ĭ_p ~ $ The outside temperature is 100F.It is a property that refers to the amount of energy required to increase the temperature of an object. If the inside wall temperature of the pipe is maintained at 550F, calculate the heat loss per foot of length. layer of asbestos insulation (k a= 0.14 Btu/hr-ft- oF) as shown in Figure 5. outside diameter (OD) is covered with a 3 in. The evaluation of heat transfer through a cylindrical wall can be extended to include a composite body composed of several concentric, cylindrical layers, as shown in Figure 4.Ī thick-walled nuclear coolant pipe (k q= 12.5 Btu/hr-ft-F) with 10 in. The thermal conductivity of the stainless steel is 108 Btu/hr-ft-> oF.Ĭalculate the heat transfer rate through the pipe.Ĭalculate the heat flux at the outer surface of the pipe.Ī 10 ft length of pipe with an inner radius of 1 in and an outer radius of 1.25 in has an outer surface temperature of 250F. The temperature of the inner surface of the pipe is 122 oF and the temperature of the outer surface is 118 oF. This expression for log mean area can be inserted into Equation 2-5, allowing us to calculate the heat transfer rate for cylindrical geometries.Ī stainless steel pipe with a length of 35 ft has an inner diameter of 0.92 ft and an outer diameter of 1.08 ft. Heat flows from the warmer fluid to the cooler fluid, at a rate proportional to the difference between the fluid entrance temperatures. Substituting the expression 2 p>rL for area in Equation 2-7 allows the log mean area to be calculated from the inner and outer radius without first calculating the inner and outer area. Related terms: Energy Engineering Nanoparticles Heat Exchanger. From: Transport Phenomena in Heat and Mass Transfer, 1992. For a problem involving cylindrical geometry, it is necessary to define a log mean cross-sectional area (A lm). The effectiveness is defined as the ratio of the actual heat transfer rate to the thermodynamically limited maximum heat transfer rate in a counter flow heat exchanger of infinite surface area. Neither thearea of the inner surface nor the area of the outer surface alone can be used in the equation. The development of an equation evaluating heat transfer through an object with cylindrical geometry begins with Fouriers law Equation 2-5.įrom the discussion above, it is seen that no simple expression for area is accurate. The surface area (A) for transferring heat through the pipe (neglecting the pipe ends) is directly proportional to the radius (r) of the pipe and the length (L) of the pipe.Īs the radius increases from the inner wall to the outer wall, the heat transfer area increases. Figure 3 is a cross-sectional view of a pipe constructed of a homogeneous material. Across a cylindrical wall, the heat transfer surface area is continually increasing or decreasing. Heat transfer across a pipe or heat exchanger tube wall is more complicated to evaluate. Heat transfer across a rectangular solid is the most direct application of Fouriers law. This may seem strange, considering that the rate of heat transfer across a heating surface is directly proportional to the temperature difference. So, superheated steam is not as effective as saturated steam for heat transfer applications. Heat Transfer Engineering | Thermodynamics Provides lower rates of heat transfer whilst the steam is superheated. Conduction - Cylindrical Coordinates - Heat Transfer
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