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Fouriers Law of Heat Conduction and Newtons Law of Cooling - Assignment Example

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The assignment "Fourier’s Law of Heat Conduction and Newton’s Law of Cooling" explains the difference between heat conduction, heat convection, and heat radiation, calculation, the thermal resistance, the role of material types, energy cost, and economic lagging in reducing the heat energy. …
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Fouriers Law of Heat Conduction and Newtons Law of Cooling
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Explain and Fourier’s law of heat conduction Answer If we consider an object in which heat is flowing in an X direction. Heat must flow from a higher to a lower temperature. Assuming that T1 is greater than T2. The temperature difference is given by T1-T2=ΔT and this creates Q which is the heat quantity. From the second law of dynamics, the ΔT is a drop in temperature. Thermal resistance R between the ends opposes the flow of heat. If we combine these quantities using ohms law; Fourier’s law of heat conduction states that the rate of heat flow through a homogenous solid is directly proportional to the area of the section at right angles to the direction of heat flow and to the temperature difference along the path of heat flow . b. Explain with examples the difference between heat conduction, heat convection and heat radiation Answer Heat conduction refers to the transfer of heat between substances that are in direct contact with each other. It occurs when a substance is heated and particles gain more energy hence increasing the vibration. An example of heat conduction is a pot placed on a hot burner. Heat convection refers to the transfer of heat through liquids and gases. An example is water heating in a boiler. Heat radiation refers to the transmission of heat through empty space by thermal radiation. Example of radiation is heat released from the filament of bulb. c. A wall with an area Of 35 m2 is made up of six layers. On the inside is plaster 20mm thick, then there is the brick 100 mm thick, followed by the insulation of 60 mm thick, then the brick of 100 mm thick, then there is the insulation of 65 mm thick and finally brick 100 mm thick. Calculate the thermal resistance, the heat transfer between the layers and the overall heat transfer coefficient given that the thermal conductivity of plaster is 20 W/ m K, the thermal conductivity of the brick is 0.6 W/m K and the thermal conductivity of the insulation is 0.08 W/ m K. The inner surface temperature of the wall is 22°C and the outer is -4°C. Solution Thermal resistance The overall heat transfer coefficient 2. Explain and state Newton’s law of cooling Answer The concept of temperature difference occurs in situations where there is flow of energy from a system to the surrounding environment. This occurrence leads to both heating and the opposite would lead to cooling of the body. This can be explained using Newton’s law of cooling. Newton’s law of cooling states that the rate of change of temperature of an object is proportional to the difference between its own temperature and the ambient temperature. b. The figure below shows a cross section through an insulated heated pipe which is made of steel (thermal conductivity, k = 45 W/m K) with an inner radius (r1) of 800 mm and an outer radius (r2) of 1300 mm. The pipe is coated with two layers of insulating materials, the first layer has a thickness of 200 mm and thermal conductivity, k =25.9 W/m K, the second layer has a thickness of 400 mm and the thermal conductivity, k = 35 W/m K. Air at Tfi = 310 °C flows through the pipe and the convective heat transfer coefficient from the air to the outside has a value of h1= 120 VV/m2 K, The outside surface of the pipe is surrounded by air which is at 50 (rf2) and the convective heat transfer coefficient on this surface has a value of h2= 90 W/m2 K. Calculate, (a) the thermal resistance (b) the heat transferred (c)temperature T1, 12, T3 and T4 (d) comment on your results. Heat transferred Temperature T1, 12, T3 and T4 The temperatures increase as the size of the radius increases. 3. Evaluate and justify the role of material types, energy cost and economic lagging in reducing the heat energy. Answer Economic lagging is used to contain heat in applications which generate, transfer and transmit heat. It actually comes in handy in industrial appliances such as boilers, steam and hot fluid plates, furnaces, ovens heaters and storage tanks. The materials used in lagging include those that appear low on the thermal conductivity scale and they operate by dividing the airspace into very small pockets of air therefore minimizing conduction and radiation and eliminating convection completely. Properly designed insulation systems reduce heat loss, conserve energy through reduction of heat stored in the system, maintenance of the desired processes and temperatures and assisting in heat transfer within the system. Lagging materials such as silica, alumina silica and high alumina are used for low scale lagging while materials such as carbon steel which has a temperature range of 0-500 degrees Fahrenheit and insulating brick which has a temperature range of 30-2800 degrees Fahrenheit are used in large scale economic lagging. These material types are all based on a single objective of reducing the amount of energy used. b. Two surfaces A and B close to each other and are exchanging heat with each other. Surface A is coated with a white paint and is maintained at a temperature 2000 C. It is located directly opposite to surface B which is considered a black body and is maintained at 8000 C.Calculate the amount of heat energy that needs to be removed from surface A per unit area to maintain its constant temperature. Solution 4. The Reynolds number (Re) is a function of density viscosity and velocity of a fluid and a characteristic length. Establish that by dimensional analysis and suggest possible methods to obtain C and d. Solution Reynolds number is dimensionless C and d can be obtained through experimental analysis and through use of dimensionless fluxes in forced heat transfer. b. Discuss the importance of applying dimensional analysis in engineering applications such as hat transfer Answer Dimensional analysis is used in engineering works to facilitate the correlation and interpretation of experimental data that needs to be analyzed to understand the functioning of a system. It is used to provide a means of combining the many parameters that may be obtained from an experiment into lesser number of dimensionless groups which are easier and less complex to work with. Dimensional analysis also reduces the amount of experimental work that may be needed to determine the effect of parameter variation on the dependent parameter in the experimental data. 5. In an experimental result showed that air at 300k and bulk velocity at 10m/s flows over a flat horizontal plate with a temperature 1000k at all points on its surface. Given that calculate the heat transfer rate from one side over the first 100 mm and the first 200 mm Solution First we calculate Reynolds number at each length L=100 mm We solve for For L=200 mm The value of Reynolds number increases with distance and hence increases the rate of heat transfer. References Munson, B. R., Young, D. F., Okiishi, T. H.,(2002) “Fundamentals of Fluid Mechanics”, 4th Edition, Chapter 7, Wiley. Fox, R. W., McDonald, A. T., Pritchard, P. J.,(2004) “Introduction to Fluid Mechanics”, 6th Edition, Chapter 7, Wiley. Read More
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