Create a 5 pages page paper that discusses fouriers law of heat conduction and newtons law of cooling.
Create a 5 pages page paper that discusses fouriers law of heat conduction and newtons law of cooling. Fourier’s law of heat conduction states that the rate of heat flow dQ/DT 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 dT/dx.
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. An example of radiation is heat released from the filament of the 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 is 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
R_plastic=∆x/kA=0.02/20X35=0.000028571
R_BRICK=∆x/kA=0.1/0.6X35=0.004761904
R_insulation=∆x/kA=0.065/0.08X35=0.023214285
R_BRICK=∆x/kA=0.1/0.6X35=0.004761904
R_insulation=∆x/kA=0.065/0.08X35=0.023214285
R_BRICK=∆x/kA=0.1/0.6X35=0.004761904
Rtotal=0.000028571+0.004761904+0.023214285+0.004761904+0.023214285+0.004761904
=0.10359999 W/M K
The overall heat transfer coefficient
Q=∆Toverall/Rtotal=(22–4)/0.10359999=250.9652 W.M^2 K
Newton’s law of cooling
Answer
The concept of temperature difference occurs in situations where there is a flow of energy from a system to the surrounding environment. This occurrence leads to both heating and the opposite would lead to a cooling of the body. This can be explained using Newton’s law of cooling.
