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XX = 23 According to Fourier's law, the heat transfer per unit area through a material is proportional to the temperature gradient. This can be
XX = 23
According to Fourier's law, the heat transfer per unit area through a material is proportional to the temperature gradient. This can be written mathematically as: Q=kLT2T1 Where T1[C] is the temperature +at face 1,T2[C] is the temperature at face 2,L[m] is the thickness, k[W/(mK)] is a material constant called the thermal conductivity, and Q[W/m2] is the heat flux from face 1 to face 2 . (a) At each material interface the heat flux ( Q ) from each side must be equal. Using this property, (i) write the system of equations for the temperature at each interface of the wall shown Figure 1 below (brick-air, air-foam, foam-concrete) then (ii) express these equations in matrix form. [12] For example, at the interface between the foam and concrete: Q3=kfLfTf,cTa,fSinceQ3=Q4kfLfTf,cTa,fQ4=kcLcTinTf,=kcLcTinTf,c Where Tf,c is the temperature at the foam-concrete interface, Ta,f is the temperature at the air-foam interface, and Tin is the internal temperature. kf and kc are the thermal conductivities of the foam and concrete respectively. Hint: It can simplify the calculations to define the conductance of a layer i as: Ci=Liki (b) If the outside temperature is 20C and the inside temperature is +20C, use matrix methods to calculate the temperature at the interface between each layer (Ti). Plot" the variation of temperature as a function of thickness through the wall. [15] (c) What is the heat flux through the wall (between inside and outside) and in what direction does the heat flow? [3] XX are the last two digits of your URN. For example, if your URN is 6835725,XX=25 and the thickness of the air gap =25+10= 35mm. Write your value for XX at the top of your solution. "You can use software (Excel, Matlab, etc) to create the plot if you wish. + Note Kelvin (K) and Celsius (C) have the same scale, so Eqn (1) is dimensionally consistent. According to Fourier's law, the heat transfer per unit area through a material is proportional to the temperature gradient. This can be written mathematically as: Q=kLT2T1 Where T1[C] is the temperature +at face 1,T2[C] is the temperature at face 2,L[m] is the thickness, k[W/(mK)] is a material constant called the thermal conductivity, and Q[W/m2] is the heat flux from face 1 to face 2 . (a) At each material interface the heat flux ( Q ) from each side must be equal. Using this property, (i) write the system of equations for the temperature at each interface of the wall shown Figure 1 below (brick-air, air-foam, foam-concrete) then (ii) express these equations in matrix form. [12] For example, at the interface between the foam and concrete: Q3=kfLfTf,cTa,fSinceQ3=Q4kfLfTf,cTa,fQ4=kcLcTinTf,=kcLcTinTf,c Where Tf,c is the temperature at the foam-concrete interface, Ta,f is the temperature at the air-foam interface, and Tin is the internal temperature. kf and kc are the thermal conductivities of the foam and concrete respectively. Hint: It can simplify the calculations to define the conductance of a layer i as: Ci=Liki (b) If the outside temperature is 20C and the inside temperature is +20C, use matrix methods to calculate the temperature at the interface between each layer (Ti). Plot" the variation of temperature as a function of thickness through the wall. [15] (c) What is the heat flux through the wall (between inside and outside) and in what direction does the heat flow? [3] XX are the last two digits of your URN. For example, if your URN is 6835725,XX=25 and the thickness of the air gap =25+10= 35mm. Write your value for XX at the top of your solution. "You can use software (Excel, Matlab, etc) to create the plot if you wish. + Note Kelvin (K) and Celsius (C) have the same scale, so Eqn (1) is dimensionally consistentStep by Step Solution
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