(a) A furmace wall is a made up of steel plate 10mm thickness (k15kcot/mhC lined en inside with silica tricks 150mm thickness (k=1.75kral/mhrr2C) and on the eutside with magnesia bricks 200mm thickness ( k=4.5kcal/mhrm2C. The inside and cutside surfaces of the wall are at temperatures 650C and 125C, respectively. (i) Make calculations for the heat loss from unit area of the wall. [6 Marks] (ii) It is required that the heat loss be reduced to 2500keal/hr by means of air gap benween steel and magnesia bricks. Estimate the necessary width of air gap if thermal conductivity for air is 0.03kcat/mhrC [4 Marks] (b) Consider a plane wall of thiekness L as shown in Figure Q1b. Let an clectric current pass through the plane wall causing a uniform heat generation, Se, unit of heat per unit volume per unit time. The two surfaces are maintained at temperature T1 and T2 (FigQIb). One-dimensional steady state heat conduction occurs in the x-direction which is normal to the wall. The point x=0 is positioned on the left surface of the wall. Assume steady state and one-dimensional heat conduction and assume uniform volumetric beat generation ( S e per unit volume) within the solid and thermal conductivity k is constant. Fourier's Law of heat conduction, qx=kAdxdT Formulate the following differential equation describing the temperature (I) distribution by making an energy shell balance on an element of thickness dx at a distance x and cross-sectional area, A. (Given steady-state heat balance here: Input - Output +Generation =0 ) dx2d2T+kSc=0 [7 Marks] Figure Q1b: An elementary volume in a plane wall