A large plane wall of thickness 2 L. experiences uniform heat generation (Fig. 2-62). Determine the expression for the variation of temperature within the wall, if (a) T₁ > T₂ and (b) T₁ = 7₂. Plane wall +L
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- We performed the experiment to measure the thermal conductivity of 2 materials (Brass & Steel) in the laboratory and measured the following tabulated values: Material 1 - BRASS (Diameter = 25mm) Power Temperature (°C) Q' (W) 2 3 4 6 7 8 1 5 9 14.6 78.9 77.5 76 50.2 46.7 42.4 36.1 34.6 33.6 Material 2 - STEEL (Diameter = 25mm) Power Temperature (°C) 7 Q' (W) 14.25 2 3 1 9 88.6 87.4 85 34.1 33.4 32.7 CALCULATE THE FOLLOWING: MATERIAL 1 - BRASS Calculation for Brass Quantities Calculated Values Power (Q') W Area of cross section (A) m2 Difference in Temperature between two points (AT) °C Difference in distance between two points (Ax) m Thermal conductivity of brass (k,) W/m'C MATERIAL 2 - STEEL Calculation for Steel Quantities Calculated Values Power (Q') W Area of cross section (A) m? Difference in Temperature between two points (AT) "C Difference in distance between two points (Ax) m Thermal conductivity of steel (k,) W/m°CQ1 Passage of an electric current through a long conducting rod of radius r; and thermal conductivity k, results in uniform volumetric heating at a rate of ġ. The conduct- ing rod is wrapped in an electrically nonconducting cladding material of outer radius r, and thermal conduc- tivity k, and convection cooling is provided by an adjoining fluid. Conducting rod, ġ, k, 11 To Čladding, ke For steady-state conditions, write appropriate forms of the heat equations for the rod and cladding. Express ap- propriate boundary conditions for the solution of these equations.Consider a wall of thickness 50 mm and thermal conductivity 14 W/m.K, the left side (x-0) is insulated. Heat generation (q) is present within the wall and the one dimensional steady-state temperature distribution is given by T(x) = ax +bxtc [°C] , where c 200 °C, a = -1285 °C/m , b=needs to he determined, andx is in meters. What is the heat fluxes at the right side, x = L, (kW/m )? 9, K 4L) Insulation
- A plane wall is insulated on its left side (? = 0). The wall generates energy uniformly at arate of ?̇ [W m 3⁄ ] and has thermal conductivity ?. On its right side (? = ?), the wall is exposedto a fluid at temperature ?" with convection coefficient ℎ.a) Draw a schematic of this plane wall. Make the schematic large enough that you cansketch the temperature profile within the wall after solving for it in later steps.b) Write out the full form of the Heat Diffusion Equation (HDE) in the appropriatecoordinate system for this physical scenario. Simplify the HDE and write out theappropriate boundary conditions in their general form (e.g., ?| #$% = ?& ).c) Derive an expression for the steady-state temperature distribution ?(?) within thewall. You may start from the general solution provided in Appendix C of the Bergmantextbook; or you may derive the solution directly from the differential equation andthe boundary conditions. Annotate your schematic by sketching the temperatureprofile…The wall (thickness L) of a furnace is comprised of brick material (thermal conductivity, k = 0.2 Wm¯' K'). Given that the atmospheric temperature is 0°C at both sides of wall, the density (p) and heat capacity (c) of the brick material are 1.6 gm cm³ and 5.0 J kg K¯l respectively. du Solve pc = k- subject to initial conditions as u(x,0) = x²(L – x). ốt Consider the case 2=- p² only.The initial temperature distribution of a 5 cm long stick is given by the following function. The circumference of the rod in question is completely insulated, but both ends are kept at a temperature of 0 °C. Obtain the heat conduction along the rod as a function of time and position ? (x = 1.752 cm²/s for the bar in question) 100 A) T(x1) = 1 Sin ().e(-1,752 (³¹)+(sin().e (-1,752 (²) ₁ + 1 3π TC3 .....) 100 t + ··· ....... 13) T(x,t) = 200 Sin ().e(-1,752 (²t) + (sin (3). e (-1,752 (7) ²) t B) 3/3 t + …............) C) T(x.t) = 200 Sin ().e(-1,752 (²t) (sin().e(-1,752 (7) ²) t – D) T(x,t) = 200 Sin ().e(-1,752 (²)-(sin().e (-1,752 (²7) ²) t E) T(x.t)=(Sin().e(-1,752 (²t)-(sin().e(-1,752 (²) t+ t + ··· .........) t +.... t + ··· .........) …..)
- A piece of beef steak 7 cm thick will be frozen in the freezer room -40 ° C. the product has a moisture content of 73%, a density of 970 kg / m cubic, and a thermal conductivity (frozen) of 1.1 W / (mK). Estimate the freezing time using the Plank equation. This product has an initial freezing temperature of -1.75 ° C, and the movement of air in the freezing room gives a convective transfer coefficient of 10 W / (m squared K)You have smooth slabs of Titanium (Ti), polycrystalline silica (SiO2), and atactic PS. Rank them from highest to lowest: Thermal conductivity (k) Heat capacity (Cp) Energy band gap (Eg) Thermal expansion coefficient (a) Transparency ResistivityBoth ends of a 32 cm long rod are maintained a constant temperature of 100 °C. Dimensions and thermal parameters of the rod are as follows: Diameter D = 2 cm Convection coefficient h = 10 W/m2K Ambient temperature T¥ = 20 °C Thermal conductivity k = 10 W/mK What is the midpoint temperature of the rod?
- (a) Consider nodal configuration shown below. (a) Derive the finite-difference equations under steady-state conditions if the boundary is insulated. (b) Find the value of Tm,n if you know that Tm, n+1= 12 °C, Tm, n-1 = 8 °C, Tm-1, n = 10 °C, Ax = Ay = 10 mm, and k = = W 3 m. k . Ay m-1, n m, n | Δx=" m, n+1 m, n-1 The side insulatedThermal diffusivity is a(A) Function of temperature(B) Physical property of a substance(C) Dimensionless parameter(D) All of these2. The slab shown is embedded in insulating materials on five sides, while the front face experiences convection off its face. Heat is generated inside the material by an exothermic reaction equal to 1.0 kW/m'. The thermal conductivity of the slab is 0.2 W/mk. a. Simplify the heat conduction equation and integrate the resulting ID steady form of to find the temperature distribution of the slab, T(x). b. Present the temperature of the front and back faces of the slab. n-20- 10 cm IT- 25°C) 100 cm 100 cm