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 +bx+c [°CJ, where c = 200 °C, a = -1144 °C/m is the heat fluxes at the right side, x L, (kW/m)? ,b=needs to he determined, and x is in meters. What 9, K 4L) Insulation
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- Find the two-dimensional temperature distribution T(x,y) and midplane temperature T(B/2,W/2) under steady state condition. The density, conductivity and specific heat of the material are p= 62400 kg/m', k-400 W/m.K, and cp=2500 J/kg.K, respectively. A uniform heat flux q%=1000 W/m² is applied to 2. the upper surface. The right and left surfaces are also kept at 0°C. Bottom surface is insulated. 9% (W/m³) y4 T = 0 °C T = 0°C W= 520 cm B=1560 cmThe steady-state temperature distribution in a one-dimensional wall of thermal conductivity k=83 (W/m) °C) and thickness 300 mm is observed to be T(°C) = ax2+bx+c, where a = -2500 °C/m2 , b = 500 °C/m, c = 250 °C and x is in meters. a) What is the heat generation rate ?̇ (W/m3 ) in the wall ? b) Find the maximum and minimum temperatures in the wall. c) Find the amount of heat transferred to the right side of the wall (for 1 m2 surface area.)Find the two-dimensional temperature distribution T(x,y) and midplane temperature T(B/2,W/2) under steady state condition. The density, conductivity and specific heat of the material are p=(1200*32)kg/mº, k=400 W/m.K, and cp=2500 J/kg.K, respectively. A uniform heat flux 9" =1000 W/m² is applied to the upper surface. The right and left surfaces are also kept at 0°C. Bottom surface is insulated. 9" (W/m) T=0°C T=0°C W=(10*32)cm B=(30*32)cm
- a. A slab of thermal insulator is 100 cm² in cross section and 2 cm thick. Its thermal conductivity is 2.4 x 10 -+ cal/(s cm C"). If the temperature difference between opposite faces is 180 F', how much heat flows through the slab in one day?Find the two-dimensional temperature distribution T(r,z) under steady state condition. Where, To=20 oC, TL=8200 oC. The density, conductivity and specific heat of the material are ρ =800 kg/m3, k=200 W/m.K, and cp=2500 J/kg.K, respectively. Also, r1=820 cm and L=1640 cm.Find the two-dimensional temperature distribution T(x,y) and midplane temperature T(B/2,W/2) under steady state condition. The density, conductivity and specific heat of the material are ρ =1200 kg/m 3, k=400 W/m.K, and cp=2500 J/kg.K, respectively. A uniform heat flux q =1000 W/m 2 is applied to the upper surface. The right and left surfaces are also kept at 0oC. Bottom surface is insulated.
- 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°CQi: (50 marks) Find the total heat flux of the composite wall when: B KA = KC = KF = 15 m. K KB = KD = 10 m. K KE = KG = 20 %3D m. K D. Height of B = C = D 4 cm 3 cm 4 cm 6 cm Height of F = G AT = 30 KA finned surface has been added to cool an electronic part. The surface temperature of the electronic part in contact with the fins is 60 ° C, the convection coefficient between the environment and the finned surface is 40 W / m².K and the heat conduction coefficient for the fin material (aluminum) is 180 w / m.K. a. If the electronic part has wings or no blades, the amount of heat per unit time (W) thrown into the environment, b. Find out wing effectiveness and efficiency?
- 3. Find the two-dimensional temperature distribution T(r,z) under steady state condition. Where, To=20 °C, TL=5200 °C. The density, conductivity and specific heat of the material are p 800 kg/m³, k=200 W/m.K, and cp=2500 J/kg.K, respectively. Also, r= 520 cm and L=1040cm. To TL L Tounder steady-state conditions. If you are given T1 = 200 °C and T2 = 164 °C, determine: a) the conduction heat flux, q,.cond, in m2 W from x = 0 to x = L b) if the dimensions of the triangle ares 15 mm and h 13 mm, calculate the heat transfer due to convection, q,y, in W at x = L Finsulation T2 T T = 20°C h = 500 W/m2.K Triangular Prism x L x 0 L= 50 mm k = 100 W/m-KConduction Heat Transfer X material (a) material (b) . both (a) and (b) have the same thermal conductivity the temperature distribution is independent of thermal conductivity it's not that simple 9. Fin efficiency is defined as: • tanh (mL) (hP/k Ac)1/2 (heat transfer with fin) / (heat transfer without fin) (actual heat transfer through fin) / (heat transfer assuming all fin is at T = Tb) (Tx=L-Tf)/(Tb-Tf) 10. For an infinite fin, the temperature distribution is given by: (T-Tf)/(Tb-Tf)= e-mx. The heat flow through the fin is therefore given by: k (Tb-Tf)/L ● zero, because the fin is infinite ● infinite because the fin is infinite ● (Tb-Tf) (hP/k Ac)1/2 ● (Tb – Tf) (h P / k Ac)1/2 tanh (mL) 11. The Biot number, Bi, is defined as: • Bi=hk/L • Bi=hL/k • Bi= k/LH • Bi=qL/k • Bi=p UL/k 12. For a plate of length L, thickness, t, and width, W, subjected to convection on the two faces of area L x W. What is the correct length scale for use in the Biot number? . L ● W ● t • t/2 • L/2 13. If Bi…