1.21 The thermal conductivity in a particular plane wall depends as follows on the wall temperature, T (°C): k = A + BT, where A and B are constants. The temperatures are T₁ and T2 on either side if the wall, and its thickness is L. By integration of Fourier's law, develop an expression for q. At what temperature should k be evaluated to avoid the need for integration?
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- 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?A house wall is composed of 3 layers in series to each other. There is an Aluminum layer that is 4 cm thick, Copper layer that is 6 cm think and Gold layer that is 3 cm thick. What is the total conductivity of the wall and its R-value in units of ft2.F.hr/Btu? [KAl = 2.37 W/cm.C, KCu = 4.01 W/cm.C, KAu = 3.17 W/cm.C]Give True or False for the following: 1.In liquids and gases, heat transmission is caused by conduction and convection 2.The surface geometry is the important factor in convection heat transfer 3. The heat transfer by conduction from heated surface to the adjacent layer of fluid, 4. The heat transfer is increased in the fin when &> 1 5.The unit of the thermal diffusivity is m²/s 6. Temperature change between the materials interfaces is attributed to the thermal contact resistance 7. A material that has a low heat capacity will have a large thermal diffusivity. 8. Heat conduction flowing from one side to other depends directly on thickness 9.Fin efficiency is the ratio of the fin heat dissipation with that of no fin 10.The critical radius is represented the ratio of the convicted heat transfer to the thermal conductivity
- Conduction 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…CALCULATE THE FOLLOWING: We performed the experiment to measure MATERIAL 1 - BRASS the thermal conductivity of 2 materials (Brass & Steel) in the laboratory and measured the following tabulated values: Calculation for Brass Quantities Values Material 1- BRASS Calculated Power (Q') W (Diameter = 25mm) Area of cross section (A) m2 Power Difference in Temperature Temperature (°C) between two points (AT) °C Difference in distance between two points (Ax) m Q' 1 2 3 (W) Thermal conductivity of brass (kp) W/m°C 4 5 6 8 9. 14.65 79 77.4 76 50.1 46.5 42.4 36 34.5 33.6 MATERIAL 2 - STEEL Calculation for Steel Material 2 - STEEL %3D Quantities Values Calculated (Diameter = 25mm) %3D Power (Q') W Power Area of cross section (A) m2 Temperature (°C) Difference in Temperature between two points (AT) °C Q' 1 |(W) Difference in distance between two points (Ax) m 2 3 7 8 14.2 88.6 87.5 85 33.9 33.6 32.4 Thermal conductivity of steel (kg) W/m°CCompletely solve and box the final answer. Write legibly 1. How much heat will be transferred in KW if a hot gas at 175ºC crosses a wall 100-mm thick and cross-sectional dimensions of 25 cm x 52 cm. It leaves the wall at 52ºC. The thermal conductivity of the wall is 12.4 W/m-ºK?
- Jes 1:01 i docs.google.com/forms what is the Physicists recognize four fundamental forces.* Flectrical force O - Gravitational force -Strong nuclear force Weak nuclear force O -Muscular force Force of friction Ff is described by • Ff = µN * O p synovlal fluld in the Joints P -coefficient of fraction between two surface in The temperature of the human body is normally about 98.6°F.calculate the temperature of the body "C* oC = (5/9)x(oF-32) =5/9(98.6- 32)-37 oc oc=(9/5)x(of -32) =9/5(98,6-32) =13,3 Under resting conditions the body energy is being used as follows * O 27% by the liver and splee O 20-%by the skeletal musules Under resting conditions the body energy is being used as follows. O 19% by the brain. 15% by the kidney. We can write the first law of thermodynamics as: * O AU-AQ -A.. O AU-AQ +AW. 1kcal = J O 1 Kcal =4184j. O O O OThe left side of this equation tells how much energy Q the cylinder gives to the water while it cools. The right side of this equation tells how much energy Q the water and aluminum cup absorb from the cylinder to warm up. Because it is the same energy, they are equal. What is known in this equation? Mcyl 411.7 g, malum 46.5 g, malum+water = 175 g Can you find: mwater =? g Twater = Talum = 20°C (water and cup of room temperature) 90°C, T; = 35°C (hot cylinder and cool "cylinder+cup+water" temperatures) Tcyl kCal Calum = 0.22, Cwater 1 (specific heat of water and aluminum, measured in units kg-°C What are we looking for is Ccul - How we find it? Plug all the numbers into the equation (1), Ccul will be one unknown which you can calculate from the equation. Important, convert all the masses from grams to kilograms! After you find Ccyl, compare it to known value for the copper 0.093(our cylinder is made out of copper). |Ceyl -0.093| % : · 100% 0.093The amount of heat conducted through a wall of length r is given by Fourier's Law:. CONDUCTION RATE EQUATION T FOURIER'S LAW q, = -k A dT dx T, >T, where q, is the heat flux, k is a proportionality factor, Ais the wall's cross-sectional area, and 4 is the temperature gradient throughout the wall. Our friend Matt Labb wants to find T2 (temperature of the wall's rightmost edge) given q,, k, and A. Is this possible? If so, briefly explain how to find Tp. If not, briefly explain why.
- Let's assume that the outdoor temperature in your region was 1 C on 26.12.2002. Let's assume that you use a 2088 W heater in the room in order to keep the indoor temperature of the room at 20 ° C. In the meantime, a 68 W light bulb for lighting, a computer you use to solve this question and load it into the system (let's assume it consumes 217 W of energy), you and your two friends (three people in total) are in the room to assist you in solving the questions. A person radiates 45 J of heat per second to his environment. When you consider all these conditions, calculate the exergy destruction caused by the heat loss from the exterior wall of your room.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°CA plane wall is insulated on its left side (x = 0). The wall generates energy uniformly at a rate of q [W/m³] and has thermal conductivity k. On its right side (x = L), the wall is exposed to a fluid at temperature Too with convection coefficient h. b) a) Draw a schematic of this plane wall. Make the schematic large enough that you can sketch the temperature profile within the wall after solving for it in later steps. Write out the full form of the Heat Diffusion Equation (HDE) in the appropriate coordinate system for this physical scenario. Simplify the HDE and write out the appropriate boundary conditions in their general form (e.g., Tlx-o = T₁). c) Derive an expression for the steady-state temperature distribution 7(x) within the wall. You may start from the general solution provided in Appendix C of the Bergman textbook; or you may derive the solution directly from the differential equation and the boundary conditions. Annotate your schematic by sketching the temperature profile…