The below shown graph is drawn from the tabulated values of steel which we measured during the experiment of thermal conductivity: (Consider the value of heater power (Q') and the area of cross section (A) of the material from the tabulated values)
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- Material 1- BRASS The below shown graph is drawn from the tabulated values of steel which we measured during the experiment of thermal (Diameter = 25mm) conductivity: (Consider the value of heater power (Q') and the area of cross section (A) of the material from the tabulated values) Power X-axis 1 unit =1 cm y-axis 1 unit = 10°C Temperature (°C) 50- Q' 1 2 3 4 5 6 7 8 (W) 14.65 79 77.4 76 50.1 46.5 42.4 36 34.5 : 20- 10- Distance X in cm 123 10 Material 2 - STEEL Calculate the following: Thermal conductivity of (Diameter = 25mm) steel Calculated Value quantities Power Difference in temperature (AT) C° (with sign) Difference in distance Temperature (°C) between test points (Ax) cm Slope from the graph °C/m (with sign) Q' 1 (W) 2 3 7 8 Thermal conductivity of steel ksb W/m°C 14.2 88.6 87.5 85 33.9 33.6 32.4CALCULATE 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°CThe following table shows data collected to measure the thermal conductivity of steel specimen of 30 mm diameter: Specimen Length (mm) t1 (C) Heater Heater Qw t lw ( C) t 2w ( C) H K (V) (A) (W/m. ( C) K) (W) (kg/s) (W) Steel 60 100 0.65 20 23 65 49 i. Use the data provided above to complete the missing information in the table. (use V=0.5 L , t=123s) ii. Plot the variation of temperature with distance using the data presented in the table. Explain the relationship you obtain. Additional information: = 998 kg/m3 p water Cp = 4180 J/Kg.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 ResistivityA 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]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°C
- Complete the following table for refrigerant-134a. Show your work and explain how you got your values. T (oF) P psia) h (Btu/Ibm) X Phase description 80 78 15 0.6 10 70 180 129.46 110 1.0Completely 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?The below shown graph is drawn from the tabulated values of steel which we measured during the experiment of thermal conductivity: (Consider the value of heater power (Q') and the area of cross section (A) of the material from the tabulated values) X-axis 1 unit = 1 cm y-axis 1 unit = 10°C 80- 70- 60- 50- 40- 30- 20- 10- 6. Distance X in cm 4. 5. 8. 10 Temperature T degree Celsius
- The 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% 0.093Jes 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 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.