Figure Q2 shows discrete locations of sampled soil electrical conductivity (dS/m). Estimate the magnitude of electrical conductivity at points P, R and T using inverse distance weighting 24 24 Estimate the accuracy of the estimation based on R² if the actual values at P is 5.2 dS/m, at 11 dS/m, and 6.9 dS/m at R 21 12 18 15 12 11 9 6 3 3 1.5 23 R 0 3 6 9 50 17 T 12 12 15 18 Figure Q2 24 21 24
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- 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°CGive 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 conductivityi. Derive the linear and quatratic approximation of the below resistance temperature readings(temperature from 40 °c to 80 °c). (. ii. find out the resistance of RTD at a temperature of 200°c using linear approximation? ( Temperature 40 45 50 55 60[TO] 65 70 75 80 Resistance ohm 115.239 120.521 167.096 254.966 384.13 554.587 766.34 1019.38 1313.724
- In the Fig. 2 below, let Ki = K2 = K and ti = t=t. %3D T -T X Fig. 2 (a) Let T= 0 °C and T= 200 °C. Solve for T: and unknown rates of heat flow in term of k and t. MEC_AMO_TEM_035_02 Page 2 of 11 Finite Element Analysis (MECH 0016.1) – Spring - 2021 -Assignment 2-QP (b) Let T- 400 °C and let fs have the prescribed value f. What are the unknowns? Solve for them in term of K, t, and f.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°Cexplains how the two methods of parameter estimation, namely the method of parameter estimation and the method of moments and percentile matching, are used to fit distributions to data for actuarial calculations.
- O Estimate the MRR (in cc/sec) of an alloy containing 18% cobalt, 62% nickel and 20% chromium during ECM with current of 500 ampere. The density of the alloy is 8.28 g/cc. The following data is available. Metal Gram atomic weight Velency Cobalt 58.93 Nickel 58.71 2 Chromium 51.99 Assume Faraday's constant is 96500 coulombs/ mole.The stress profile shown below is applied to six different biological materials: Log Time (s] The mechanical behavior of each of the materials can be modeled as a Voigt body. In response to o,= 20 Pa applied to each of the six materials, the following responses are obtained: 2 of Maferial 6 Material 5 0.12 0.10 Material 4 0.08 Material 3 0.06 0.04 Material 2 0.02 Material 1 (a) Which of the materials has the highest Young's Modulus (E)? Why? Log Time (s) (b) Using strain value of 0.06, estimate the coefficient of viscosity (n) for Material 6. Stress (kPa) StrainAn experimental data set fits the equation: y- yo : = 1 - e 1-e Yoo - Yo shown in the table and also shown graphically. Using this data, determine the time constant, t. Show your work and justify your answer. 125 0.00 25.00 0.50 47.12 105 2.00 88.21 85 4.00 111.47 5.25 117.76 65 7.00 121.98 8.00 123.17 45 9.50 124.13 25 6 10.00 124.33 0.00 2.00 4.00 6.00 8.00 10.00 12.00 14.00 15.00 124.94
- 45 TV $55 D30 215 4-018 R10 15 CHO ED FV 50 21 RSV 41 425 100 142 21 5. 11 85The heat transfer conducted through material is calculated from the equation: Q = KX AXTD/L Where K: Conductivity of material A: Area of heat transfer TD: Temperature difference across material L: Thickness of material A student measures the area, thickness and temperature difference and assumes that the error in conductivity is negligible. The student also estimates the uncertainty range for each variable. In estimating the maximum possible value of Q, the student should use the following formula: A B Q max= K x A max x TD max / L max Q max= K x A max x TD max / L nom Q max= Q nominal + dQ/dLmin Q max= K x A max x TD max / L min100 80 60 40 20 0.002 0.004 0.006 0.008 0.01 0.012 Strain, in/in. FIGURE P1.17 1.18 Use Problem 1.17 to graphically determine the following: a. Modulus of resilience b. Toughness Hint: The toughness (u) can be determined by calculating the area under the stress-strain curve u = de where & is the strain at fracture. The preceding integral can be approxi- mated numerically by using a trapezoidal integration technique: u, = Eu, = o, + o e, - 6) %3D c. If the specimen is loaded to 40 ksi only and the lateral strain was found to be -0.00057 in./in., what is Poisson's ratio of this metal? d. If the specimen is loaded to 70 ksi only and then unloaded, what is the permanent strain? Stress, ksi