Lab1_COLLINS

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Dec 6, 2023

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CE343 LAB01 Elijah Collins Exer. 1 1. From the measurements of D in Table 1.2, determine, using a spreadsheet package, D and eD for a 95% confidence level, and express your result in the form, D = D ± eD. Answer: Dbar = 3.218125 cm E D =t sd / √ N E d =2.45/ √7 = 0.926 cm D = 3.22 ± 0.926 cm 2. If the manufacturer’s specification were that D = 3.5 cm, is there any (statistical) reason to doubt this specification? Answer: No, while the average is below the manufacturer’s specifications, the allowable error is within the range that has been specified by the manufacturer.

We can also say that it is statistically insignificant because Dbar-Dtrue < e d 3. Determine eW,D for the tabulated data (assume that ρg = 50 kN/m3). e W,D = 50,000(pi/6[3*3.22^2*0.93) = 252.13 N 4. Show how Eqn. 1.6 was obtained. (assume that ρ= 5 g/cm3, eρ = 0.1 g/cm3, and g = 980 cm/s2). the uncertainty in W , e W,x due to a specific independent variable, x, is simply the product of the partial derivative of W with respect to x and the uncertainty in x. Thus, eW,ρ = (∂W/∂ρ)eρ = [(π/6)gD3]eρ Determine eW,total for the tabulated data: eW total = 73,800 g*cm/s^2 work is shown below

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Exer. 2 Inspect the laboratory pipe and open-channel flow systems shown in Figs. 1.1 and 1.2 and become familiar with the various features. Other features not discussed above may also be pointed out. Examine another pipe system (e.g., that attached to the wall) and another open-

channel system (e.g., the smaller system beside that in Fig. 1.2), and identify similar features as those discussed above. What features in these other systems can you identify that are similar to those in the systems in Figs. 1.1 and 1.2. Answer: The pipe system situated in the basement of a building mirrors the features depicted in Figures 1.1 and 1.2. This crucial network comprises components such as strategically placed pipe bends, pumps for pressure enhancement akin to those in Figure 1.1, and flange joints facilitating maintenance, resembling the examples shown. Flow control is achieved through valves similar to those in both figures, while drainpipes prevent flooding similar to the depicted drain pipe. Bleed valves release air pockets, optimizing flow like their counterparts, and reservoirs, as in both figures, ensure a steady supply. This alignment underscores the essential nature of these components in efficiently delivering water, heating, and other fluids throughout the building's infrastructure, maintaining functionality and comfort. Exer. 3 Plot the measured values of W and D in Table 1.3 with linear scales. Design your plot with the tips of Sec. 1.4.1 in mind; in particular, use only symbols for the data points. 0 10 20 30 40 50 60 70 80 90 0 0.05 0.1 0.15 0.2 Weight (N) Diameter (M) Weight Against Diameter

Exer. 4 Plot W against D3 on linear scales. Exer. 5 Using linear regression, determine the slope and intercept in the relation between W and D3. Draw the fitted line through the data points in the plots W vs D and W vs D3 (the fitted lines should not contain any symbols.) W vs D^3 = y = 12650x + 0.9949 y = 12650x + 0.9949 0 10 20 30 40 50 60 70 80 90 0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 Weight (N) Diameter^3 (m^3) Weight against Diameter^3 y = 12650x + 0.9949 0 10 20 30 40 50 60 70 80 90 0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 Weight (N) Diameter^3 (m^3) Weight against Diameter^3

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Exer. 6 Using linear regression, determine the slope and intercept, this time forcing the best-fit line through the origin and compare it with your previous results. Redraw your plots with the revised fitted line. y=12869x It changed the slope and got rid of the intercept. y = 596.91x - 37.049 -20 -10 0 10 20 30 40 50 60 70 80 90 0 0.05 0.1 0.15 0.2 Weight (N) Diameter (M) Weight Against Diameter y = 285.33x 0 20 40 60 80 100 0 0.05 0.1 0.15 0.2 Weight (N) Diameter (m) Weight Against Diameter, linear scales, forced intercept at 0

Exer. 7 1. Plot the given data in W vs D on log-log scales by hand with the log-log graph paper provided, and fit a straight line through the data points by eye, and determine the ‘slope’ and ‘intercept’. Note that determining manually the slope and intercept of a log-log plot will require transformation of values. See plot below: The slope appears to be about 3 and the intercept should be around 4 as well 2. Plot W vsD on log-log scales using the computer. Do a linear regression analysis of the transformed data (depending on the method chosen, you may need to transform explicitly your variables by first taking logarithms). (If you use trendlines in charts for fitting pur- poses, be careful about the type of trendline used). Is the ‘slope’ what you expected? Draw the fitted line through the data points on the computer. The slope appears to be as expected. We assume it will be the power that our diameter is raised to. y = 12869x 0 10 20 30 40 50 60 70 80 90 0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 Weight (N) Diameter^3 (m^3) Weight Against Diameter^3, with forced intercept at 0

3. Assuming that the acceleration due to gravity, g = 9.8 m/s, what is the density of the sphere material from your regression results? What would be your estimate of the weight of sphere of the same material, but with a diameter of 0.1 m? How about a sphere of diameter 0.25 m? Answer: Density based on the regression analysis would be 2630 kg/m^3 0.1 m would be about 13.5 N and 0.25 N would be 211 N but we have no data in that range so this would be unwise to extrapolate data that is outside of our given data set. Density – Mass/ pi/6 D^3 y = 2.9573x + 4.0933 0 0.5 1 1.5 2 2.5 -1.4 -1.2 -1 -0.8 -0.6 -0.4 -0.2 0 Log Weight (N) Log Diameter (m) Weight log, W, (N) vs Diameter Log D (m)

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