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ChemMaterials Analysis

by: Landen Tromp

ChemMaterials Analysis ECM 005

Landen Tromp
GPA 3.8

Bruce Gates

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Bruce Gates
Class Notes
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This 5 page Class Notes was uploaded by Landen Tromp on Tuesday September 8, 2015. The Class Notes belongs to ECM 005 at University of California - Davis taught by Bruce Gates in Fall. Since its upload, it has received 43 views. For similar materials see /class/187478/ecm-005-university-of-california-davis in Engineering Chemical Materials at University of California - Davis.

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Date Created: 09/08/15
ECM 5 Problem Set 2 Solutions Prepared by Joey Kistler 1122013 1 Total height of water above orifice height of the water above the base of the tank read from scale distance between the narrowest point of the orifice and the base of the tank a Sample Data for Tank 2 Tank Height 70 cm Tank Extension Height 104 cm Plate Height 11 cm Total Height 70 104 11 815 cm Height cm Trial 1 Time s Height cm Trial 1 Time s 815 0 475 5810 795 314 455 6215 775 606 435 6645 755 921 415 7061 735 1237 395 7500 715 1548 375 7942 695 1874 355 8434 675 2191 335 8895 655 2530 315 9414 635 2886 295 9879 615 3207 275 10421 595 3560 255 10987 575 3930 235 11582 555 4289 215 12181 535 4637 195 12825 515 5023 175 13548 495 5397 155 14327 Tank 2 Trial 1 Data Total Water Height cm l N w Jgt Ln m l 00 w o o o o o o o o o o 0 2000 4000 6000 8000 10000 12000 14000 Time s 2 I calculated the AhAt for 10time intervals as mentioned in the problem statement Because the entire set of data took almost 4 hours to collect I chose the time intervals to span the entire data set Mytime intervals are Oto 10 min Oto 25 min Oto 50 min Oto 75 min Oto 100 min Oto 125 min Oto 150 min Oto 175 min Oto 200 min Oto 240 min Sample Calculations ForOto 10 min used points 0 815 and 606 775 AhAt 775 815606 0 00066 cms MM 303 s Oto 25 min 0815 and 1548 715 AhAt 715 8151548 0 00065 cms Atm 774 s Oto 50 min 0815 and 2886 635 AhAt 635 8152886 0 00062 cms Atm 1443 s Oto 100 min 0815 and 6215 455 AhAt 455 8156215 0 00058 cms Atm 3108 s Oto 175 min 0815 and 10421 275 AhAt 275 81510421 0 00052 cms Atm 5211 s Oto 240 min 0815 and 14327 155 AhAt 155 81514327 0 00046 cms Atm 7164 s 3 Graph the average rate of change of h as a function of the average time Below are the data calculated in response to question 2 70001 70002 70003 70004 AhAt cms 70005 70006 9 70007 0 1000 2000 3000 4000 5000 6000 7000 8000 MM 5 This plot shows howthe value of AhAt approaches a limitingvalue as At approaches zero this is of course dhdtevaluated atthe initial height 4 To estimate the average rate of change of h at each ofthe 10 different times we need to estimate dhdt at each time we approximate this as AhAt choosing a small enough increment in time to make the approximation a good one A good way to estimate this is by using two data points that create an interval where this height is the average height ofthe interval The smaller the interval the better the estimate will be So I will choose 1 point before and after the desired height to use for the interval Sample Calculation For a height of 775 cm I use the points 760 765 and 461 785 AhAt 765 785760 461 00067 cms For a height of 715 cm I use the points 1712 705 and 1392 725 AhAt 705 7251712 1392 00063 cms 70001 70002 70003 70004 AhAt cms 70005 70006 70007 70008 h cm Note in this graph that all the values of AhAt are negative fwe wanted to represent the average rate of fall of the liquid level we would instead plot iAhAt these values are positive 5 To estimate the average rate of change of h at certain heights I have used very small intervals of time As these time intervals get smaller and smaller my estimate will approach the exact derivative dhdt at the particular value of h By doing this we recognize what is meant by the derivative of the function as u 111 l39iiiim it r it This states that as At becomes smaller and smaller and approaches 0 the value of AhAt approaches the derivative of the function dhdt described below


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