科技与专业英语期末考试A标准答案及评分标准

试卷类型 A 2009 年 12 月 30 日 苏州科技学院试题标准答案及评分标准 课课程名称 科技与程名称 科技与专业专业英英语语 使用专业给排水工程命题教师陈新课程所在教研室给排水教研室 STANDARDS 25 SCORES OF EVERY STEP a to g TOP SCORE IS 100 IF GOT MORE THAN 100 a DETERMINE the effective size and the uniformity coefficient of the sand Plot the cumulative percent passing versus the corresponding sieve size List the cumulative percent passing versus the corresponding sieve size in the following table Cumulative weight passing US sieve size Designation Size of opening mm geometric mean size mm dg 2 ABCDE 1400 1050 41 50 15 00 2 1000 1490 1251 641 54 10 811 11 6 700 2100 1771 324 010 02 520 05 0 500 2970 2501 169 521 08 232 016 4 400 4200 3531 818 540 618 549 537 0 300 5900 4981 431 061 032 062 354 0 200 8400 7041 249 078 358 178 368 0 161 1901 000163 290 076 388 549 0 121 6801 414282 896 090 094 492 0 82 3802 000489 099 096 797 898 0 63 3602 828898 099 999 099 099 0 44 7604 00016100100100100100 d10 d15 9 d50 and d60 read from the graphs and list them in the following table The uniformity coefficient UC and the geometric standard deviation are calculated and list in the following g table The geometric mean size is calculated and list in the above table 6050 1015 9 g dd UC dd ABCDE d100 310 210 320 140 26 d15 90 380 270 390 180 29 d500 850 480 760 430 53 d601 10 580 860 550 68 UC3 552 762 73 92 6 g 2 241 781 952 391 83 第1页共7页 0 10 20 30 40 50 60 70 80 90 100 0 1110 Size of opening mm Cumulative weight passing 系列1系列2系列3系列4系列5 第2页共7页 b The amount of stock sand needed to obtain one ton of filter sand is listed in the following table Cumulative weight passing Size of opening mm ABCDE 0 1050 41 50 15 00 2 0 1491 54 10 811 11 6 0 2104 010 02 520 05 0 0 2979 521 08 232 016 4 0 42018 540 618 549 537 0 0 59031 061 032 062 354 0 0 84049 078 358 178 368 0 1 19063 290 076 388 549 0 1 68082 8 78 8 96 0 86 0 90 0 87 5 94 4 74 4 92 0 87 0 2 38089 099 096 797 898 0 3 36098 099 999 099 099 0 4 760100100100100100 depth1 271 161 141 351 15 c The US standard sieve size is 12 d Determine the velocity using Stokes law 22 2 2 62 6 3 2 3 1818 9 81 2 65 1 18 1 003 10 2 Check the Reynolds nu 0 210 10 0 0395 0 03950 2 mber 0 85 1 1 003 10 10 10 7 03 PWpPWp p t W pp R gdgd v m sm m s ms m smv d N ms The use of Stokes law is not appropriate for Reynolds number 1 0 Therefore Newton s law must be used to determine the velocity in the transition region The drag coefficient term in Newton s equation is dependent on the Reynolds number which is a function of the settling velocity Because the settling velocity is not known an initial velocity must be assumed The assumed velocity is used to compute the Reynolds number which is used to determine the drag coefficient which is used in the Newton equation to calculate the settling velocity A solution is achieved when the initial assumed settling velocity is approxi mately equal to the settling velocity resulting from Newton s equation The solution process is iterative as illustrated below For the first assumed velocity se the Stokes law settling velocity calculated above Using the resulting Reynolds number also determined previously compute the drag coefficient 第3页共7页 7 243243 0 34 8 037 03 0 3484 d R R C NN Use the drag coefficient in Newton equation to determine the particle settling velocity 23 4 9 81 2 65 1 0 0 03050 039 210 10 4 33 5 4 8 8 PW p tp dw m sm g vd C m sm s Because the initial assumed velocity does not equal the velocity calculated by Newton equation iteration is necessary For the second iteration assume a settling velocity value of 0 0250 m s and calculate the Reynolds number Use the Reynolds number to determine the drag coefficient and use the drag coefficient in Newton equation to find the settling velocity 62 23 3 0 02500 210 10 4 449 4 4494 449 243243 0 340 34 4 9 81 2 65 1 0 210 10 4 0 85 33 1 1 0 02520 003 10 7 156 7 1 0 5 250 6 R R PW p tp w pp R d d NN m sm g vd m s m smv d N ms C C m s For the 3rd iteration assume a settling velocity value of 0 0260 m s and calculate the Reynolds number Use the Reynolds number to determine the drag coefficient and use the drag coefficient in Newton equation to find the settling velocity 62 23 3 0 02600 210 10 4 627 4 6274 627 243243 0 340 34 4 9 81 2 65 1 0 210 10 4 0 85 33 1 1 0 02560 003 10 6 922 6 9 0 2 260 2 R R PW p tp w pp R d d NN m sm g vd m s m smv d N ms C C m s For the 4th iteration assume a settling velocity value of 0 0253 m s and calculate the Reynolds number Use the Reynolds number to determine the drag coefficient and use the drag coefficient in Newton equation to find the settling velocity 第4页共7页 62 23 3 243243 0 340 34 0 02530 210 10 4 503 4 5034 50 0 85 1 1 0 4 9 81 2 3 65 1 0 21 03 10 0 10 4 7 0 0 84 7 084 253 3 0 3 R R PW p pp R d d tp w m smv d NN m sm g vdm N ms s C C Because the initial assumed velocity IS equal to the velocity calculated by Newton equation iteration is FINISHED and the terminated velocity of the sand is 0 0253m s The backwash velocity can be calculated by the following equation 4 54 54 0 0253 0 404 1 10 0 0246 min1 476 p t vvm sm smm h e The depth of sieved material would Use the drag coefficient in Newton equation to determine the particle settling velocity have to be placed in the filter to produce 600mm of usable filter sand are listed in the following table Cumulative weight passing Size of opening mm ABCDE 0 1050 41 50 15 00 2 0 1491 54 10 811 11 6 0 2104 0 4 83 10 0 10 42 2 5 2 78 20 0 21 19 5 0 5 43 0 2979 521 08 232 016 4 0 42018 540 618 549 537 0 0 59031 061 032 062 354 0 0 84049 078 358 178 368 0 1 19063 290 076 388 549 0 1 68082 8 95 17 96 0 89 58 90 0 97 22 94 4 78 81 92 0 94 57 depth630670617761634 f Plot the cumulative percent passing versus the corresponding sieve size List the cumulative percent passing versus the corresponding sieve size in the following table Cumulative weight passing US sieve size Designation Size of opening mm ABCDE 500 2979 521 08 232 016 4 400 42018 540 618 549 537 0 300 59031 061 032 062 354 0 200 84049 078 358 178 368 0 161 19063 290 076 388 549 0 121 68082 896 090 094 492 0 第5页共7页 第6页共7页 0 10 20 30 40 50 60 70 80 90 100 0 1110 Size of opening mm Cumulative weight passing 系列1系列2系列3系列4系列5 g list the given parameter 33 233 3 2 62 2 2 2 67 10 0 40 0 6 9 81 0 85 1 003 10 10 160min1602 67 10 60 2 0 s vm slm m qL mmms ms gm s msC set up computation table to determine the summation term 12 243 0