2-4_2007_jun_a

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JAN一月FEB二月MAR三月APR四月MAY五月JUN六月JUL七月A...

JAN一月FEB二月MAR三月APR四月MAY五月JUN六月JUL七月A...

注释: Notice:
北京达富保留根据实际情况调整培训课程的权力 Beijing DAFU reserves the right to cancel or add training courses. 新人培训时间 产品培训时间 每周二/周六下午1:30 上午10:30
图例: Symbles: 新人培训 关帐 承保 回执
产品培训 总监学院 超级星期天 总监例会
产品培训 总监专门培训 超级星期天大型 总监例会
P LANNER
OCT十月 NOV十一月
2007 计划表
DEC十二月
1 2新人培训 3 4 1 5 2 6新人培训 3新人培训 7 4 8 回执 5 9新人培训 6新人培训 10 7 11 8 回执 12 9 13新人培训 10新人培训 14总监学院 11 15产品培训 12 16新人培训 13新人培训 17 14 18总监例会 15 19 16 20新人培训 17新人培训 21 18 22 19 23新人培训 20新人培训 24 21 25 关帐 22总监例会 26 23 27新人培训 24新人培训 28超级星期天 25超级星期天 29 26 关帐 30新人培训 27新人培训 31 承保 28 29 30 承保
1 2 3新人培训 4 5 6新人培训 7 8 回执 9 10新人培训 11 12 13新人培训 14 15 16 关帐 17新人培训 18 19 20新人培训 21 22 23 24新人培训 25 26 27新人培训 28 承保
1 2 3新人培训 4 5 6新人培训 7 8 回执 9 10新人培训 11 12 13新人培训 14 15 16 17新人培训 18 19 20新人培训 21 22 23 24新人培训 25 26 关帐 27新人培训 28 2月 1 2 新人培训 3 4 5 6新人培训 7 8 9新人培训 10 11 12 13新人培训 14 15 16新人培训 17 18 19 20新人培训 21 22 23新人培训 24 25 关帐 26 27新人培训 28 29 30新人培训 31 承保 FEB二月 MAR三月 APR四月 1 2 3新人培训 4 5 6 回执 7新人培训 8 9 10新人培训 11 12 13 14新人培训 15 16 17新人培训 18 19 20 21新人培训 22 23 24新人培训 25 关帐 26 27 28新人培训 29 30 承保 新人培训 MAY五月 JUN六月 JUL七月 AUG八月 SEP九月 1 2 1新人培训 3新人培训 2 4 1 3 5 2 4 1 6 回执 3 5新人培训 2新人培训 7新人培训 4新人培训 1新人培训 6 3 8 5 2 7 4产品培训 9产品培训 6产品培训 3 产品培训 8 回执/新人培 5新人培训 10新人培训 7新人培训 4新人培训 9 6 11 8 回执 5 10 7 12 9 6 11 8 回执 13 10 7 回执 12新人培训 9新人培训 14新人培训 11新人培训 8新人培训 13 10 15 12 9 14产品培训 11 16 13 10 15新人培训 12新人培训 17新人培训 14新人培训 11新人培训 16 13 18 15 12 17 14 19总监例会 16 13 18 15 20 17 14 19新人培训 16新人培训 21新人培训 18新人培训 15新人培训 20 17 22 19 16 21产品培训 18 23 20 17 22新人培训 19新人培训 24新人培训 21新人培训 18新人培训 23 20 25 关帐 22 19 24 21总监例会 26 23总监例会 20总监例会 25 关帐 22 27 24 21 26新人培训 23新人培训 28新人培训 25新人培训 22新人培训 27总监学院 24 29超级星期天 26超级星期天 23超级星期天 28产品培训 25 关帐 30 27 关帐 24 29新人培训 26新人培训 31 承保/新人 28新人培训 25 关帐/新人 30 27 29 26 31 承保 28 30 27 29 承保 31 承保 28 承保 30新人培训 29新人培训 31超级星期天 30

P204萃取硫酸体系中钒的性能研究_胡建锋 (1)

P204萃取硫酸体系中钒的性能研究_胡建锋 (1)
8 min 。
- 1 - 1
; 反萃
在萃取前都要进行皂化 , 以提高萃取率 。 皂化过程 可看成酸碱中和反应 , P204 中 = P ( O) OH 活性基 团中的氢被钠离子取代 。 当 P204 与 NaOH 的克分 子数相等时 , 皂化度为 100 % 。 有机相组成 : 20 %P204 , 3 % 仲辛醇 , 77 % 煤 油。 用 200 g・ L
3
Ξ
摘要 : 用 P204 对硫酸体系钒溶液进行液2液萃取 , 研究了 P204 对钒的萃取性能 。 通过调节 P204 浓度 、 平衡 pH 值及皂化度考察对钒萃取率的 影响 , 得到萃取等温线 。 试验确定了最佳工艺参数 : P204 体积浓度取 15 %~20 % , 萃取平衡 pH 值在 1. 5 左右 , 萃取前必须皂化 。 在最佳条件 下 , 萃取含钒 4. 5 g・ L - 1 溶液的一次萃取率可达 80 %(OΠ A = 1∶ 1) 。
协同萃取五价钒的机制 。 本文研究 P204 萃取剂在 硫酸体系下对钒的萃取性能 , 参考皂化度 、P204 浓度 、 平衡 pH 值等条件对萃取率的影响 , 同时考 察 P204 的循环使用次数 , 饱和吸附量等因素 , 揭 示 P204 萃取钒的性能与规律 。
1 实 验
萃取剂 : P204 ( D2EHPA) ( 学名 : 二2 ( 22乙基己 基) 磷酸 ) , 颜色淡黄色 , 分子量 322. 43 , 密度为
图2 萃取等温线
Fig. 2 Extraction isothermal lines
图3 饱和吸附量平衡曲线
Fig. 3 Equilibrium line of saturation adsorption
3 期 胡建锋等 P204 萃取硫酸体系中钒的性能研究 369

Agilent_TMToolkit_2_1_SP1_Junuary2007_ReadMe

Agilent_TMToolkit_2_1_SP1_Junuary2007_ReadMe

Agilent T&M Toolkit 2.1 with Test Automation Service Pack 1 Read MeAgilent T&M Toolkit v2.1 SP1For Visual Studio 2005Agilent T&M Toolkit SP1 is a Service Pack of bug fixes for Toolkit v2.1.Bugs fixed in SP1Defect Description28488Memory leak occurs when using FFT, DSP functions.28531Error report dialog shows up when create a new instrument session due to wrong version of EnvDTE was shipped with TK 2.1.28520When using negative numbers with the strip chart an exception is thrown when using autoscale.Frequently Asked Questions1.Where do I go for instrument drivers and other related information? The Driver WrapperWizard will walk you through the steps needed to use a driver in Toolkit, including where to getdrivers. You can download drivers for Agilent/HP instruments from:/find/drivers2.How do I convert my installation from a time-limited evaluation to full-use product? TheT&M Toolkit uses a Product Key to enable the product for full use. The Product Key is part of themedia and manuals kit sent to you when you purchase Toolkit. Once the product is purchased, you can do one of the following:o During the 30-day evaluation period, enter the Product Key on the T&M Toolkit > Product Key menu item while Visual Studio 2005 is running. Using the same menu item, you canreview the Product Key at any time.o After the 30-day evaluation period, Toolkit will allow you to enter your Product Key in a dialog box that appears when you try to start the software.3.Will my Product Key from Toolkit or Test Automator 2.0 work with the 2.1 versions? Yes itwill. Toolkit 2.1 is a no-cost upgrade for existing Toolkit 2.0 users and the Product Key from either will work with both products.4.To install on Windows XP under a non-administrator account. Please follow theseinstructions. This is not an issue on Windows 2000.1.On the Start menu, choose Run.2.Type the following command. Substitute your own domain name & administrator accountname in the italicized text.Change the "D:" to the appropriate drive letter or path to theinstallation program.runas /user:<domain name>\<admin user name>"D:\Toolkit\setup.exe /runas"Thank you for your interest in Agilent Technologies T&M Toolkit with Test Automation v2.1. We invite you to visit our web site at /find/toolkit for additional product information. For more information on Toolkit 2.1 see the Toolkit 2.1 Help. Instrument Drivers can be found at /find/drivers.Microsoft and Visual Studio are U.S. registered trademarks of Microsoft Corporation.Adobe is a trademark of Adobe Systems Incorporated.Copyright Agilent Technologies 2007。

2025届郑州二中学七年级数学第一学期期末达标检测试题含解析

2025届郑州二中学七年级数学第一学期期末达标检测试题含解析

2025届郑州二中学七年级数学第一学期期末达标检测试题注意事项1.考试结束后,请将本试卷和答题卡一并交回.2.答题前,请务必将自己的姓名、准考证号用0.5毫米黑色墨水的签字笔填写在试卷及答题卡的规定位置.3.请认真核对监考员在答题卡上所粘贴的条形码上的姓名、准考证号与本人是否相符.4.作答选择题,必须用2B 铅笔将答题卡上对应选项的方框涂满、涂黑;如需改动,请用橡皮擦干净后,再选涂其他答案.作答非选择题,必须用05毫米黑色墨水的签字笔在答题卡上的指定位置作答,在其他位置作答一律无效.5.如需作图,须用2B 铅笔绘、写清楚,线条、符号等须加黑、加粗.一、选择题:本大题共12个小题,每小题3分,共36分.在每小题给出的四个选项中,只有一项是符合题目要求的.1.下列各数中,绝对值最大的是( )A .2B .﹣1C .0D .﹣32.某车间有22名工人,生产某种由一个螺栓套两个螺母的产品,每人每天生产螺母2000个或螺栓1200个,若分配x 名工人生产螺栓,其他工人生产螺母,恰好使每天生产的螺栓和螺母配套,则下面所列方程中正确的是( ) A .20001200(22)x x =-B .12002000(22)x x =-C .212002000(22)x x ⨯=-D .200021200(22)x x =⨯-3.将一副三角尺按不同位置摆放,摆放方式中∠α 与∠β 互余的是( )A .B .C .D .4.8的倒数是( )A .﹣8B .8C .18D .﹣185.我国古代数学名著《孙子算经》中记载了一道题,大意是:求100匹马恰好拉了100片瓦,已知1匹大马能拉3片瓦,3匹小马能拉1片瓦,问有多少匹大马、多少匹小马?若设大马有x 匹,那么可列方程为( )A .3x 3100-x 100+=()B .1x 1003+=C .3x 100-x 100+=()D .13x 100-x 1003+=() 6.一个角的余角是它的补角的25,这个角的补角是( ) A .30° B .60° C .120° D .150°7.在大课间活动中,同学们积极参加体育锻炼.小丽在全校随机抽取一部分同学就“一分钟跳绳”进行测试,并以测试数据为样本绘制如图所示的部分频数分布直方图(从左到右依次分为六个小组,每小组含最小值,不含最大值)和扇形统计图,若“一分钟跳绳”次数不低于130次的成绩为优秀,全校共有1200名学生,根据图中提供的信息,下列说法不正确...的是( )A .第四小组有10人B .本次抽样调查的样本容量为50C .该校“一分钟跳绳”成绩优秀的人数约为480人D .第五小组对应圆心角的度数为45︒8.如图所示,已知AOB ∠与BOD ∠互为余角,OC 是BOD ∠的平分线,20AOB ∠=︒,则COD ∠的度数为( )A .70︒B .35︒C .50︒D .20︒9.下列各式中,运算正确的是( )A .22m n mn +=B .21526a a +=C .2(4)24x x --=-+D .23(32)a a -=--10.如图,在一个长方形中放入三个正方形,从大到小正方形的边长分别为a 、b 、c ,则右上角阴影部分的周长与左下角阴影部分周长差为( )A .+a bB .b c +C .2aD .2b11.若与是同类项,则的值是()A.0 B.1 C.2 D.312.如图所示,已知∠AOB=90°,∠BOC=30°,OM平分∠AOC,ON平分∠BOC,则∠MON的度数为()A.30B.45︒C.60︒D.75︒二、填空题(每题4分,满分20分,将答案填在答题纸上)13.如图所示的整式化简过程,对于所列的每一步运算,第2步依据是______(填“运算率”)14.已知P点坐标为(2﹣a,3a+6),且点P在x轴上,则点P的坐标是____.20,若该彩电的进价为3000元,则标价是15.元旦当天,怡佳商场把品牌彩电按标价的8折出售,仍然获利00___________元.16.如图,从点O引出的射线(任两条不共线)条数与角的总个数有如下关系:从点O引出两条射线形成1个角;如图1从点O引出3条射线共形成3个角;如图2从点O引出4条射线共形成6个角;如图3从点O引出5条射线共形成10个角;(1)观察操作:当从点O引出6条射线共形成有________个角;(2)探索发现:如图4当从点O引出n条射线共形成________个角;(用含n的式子表示)(3)实践应用:8支篮球队进行单循环比赛(参加比赛的每两支球队之间都要进行一场比赛),总的比赛场数为__________场.如果n支篮球队进行主客场制单循环赛(参加的每个队都与其它所有队各赛2场)总的比赛场数是______场.17.某种商品每件的进价为80元,标价为120元,后来由于该商品积压,将此商品打八折销售,则该商品每件销售利润为__元三、解答题 (本大题共7小题,共64分.解答应写出文字说明、证明过程或演算步骤.)18.(5分)观察下列等式:第1个等式:11111212a ==-⨯; 第2个等式:21112323a ==-⨯; 第3个等式:31113434a ==-⨯; 第4个等式:41114545a ==-⨯;…… 解答下列问题:(1)按以上规律写出第5个等式:5a = = ;(2)求1232019a a a a ++++的值; (3)求1111366991220162019=++++⨯⨯⨯⨯的值. 19.(5分)如图,已知∠BOC =2∠AOC ,OD 平分∠AOB ,且∠COD =20°,求∠AOB 的度数.20.(8分)在如图的方格纸中,ABC 的三个顶点都在格点上.(1)画出ABC 向下平移3个单位后的111A B C △;(2)若222A B C △与ABC 关于点O 成中心对称,请画出222A B C △.21.(10分)如图,第一个图形是一个六边形,第二个图形是两个六边形组成,依此类推:(1)写出第n个图形的顶点数(n是正整数);(2)第12个图有几个顶点?(3)若有122个顶点,那么它是第几个图形22.(10分)画出如图由11个小正方体搭成的几何体从不同角度看得到的图形.23.(12分)如图,A、B、C和D、E、F分别在同一条直线上,且∠1=∠2,∠C=∠D,试完成下面证明∠A=∠F 的过程.证明:∵∠1=∠2(已知),∠2=∠3()∴(等量代换)∴BD//CE()∴∠D+∠DEC=()又∵∠C=∠D(已知)∴∠C+∠DEC=180°()∴()∴∠A=∠F()参考答案一、选择题:本大题共12个小题,每小题3分,共36分.在每小题给出的四个选项中,只有一项是符合题目要求的. 1、D【解析】试题分析:∵|2|=2,|﹣1|=1,|0|=0,|﹣3|=3,∴|﹣3|最大,故选D .考点:D .2、C【分析】分配x 名工人生产螺栓,则有22-x 名工人生产螺母,根据一个螺栓套两个螺母可知螺母的数量是螺栓的两倍,要想数量相等则螺栓需要乘2,以此列出方程即可.【详解】根据题意分配x 名工人生产螺栓,每人每天生产1200个,则每天能生产1200x 个.分配22-x 名工人生产螺母,每人每天生产2000个,则每天能生产2000(22-x)个.数量要相等则螺栓需要乘2,则方程为: 212002000(22)x x ⨯=-.故选C .【点睛】本题考查一元一次方程的应用,关键在于读懂题意,找到数量关系.3、C【分析】根据图形,结合互余的定义判断即可.【详解】解:A 、∠α与∠β不互余,故本选项错误;B 、∠α与∠β不互余,故本选项错误;C 、∠α与∠β互余,故本选项正确;D 、∠α与∠β不互余,∠α和∠β互补,故本选项错误;故选:C .【点睛】本题考查了余角和补角的应用,掌握余角和补角的定义是解题的关键.4、C【分析】根据乘积为1的两个数互为倒数进行求解即可得.【详解】解:因为8×18=1,所以8的倒数是18, 故选C .【点睛】本题考查了倒数的概念,熟练掌握倒数的概念是解题的关键.5、D【分析】设大马有x 匹,则小马有(100-x )匹,根据等量关系:大马拉瓦数+小马拉瓦数=100,根据等量关系列出方程.【详解】解:设大马有x 匹,则小马有(100-x )匹,由题意,得:13x 100-x 1003+=(). 故选D .【点睛】本题考查了用一元一次方程解实际问题,关键找到大小马的总数和大小马拉的瓦总数两个等量关系.6、D【分析】首先根据余角与补角的定义,设这个角为x °,则它的余角为(90°﹣x ),补角为(180°﹣x ),再根据题中给出的等量关系列方程即可求解.【详解】设这个角的度数为x ,则它的余角为(90°﹣x ),补角为(180°﹣x ),根据题意得: 90°﹣x 25=(180°﹣x ) 解得:x =30°. 当x =30°时,这个角的补角是:180°﹣30°=150°. 故选D .【点睛】本题考查了余角与补角,属于基础题中较难的题,解答此类题一般先用未知数表示所求角的度数,再根据一个角的余角和补角列出代数式和方程求解.7、D【分析】结合条形图和扇形图,求出样本人数,进行解答即可.【详解】根据直方图可知第二小组人数为10人,根据扇形图知第二小组占样本容量数的20%,则抽取样本人数为1020%50÷=人,故B 选项正确;所以,第四小组人数为50410166410-----=人,故A 选项正确; 第五小组对应的圆心角度数为636043.250︒⨯=︒,故D 选项错误; 用样本估计总体,该校“一分钟跳绳”成绩优秀的人数约为1064120048050++⨯=人,故C 选项正确; 故选:D .【点睛】本题综合考查总体、个体、样本、样本容量,以及扇形统计图和频数(率)分布直方图.准确理解总体、个体、样本、样本容量、扇形统计图和频数(率)分布直方图等的相关概念是关键.8、B【分析】根据余角的性质以及角平分线的性质求解即可.【详解】∵AOB ∠与BOD ∠互为余角,20AOB ∠=︒∴99207000AOB BOD ∠=∠=︒-︒=︒-︒∵OC 是BOD ∠的平分线 ∴11703522BOD COD ∠==⨯︒=∠︒ 故答案为:B .【点睛】本题考查了角度的问题,掌握余角的性质以及角平分线的性质是解题的关键.9、D【分析】利用合并同类项、去括号、添括号对各项进行判断即可.【详解】解:A 、2m 和n 不是同类项,不能合并,故选项错误;B 、21a 和5不是同类项,不能合并,故选项错误;C 、2(4)28x x --=-+,故选项错误;D 、23=32=(32)a a a -=-+--,故选项正确.故选D.【点睛】本题考查了合并同类项和去(添)括号,解题的关键是掌握同类项的概念和去(添)括号的法则,难度不大. 10、D【分析】设重叠部分的小长方形的长与宽分别为,x y ,如图,在图上依次表示阴影部分的各边的长,从而利用周长公式可得答案.【详解】解:设重叠部分的小长方形的长与宽分别为,x y ,如图,在图上依次表示阴影部分的各边的长,所以右上角阴影部分的周长与左下角阴影部分周长差为:()()()()2222a b x c b c y b x a y +--++-----22222222222a b x c b c y b x a y =+--++--+-+2b =.故选D .【点睛】本题考查的是整式的加减,列代数式,去括号,掌握列代数式与去括号是解题的关键.11、C【解析】利用同类项定义列出方程组,即可求出值. 【详解】∵与是同类项, ∴, 则a−b=2,故选:C.【点睛】此题考查同类项,解题关键在于掌握其定义.12、B【分析】根据题意计算出∠AOC ,∠MOC ,∠NOC 的度数,再根据MON MOC NOC ∠=∠-∠计算即可.【详解】解:∵∠AOB=90°,∠BOC=30°,∴∠AOC=∠AOB+∠BOC=120°,又∵OM 平分∠AOC ,ON 平分∠BOC ∴111206022MOC AOC ∠=∠=⨯︒=︒ 11301522NOC BOC ∠=∠=⨯︒=︒ ∴601545MON MOC NOC ∠=∠-∠=︒-︒=︒,故答案为:B .【点睛】本题考查了基本几何图形中的角度计算,掌握角度的运算法则是解题的关键.二、填空题(每题4分,满分20分,将答案填在答题纸上)13、加法交换律【解析】直接利用整式的加减运算法则进而得出答案.【详解】原式=2a 2b+5ab+a 2b-3ab=2a 2b+a 2b+5ab-3ab=(2a 2b+a 2b )+(5ab-3ab )=3a 2b+2ab .第②步依据是:加法交换律.故答案为:加法交换律.【点睛】此题主要考查了整式的加减运算,正确掌握相关运算法则是解题关键.14、(4,0)【分析】根据x 轴上点的纵坐标为0列方程求出a ,再求解即可.【详解】∵P 点坐标为(2a -,36a +),且点P 在x 轴上,∴360a +=,解得2a =-,()2224a -=--=,所以,点P 的坐标为(4,0).故答案为:(4,0).【点睛】本题考查了点的坐标,熟记x 轴上点的纵坐标为0是解题的关键. 15、1【分析】设标价为x 元,根据题意列出方程,解方程即可.【详解】设标价为x 元,根据题意有80%300020%3000x -= 解得4500x =故答案为:1.【点睛】本题主要考查一元一次方程的应用,能够根据题意列出方程是解题的关键.16、15 ()12n n - 28 n (n -1)【分析】(1)现察图形可知,2条射线组成1个角,3条射线就可以组成2+1=3个角,4条射线可以组成3+2+1=6个角,依此可得6条射线组成角的个数是1+2+3+4+5然后计算即可;(2)根据(1)的规律可知:n条射线组成角的个数是1+2+3+…+(n-1),然后计算即可;(3)将每只球队当作一条射线,每场单循环赛当作一个角,然后利用(2)的规律解答即可;【详解】解:(1)现察图形可知,2条射线组成1个角,3条射线就可以组成2+1=3个角,4条射线可以组成3+2+1=6个角,依此可得6条射线组成角的个数是1+2+3+4+5=15;(2)根据(1)的规律可知:n条射线组成角的个数是1+2+3+…+(n-1)=()12n n-;(3)将每只球队当作一条射线,每场单循环赛当作一个角,所以8支篮球队进行单循环比赛相当于8条射线可以组成的角,即比赛场数()8812-=28;如果n支篮球队进行主客场制单循环赛(参加的每个队都与其它所有队各赛2场)总的比赛场数是()12n n-×2= n(n-1).故答案为(1)15,(2)()12n n-,(3)28, n(n-1).【点睛】考查了数角的个数、归纳总结规律以及迁移应用规律的能力,根据题意总结规律和迁移应用规律是解答本题的关键.17、1【分析】设该商品每件销售利润为x元,根据进价+利润=售价列出方程,求解即可.【详解】设该商品每件销售利润为x元,根据题意,得80+x=120×0.8,解得x=1.答:该商品每件销售利润为1元.故答案为1.【点睛】此题考查一元一次方程的应用,正确理解题意找到等量关系是解题的关键.三、解答题(本大题共7小题,共64分.解答应写出文字说明、证明过程或演算步骤.)18、(1)15,16;(2)20192020;(3)2242019【分析】(1)分子是1,分母是两个连续奇数的乘积,等于分子是1,两个连续数为分母的分数差,由此规律解决;(2)利用发现的规律拆项相互抵消计算即可.(3)利用发现的规律拆项相互抵消计算即可.【详解】解:(1)第1个等式:11111212a ==-⨯;第2个等式:21112323a ==-⨯; 第3个等式:31113434a ==-⨯;第4个等式:41114545a ==-⨯;…… 第5个等式:51115656a ==-⨯; 故答案为:156⨯;1156-; (2)12320191111111112233420192020a a a a ++++=-+-+-++- 211200=- 20192020=; (3)1111366991220162019++++⨯⨯⨯⨯ 11111113366920162019⎛⎫=⨯-+-++- ⎪⎝⎭111332019⎛⎫=⨯- ⎪⎝⎭167232019=⨯ 2242019=. 【点睛】 此题考查数字的变化规律,找出算式之间的联系,发现规律解决问题.19、120°【分析】此题可以设∠AOC=x ,进一步根据角之间的关系用未知数表示其它角,再根据已知的角列方程即可进行计算.【详解】解:设∠AOC =x ,则∠BOC =2x .∴∠AOB =3x .又OD 平分∠AOB ,∴∠AOD =1.5x .∴∠COD =∠AOD ﹣∠AOC =1.5x ﹣x =20°.∴x =40°∴∠AOB =120°.【点睛】此题考查角平分线的定义及角的计算,设出适当的未知数,运用方程求出角的度数是解题的关键.20、(1)见解析;(2)见解析【分析】(1)分别作出A、B、C的对应点A1、B1、C1即可;(2)分别作出A、B、C的对应点A2、B2、C2即可.【详解】(1)如图所示:(2)如图所示:【点睛】考查了作图-平移变换和画中心对称图形,解题关键是正确确定对应点的位置.21、(1)4n+2;(2)50;(3)第30个图形(1)由题意可知第1个图形的顶点数为4+2,第2个图形的顶点数为2×4+2,第3个图形的顶点数为3×4+2,…,【分析】即可得出第n个图形的顶点数为4n+2;(2)根据题意将n=12代入4n+2,即可得出第12个图有几个顶点;(3)根据题意由4n+2=122,解出n的值即可得出结果.【详解】解:(1)第1个图形的顶点数为:4+2,第2个图形的顶点数为:2×4+2,第3个图形的顶点数为:3×4+2,…,第n个图形的顶点数为:n×4+2=4n+2;(2)第12个图的顶点数为:4×12+2=50,∴第12个图有50个顶点;(3)4n+2=122,解得:n=30,∴若有122个顶点,那么它是第30个图形.【点睛】本题考查图形的变化规律以及解一元一次方程等知识.根据题意认真观察并得出规律是解题的关键.22、见解析;【解析】利用组合体从不同的角度观察得出答案即可.【详解】解:如图所示:.【点睛】此题主要考查了三视图的画法,正确根据观察角度得出图形是解题关键.23、对顶角相等;∠1=∠3;同位角相等,两直线平行;180°;两直线平行,同旁内角互补;等量代换;DF∥AC;同旁内角互补,两直线平行;两直线平行,内错角相等.【分析】根据已知条件和对顶角相等得出∠1=∠3,从而可得BD//CE,再根据两直线平行同旁内角相等和等量代换可得∠C+∠DEC=180°,从而可得DF∥AC,继而证明∠A=∠F.【详解】证明:∵∠1=∠2(已知),∠2=∠3(对顶角相等),∴∠1=∠3(等量代换),∴BD∥CE(同位角相等,两直线平行),∴∠D+∠DEC=180°(两直线平行,同旁内角互补),又∵∠C=∠D(已知),∴∠C+∠DEC=180°(等量代换),∴DF∥AC(同旁内角互补,两直线平行),∴∠A=∠F(两直线平行,内错角相等).故答案为:对顶角相等;∠1=∠3;同位角相等,两直线平行;180°;两直线平行,同旁内角互补;等量代换;DF∥AC;同旁内角互补,两直线平行;两直线平行,内错角相等.【点睛】本题主要考查了平行线的判定与性质,平行线的判定是由角的数量关系判断两直线的位置关系,平行线的性质是由平行关系来寻找角的数量关系.。

9不同场所暗娼的高危行为和HIVSTI感染状况调查

9不同场所暗娼的高危行为和HIVSTI感染状况调查
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性,说明当地暗娼中HIV的流行尚处于较低水平。 从1997年开始,设在贵阳市的暗娼监测哨点问断地 本调查前对当地暗娼的活动场所和人数绘制分 布图发现,发廊和街头暗娼的比例相对较低。为了了 解不同场所中暗娼的行为和HIv/STIs感染状况,在 调查时增加了发廊和街头暗娟的样本人数,故其构成 不能反映当地实际暗娼人群的场所构成。对于宾馆、 酒店及娱乐场所中的暗娼采用两阶段概率抽样的方 法,发廊和街头的暗娼基本都纳入到调查对象中,这 3部分来源暗娼的调查结果能够反映3类不同场所 中暗娼的情况。 本次调查是选择吸毒人群HIV感染率较高的地 区进行的,仅从来自发廊的暗娼中检出1例HIV阳 检出HIV阳性者,HIV阳性率基本约在1%以下”J。 与国内研究”1有类似的发现,来自街头等低档 性服务场所的暗娼,年龄偏大,文化程度偏低,首次性 交年龄和首次从事性交易年龄偏大,从事商业性性行 为的时间较长。无论是与固定性伴还是非固定性伴 发生性关系时,使用安全套的比例均较低。STIs感 染率也最高。因此,对来自街头、发廊等低档场所的 暗娼进行HIV和STI监测,能够更早、更快地发现艾 滋病的流行。 目前,我国梅毒的流行呈上升趋势,病例报告数 据显示,从1991~2005年平均每年增长71.9%L4 J。

Acomparative study onthehydro-mechanical behavior of compacted

Acomparative study onthehydro-mechanical behavior of compacted

A comparative study on the hydro-mechanical behavior of compacted bentonite/sand plug based on laboratory and field in filtration testsQiong Wang a ,Anh Minh Tang a ,Yu-Jun Cui a ,c ,⁎,Jean-Dominique Barnichon b ,Wei-Min Ye ca Ecole des Ponts ParisTech,Franceb Institut de Radioprotection et de SûretéNucléaire (IRSN),France cTongji University,Chinaa b s t r a c ta r t i c l e i n f o Article history:Received 1December 2012Received in revised form 15May 2013Accepted 19May 2013Available online 25May 2013Keywords:Small-scale test In situ experimentBentonite/sand mixture Technological void Swelling pressure Swelling stainSEALEX is a research project aiming at identifying the key factors that affect the long-term performance of bentonite-based sealing systems with an initial technological void.In this context,a series of in situ experiments have been performed in field conditions.Meanwhile,a small-scale test (1/10)was carried out in controlled conditions in the laboratory,aiming at providing useful information for analyzing the in situ tests in terms of sat-uration time and sealing effectiveness.In this paper,the results of the small-scale test are presented along with the results from the first in situ test (PT-N1).It was observed that during the saturation process,the evolution of the injected water volume followed a hyperbolic relationship with time in both the laboratory and field condi-tions.In the laboratory conditions,a decrease in axial swelling pressure occurred due to filling of the technolog-ical void.By contrast,this decrease has not been observed in the field parison of the injected water and the axial swelling pressure between the two different scales enabled the de finition of a same time upscaling ratio of 2.5(in situ experiment/small-scale test).Accordingly,the saturation duration of the in situ experiment was estimated to be equal to two years.For the small-scale test,a swelling strain evolution rate of 0.588mm/day was identi fied in the case of in filtration from two sides of the sample.This is useful when predicting the evolution of swelling strain in the case of failure of the sealing plug.After filling of an additional 20%void,a swelling pressure of 0.18MPa was obtained,indicating the favorable sealing capacity of the material after filling the technological void.©2013Elsevier B.V.All rights reserved.1.IntroductionIn the design of deep geological repository for high level long lived radioactive wastes,compacted bentonite-based materials are often considered as buffer/sealing materials.These materials are expected to exhibit a swelling pressure high enough to ful fill their buffer/sealing functions.Numerous laboratory studies have been conducted to assess the performance of buffer/sealing materials (e.g.Delage et al.,1998;Lloret et al.,2003;Romero et al.,2005;Lloret and Villar,2007).Vari-ous experiments were also performed in the underground research laboratories (URL)(TSX at Manitoba,Canada;FEBEX at Grimsel,Switzerland;RESEAL at Mol,Belgium;KEY at Bure,France,etc.).Re-cently,IRSN (Institut de Radioprotection et de SûretéNucléaire,France)has launched the SEALEX project aiming at identifying and quantifying the key factors related to the long-term performance of bentonite-based sealing systems taking into account an initial technological void.This project consists of a series of in situ experiments in the Tournemire URL,and a small-scale test (1/10)in the laboratory.The in situ experimental program was purposefully built allowing systematical exploration of the effects of technical speci fications,design,construction,defect,etc.,by changing a single parameter each time.As a reference case (see Barnichon and Deleruyelle,2009;Barnichon et al.,2012for more details),the first test PT-N1with a clay core made up of pre-compacted monolithic disks of MX80bentonite/sand mixture (70/30in dry mass)has been conducted in the URL of Tournemire.Due to the low permeability of this material,saturation is expected to be reached in several years (see Barnichon et al.,2012).During the saturation process,the injected water volume,total pres-sure,pore water pressure and relative humidity changes have been monitored at several positions within the plug.After the saturation stage,hydraulic tests will be performed to determine the overall hy-draulic properties (permeability,occurrence of leakage)of the sealing system.In addition to this reference case,three other tests are designed to quantify the impact of the technical speci fication and design of the sealing plug by changing the intra-core geometry (jointed instead of monolithic disks),core composition (MX80/sand ratio)and core condi-tions (compacted in field instead of pre-compacted).Moreover,to in-vestigate the effect of altered conditions,an additional test is designedEngineering Geology 162(2013)79–87⁎Corresponding author at:Ecole des Ponts ParisTech,6-8av.Blaise Pascal,CitéDescartes,Champs-sur-Marne,77455Marne la Vallee,France.Tel.:+33164153550;fax:+33164153562.E-mail address:yujun.cui@enpc.fr (Y.-J.Cui).0013-7952/$–see front matter ©2013Elsevier B.V.All rights reserved./10.1016/j.enggeo.2013.05.009Contents lists available at SciVerse ScienceDirectEngineering Geologyj o u rn a l ho m e p a g e :w w w.e l s e v i e r.c o m /l o c a t e /e n g g e oto simulate an incidental decrease of swelling pressure caused by failure of the confining structure.Based on the design of the in situ experiments,a laboratory small-scale test(1/10)was performed,focusing on the recovery capacity of the bentonite-based seal with technological voids.The ma-terial identical to that used in test PT-N1was used(MX80bentonite/ sand mixture).A confining cell of stainless steel was used to simulate the constant-volume boundary conditions.After the initial saturation process as in the PT-N1in situ experiment,the seal evolution upon a confinement failure was simulated by allowing a given amount of free swell.This free swell was followed by a last stage of wetting under con-stant volume conditions.To assess the sealing capacity,the injected water volume,axial swelling pressure and swelling strain were moni-tored in different stages.It was expected to obtain useful information from the laboratory small-scale test for analyzing thefield tests in terms of saturation time and sealing effectiveness.In this paper,the results of the small-scale test are presented along with the results from the in situ test(PT-N1).An upscaling ratio was obtained by comparing the injected water volume and the axial swelling pressure evolution between the laboratory andfield condi-tions.The time needed to reach the stabilization of axial swelling pressure for the in situ test(PT-N1)as well as the evolution of swell-ing strain and swelling pressure in the case of failure of the confining structure were estimated accordingly.2.Materials and methods2.1.MaterialsThe soil studied is a compacted MX80/sand mixture with a pro-portion of70/30in dry mass.The bentonite is from Wyoming,USA, with a high content of montmorillonite(80%).It has a liquid limit of 575%,a plastic limit of53%and a unit mass of2.77Mg/m3.The cation exchange capacity(CEC)is76meq/100g(83%of Na+).The quartz sand used in the mixture comes from Eure and Loire(France)with a unit mass of2.65Mg/m3.It was sieved at2mm prior to being mixed with the bentonite.The water used has the same chemical composition as the pore water of the Callovo-Oxfordian claystone from the ANDRA URL in Bure(France),namely synthetic water(Wang et al.,2012,2013).It was obtained by mixing the corresponding chemical compounds (see Table1)with distilled water using a magnetic stirrer until full dissolution.2.2.SEALEX in situ test(PT-N1)As mentioned above,the in situ experiment(PT-N1)has been conducted in the Tournemire URL excavated in Toarcian claystone.A horizontal borehole(0.60m in diameter)was drilled for this purpose. Fig.1shows the layout of the experiment.A seal made up of compacted MX80/sand mixture was sandwiched between two porous plates, allowing water inflow from two water reservoirs(i.e.upstream and downstream).The14.33%annular technological void(volume of void/ volume of borehole)was defined by adopting a smaller initial diameter (0.555m)of the pre-compacted seal as compared to the borehole diameter(0.60m).The upstream plate is in direct contact with the host-rock while the downstream one is retained by a confining system ensuring a constant-volume condition.A packer-like device was used to prevent water leakage from the interface between the confining plug and host-rock.The clay seal in test PT-N1is made up of8monolithic pre-compacted disks(0.555m in diameter and0.15m thick)of MX80/ sand mixture with an initial dry density of1.97Mg/m3(Figure2). The disks were arranged in vertical slices giving rise to the geometry of seal as shown in Fig.2.The bricks were obtained through uniaxial compaction of the mixture at its initial water content of11%.The ini-tial dry density(1.97Mg/m3)of the bricks was selected based on the consideration of the14.33%technological void and the need to have a final dry density of1.67Mg/m3after saturation of the plug andfilling of the initial technological voids.Three types of sensors were installed within the compacted blocks to monitor the swelling pressure,pore pressure and relative humidity.For clarity,only the distribution of sensors for swelling pressure measurement is shown in this paper(Figure3a).Three total pressure sensors were installed on the surface of the column at section0.60m(from the downstream saturation system,L-01,L-02, L-03)to measure the radial swelling pressure;two total pressure sensors were installed at section0and1.20m to measure the axial swelling pressure(A-01,A-02).For each sensor,a hole as shown in Fig.3b was prepared at their pre-assigned positions before the assem-blage of blocks,keeping the hole to a minimum size.Wireless sensors (d=32mm)were used to limit preferentialflow along cables and a wireless transmitter was installed at each measurement section.Data were recorded automatically by a data acquisition system.Regarding the test operational phases,a volume of water of49l wasfirst injected,which corresponded to the volume of the techno-logical void adopted.This process ended in one hour.Afterwards, the water supply was stopped because the side packer was not prop-erly inflated;it restarted after20days under a water pressure of 0.1MPa.During the saturation process,the swelling pressure,pore pressure,water content or water saturation within the plug were monitored.The injected water volumes at both upstream and down-stream chambers were also measured.When the saturation process is completed,hydraulic tests will be performed to determine the overall hydraulic properties(permeability,occurrence of leakage)of the corresponding sealing systems.boratory small-scale testThe experimental devices used for the laboratory small-scale test (1/10)are shown in Fig.4.A stainless steel cell of60mm in inner diameter and200mm long was used.As in the in situ test,an annular technological void was defined by adopting a smaller initial diameter (55.5mm)for the pre-compacted sample as compared to the diame-ter of the hydration cell(60mm).Note however that the hydration cell was placed in the vertical direction(see Figure4)and it was then different from the in situ test which is performed in a horizontal borehole(see Figures1and2).Water supply was conducted through the water inlets in the bottom base which was connected to burettes. This allowed measurement of the total amount of water taken up by the sample.A piston of60mm diameter was used to simulate the confining structure.On the bottom of the piston,there was drainage with two inlets(upside inlet in Figure4)and a porous stone of 50mm diameter,allowing water/airflow.A mechanical press was used to restrain the axial deformation and a force transducer was used to monitor the axial swelling pressure.A displacement transducer fixed on the piston allowed monitoring of the axial displacement to an accuracy of1μm.The axial pressure and axial displacement were recorded automatically to a data logger,while the inlet water volume was measured manually by determining the water level in the burettes.Table1Chemical composition of the synthetic water.Compound NaHCO3Na2SO4NaCl KCl CaCl2.2H2O MgCl2.6H2O SrCl2.6H2O Mass(g)per liter of solution0.28 2.2160.6150.075 1.082 1.3560.05380Q.Wang et al./Engineering Geology162(2013)79–87Note that in this small-scale test,the radial swelling pressure was not measured.A monolithic cylindrical sample(55.5mm in diameter,120mm high)was used in the test.It was statically compacted in a mold to the same dry density as in the in situ test(1.97Mg/m3).In order to ensure the homogeneity of the specimen,the compaction was carried out in two layers.The surface of thefirst compacted layer was care-fully scarified before the second layer was added to ensure a good junction between them.Fig.5a shows the pre-compacted specimen with the hydration cell.After compaction,the specimen(55.5mm in diameter)was placed at the center of the cell(60mm in inner diameter),leaving an annular void(2.25mm)between the specimen and cell wall(Figure5b).An initial axial stress of0.1MPa was applied on the specimen before hydration in order to ensure good contacts between the load cell and the piston,between the piston and the sample,between the sample and the cell bottom,as well as satisfac-tory load measurement.Then,the upside inlets(see Figure4)were sealed and vacuum was applied to evacuate all air in the voids(tech-nological void mainly).The synthetic water wasfinally injected from the bottom.As described in Fig.6,hydration was carried out in three stages. First,the axial deformation was restrained and water was injected to the sample;during this stage(Stage1.initial saturation phase), the evolution of the vertical swelling pressure was monitored.Once the hydration ended,the confining pressure in the axial direction was removed by unloading,allowing a free swell of20%(Stage2. recovery of the void phase).To reduce the test duration in this stage,two-side infiltration was applied by injecting water from both the bottom and the top,while recording changes in axial swelling strain over time.This stage aimed at simulating the case of a satura-tion defect or a failure of confining structure that may occur during the long-term lifespan of the disposal system.The free swell of20% represents the sealing capacity required afterfilling the technological void.When the axial swelling strain reached the desired value of20%, the piston was automatically blocked thanks to a reserved distanceof Fig.2.Geometry of the clay plug and the pre-compacted blocks.(a)Distribution of the total pressure sensorsFor total pressure sensorA-01 (d = 32 mm)0.555 m(b) Hole machined for installation of wireless sensorFig.3.Distribution and installation of the total pressuresensors.yout of the SEALEX in situ experiment(after Barnichon and Deleruyelle,2009).81Q.Wang et al./Engineering Geology162(2013)79–8724mm (corresponding to 20%swelling strain)between the piston and the load cell (Figure 7);the evolution of swelling pressure was monitored again (Stage 3.con finement phase).3.Experimental results 3.1.In situ test (PT-N1)Fig.8shows the evolution of injected water volume over time.As mentioned above,a volume of 49l was first injected to fill the tech-nological void and the injection was stopped for 20days due to a technical problem related to the packer.After resuming the injection,it was observed that the increase rate of water volume was followed by an asymptotic curve with a decreasing rate.After 367days,the total injected water volume was 71.39l (Figure 8).Examination of the curve shows that the shape of water volume versus time (after rejection)can be described by a hyperbolic function.Fig.9presents the time/water volume (day/l)versus time.A good linear relationship is obtained,con firming that the water volume –time curve is of a hyperbolic shape.Thereby,the following equation can be adopted for this relationship:t¼a þbt 1where t is time,V is injected water volume,a and b are the intercept and the slope of the straight line,respectively (Figure 9).According to this hyperbolic relationship,the maximum water volume corresponds to 1/b (Eq.(2)),equal to 72.46l.This is to be compared with the total volume of voids including the technological void and the soil porosity:69.1l.V max¼lim t →∞V t ðÞ¼limt →∞1¼12During hydration,both the axial and radial swelling pressureswere recorded by the total pressure sensors (see Figure 3)and the results are shown in Fig.10.The data by the sensor located at 1.20m section for the axial swelling pressure measurement were unfortunately not available;only the axial swelling pressure values at 0m section are presented.This pressure increased at an almost constant rate and reached 1.63MPa after 367days.For the radial swelling pressure,the evolution rates were very different for the three sensors (see Figure 3);the pressure values reached were 1.78MPa,0.56MPa and 1.05MPa for sensors L-01,L-02and L-03,respectively.This indicates the heterogeneous radial swelling under the in situ conditions.3.2.Small-scale testFig.11shows the measured water in flow over time.Once the water supply was connected to the bottom inlet,water volume increased rapidly and reached 49ml in a few minutes.This value corresponded exactly to the volume of technological void (49ml).Fig.4.Experimentaldevices.Fig.5.Sample preparation.82Q.Wang et al./Engineering Geology 162(2013)79–87Afterwards,water volume increased gradually to reach a maximum value of 70.6ml.No more water could in filtrate after about 200days.The total volume of void (including the technological void V tech and the void inside the soil V v −s )that could be filled with water was esti-mated at 69.1ml.The result shows that a little more water was injected with respect to the estimated one (70.6ml against 69.1ml),but the dif-ference is quite small.For further analyzing the evolution of water volume,the time/water volume (day/ml)is plotted versus time in Fig.12.As in the case of in situ test,a straight line is obtained justifying a hyperbolic relationship between the water volume and the elapsed time (Eq.(1)).The value of 1/b corresponds to the maximum volume of water:1/b =71.43ml,which is very close to the measured water volume (70.6ml).Figs.13and 14depict the evolution of swelling pressure in the first stage (i.e.initial saturation phase).Once water was injected into the specimen,the axial swelling pressure increased very quickly (Figure 14).After about 2days,the swelling pressure reached a first stability stage (1.30MPa)and it restarted to increase on the 4th day.When the swelling pressure reached 1.45MPa after about 12days,a signi ficant decrease of swelling pressure occurred and a minimum value of 0.70MPa was reached.Afterwards,the swelling pressure increased again after about 33days (Figure 13),but at a slower rate.In addition,the evolution curve shows fluctuating pattern.The swelling pressure reached a mean value of 1.80MPa after 300days and then fluctuated in the range of 1.75–1.95MPa.The axial deformation during the swelling pressure development was also recorded,and shown in Fig.13.It followed the same trend as the swelling pressure.Note however that the variation of displace-ment was smaller than 0.2mm.It represents 0.16%of the specimen height (120mm),suggesting a satisfactory control of axial displace-ment in this stage.According to the data obtained in the first stage (Figure 13),no obvious swelling pressure increase occurred during a period of 50days from day 300to day 350.Thus,it was decided to start the sec-ond stage.For this purpose,the con fining pressure was removed on day 350,allowing the free swell.Changes in axial swelling strain were recorded and presented in Fig.15.The uplifting of load cell led to an in-stantaneous rebound of 1.1%(1.4mm/120mm).Afterwards,the axial swelling strain increased almost linearly at a rate of 0.145mm/day.Following this rate,20%of swelling strain was expected to be reached after 157days.In order to reduce the test duration,two-side in filtration was applied on day 364.This resulted in an increase of the swelling strain rate to 0.588mm/day,which is four times faster than that with one-side in filtration.The expected value of 20%(24mm)was reached on day 400.The piston was then re-blocked automatically to start Stage 3.Note that the measured axial swell was 24.4mm.The evolution of swelling pressure was then measured again and the results are presented in Fig.16.Small fluctuation was observed and this fluctuation can be attributed to the daily temperature varia-tions.As expected,the evolution curve follows a hyperboliccurveFig.6.A schematic description of the three stages of the small-scale test.LVDTFig.7.Lifting of load cell for free swell.83Q.Wang et al./Engineering Geology 162(2013)79–87with a decreasing rate over time.It reached stabilization on day 520with a final swelling pressure of 0.18MPa.parison and discussionIt was observed that more water in filtrated into the soil than that calculated by considering the technological void and the soil porosity in both the in situ test (71.39l)and laboratory small-scale test (70.60ml).Even though the water volume has not yet reached stabi-lization in the in situ conditions,the discrepancy is found to be larger than in the small-scale test.This can be related to the natural condi-tions of the in situ test,where some water intake by the host-rock did occur.With a well-controlled condition in the small-scale test,the larger in filtrated water obtained may be related to the low water density (1.00Mg/m 3)considered in the determination of soil void ratio.Indeed,for high plasticity materials as the MX80bentonite,the water density can be much higher than 1.00Mg/m 3(Marcial,2003;Villar and Lloret,2004;Lloret and Villar,2007;Jacinto et al.,2012).This is in agreement with the observation from the KBS-3H mock-up test (Börgesson et al.,2005).Regarding the evolution of water volume,a hyperbolic relation-ship between the injected water volume and elapsed time was obtained in both tests.Accordingly,the maximum water volume was estimated at 72.46l and 71.43ml for the in situ and small-scale tests,respectively.To compare the evolution curve at different scales,the water volume was normalized by using these two values.The nor-malized water volume is equal to the ratio of water volume at time t (V t )to the maximum water volume that can be injected (72.46l and 71.43ml for the in situ and small-scale test,respectively).In terms of time scale,an upscaling ratio of 2.5(in situ test/small-scale test)was found from the normalized water volume shown in Fig.17,where very similar evolution curves (normalized water volume versus normalized time)were obtained for the two tests (in situ and small-scale tests).This upscaling ratio is much smaller than that estimated based on the consolidation theory by considering the experiment scale and hydration conditions:according to the in filtration length,the hydration rate of the in situ test should be 100times lower than in the small-scale test (1/10).As the two-side in filtration applied in the in situ test increased the hydration rate by 4times,the upscaling ratio should be equal to 25(in situ test/small-scale test),still tenElapsed time (day)I n j e c t e d w a t e r v o l u m e (L )Fig.8.Injected water volume versus time.Elapsed time (day)T i m e /w a t e r v o lu m e (d a y /L )Fig.9.Time/water volume versus elapsed time after water rejection.Elapsed time (day)S w e l l i n g p r e s s u r e (M P a )Fig.10.Evolution of swelling pressure.102030405060708004080120160200Elapsed time (day)I n j e c t e d w a t e r v o l u m e (m L )V tech = 49.0 mLTotal volume of voidFig.11.Water volume injected into the specimen.84Q.Wang et al./Engineering Geology 162(2013)79–87times larger than the rate identi fied from the measurements.In fact,under the field conditions,water may fill some voids between the pre-compacted disks during the first minutes (0.15m thick,see Figure 2).This in filtration length of 0.15m leads to an upscaling ratio to 0.56.The upscaling ratio of 2.5observed is possibly related to the combined effect of these two phenomena.Using this upscaling ratio,the axial swelling pressure evolution curve obtained from the in situ test (Figure 10)was normalized and presented in Fig.18,together with the axial swelling pressure mea-sured in the first stage of the small-scale test.Except for the first 33days (see Figure 13)where signi ficant decrease of swelling pres-sure occurred in the small-scale test,the normalized curve of swelling pressure for the in situ test joins the curve of the small-scale test,con-firming the upscaling ratio of 2.5.Based on the results of small-scale test,this ratio allows the time needed to reach the stabilization of swelling pressure in the in situ test to be estimated.It can be observed in the small-scale test that the swelling pressure reached the stability after about 300days.Thus,155days more is needed to reach the maximum swelling pressure at the normalized time scale for the in situ test.Accordingly,it can be estimated that the maximum swellingy = 0.0140 x + 0.0524R 2= 0.99870.00.51.01.52.02.53.004080120160200Elapsed time (day)T i m e /w a t e r v o l u m e (d a y /m L )Fig.12.Time/water volume versus elapsed time.050100150200250300350400Elapsed time (day)S w e l l i n g p r e s s u r e (M P a )-0.100.10.20.30.40.50.6D i s p l a c e m e n t (m m )Fig.13.Evolution of axial swelling pressure and displacement in the firststage.0.00.40.81.21.605101520253035Elapsed time (day)S w e l l i n g p r e s s u r e (M P a )Fig.14.Evolution of swelling pressure during the first 33days.0.00.51.01.52.02.53.03.54.0350355360365Elapsed time (day)S w e l l i n g s t r a i n (m m )(a) Evolution of axial swelling strain during the first051015202530350360370380390400410S w e l l i n g s t a i n (m m )Elapsed time (day)One-side infiltrationTwo-side infiltrationε= 20 %(b) Evolution of axial swelling strainFig.15.Evolution of axial swelling strain during Stage 2.85Q.Wang et al./Engineering Geology 162(2013)79–87pressure in the in situ test should be reached after 388days,corre-sponding to 754days starting from the time of water injection.As regards the kinetics of the axial swelling pressure in the first 33days of small-scale test,stabilization was attained after about 2days at 1.30MPa and restarted to increase from day 4(Figure 14).Wang et al.(2013)observed similar phenomenon in swelling pres-sure tests on smaller samples (35mm in diameter,10mm in height)with the same percentage of technological void.This is related to changes in microstructure of soil.With the progress of hydration,the ef-fect of microstructure changes is reduced,resulting in the re-increase in swelling pressure (Cho et al.,2000;Baille et al.,2010).When the swell-ing pressure reached 1.45MPa on day 12,a signi ficant decrease of swelling pressure occurred,reaching a minimum value of 0.70MPa.This decrease can be attributed to the filling of the technological void.Afterwards,the swelling pressure increased again from day 33.How-ever,the evolution curve showed fluctuation due to the reorganization of soil microstructure under the effect of technological void.On the contrary,the axial swelling pressure measured in the in situ test increased constantly without any fluctuation.This can be re-lated to a coupled effect of large-scale and pressure sensor location.Indeed,in the small-scale test,the axial swelling pressure was mea-sured on the whole cross section and any changes in axial pressure could be monitored.However,in the in situ test,the total pressure sensor was installed in the center of the cross section (Figure 3)and the axial swelling pressure herein corresponded to the local one.Therefore,the axial swelling pressure changes that occurred in the zone of technological void could not be detected by this sensor.After the axial swelling pressure of small-scale test reached stabi-lization,removing the axial con fining restriction led to a very small rebound of 1.1%(1.4mm/120mm).This observation provides valu-able information for the con finement removal phase of the in situ test.During the free swelling process (Stage 2),a swelling strain evo-lution rate (swelling strain/time)of 0.588mm/day was observed under two-side in filtration bined with the upscaling ratio,this result (0.588mm/day)allows prediction of the swelling strain evolution in the in situ test when simulating an incidental decrease of the swelling pressure caused by a failure of the concrete con fining structure.As the piston was re-blocked automatically in the small-scale test (Stage 3),swelling pressure developed again.This indicates the favor-able sealing capacity after filling of the technological void.If a saturation defect or a con fining structure failure occurs in the field,a 20%addi-tional void could be sealed.A final swelling pressure of 0.18MPa was attained,which is in accordance with the swelling pressures measured in the laboratory on smaller samples (Wang et al.,2013):after the 20%free swell (24.5mm),the dry density decreased to a final value of 1.39Mg/m 3;this corresponds to a swelling pressure of 0.23MPa.5.ConclusionIn the context of SEALEX project,a laboratory small-scale test (1/10)was carried out to investigate the recovery capacity of bentonite-based plug with technological void.By comparison with the first results from the in situ test PT-N1,the phenomena identi fied in the laboratory were used for interpreting and estimating the results from the in situ test.During the saturation process,a hyperbolic relationship between the injected water volume and elapsed time was obtained in both laboratory and field tests.However,a little more water was injected as compared to the water volume estimated by considering the total rger discrepancy was found for the in situ test due to the effect of natural conditions.Elapsed time (day)S w e l l i n g p r e s s u r e (M P a )23242526D i s p l a c e m e n t (m m )Fig.16.Evolution of axial swelling pressure during Stage 3.0.70.80.91.01.1050100150200N o r m a l i z e d v o l u m eNormalized time (day)In-situ test PT-N1Small-scale test (1/10)Fig.17.Normalized water volume versus normalized time.050100150200250300350400S w e l l i n g p r e s s u r e (M P a )Elapsed time (day)Fig.18.Swelling pressure versus normalized time.86Q.Wang et al./Engineering Geology 162(2013)79–87。

简单的英文各种缩写

简单的英文各种缩写
英文缩写(英语星期月份等)
星期
星期一: Mon.=Monday
星期二: Tues.=Tuesday
星期三:Wed.=Wednesday
星期四: Thur.=Thurday
星期五: Fri.=Friday
4月21日写作:April 21(st)读作:the twenty-first of April(英)/April twenty-first(美)
四、年月日
表示某年某月某日,按"月份 日期的序数词 年份"写出,年份前常用逗号。如:1993年9月2日写作:September 2(nd), 1993
FDA(Food and Drug Administration)美国食品药物管理局
例句:This new drug has to be approved by the FDA.(这新药要经 FDA 批准)
USDA(United States Department of Agriculture)美国农业部
AAUP (American Association of University Professors) (美国大学教授协会)
例句:The AAUP has many branches at different colleges and universities in the U.S. (AAUP 在美国各大学有很多分会)
EXP (export)出口
MAX (maximum)最大的、的最大限度的
MIN (minimum)最小的,最低限度
DOC (document)文件、的单据
INT (international)国际的
EMS (express mail special)特快传递

干燥综合症评分最终版

干燥综合症评分最终版

Articular domain [2]
Please be careful of not rating articular involvement not related to the disease, such as osteoarthritis
No activity
Absence of currently active articular involvement
ESSPRI的实验结果
1.多变量回归分析证明,干燥症状、肢体痛、疲劳与PGA符合度好,但是 精神疲倦与PGA符合度较差。
2.患者认为最需要改善的症状:干燥症状( 38.7%)、疲劳( 32.2% )、 肢体痛(19.6%),最不重要的为精神疲倦
3.在干燥症状中,口干、眼干被认为最重要
4.ESSPRI最终评分为干燥症状、疲乏、肢体痛三个积分的平均值,范 围为0——10分
High activity
≥ 6 synovitis among a 28 count
Cutaneous domain [3]
Please be careful of rating as “No activity” stable long lasting features that are related to damage rather than
Moderate activity Limited cutaneous vasculitis, including urticarial vasculitis2,
or purpura limited to feet and ankle,
High activity
or subacute cutaneous lupus Diffuse cutaneous vasculitis, including urticarial vasculitis2,

常用英文缩写

常用英文缩写

常用英文缩写(英语星期月份等)星期星期一:Mon.=Monday星期二:Tues.=Tuesday星期三:Wed.=Wednesday星期四:Thur.=Thurday星期五: Fri.=Friday星期六:Sat.=Saturday星期天:Sun.=Sunday月份一月份=JAN. Jan.=January二月份=FEB. Feb.=February三月份=MAR. Mar.=March四月份=APR. Apr.=April五月份=MAY May=May六月份=JUN. Jun.=June七月份=JUL. Jul.=July八月份=AUG. Aug.=August九月份=SEP. Sept.=September十月份=OCT. Oct.=October十一月份=NOV. Nov.=November十二月份=DEC. Dec.=December注意:“.”不能省略这里给大家个例子,比如今天2007年3月20日Mar.20,2007写日期时,可以用基数词(避免出现不必要的失误)1,2,3,4,5,。

28,29,30,31等。

怎样用英语表达年、月、日一、年份在英语中,年份一般用阿拉伯数字写出,其读。

写方法有以下几种:1、四位数的年份,一般前两个数为一个单位,后两个数为一个单位,依次按基数词读出。

如:1763年写作:1763读作:seventeen sixty-three或seventeen hundred and sixty-three2006年写作:2006。

读作:two thousand and six2063年写作:2063。

读作:twenty sixtythree或twenty hundred andsixty-three1050年写作:1050。

读作:ten fifty或ten hundred and fifty2、三位数的年份,可以按基数词读出,或者第一个数字为一个单位,后两个数字为一个单位,按基数词读出。

A variation of Gronwall's lemma

A variation of Gronwall's lemma

a r X i v :0704.0922v 2 [m a t h .C A ] 24 J u n 2007A variation of Gronwall’s lemmaQuang-Cuong Pham LPPA,Coll`e ge de FranceParis,France cuong.pham@ens.frFebruary 1,2008AbstractWe prove a variation of Gronwall’s lemma.The formulation and proof of the classical Gronwall’s lemma can be found in [1].We prove here a variation of this lemma,which we were not able to find in the literature.The main difference from usual versions of the Gronwall’s lemma is that −λis negative .Lemma 1Let g :[0,∞[→R be a continuous function,C a real number and λa positive real number.Assume that∀u,t 0≤u ≤t g (t )−g (u )≤ tu−λg (s )+Cds (1)Then∀t ≥0g (t )≤Cλ+e −λt(2)where [·]+=max(0,·).Proof Case 1:C =0,g (0)>0.Define h (t )by∀t ≥0h (t )=g (0)e −λtRemark that h is positive with h (0)=g (0),and satisfies (1)where the inequality has been replaced by an equality∀u,t 0≤u ≤t h (t )−h (u )=− tuλh (s )ds1Consider now the set S={t≥0|g(t)>h(t)}.If S=∅then the lemma holds true.Assume by contradiction that S=∅.In this case,let m=inf S<∞.By continuity of g and h and by the fact that g(0)=h(0), one has g(m)=h(m)and there existsǫ>0such that∀t∈]m,m+ǫ[g(t)>h(t)(3) Consider nowφ(t)=g(m)−λ t m g(s)ds.Equation(1)implies that∀t≥m g(t)≤φ(t)In order to compareφ(t)and h(t)for t∈]m,m+ǫ[,let us differentiate the ratioφ(t)/h(t).φh2=−λgh+λhφh2≥0Thusφ(t)/h(t)is increasing for t∈]m,m+ǫ[.Sinceφ(m)/h(m)=1,one can conclude that∀t∈]m,m+ǫ[φ(t)≥h(t)which implies,by definition ofφand h,that∀t∈]m,m+ǫ[ t m g(s)ds≤ t m h(s)ds(4) Choose now t0such that m<t0<m+ǫ,then one has by(3)t0m g(s)ds> t0m h(s)dswhich clearly contradicts(4).Case2:C=0,g(0)≤0Consider the set S={t≥0|g(t)>0}.If S=∅then the lemma holds true.Assume by contradiction that S=∅.In this case,let m=inf S<∞. By continuity of g and by the fact that g(0)≤0,one has g(m)=0and there existsǫsuch that∀t∈]m,m+ǫ[g(t)>0(5) Let t0∈]m,m+ǫ[.Equation(1)implies thatg(t0)≤−λ t0m g(s)dswhich clearly contradicts(5).2Case3:C=0Defineˆg=g−C/λ.One has∀u,t0≤u≤tˆg(t)−ˆg(u)=g(t)−g(u)≤ t u−λg(s)+Cds=− t uλˆg(s)ds Thusˆg satisfies the conditions of Case1or Case2,and as a consequence∀t≥0ˆg(t)≤[ˆg(0)]+e−λtThe conclusion of the lemma follows by replacingˆg by g−C/λin the above equation.References[1]I.Gikhman,A.Skorokhod.Introduction to the theory of random pro-cesses.WB Saunders Company,Philadelphia,1969.3。

21483296

21483296
维普资讯
第2 6卷
第 3期
北 京 理 工 大 学 学 报
T a sci s f e i s tt f c n l y r a t n in I t ueo h o g n o o B jg n i Te o
V0 . 6 No 3 12 .
( . t nl aoaoyo Fa e ra t tr l eigI tue f eh o g , ei 0 0 1 hn ; 1Nai a L b rtr f l o meR t dn eis a Ma a ,B i n i t o cnl y B in 1 08 ,C i j n s t T o jg a 2 Sh o o tr s c ne n n ier g eigIs tt o eh o g , eig10 8 ,C i ) .col f Ma i i c dE g e n ,B i ntue f cnl y B i 0 0 1 hn ea S e a l n i j n i T o j n a
t i i g c s s an n a e .
Ke r s p x ei y wod :e o y rs n;DOP ;itisc l a ead n ;f mertr a c O n r ial f mertr a t l n yl a ea d n y
A src: n v1 hs h rs o tiigeo yrs Ssnh s e yterat n o O O ( , btat A o e p op ou. na n px ei Wa y tei db h eci f P c n n z o D 9
1 -iy r-.x .0p op a h nnh ee1一xd )a d te d l iy ehro i h n lA. T e 0dh do9o a1 .h s h p e a trn 一0o ie n h i y d l te fbs e o g c p h

21495683

21495683
企业. 解决“ 农” 为 三 问题 争 创 辉 煌 。 关 键 词 :三农 ” “ 问题 ; 业 机 械 化 ; 机 工 业 ; 业 管 理 农 农 企 中 圈分 类 号 : 3 F 2 . S : 3 33 2 文 献 标 识码 : A
Ab ta t T ea r— c ieyid s ypa si otn u cin sr c: h go ma hn r ut ly mp ra t n t s n r f o
维普资讯
第 2期
中国农机化・H N S G IU T R LM C A IA N C IE EA RC L U A E H NZ Ⅱo
打造农 机工业“ 品"为 精 , 解决“ 三农 " 问题争创辉煌
余 泳 昌, 文 艺, 春 丽 , 天 佑 刘 冯 孙
摘要: 机工业在解决 “ 农 三农 ” 问题的过程 中发挥着不可忽视的 作 用,农业机械化 的发 展促进 了农业结 构调整和产业化进程, 加快了全面建设小康社会的步伐 。农 机工业企业应认 清形势, 迎接机 遇和挑战, 精 、 、 、 的方 向发展, 向“ 新 全 量” 树立精 品意识, 广纳人才, 加大科研合作, 博采众 长, 提高产品竞争力, 创一流 争
nzt n hsp mo dte ar utr t c r dut e t d i i a r t h gi l a s u t ajs n ao o e c ul r u e m n a
id s il a o rc s,h sq ik n d tese so uligte n ut ai t n po e s a uc e e h tp fb i n h r zi d we — f o it n a l ru d wa . h go ma hn r ne- l o s cey i a - n y T ea r- c ie e tr l n l o y p ss s o l a e ce r u d rt dn fte stain a d i r e h ud h v a la n e a ig o h i t sn u o n me tte o p ru i d c aln e d v lp i h i cin o e h p otnt a h l g , e eo n tedr t f yn e e o

BAV199DW 四极管数据手册说明书

BAV199DW 四极管数据手册说明书

QUAD SURFACE MOUNT LOW LEAKAGE DIODEFeatures∙ Surface Mount Package Ideally Suited for Automated Insertion ∙ Very Low Leakage Current∙ Totally Lead-Free & Fully RoHS Compliant (Notes 1 & 2) ∙Halogen and Antimony Free. “Green” Device (Note 3)Mechanical Data∙ Case: SOT363∙ Case Material: Molded Plastic. UL Flammability Classification Rating 94V-0∙ Moisture Sensitivity: Level 1 per J-STD-020∙ Terminals: Finish - Matte Tin Annealed over Alloy 42 Leadframe (Lead Free Plating). Solderable per MIL-STD-202, Method 208 ∙ Polarity: See Diagram∙Weight: 0.008 grams (Approximate)Ordering Information (Note 4)2. See /quality/lead_free.html for more information about Diodes Incorporated’s definitions of Hal ogen- and Antimony-free, "Green" and Lead-free.3. Halogen- and Antimony-free "Green” products are defined as those which contain <900ppm bromine, <900ppm chlorine (<1500ppm total Br + Cl) and <1000ppm antimony compounds.4. For packaging details, go to our website at /products/packages.html.Marking InformationTop ViewSOT363Internal Schematic Top ViewAC 1AC 2C 2A 1A 2C 1K52 = Product Type Marking Code YM = Date Code Marking Y = Year (ex: C = 2015)M = Month (ex: 9 = September)K 52K 52Y MY M(@T = +25°C, unless otherwise specified.)Thermal CharacteristicsElectrical Characteristics (@T A = +25°C, unless otherwise specified.)Notes: 5. Part mounted on FR-4 PC board with recommended pad layout, which can be found on our website at . 6. Short duration pulse test used to minimize self-heating effect.I , I N S T A N T A N E O U S F O R W A R D C U R R E N T (A )F V , INSTANTANEOUS FORWARD VOLTAGE (V)Fig. 2 Typical Forward Characteristics, Per Element F 0.010.1P , P O W E R D I S S I P A T I O N (m W )D T Fig. 1 Power Derating Curve, T APackage Outline DimensionsPlease see AP02001 at /_files/datasheets/ap02001.pdf for the latest version.1001,000I , L E A K A G E C U R R E N T (p A )R V , REVERSE VOLTAGE (V)Fig. 3 Typical Reverse Characteristics, Per ElementR 30020025010015050C)°I , F O R W A R D C U R R E N T (m A )F 00.20.40.60.811.21.41.61.82Suggested Pad LayoutPlease see AP02001 at /_files/datasheets/ap02001.pdf for the latest version.。

SHGOPEN 2007经典战役回顾 新科状元的对决

SHGOPEN 2007经典战役回顾 新科状元的对决

SHGOPEN 2007经典战役回顾新科状元的对决
粥脂弱
【期刊名称】《电子竞技》
【年(卷),期】2007(000)006
【摘要】在刚刚结束的shgOpen 2007上,去年的ESWC和WCG两大新科世界
冠军MIBR和PGS会师决赛,不管最后结果如何,他们都再次向世人证明了他们的实力。

而这场火星撞地球的精彩对决也一定会吊起大家的胃口,实际上本场比赛也是
异常惨烈,比分呈交替上升状。

但做为以稳定和整体配合为特点的两支战队,他们的
比赛节奏也是十分之缓慢,同时没有什么新的打法和战术,完全就是在拼发挥和应变。

而这场发生在这两者之间的战斗则可以当作一场nuke标准打法的典型教材来使用,下面就让我们从整体的角度来观察一下整场比赛的走势:
【总页数】1页(P)
【作者】粥脂弱
【作者单位】
【正文语种】中文
【中图分类】G899
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2.SHGOPEN 2007经典战役回顾冷静冷静再冷静侦查侦查再侦查 [J], Beezilbub
3.DotA经典对决回顾——AOE流 [J], flyzy
4.四保一流 GL vs Nebula经典对决回顾 [J], 大熊猫
5.War3经典战役回顾 [J], 柳子
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Part 2 Examination – Paper 2.4Financial Management and Control June 2007 Answers 1(a)Expected net present value of Proposed 1Scenario 1Scenario 2Scenario 3Number of sunny days100125150Saving (£/day)700700700Annual saving (£)70,00087,500105,000Costs(24,000)(24,000)(24,000)–––––––––––––––––––––––––Net annual savings 46,00063,50081,000–––––––––––––––––––––––––Present value of net savings at 10%460,000635,000810,000Investment500,000500,000500,000–––––––––––––––––––––––––Net present value(40,000)135,000310,000–––––––––––––––––––––––––Probability30%60%10%Expected net present value = (–40,000 x 0·3) + (135,000 x 0·6) + (310,000 x 0·1) = £100,000The ENPV is £100,000 so if the investment is evaluated on this basis, it is financially acceptable. In reaching a decision, however, the company should consider that there is a 30% chance of making a loss. This may be seen as an unacceptably high risk. Furthermore, the number of sunny days each year will not be constant, as assumed here, and may or may not be exactly 100, 125 or 150 days. It is possible the net present values of Scenarios 1 and 3 represent extremes in terms of expectations, and that the net present value of Scenario 2 may be most useful as representing the most likely outcome, even on a joint probability basis. It is also worth noting that inflation has not been taken into account and that the ever-increasing cost of energy may make the proposed investment much more financially attractive if it were factored into the analysis.WorkingsPresent values must be calculated with the before-tax cost of capital of 10%, since before-tax cash flows are being evaluated here. The present value of a perpetuity is found by dividing the constant annual cash flow by the cost of capital.Present value of net savings, Scenario 1 = 46,000/0·10 = £460,000Present value of net savings, Scenario 2 = 63,500/0·10 = £635,000Present value of net savings, Scenario 3 = 81,000/0·10 = £810,000(b)Net present value of Proposal 2Year1234££££Contribution (W1)248,400334,230449,709550,800Fixed costs (W2)(185,400)(190,962)(196,691)(202,592)––––––––––––––––––––––––––––––––––––63,000143,268253,018348,208T axation (30%)(18,900)(42,980)(75,905)(104,462)––––––––––––––––––––––––––––––––––––44,100100,288177,113243,746T ax benefits (W3)16,57512,4319,32312,970––––––––––––––––––––––––––––––––––––After-tax cash flows60,675112,719186,436256,7168% discount factors0·9260·8570·7940·735––––––––––––––––––––––––––––––––––––Present values56,18596,600148,030188,686––––––––––––––––––––––––––––––––––––£Sum of present values489,501PV of scrap value (W4)36,750––––––––526,251Initial investment221,000––––––––Net present value305,251––––––––The net present value is positive and so Proposal 2 is acceptable. Note that the after-tax cost of capital of 8% is used to discount after-tax cash flows in evaluating this investment proposal.Workings(W1)Year1234Selling price (£/unit)12·4812·9813·5014·04Variable cost (£/unit)4·204·414·634·86––––––––––––––––––––––––––––––––Contribution (£/unit)8·288·578·879·18Sales volume (units/yr)30,00039,00050,70060,000––––––––––––––––––––––––––––––––T otal contribution (£/yr)248,400334,230449,709550,800––––––––––––––––––––––––––––––––(W2)T otal fixed production overhead cost in current price terms = 6 x 30,000 = £180,000Inflating this current cost at 3% per year:Year1234Fixed costs (£/yr)185,400190,962196,691202,592(W3)Year Capital allowances Tax benefits1221,000 x 0·25 =55,25055,250 x 0·3 =16,575255,250 x 0·75 =41,43841,438 x 0·3 =12,431341,438 x 0·75 =31,07831,078 x 0·3 =9,3234By difference43,23443,234 x 0·3 =12,970–––––––––––––––221,000 – 50,000 =171,00051,299–––––––––––––––(W4)Present value of scrap value = 50,000 x 0·735 = £36,750(c)Before-tax return on capital employed of Project 3T otal cash flow over five years before advertising and depreciation = £500,000T otal depreciation over five years = 300,000 – 30,000 = £270,000T otal accounting profit over five years = 500,000 – 100,000 – 270,000 = £130,000Average annual accounting profit = 130,000/5 = £26,000 per yearAverage investment = (initial investment + scrap value)/2 = (300,000 + 30,000)/2 = £165,000ROCE = 100 x (26,000/165,000) = 15·8%The ROCE of Proposal 3 is marginally greater than the target level of 15%. ROCE cannot be recommended as an investment appraisal method, however, and the NPV of Proposal 3 should be calculated in order to determine whether it is financially acceptable.(d)G TK plc is a company with a small overdraft and no long-term debt. If the three proposals represent the total capitalinvestment needs of the company, the amount of finance needed is as follows.Finance needed Project lifeProposal 1£500,000PermanentProposal 2£221,000Four yearsProposal 3£400,000*Five years–––––––––––£1,121,000–––––––––––*It is assumed that advertising costs would be met from finance raisedEquity financeThe equity financing choices available to GTK plc are a rights issue or a placing.Rights issueIn this method of raising new equity finance, new shares are offered to existing shareholders pro rata to their existing shareholdings, meeting the requirements of company law in terms of shareholders’ pre-emptive rights. Since GTK plc has several million pounds of shareholders’ funds, it may be able to raise £1·1 million through a rights issue, but further investigation will be needed to determine if this is possible. Factors to consider in reaching a decision will include the number of shareholders, the type of shareholders (institutional shareholders may be more willing to subscribe than small shareholders), whether a recent rights issue has been made, the recent and expected financial performance of GTK plc, and the effect of a rights issue on the company’s cost of capital. A rights issue would not necessarily disturb the existing balance of ownership and control between shareholders. Approximately half of the finance needed is for a permanent investment and the permanent nature of equity finance would match this.PlacingThis way of raising equity finance involves allocating large amounts of ordinary shares with a small number of institutional investors. Existing shareholders will need to agree to waive their pre-emptive rights for a placing to occur, as it entails issuing new shares to new shareholders. The existing balance of ownership and control will therefore be changed by a placing. Since GTK plc is a listed public limited company, 25% or more of its issued ordinary share capital will be in public hands and theIdentifying alternative courses of actionOnce organisational objectives have been identified, alternative courses of action that may lead to achieving those objectives can be identified. Strategic analysis of the organisation and its environment can indicate potential courses of action. For example, a company may look at its existing products and markets, its potential markets, the threat posed by its competitors, the impact of changes in technology on its products and production processes, and so on, and decide that a key objective is the development of new products to replace existing products in existing markets that are reaching the end of their product life cycle.Evaluating alternative courses of actionAt this stage the various alternative courses of action are considered from the point of view of suitability, feasibility and acceptability. In order for this to be done, detailed information about each alternative course of action needs to be gathered and analysed.Selecting alternative courses of actionOnce the most appropriate alternative courses of action have been selected, long-term plans to implement them are formulated. Because these plans are long-term in nature, they will of necessity be less detailed than short-term plans, and will need to allow a degree of flexibility in responding to the changing organisational environment.Preparing and implementing the budgetA budget is a short-term plan formulated in financial terms and will show in detail the short-term actions the organisation willtake in working towards its long-term objectives. Once the budget has been formulated, finalised and agreed it can be implemented.Monitoring actual resultsIn order to achieve the long-term objectives that are reflected in the budget, the organisation must ensure that actual performance is proceeding according to plan. It will therefore need to monitor actual performance and results.Responding to divergences from planDivergences from planned activity, as measured by variances from budget, can lead to action if they are deemed to be significant. This action may be corrective in nature, in order to bring actual activity back into line with planned activity, or may entail revision of the budget if one of its underlying assumptions is seen as being in error.(b) A fixed budget is one prepared in advance of the relevant budget period which is not changed or amended as the budgetperiod progresses. This budget represents a periodic approach to budgeting, since a new budget is prepared towards the end of the budget period for the subsequent budget period. In this way, an organisation may set a new budget on an annual basis.A rolling budget, sometimes called a continuous budget, represents an alternative approach to periodic budgeting. Here, aportion of the budget period is replaced on a regular basis so that the overall budget period remains unchanged. For example, with a budget period of one year, at the end of each quarter a new quarter could be added to the end of the budget period and the elapsed quarter could be deleted, so that the budget was always looking one year ahead. Continuous budgeting continues to increase in popularity.A zero-based budget is a periodic budget which seeks to dispose of the incremental approach to budgeting. In the incrementalapproach, an increment is added to the relevant figure from last year’s budget, for example to take account of inflation. In this way, inefficiency can become embedded in the annual budget and profitability may suffer as a result. With the zero-based approach, each element of planned activity is required to be justified in terms of its contribution towards achieving organisational objectives. This involves the formulation of decision packages, which describe particular activities in such a way that managers can compare them in terms of their competing claims on organisational resources, and then rank them from a cost-benefit point of view. In this way, zero-based budgeting looks at each budget period with a new perspective.A fixed budget is likely to be useful in circumstances where the organisational environment is relatively stable and can bepredicted with a reasonable degree of certainty.A rolling budget is likely to be useful in circumstances where the future is less certain and more flexibility is needed in theorganisational response to its changing environment. For this reason, rolling budgets are popular with new organisations. A cash budget is often a rolling budget because of the need to keep tight control of this area of financial management. A rolling budget is also supported by the availability of cheap and powerful information processing via personal computers and computer networks.A zero-based budgeting approach tends to be most beneficial when used with services and with discretionary activities, andso is most widely used in the public sector.(c)Linear regression is a powerful way of analysing past information in order to derive linear relationships and so is ideally suitedto deriving cost equations from past accounts. Sales volume, however, is unlikely to follow a linear relationship alone. Linear regression could be used to determine the overall trend being followed by sales volume on, for example, an annual basis, but inspection of historic sales volumes is likely to show variations about the trend. These could be due to seasonal variations, or longer-term cyclical variations. Time-series analysis can extract these seasonal and cyclical variations and therefore produce forecasts of sales volumes that are likely to be more accurate in a given period than forecasts based on the underlying trend alone. In forecasting future sales volumes, therefore, both quantitative methods have their place in increasing forecasting accuracy.3(a)Discussion of performance of Woodside CharityIn a year which saw fundraising fall £80,000 short of the target level, costs were over budget in all areas of activity except overnight shelter provision. The budget provided for a surplus of £98,750, but the actual figures for the year show a shortfall of £16,980.Free meals provision cost £12,750 (14%) more than budgeted. Most of the variance (69%) was due to providing 1,750 more meals than budgeted, although £4,000 of it was due to an increase of 20p in the average cost per meal.Variable cost of overnight shelter provision was £26,620 (11%) less than budgeted. £31,000 was saved because usage of the service was 1,240 bed-nights below budget, but an adverse variance of £4,380 arose because of an increase of 50p in the average unit cost of provision.Variable advice centre costs were £16,600 (37%) above budget. This was due to increased usage of the service, which was 17% up on budget from 3,000 to 3,500 sessions, and to an increase in the average cost of provision, which rose by 17% from £15 to £17·60 per session.Fixed costs of administration and centre maintenance were £18,000 (28%) above budget and the costs of campaigning and advertising were £15,000 (10%) above budget.While investigation of some of the variances in the reconciliation statement below may be useful in controlling further cost increases, the Woodside charity appears to have more than achieved its objectives in terms of providing free meals and advice.The lower usage of overnight shelter could lead to transfer of resources from this area in the next budget to the services that are more in demand. The reasons for the lower usage of overnight shelter are not known, but the relationship between the provision of effective advice and the usage of overnight shelter could be investigated.Operating statement£Budgeted surplus (W1)98,750Funding shortfall (W3)(80,000)–––––––££18,750Favourable AdverseFree meals (W4)Price variance4,000Usage variance8,750Overnight shelter (W5)Price variance4,380Usage variance31,000Advice centre (W6)Price variance9,100Usage variance7,500Campaigning and advertising (W7)Expenditure variance15,000Fixed cost (W8)Expenditure variance18,000––––––––––––––31,00066,730(35,730)–––––––Actual shortfall (W2)(16,980)–––––––Workings(W1)Budgeted figures£Free meals provision91,250(18,250 meals at £5 per meal)Overnight shelter (variable)250,000(10,000 bed-nights at £30 – £5 per night)Advice centre (variable)45,000(3,000 sessions at £20 – £5 per session)Fixed costs65,000(10,000 x £5) + (3,000 x £5)Campaigning and advertising150,000––––––––601,250Surplus for unexpected costs98,750––––––––Fundraising target700,000––––––––(W2)Actual figures£Free meals provision104,000(20,000 meals at £5·20 per meal)Overnight shelter223,380(8,760 bed-nights £25·50 per night)Advice centre61,600(3,500 sessions at £17·60 per session)Fixed costs83,000Campaigning and advertising165,000––––––––636,980Shortfall16,980––––––––Funds raised620,000––––––––(W3)Funding shortfall – 700,000 – 620,000 = £80,000 (A)(W4)Free meals price variance = (5·00 – 5·20) x 20,000 = £4,000 (A)Free meals usage variance = (18,250 – 20,000) x 5·00 = £8,750 (A)(W5)Overnight shelter price variance = (25·00 – 25·50) x 8,760 = £4,380 (A)Overnight shelter usage variance – (10,000 – 8,760) x 25 = £31,000 (F)(W6)Advice centre price variance = (17·60 – 15·00) x 3,500 = £9,100 (A)Advice centre usage variance = (3,000 – 3,500) x 15·00 = £7,500 (A)(W7)Campaigning and advertising expenditure variance = 150,000 – 165,000 = £15,000 (A)(W8)Fixed cost expenditure variance = 65,000 – 83,000 = £18,000 (A)(b)Financial management and control in a not-for-profit organisation (NFPO) such as the Woodside charity must recognise thatthe primary objectives of these organisations are essentially non-financial. Here, these objectives relate to helping the homeless and because the charity has no profit-related objective, financial management and control must focus on providing value for money. This means that resources must be found economically in order to keep input costs as low as possible; that these resources must be used as efficiently as possible in providing the services offered by the charity; and that the charity must devise and use effective methods to meet its objectives. Financial objectives could relate to the need to obtain funding for offered services and to the need to control costs in providing these services.Preparing budgetsThe nature of the activities of a NFPO can make it difficult to forecast levels of activity. In the case of the Woodside charity, homeless people seeking free meals would be given them, and more food would be prepared if necessary, regardless of the budgeted provision for a given week or month. The level of activity is driven here by the needs of the homeless, and although financial planning may produce weekly or monthly budgets that consider seasonal trends, a high degree of flexibility may be needed to respond to unpredictable demand. This was recognised by the charity by budgeting for a fundraising surplus for unexpected costs.It is likely that forecasting cost per unit of service in a NFPO can be done with more precision if the unit of service is small and the service is repetitive or routine, and this is true for the Woodside charity. It is unlikely, though, that a detailed analysis of costs has been carried out along these lines, and more likely that an incremental budget approach has been used on a total basis for each service provided. It depends on the financial skills and knowledge available to the charity from its three full-time staff and team of volunteers.Controlling costsBecause of the need for economy and efficiency, this is a key area of financial management and control for a NFPO. The costs of some inputs can be minimised at the point of buying, for example the Woodside charity can be economical when buying food, drink, crockery, blankets, cleaning materials and so on. The costs of other inputs can be minimised at the point of use, for example the Woodside charity can encourage economy in the use of heating, lighting, water consumption, telephone usage and postage. In an organisation staffed mainly by volunteers with an unpredictable clientele, cost control is going to depend to a large extent on the way in which responsibility and authority are delegated.Collecting informationCost control is not possible without collecting regularly information on costs incurred, as well as storing and processing this information. In the Woodside charity, provision has been made in the budget for fixed administration costs and the administration duties must hopefully relate in part to this collecting of costing information. Without it, budgeting and financial reporting would not be possible. Annual accounts would be needed in order to retain charitable status and to show providers of funds that their donations were being used to their best effect.Meeting objectivesA NFPO organisation must be able to determine and demonstrate whether it is meeting its declared objectives and so needsto develop measures to do this. This can be far from easy. The analysis of the performance of the Woodside charity over the last year shows that it may be possible to measure objective attainment quantitatively, i.e. in terms of number of free meals served, number of bed-nights used and number of advice sessions given. Presumably, objectives are being met to a greater extent if more units of service are being provided, and so the adverse usage variances for free meals and advice sessions can in fact be used to show that the charity is meeting a growing need.The meaning of quantitative measures of service provision may not be clear, however. For example, the lower usage of bed-nights could be attributed to the effective provision of advice to the homeless on finding housing and financial aid, and so may also be seen as a success. It could also be due to dissatisfaction amongst the homeless with the accommodation offered by the shelter. In a similar vein, the higher than budget number of advice sessions may be due to repeat visits by homeless people who were not given adequate advice on their first visit, rather than to an increase in the number of people needing advice. Qualitative measures of objective attainment will therefore be needed in addition to, or to supplement, quantitative ones.4(a)Profit and loss accounts for TFR Ltd for the four-year periodYear Current Year 1Year 2Year 3Year 4£££££T urnover210,000255,000300,000345,000390,000Expenses168,000204,000240,000276,000312,000––––––––––––––––––––––––––––––––––––––––Net profit42,00051,00060,00069,00078,000Interest2,00011,0008,7506,5004,250––––––––––––––––––––––––––––––––––––––––Profit before tax40,00040,00051,25062,50073,750T ax10,00010,00012,81315,62518,438––––––––––––––––––––––––––––––––––––––––Profit after tax30,00030,00038,43846,87555,313Dividend15,00015,00019,21923,43827,656––––––––––––––––––––––––––––––––––––––––Retained profit15,00015,00019,21923,43827,656––––––––––––––––––––––––––––––––––––––––Equity finance200,000215,000234,219257,656285,313Debt finance nil75,00050,00025,000nilInterest cover (times)21·04·66·910·618·4Debt/equity (%)nil352110nilReturn on equity (%)1514161819ROCE (%)2118212427ROCE (%)*1916202326*Including the existing and continuing overdraft in capital employedWorkingsAnnual interest (assuming the continuing overdraft is maintained at the current level)Year 1 interest payment = 100,000 x 0·09 = 9,000 + 2,000 = £11,000Year 2 interest payment = 75,000 x 0·09 = 6,750 + 2,000 = £8,750Year 3 interest payment = 50,000 x 0·09 = 4,500 + 2,000 = £5,500Year 4 interest payment = 25,000 x 0·09 = 2,250 + 2,000 = £4,250(b)Financial implications for TFR Ltd of accepting bank loanA key consideration is whether TFR Ltd will be able to meet the annual payments of interest and capital. It is assumed, inpreparing a cash flow forecast, that there is no difference between profit and cash, and that inflation can be ignored. The annual cash surplus after meeting interest and tax payments is therefore assumed to be equal to retained profit.Year1234Retained profit15,00019,21923,43827,656Capital repayment25,00025,00025,00025,000––––––––––––––––––––––––––––Net cash flow(10,000)(5,781)(1,563)2,656––––––––––––––––––––––––––––TFR Ltd is clearly not able to meet the annual capital repayments. In order to do so, it will need to change the dividend policy it appears to have maintained for several years of paying out a constant proportion of profit after tax as dividends. One possible course of action is to cut its dividend now and then increase it in the future as profitability allows. Since TFR Ltd is owner-managed, a change in dividend policy may be possible, depending of course on the extent to which the owner or owners rely on dividend income. The annual cash flow shortfall is less than the annual dividend payment, so a change in dividend policy would probably allow the loan to be accepted.Year1234Profit after tax30,00038,43846,87555,313Capital repayment25,00025,00025,00025,000––––––––––––––––––––––––––––Available funds5,00013,43821,87530,313––––––––––––––––––––––––––––It is useful to consider key financial information after the loan has been paid off, i.e. in year 5, assuming that no further turnover growth occurs after the fourth year:Year Year 5£T urnover390,000Expenses312,000––––––––Net profit78,000Interest2,000––––––––Profit before tax76,000T ax19,000––––––––Profit after tax57,000Dividend28,500––––––––Retained profit28,500––––––––Equity finance313,813Debt finance nilInterest cover (times)39Debt/equity (%)nilReturn on equity (%)18ROCE (%)25ROCE (%)*23*Including the existing and continuing overdraft in capital employedThe effect on financial risk of taking on the loan can be examined. If the interest and capital payments are kept up, financial risk will be lower than its current level at the end of four years, all things being equal. Interest cover increases from its current level after five years, from 21 times to 39 times, but is on the low side at the end of the first year (4·6 times), although an improved level is reached at the end of the second year (6·9 times), with further increases in subsequent years. The debt/equity ratio peaks at 35% at the end of the first year and falls rapidly thereafter, at no time looking dangerous, and TFR Ltd returns to its current ungeared position after five years. The bank, as provider of debt finance, would be interested in the trend in these ratios, as well as in the ongoing cash flow position.Both return on equity (ROE) and return on capital employed (ROCE) improve with growth in turnover, but are lower than current levels in the first and second years following taking on the loan. At the end of five years ROE has improved to 18% from 15% and ROCE from 19% to 23%. Interest and capital payments would not increase with inflation.Provided TFR Ltd can meet the interest and capital repayments, business expansion using debt finance may be financially feasible. However, this analysis has ignored any potential pressure for reduction or repayment of the overdraft. An average overdraft of £20,000 is quite large for a company with an annual turnover of £210,000 and therefore cannot be ignored in any assessment of financial risk. TFR Ltd may therefore consider asking for a longer repayment period, with lower annual capital repayments, if it plans to reduce the size of the overdraft or if it is concerned about future cash flow problems. (c)TFR Ltd is owner-managed and profitable, and financed by equity apart from its large overdraft. It is currently seeking a bankloan in order to finance an expansion of business.Equity financeThe owner could inject new equity finance himself but his personal financial situation may make this impossible. There are unlikely to be any wealthy individuals willing to invest in his company because there are likely to be more attractive investments elsewhere. Investing in a UK pension fund, for example, carries a tax incentive in that the UK government increases any contributions by the amount of income tax paid. There is therefore a disincentive to invest in the shares of a small company which may be difficult to sell in the future unless another investor can be found who wishes to buy the shares.However, there is in the UK a Business Angel network which can bring potential investors and small companies together, with the added bonus that the Business Angel may have expertise and experience to offer that could be useful in a small company situation. The owner of TFR Ltd may wish to look into this possibility.There is also a UK government initiative called the Enterprise Investment Scheme, which is of potential benefit to trading companies rather than service companies. The government offers tax advantages in terms of income tax and capital gains tax in order to encourage investment by individuals in the ordinary shares of small companies.A further UK government scheme offers tax advantages to Venture Capital T rusts, who are required to invest a large part oftheir funds in the ordinary shares of small companies.Other government assistance schemesA range of other UK and EU government assistance schemes exist but almost all of these are targeted towards companies inparticular geographic locations, or within particular ranges in terms of number of employees, or with particular funding requirements, for example training.Debt financeSmall companies are faced with a risk-averse attitude from banks when they seek to raise debt finance. Banks tend to ask for personal guarantees from owners and will set interest rates at higher levels than those charged to larger companies. TFR。

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