油藏工程课程设计计算实例市公开课一等奖省赛课微课金奖课件.pptx

单击此处编辑母版标题样式,单击此处编辑母版文本样式,第二级,第三级,第四级,第五级,*,油藏工程课程设计,（计算实例）,一,.,油区基本资料,1.,油藏基本参数,(,以,2,班,18,号同学为例做详细实例分析,),含油面积：,地层压力梯度,:,0.1MPa/10m,地温梯度,:,3.7/100m,由上表可计算得,(,厚度加权法,):,油藏有效厚度之和为,:,序 号,1,2,3,4,5,6,7,8,9,10,有效厚度,m,2.61,2.85,3.31,3.27,4.05,3.15,3.42,4.28,4.38,4.23,孔隙度,0.345,0.3552,0.3378,0.3352,0.3366,0.3328,0.3358,0.3367,0.3427,0.3438,渗透率,mD,202.5,364.3,451.4,582.1,605.4,668.8,516.9,402.6,348.2,284.9,第1页,平均孔隙度为,:,平均渗透率为,:,2,.,流体物性,地面条件下,:,地层条件下,:,地层泡点压力,:,原油体积系数,:,水体积系数,:,第2页,3.,油水相渗关系,:,S,w,k,ro,K,rw,0.320,0.6500,0.0000,0.352,0.5861,0.0017,0.384,0.5244,0.0059,0.416,0.4651,0.0120,0.448,0.4082,0.0198,0.480,0.3538,0.0292,0.512,0.3021,0.0402,0.544,0.2532,0.0527,0.560,0.2300,0.0595,0.576,0.2072,0.0666,0.592,0.1853,0.0741,0.608,0.1644,0.0818,0.624,0.1439,0.0900,0.640,0.1251,0.0984,0.656,0.1062,0.1072,0.672,0.0895,0.1162,0.688,0.0728,0.1257,0.704,0.0581,0.1353,0.736,0.0316,0.1557,0.768,0.0112,0.1773,0.80,0.0000,0.,Swi=0.32 Sm=0.8 Sor=0.2,第3页,4,.,油水井操作条件,:,注采压差,P=,3MPa;,排状注水排距与井距之比为,2:1;,早期采油速度为,3;,油水井正常生产时间为,300,天,/,年,;,5.,井眼半径,:,0.1m,6.,惯用经济指标：,钻井成本,:,1000,元,/,米,;,注水单价,:6,元,/,米,3,;,输油单价,:,60,元,/,吨,;,生产维护费,:,150,元,/,吨,;,作业费用,:,0,元,/(,井,年,),;,地面工程建设费,0.5,钻井费用,;,原油商品率,:,95;,原油价格,:,1200,元,/,吨贷款利率,:,5.48;,存款利率,:,1.98;,第4页,二,.,设计内容,1.,分析油藏纵向非均质性,并计算储量,惯用一个处理纵向非均质性方法是看反应渗透率非均质程度参数,变异系数,(V),所以地层非均质性不大。,储量为：,带入详细数据得,第5页,2.,依据油藏特点论证划分开发层系可行性,划分开发层系,是将特征相近油层合在一起,用一套生,产井网单独开采,依据开发层系划分标准,油层特征相近油层组合在同一开发层系,以确保各油层对注水方式和井网含有共同适应性,降低开发过程中层间矛盾,单层突进,.,油层性质相近包含,:,沉积条件,渗透率,油层分布面积,层内非均质程度等,.,依据油区各小层基本资料,该油藏各小层渗透率多在 到 间分布,属于高渗油藏,且各小层孔隙度都比较靠近,改变不大,均质程度很好,所以将此,10,小层作为一个开发层系进行开发,.,3.,注水方式选择,选择,排状注水方式,进行开发,.,采油速度要到达,3,，,而排,状注水适合用于高速开采油田,.,第6页,4.,利用排状注水开发指标计算方法,计算对应开发指标,.,计算初始产量,并依据油田产量要求确定油井数及水井数,.,初始产量为,设井距为,X,m,则排距为,截面积,;,由相渗关系数据表知,束缚水饱和度 对应 为,0.6500,化为地面产量为,早期采油速度为,5%,故初始年采油量为,第7页,所以油井数为,取套数,n=241口,.,因为油井与水井数比为,1:1,故水井数量,n=241,口,.,计算井距与排距,设井距为,X,则排距为,2X,如右图,:,（,2x+2x,）,xn=S,则,排距,求前缘含水饱和度,依据相渗关系表中数据,计算含水率,及 随,改变率,.,X,2X,2X,第8页,所用公式,:,其中,全部计算结果数据见下表,:,第9页,SW,Kro,Krw,fw(Sw),fw(Sw),0.32,0.65,0,0.0000,0,0.352,0.5861,0.0017,0.2086,7.8997,0.384,0.5244,0.0059,0.5056,7.6941,0.416,0.4651,0.012,0.70104.8364,0.448,0.4082,0.0198,0.8151,2.8332,0.48,0.3538,0.0292,0.8824,1.6957,0.512,0.3021,0.0402,0.9236,1.0535,0.544,0.2532,0.0527,0.9498,0.7411,0.56,0.23,0.0595,0.9592,0.5347,0.576,0.2072,0.0666,0.9669,0.4381,0.592,0.1853,0.0741,0.9732,0.3582,0.608,0.1644,0.0818,0.9784,0.2965,0.624,0.1439,0.09,0.98270.245,0.64,0.1251,0.0984,0.9862,0.2033,0.656,0.1062,0.1072,0.9892,0.1685,0.672,0.0895,0.1162,0.9916,0.1391,0.688,0.0728,0.1257,0.9937,0.1156,0.704,0.0581,0.1353,0.9953,0.0855,0.736,0.0316,0.1557,0.9978,0.0626,0.768,0.0112,0.1773,0.9993,0.0348,0.8,0,0.2,1,0.0219,第10页,应该绘制曲线,:,1,、,关系曲线,Swi,Swf=0.384,第11页,2,、,关系曲线,第12页,依据表中数据 和 及,在坐标纸中以 为横,坐标,以 为纵坐标,分别绘出 关系曲线和,关系曲线,在含水率与含水饱和度曲线图中,经过束缚水饱和,度 点对 先作切线,得到切点所对应含水饱和,度即为水驱油前缘含水饱和度,所以求得,前缘含水饱,和度,(,见图,).,计算见水前开发指标,求非活塞因子,B,由数据知,则,束缚水饱和度为,残余油饱和度为,最大含水饱和度,第13页,第14页,积分,以一年为一个点计算见水前开发指标,(,以下,q,均为单井日产量,),公式,:,其中,为初始产量,为油水前缘位置,.,第15页,第一年,:,未见水,.,第二年,:,将第一年中求得 有因次化,作为第二年初始产量,.,第16页,同理求得,:,故在第二年见水,.,无因次见水时间,依据 定义,将 有因次化得见水时间,.,设见水时刻为第二年第,X,天,则,6,第17页,故见水时间,t=300+218=518,即在第,518,天见水,.,见水时刻无因次产量为,将其有因次化为,计算见水后开发指标,公式,:,第18页,第二年,(,之后,),将 作为此,82,天产量,.,对应 曲线,找到,5.5315,对应 为,0.416,由数据表中数据可计算,第19页,无因次开发时间,第三年,:,将第二年中计算所得年末产液量作为第三年产量,用一样步骤计算,及对应,有因次产量,直 到,若在 曲线图中查到 不是数据表中含水饱和度,则利用插值法求得对 应,计算结果以下,:,第20页,A,B,C,近似认为,若 在 与 之间,即可求出 ，然后可得出对应,用插值法求,S,we,：,第21页,第22页,将计算结果汇总,制成开发指标汇总表,其中,:,日产液,=,单井日产量,井数（,以第,0,年为例,）,采出程度,=,累计产油量,地质储量,含水率,=,上一年年末来计算求得含水率,.,在第二年见水,故将第二年分成两部分,:,见水前,和,见水后,对应表中第,1,年,.,第23页,见水前,后开发指标汇总表,见水前,见水后,第24页,经济评价,原油销售额,=,年产油量,原油商品率,原油价格,原油商品率,:95;,原油价格,:1200,元,/,吨,;,生产维护费,=,年产油量,生产维护费,生产维护费,:150,元,/,吨,;,注水费,=,年注水量,注水单价,注水单价,:6,元,米,3,;,输油费,=,年产油量,输油单价,输油单价,:60,元,/,吨,;,作业费,=,n,（油井数）,作业费用,作业费用,:0,元,/(,井,.,年,);,钻井及地面建设费用,=,钻井成本,(,油井数,+,水井数,),井深,（,1+0.5,）,钻井成本,:1000,元,/,米,;,纯收入,=,收入,-,支出,第,i,年累计现金流量,=(,第,i-1,年累积现金流量,)(1+,利率,),计息周期,+(,第,i,年纯收入,),当累积现金流量为,负,时，利率用贷款利率,;,贷款利率,:5.48,当累积现金流量为,正,时，利率用存款利率,;,存款利率,:1.98,第25页,经济评价结果表,第26页,依据计算结果中含水率在,70%98%,间数据,进行以下,2,个方面计算,利用水驱特征曲线方法,计算该油藏可采储量和,采收率,.,第五型水驱规律曲线方程为,依据见水后开发指标中累积产油量,含水率资料,可求,出水油比,R,及,.,结果见下表,第27页,依据上表绘图,:,如图所表示,:,将曲线向左延伸,在横坐标,取得 这就是该油田可采储量,由此得,其最终水驱采收率为,:,绘成直线,第28页,利用对应油藏工程方法,判断该油藏递减类型,确定对应递减规律,并计算该油藏可采储量和采收率,(,以概算指标中含水量为,98%,对应日产油量作为极限油量,),利用图解法判断是否属于指数递减或调和递减,假如都不是,则是双曲递减,.,由开发指标汇总表可知第八年末,(,即第九年初,),含水率已到达,97.8%,经过深入计算得在第九年末,对应图中,有因次化,.,得第九年年末产液量为 ，,产油量为,则年产油量为,即对应含水率,98%,极限产量为,指数递减,半对数直线关系为,调和递减,半对数直线关系为,第29页,所以,将时间,t,产量,累积产量 及 计算并列数据表下：,依据数据表,分别绘出,以下,:,其中 为递减期,累积产油量,.,绘成直线,绘成直线,第30页,比较两图可知,递减类型更符合调和递减,基本为一直线,利,用数据表中,t=3,和,t=6,时两点可求得调和递减类型斜率为,即,令,可求得对应截距为,2.802,即,所以可得递减早期产量为,则初始递减率为,方程为,含水率为,98%,对应极限产量为 代入上式,得到,:,第31页,即递减期最大累计产油量为,而总最大累积产油量为,可得采收率为,第32页,