登录
首页 » Fortran » two-FVM

two-FVM

于 2021-03-23 发布 文件大小:49KB
0 236
下载积分: 1 下载次数: 292

代码说明:

  本程序是采用有限体积法求解二维对流扩散方程的经典程序。程序里包含各种离散方法,划分网格方法等。适合CFD初学者阅读。(The program is two-dimensional finite volume method for solving convection-diffusion equation of the classic program. Program includes a variety of discrete methods, meshing methods. CFD for beginners)

文件列表:

2dc
...\grid.f,16503,2011-09-05
...\grid.inp,355,2011-09-05
...\pcol.f,48078,2011-09-05
...\pcol.inp,748,2011-09-05
...\pipe
...\....\htm" target=_blank>dcpipe,755,2011-09-05
...\....\htm" target=_blank>dgpipe,247,2011-09-05
...\....\grid.f,16503,2011-09-05
...\....\pcol1.f,50268,2011-09-05
...\plot.f,49706,2011-09-05
...\plot.inp,380,2011-09-05
...\psc.f,11130,2011-09-05
...\psc.inp,405,2011-09-05
...\pscus.f,10933,2011-09-05
...\pscus.inp,328,2011-09-05
...\htm" target=_blank>readme,6588,2011-09-05

下载说明:请别用迅雷下载,失败请重下,重下不扣分!

发表评论

0 个回复

  • Meshless_SPH
    无网格算法中的光滑粒子水动力学sph算法源程序(溃坝)(Meshless algorithm smoothed particle hydrodynamics sph algorithm source (dam))
    2021-01-21 21:38:45下载
    积分:1
  • ADINA-Program
    有限元源程序,结构分析软件,可以用于结构工程、近海工程等领域。(Adina FEM Program, can be used in structural engineering, offshore engineering, and so on)
    2012-12-26 20:47:21下载
    积分:1
  • 11087 统逆序对
    说明:  Description 设a[0…n-1]是一个包含n个数的数组,若在ia[j],则称(i, j)为a数组的一个逆序对(inversion)。 比如 有5个逆序对。请采用类似“合并排序算法”的分治思路以O(nlogn)的效率来实现逆序对的统计。 一个n个元素序列的逆序对个数由三部分构成: (1)它的左半部分逆序对的个数,(2)加上右半部分逆序对的个数,(3)再加上左半部分元素大于右半部分元素的数量。 其中前两部分(1)和(2)由递归来实现。要保证算法最后效率O(nlogn),第三部分(3)应该如何实现? 此题请勿采用O(n^2)的简单枚举算法来实现。 并思考如下问题: (1)怎样的数组含有最多的逆序对?最多的又是多少个呢? (2)插入排序的运行时间和数组中逆序对的个数有关系吗?什么关系? 输入格式 第一行:n,表示接下来要输入n个元素,n不超过10000。 第二行:n个元素序列。 输出格式 逆序对的个数。 输入样例 5 2 3 8 6 1 输出样例 5(Set a[0... N-1] is a n array containing n numbers. If there is a [i] > a [j] i n the case of I < j, then (i, j) is a n inversion pair of a array. For example, has five reverse pairs. Please use the idea of "merge sorting algorithm" to achieve the statistics of inverse pairs with O (nlogn) efficiency. The number of inverse pairs of a sequence of n elements consists of three parts: (1) The number of reverse pairs in the left half, (2) the number of reverse pairs in the right half, (3) the number of elements in the left half is greater than that in the right half. The first two parts (1) and (2) are implemented by recursion. To ensure the final efficiency of the algorithm O (nlogn), how should the third part (3) be implemented? Do not use O (n ^ 2) simple enumeration algorithm to solve this problem.)
    2019-01-07 23:52:06下载
    积分:1
  • fdk-f
    基于FDK的锥束CT全三维重建算法,Matlab语言版本。(FDK cone-beam CT based full three-dimensional reconstruction algorithm, Matlab language version.)
    2011-10-13 12:54:43下载
    积分:1
  • UPWIND
    fortran code for LINEAR_WAVE PROPAGATION by upwind method.
    2013-03-26 18:53:58下载
    积分:1
  • wave_propagation
    西安电子科技大学吴振森教授的波传播与电磁散射课件,里面有很多fortran计算的代码,此课程为博士课程,供专业人士学习.(The courseware of the Xidian University Professor Wu Zhensen wave propagation and electromagnetic scattering, which have many fortran code, this course is Dr. courses for professionals to learn.)
    2012-09-15 11:17:28下载
    积分:1
  • SVDCMP
    奇异值快速分解程序,相比其他奇异值分解程序,速度快,解的精度高。(Fast singular value decomposition process, compared to other singular value decomposition process, fast, high precision solution.)
    2020-08-14 15:28:27下载
    积分:1
  • 雷达sra成像 C语言
    描述了雷达sra成像等,用c语言开发 ,包含有fft等程序(Describes the radar imaging Sra, using c language development, contains the procedures fft)
    2008-12-14 20:48:52下载
    积分:1
  • Least-squares
    在visual c++6.0的环境下,使用C++语言编写的最小二乘法程序。(In visual c++6.0 environment, using C++ language least squares procedure.)
    2011-06-24 14:33:54下载
    积分:1
  • C3.17
    说明:  有两个ARMA过程,其中信号1是宽带信号,信号2是窄带信号,分别用AR谱估计算法、ARMA谱估计算法和周期图法估计其功率谱。(There are two ARMA processes, in which signal 1 is a wideband signal and signal 2 is a narrowband signal. The power spectrum is estimated by AR spectrum estimation algorithm, ARMA spectrum estimation algorithm and periodogram method respectively.)
    2020-07-27 09:27:39下载
    积分:1
  • 696516资源总数
  • 106914会员总数
  • 0今日下载