N5K 2K-update_for partner training_20091223.pptx
Autodesk Nastran 2022 用户手册说明书
MPA, MPI (design/logo), MPX (design/logo), MPX, Mudbox, Navisworks, ObjectARX, ObjectDBX, Opticore, Pixlr, Pixlr-o-matic, Productstream,
Publisher 360, RasterDWG, RealDWG, ReCap, ReCap 360, Remote, Revit LT, Revit, RiverCAD, Robot, Scaleform, Showcase, Showcase 360,
TrueConvert, DWG TrueView, DWGX, DXF, Ecotect, Ember, ESTmep, Evolver, FABmep, Face Robot, FBX, Fempro, Fire, Flame, Flare, Flint,
ForceEffect, FormIt, Freewheel, Fusion 360, Glue, Green Building Studio, Heidi, Homestyler, HumanIK, i-drop, ImageModeler, Incinerator, Inferno,
Autodesk Nastran 2022
Reference Manual
Nastran Solver Reference Manual
N5K-N2K配置模板
N5K-3# write erase
Warning: This command will erase the startup-configuration.
Do you wish to proceed anyway? (y/n) [n] y
N5K-3# reload
WARNING: There is unsaved configuration!!!
N5K-1(config-if)# fex associate 100
N5K-1(config-if)# vpc 100
N5K-1(config)#fex 101
N5K-1(config-fex)# pinning max-links 1
N5K-1(config-fex)# description "SECOND_FEX_IN_1 "
N5K-1(config-vrf)#interface mgmt0
N5K-1(config-if)#ip address 10.75.58.24/24
N5K-1(config-if)#no shutdown
3、启用SSH：
N5K-1(config)# no ssh server enale
N5K-1(config-fex)# description "N5k1 FEX 100 "
N5K-1(config-fex)# interface e1/7-8
N5K-1(config-if)# switchport mode fex-fabric
N5K-1(config-if)# fex associate 100
N5K-1(config)# spanning-tree vlan 101-104 root secondary(备份根桥)
Nexus系列交换机介绍_VMware vSphere企业级网络和存储实战_[共2页]
![Nexus系列交换机介绍_VMware vSphere企业级网络和存储实战_[共2页]](https://img.taocdn.com/s3/m/c618daa67fd5360cbb1adb94.png)
第5章 部署Nexus N5K&N2K交换机1993年，Cisco公司推出了一个交换机的品牌——Catalyst，这个品牌在市场上处于绝对的领导地位。
经过15年的时间，到2008年1月，Cisco公司又对外宣布了另一个交换机的品牌——Nexus。
Nexus交换机是适应未来云计算、虚拟化、整合化数据中心的新一代交换机产品，对以前的Catalyst交换机有重大改进和扩展，是业界最先进的产品。
本章介绍Cisco Nexus N5K&N2K交换机的基本配置以及在VMware vSphere环境的应用。
本章要点Nexus N5K&N2K交换机介绍配置使用Nexus FEX配置使用Nexus Vpc虚拟化架构使用N5K&N2K5.1 Nexus N5K&N2K交换机介绍Cisco Nexus交换机是适应未来数据中心的高密度交换机，具有比Catalyst交换机高得多的极高的性能，端口交换延迟很低，同时支持国际标准不丢包的以太网技术。
5.1.1 Nexus系列交换机介绍作为全新的品牌，Nexus系列交换机适应未来数据中心的先进架构，与Catalyst交换机及其他厂家等交换机有本质上不同。
Nexus交换机不简单是传输数据，而是一种全新的架构设计——计算机总线的延伸，目前只有Brocade交换机能部分支持，其他厂商等无法支持这些特性。
Nexus交换机具有以下特点。
1．高性能/低延迟/不丢包的以太网高性能/低延迟/不丢包的结合使得Nexus交换机具有一个独特的优势——计算机总线的延伸，相当于把彼此通信的众多计算机的总线直接连接起来。
z高性能相当于修了一条非常宽的路，宽到有多少车都可以直接上来；z低延迟相当于每辆车的速度都很快；z不丢包相当于车上的货物在道路上不丢失，以前的所有交换机都做不到这点，相当于目的地有个货物检查员来检查车上的货物，一旦发现丢了，再重发货，效率会降低。
Degree Works 5.0.2 Upgrade Functional User Guide a
Degree Works 5.0.2 Upgrade Functional User Guide and FAQFunctional User Guide to Degree Works 5.0.2 Upgrade Degree Works 5.0.2 delivers updates to the look, feel, and functionality of the Degree Works Application Suite for functional users. This document highlights some of the most pertinent changes. Ahead of installation and testing, please review the User Guides associated with this release, accessible through the Degree Works 5.0.2 Content Pack.To access the Degree Works 5.0.2 Content Pack, you must first be logged in to the Ellucian Customer Center. If you do not have an Ellucian account, you may request one by completing the login request form.Responsive DashboardWith Degree Works 5.0.2, Ellucian has introduced the Responsive Dashboard as a new visual metaphor based on “visual cards” to enhance the end-user experience when interfacing with Degree Works. For clarity, the Dashboard available in versions before Degree Works 5.0.2 is labeled the “Dashboard” and this new interface is labeled the “Responsive Dashboard.”The Dashboard remains available in Degree Works 5.0.2 and maintains the same look and functionality. The Responsive Dashboard mirrors the functionality and information displayed in the Dashboard, although the look is markedly different.In Degree Works 5.0.2, the Dashboard and the Responsive Dashboard can be utilized and viewed simultaneously.Transit and Web TransitWith Degree Works 5.0.2, Web Transit is available. Users with appropriate permissions will be able to access Web Transit with their SSO credentials.Users can continue to access Transit as a PC Tools application as well. Confirm access through your desktop with your Degree Works credentials.Web Transit and PC Tools Transit can be utilized simultaneously in Degree Works 5.0.2.New KeysWith the Degree Works 5.0.2 release, there are multiple updates to the keys that should be assigned to users to maintain Degree Works functionality. Review these carefully ahead of an upgrade. More information is available in the Degree Works Installation Guide 5.0.2beginning on page 31 and in the Degree Works Technical Guide 5.0.2 beginning on page 231.To access the Degree Works Installation Guide 5.0.2 and the Degree Works Technical Guide 5.0.2, you must first be logged in to the Ellucian Customer Center. If you do not have an Ellucian account, you may request one by completing the login request form.Functional users should be aware of the keys below ahead of testing an upgraded environment:SDFINDIDUsers must have this key to be able to search for a student in the Dashboard and the Responsive Dashboard.SDAUDITUsers must have this key to be able to view a student’s audit in the Dashboard and the Responsive Dashboard.TRANSITUsers must have this key to access Web Transit.TRANALLUsers must have this key to run jobs in Web Transit.COMPOSERAs of Degree Works 5.0, to access Composer a user or user group must have the COMPOSER key.DEBUGAs of Degree Works 5.0.1, any user with the DEBUG key can retrieve debug directly in any Degree Works application.SHPCFG FileAs of Degree Works 5.0.1, the SHPCFG file is located in Shepentry. Users no longer require command line access to retrieve this file. To access, navigate to “Settings” and look undercore.security.rules.shpcfg.bannerextract.configBeginning with Degree Works 5.0.1, the bannerextract.config file is located in Shepentry. Users no longer require command line access to retrieve this file. To access, navigate to “Settings” and look under integration.banner.extract.config.AuthenticationWhile it may require no action from you, please be aware that in Degree Works 5.0.2, authentication occurs through the Gateway and not directly in the application (except PC Tools Transit). Use your Single Sign On (SSO) credentials for all Degree Works applications (except PC Tools Transit).On the Dashboard, the “Change Password” option is disabled as of Degree Works 5.0.1; the link leads to a blank page. If you would like to remove the link, you can do so in localizations, accessible in Composer.LocalizationsLocalizations for Degree Works 5.0.2 are stored and managed in the Composer application. To access Composer, be sure to add the COMPOSER key to the user or user group.For more information about using the Composer application, refer to the Degree Works Composer Administration Guide 5.0.2.To access the Degree Works Composer Administration Guide, you must first be logged in to the Ellucian Customer Center. If you do not have an Ellucian account, you may request one by completing the login request form.Frequently Asked Questions: Degree Works 5.0.2 Am I required to use the Responsive Dashboard?Degree Works 5.0.2 allows for simultaneous use of the Dashboard and the Responsive Dashboard.The two dashboards stay in sync and both accurately reflect any changes made through administrative applications (i.e., Scribe, Shepentry, etc.) or within the other dashboard (i.e., exceptions, notes, etc.).User access is the same for each dashboard.Am I required to use Web Transit?Degree Works 5.0.2 allows for simultaneous use of PC Tools Transit and Web Transit. Each application requires a distinct set of keys, however.Information about the keys and groups required to access these applications and to run Transit jobs is available in Degree Works Technical Guide 5.0.2.To access the Degree Works Technical Guide 5.0.2, you must first be logged in to the Ellucian Customer Center. If you do not have an Ellucian account, you may request one by completing the login request form.Once you have fully transitioned to Web Transit, it is recommended to remove ucx12job from cron jobs.Has the Student Educational Planner (SEP) changed?The Student Educational Planner in the Responsive Dashboard is labeled SEP4.The Student Educational Planner in the Dashboard is now labeled SEP3.While the look and feel of SEP4 is distinct from SEP3, these planners mirror the same student information and plans. Changes made in SEP3 will be reflected in SEP4 and vice versa.How do I access Template Management for the Student Educational Planner (SEP)?In Degree Works 5.0.2, Template Management is only accessible through the Dashboard.Updates to Template Management through the Dashboard will be reflected in both SEP3 and SEP4.Where is my SHPCFG file?As of Degree Works 5.0.1, the SHPCFG file is located in Shepentry. You no longer need command line access to retrieve this.As a best practice, you may consider saving a copy of this locally as well as in the Degree Works application.I updated my SHPCFG file through Shepentry; why haven’t permissions updated?This is a Known Issue with Ellucian; the SHPCFG file will save in Shepentry with parsing errors.For more information, see CR-000166563 in the Ellucian Customer Center.If you do not have an Ellucian account, you may request one by completing the login request form.Why can’t I create a new group in Shepentry?This is a Known Issue with Ellucian; attempting to create a new group in Shepentry results in an error.For more information, see CR-000167483 in the Ellucian Customer Center.If you do not have an Ellucian account, you may request one by completing the login request form.Where are my localization files?In Degree Works 5.0.2, localizations are located in the Composer application. You no longer need command line access to view these.As a best practice, you may consider saving a copy of these files locally as well as in the Degree Works application.Why can’t I access Composer?In order to access Composer, you must have the COMPOSER key added to your key ring.How do I use Composer?For more information about using the Composer app, refer to the Degree Works Composer Administration Guide 5.0.2.To access the Degree Works Composer Administration Guide, you must first be logged in to the Ellucian Customer Center. If you do not have an Ellucian account, you may request one by completing the login request form.Where is Scribe?All Degree Works 5.x versions use Scribe on the Web, accessed via URL. PC Tools Scribe is obsolete.How do I log in?You will authenticate with your Single Sign On credentials. If you are unable to authenticate, contact your campus networking team.PC Tools Transit log in still authenticates with the Degree Works ID.Why can’t I log in to Scribe/DWShell/Composer?If you can access the Dashboard but not other applications, you may be assigned to the incorrect user class. Your Degree Works SuperUSER may need to verify your user class by checking Shepentry and reviewing the extract files.Why am I seeing a 500 error?There are several technical reasons for this error. We recommend working with the Ellucian Action Line to better diagnose the issue. You will most likely need to work with your campus networking team as well.To open a case with the Ellucian Action Line, contact us at ****************.Why am I seeing a 404 error?There are several technical reasons for this error. We recommend working with the Ellucian Action Line to better diagnose the issue.To open a case with the Ellucian Action Line, contact us at ****************.What do I do if the Self Service Banner link to Degree Works doesn’t work?You may need to update the Web Tailor settings to reflect the current Degree Works Dashboard link.。
Infoprint 250 導入と計画の手引き 第 7 章ホスト
SUBNETMASK
255.255.255.128
Type of service...............: TOS
*NORMAL
Maximum transmission unit.....: MTU
*LIND
Autostart.....................:
AUTOSTART
*YES
: xx.xxx.xxx.xxx
: xx.xxx.xxx.xxx
*
(
)
IEEE802.3
60 1500
: xxxx
48 Infoprint 250
31. AS/400
IP
MTU
1
1
IPDS TCP
CRTPSFCFG (V3R2)
WRKAFP2 (V3R1 & V3R6)
RMTLOCNAME RMTSYS
MODEL
0
Advanced function printing............:
AFP
*YES
AFP attachment........................:
AFPATTACH
*APPC
Online at IPL.........................:
ONLINE
FORMFEED
*CONT
Separator drawer......................:
SEPDRAWER
*FILE
Separator program.....................:
SEPPGM
*NONE
Library.............................:
V5R2M0 Electronic Service Agent for iSeries 常见问题解答
Frequently Asked Questions forV5R2M0 Electronic Service Agent for iSeriesQ1: Where can I get media for the Electronic Service Agent product (5798-RZG)?A1: New systems leaving IBM from manufacturing will have Electronic Service Agent preloaded on the iSeries system. OS/400 upgrades will have the product included with theOS/400 software order on the B29XX_07 standard set products CD-ROM where XX is the language id.For systems where the product is not preloaded and the install media is not readily accessible, a separate Electronic Service Agent V5R2M0 CD-ROM will be available at V5R2 GA. IBM Service Representatives (SSRs) can order this CD-ROM (SK3T-4116-01) through IBM Publications.Q2: Are there any migration instructions if I am upgrading my iSeries system from V4R5 orV5R1 to V5R2?A2: Yes. If you SLIP install V5R2, the Service Agent inventory collection task created within Management Central will exist along with the QYSDDIAL TCP/IP connection profile, so you will not have to recreate them. However, you will need to accept the license agreement for inventory collection via the Extreme Support wizard. For hardware problem reporting, you will need to obtain an activation password from your IBM Service Representative (SSR) and accept the license agreement.Q3: Why does LODRUN fail when I try to install Service Agent?A3: The methods to install Service Agent from CD-ROM are the Restore Licensed Program command (RSTLICPGM) or Install Licensed Programs (GO LICPGM option 11).Message CPF3717 is generated if LODRUN is attempted.Q4: Why do I need to use a user profile for Inventory setup that has a *SECOFR user class but not the actual QSECOFR user profile?A4: The operating system has a submit job restriction that does not allow batch jobs to be submitted in the name of system supplied user profiles, such as QSECOFR. However, for Management Central purposes, the QSECOFR user profile must be enabled.If a collection task is created by QSECOFR, the inventory collection task will be started by the QSECOFR user profile and the data will be collected normally. After collection, Management Central will attempt to submit a QYIVRIPS job to be run in the QSVCDRCTR subsystem to send the inventory data. The operating system will not allow the QYIVRIPS job to be submitted because of the submit job restriction. Message CPD1617 (‘Value specified for USER parameter not correct.’) will be generated.Q5: How do I activate Service Agent functions?A5: For existing Service Agent customers, hardware problem reporting is activated by entering GO SERVICE on a command line. You will need an activation password from your IBM ServiceRepresentative (SSR). System inventory collection and transmission is activated using iSeries Navigator and the Extreme Support Wizard under Management Central.Q6: In prior releases of Service Agent, I needed to set up a locale object to make sure the time information for the scheduled task and history panel was correct. Is that still necessary for Service Agent?A6: Although the use of an OS/400 locale object is no longer necessary for Service Agent, we recommend you make sure the system value for time (QTIME) and the system value QUTCOFFSET are set correctly for your time zone. Changes may need to be made during the calendar year for the beginning and ending of Daylight Savings Time to ensure continuity of the task operations and history panel information.Q7: In V5R2 iSeries Navigator, what additional options are available to the Electronic Service Agent object under Management Central’s Extreme Support?A7: The additional V5R2 options are listed below. To view all of the options, expand Extreme Support and click Agents. Right click on the Electronic Service Agent object.!Using the “Configure Electronic Services” option you can register existing IBM ids to the Service Agent Information site to view your system’s inventory information. If you do not!!The history panel can be refreshed using F5 or from a drop-down option from the history panel.!You can start and stop job logging of all jobs that are run by user id QSVCDRCTR.Q8: Are there new or updated inventory selections for Service Agent?A8: Yes. V5R2 Service Agent sends Group PTF information (part of the PTF inventory) as well as Software Audit information (part of Software Inventory). Service Agent can also send the System Registration information. The System Registration wizard is located in the Configuration and Service section of iSeries Navigator. Sending this information and registering your iSeries system with IBM will help make future service calls to IBM faster and more productive.Q9: Does Service Agent send all of my system inventory to IBM each time it transmits?A9: No. After the first transmission of all of your selected inventories, Service Agent will only transmit system inventory if any part of that inventory has changed since the last time Service Agent transmitted. In V5R2, for PTF inventory and Software Resource inventory, Service Agent will only transmit the specific changes, and not the entire inventory.Q10: When I tried to do Inventory -> Collect, I did not get the option to send the Service Agent information. Has there been a change?QQ11: How can I set up Service Agent to report problems from a central system to IBM for other iSeries systems or secondary partitions in my network?A11: For hardware problem reporting, the products needed for this function are Electronic Service Agent (5798RZG) and System Manager for iSeries (57xxSM1) which is fee based software.System Manager for iSeries gives your central system or main partition the ability to report hardware failures on behalf of the iSeries systems or other partitions in your network.The link below to an IBM Software Knowledge Base document provides the information toQ12: How can I set up Service Agent to send inventory information from a central system to IBM for other iSeries systems or secondary partitions in my network?A12: For inventory transmission, Management Central allows collection from other iSeries systems or secondary partitions and transmission from the central system. You must have Service Agent installed on the secondary partitions or iSeries systems. The OS/400 release running on the central system determines the types of inventory to be sent since new collection capabilities have been added over the past two OS/400 releases.Q13: During Service Agent setup, I am having problems starting the Management Central server jobs and connecting to Management Central. What should I do?A13: We recommend this excerpt from the Management Central FAQs.You will need to perform some troubleshooting. The problem is most likely a TCP/IP configuration problem on the host. The following TCP/IP related questions should be answered.• Has TCP/IP been configured (CFGTCP) on your iSeries systems in the network?• Has TCP/IP been started (STRTCP) on your iSeries systems in the network?• Can you 'ping' the host from the client or another iSeries?• Can you 'ping' the client from the iSeries?• Can you 'ping' other iSeries systems in your network from the client and from the iSeries?• If you are using a Domain Name Server (DNS), both short and long names for the system in your network must be there and entered correctly.• If you are not using a DNS, the host table for both short and long name(CFGTCP, option 10) must be there and entered correctly. For example, a commonpractice is to define a long name that conveys the networkorganization and to also define a short name such as SYS1 that is easier to remember.!Is the domain information listed in CFGTCP, option 12 specified correctly?Host name (ex: SYS1) and domain name (ex: )• Is there a port conflict? Management Central uses port 5555 (NETSTAT, option 3.) You can also view the Service Entry Table (CFGTCP, opt 21, opt 1) to identify portassignments. Management Central has three port numbers assigned to it. Non-SSLservice as-mgtctrl is port 5555, SSL service as-mgtctrl-cs is port 5577 and SSL serviceas-mgtctrl-ss is port 5566. If these are not defined in the service table, add them.The protocol is TCP.• Starting in V5R1, Management Central added a java server job (QYPSJSVR) to support java function on the GUI. In order for this java server to run, the QSECOFR user profile must NOT be disabled.If the above troubleshooting steps do not solve the problem, contact your IBM Software Support Center.Q14: Are there problem determination steps for Service Agent inventory problems?A14: Yes. The following diagram steps through a general problem determination process.Q15: When I take option 19 to Display Service Agent audit log using the Service Agent Main Menu (GO SERVICE), my screen does not automatically go to the bottom of the file for the most recent entries as in past releases. Has there been a change?A15: Yes. The audit log file has been changed for easier code page translation and is displayed using SQL. On the ‘Position to line’ field, you can enter the letter B to get to the bottom of the file for the most recent audit log entries.Q16: When I look at the job schedule entries (WRKJOBSCDE) on my system, there is a QSDAUTOTST job schedule entry. What is that entry used for?A16: QSDAUTOTST is the Service Agent automated operational test which is an integral part of V5R2. This scheduled job will send a test to the service provider every 91 days for verification purposes. Day of the week and time settings can be modified using option 1 ‘Change Service Agent Settings’ from the Service Agent Main Menu (GO SERVICE).Q17: The V5R2 AUTOPTF function has been requesting a different informational APAR than in prior releases. Has there been a change?A17: Yes. The new informational APAR used for AUTOPTF at V5R2 is II12806.Q18: Why is there a new JV1 requirement for V5R2 systems running Service Agent Inventory Collection and Transmission?A18: The V5R2 IBM Developer Kit for Java has been updated. To comply with licensing requirements and to ensure V5R2 Service Agent will function, the update must be applied to user systems.The OS/400 product 5722JV1 (IBM Developer Kit for Java) and its option 5 (Java Developer Kit 1.3) need to be restored from the operating system CD B29XX_08, where XX is the two digit language identifier. No IPL or activation of Java code is necessary.This change affects new and existing V5R2 Electronic Service Agent inventory configurations.。
中控考勤机iFace系列脱机通讯开发包开发手册
iFace 系列脱机通讯开发包 开发手册
日期: 2012 年7月
2012 ZKSoftware Inc.。版权所有，翻印必究。 ·本说明文件中的信息如有更改，恕不另行通知。 ·ZKSoftware Inc.为中控科技的司标，Biokey 为中控科技的核心技术徽标，ZKSoftware ,Biokey 已经在中国 及美国注册。 ·本说明文件中述及的其他商标和产品名称是指拥有相应商标和产品名称的公司或者制造的产品。 ZKSoftware Inc.对其他公司的商标和产品名称不拥有任何所有权。 ·开发中使用问题，非开发包问题，请自行解决。 ·使用该开发包，由于人为或程序本身造成数据丢失，本公司概不负责。 ·该文档基于 6.2.4.1 版 SDK
目录
1. SDK 介绍 .................................................................................................................................................................... 1 2. 快速开始 ................................................................................................................................................................... 2 2.1 常用词介绍....................................................
EMC_Network培训
8
与传统NetWorker比较
NetWorker Fast Start (7.4SP2)
预打包的solution 定制环境 特定配置
NetWorker 7.4
复杂环境 通用配置
手动选择并安装应用模块 手动定义备份到磁盘的设备 手动定义 clients, groups, schedules, etc
9
文档
除了 Install Guide and Release Notes外,没有新的文档
– NW Fast Start 没有增加和改变传 统NetWorker的功能
新的安装简介
一个订货号订购所有的模块 一个许可证激活所有的功能 自动安装所有备份服务器模块,减少人工干预
不同的定货号针对不同的模块 每个特定功能模块都需要一个许可证 安装时较多的可选项
自动检测应用程序并安装相应的模块
自动生成备份到磁盘的目录 自动生成 NetWorker clients, groups, schedules, etc 最短时间开始备份作业
➢ 临时启用: 45 天
▪ 通过 EMC PowerLink获取
2) 购买: 90 天测试和评估结束以后
▪ 软件注册和永久许可 ▪ PowerLink 注册和授权程序 ▪ 需要NetWorker服务器ID来获取永久授权码
Test drive NetWorker Free for 90 days
Evaluation 30 days
11
Today’s Topic
BuRA Partner Training
NetWorker 配置简化更新
N5K ISSU升级步骤
2、show version 查看升级前系统版本；
3、dir 查看升级前系统版本文件；
4、show fex 查看升级Fra bibliotekfex在线状态；
5、copy 目标升级版本到bootflash：
copy usb1:kickstart.bin bootflash:kickstart.bin
交换机reload：Y
10、确认fex正常升级及在线：show fex
11、变更失败，回退；
boot system bootflash：system_old.bin
reload
12、查看回退后版本状态及fex状态；
show version
show fex
8、show install all impact kickstart bootflash：kickstart.bin system bootflash:system.bin
预安装输出信息，验证目标版本升级安装的兼容性及影响
9、版本升级
install all kickstart bootflash：kickstart.bin system bootflash:system.bin
copy usb1:system.bin bootflash:system.bin
6、确认设备是否需要配置备份：
copy running-config usb1：n5k_backup_20160602.cfg(此条命令需要验证，如不成功，更换usb1为tftp等)
7、dir 确认目标系统版本copy成功；
N5K VPC功能配置及测试要点
N5K VPC功能配置及测试2013年01月目录一、概述 (3)1.1 测试目的 (3)1.2测试环境： (3)1.3配置前准备 (4)1.3.1配置管理地址 (4)1.3.2安装存储license (4)1.3.2N5K操作系统升级 (4)三、VPC功能测试 (5)3.1 测试目的 (5)3.2 测试拓扑 (6)3.3 测试配置 (6)3.4测试结果 (7)四、VPC兼容性测试 (11)4.1 测试目的 (11)4.2 测试拓扑 (11)4.3 测试配置 (11)4.4测试结果 (12)五、EVPC功能测试 (15)5.1 测试目的 (15)5.2 测试拓扑 (16)5.3 测试配置 (16)5.4测试结果 (17)一、概述1.1 测试目的Nexus 5500交换机作为业界第一款统一端口交换机，同时支持ethernet、 Fibre Channel和Fibre Channel over Ethernet (FCOE)协议。
本次测试的目的是为了测试Nexus 5500交换机的功能，以验证是否适合作为服务器区的汇聚交换机，同时实现服务器区的以太网汇聚和SAN存储汇聚，从而逐渐向两网融合的方向演进。
.1.2测试环境：1．主机：1台IBM P5502．系统：AIX3．FC HBA：emulex 4Gb double-port4．盘柜：HDS 23005．网络设备：两台Nexus 5596，一台Nexus 2148，一台Nexus 2248，一台Catalyst 3560G 以下是这次测试方案的拓扑。
1.3配置前准备1.3.1配置管理地址通过Console线连接到两台Nexus 5596的管理口上，分别配置两台设备的管理IP地址在同一网段。
在本例中Nexus 5596-1的管理ip是168.7.63.11，Nexus 5596-2的管理ip是168.7.63.12。
如下图所示，管理站和A、B控制器正常连接。
7k5k_Nexus update_Aug
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chemical-reaction-engineering-3ed-edition作者-octave-Levenspiel-课后习题答案
chemical-reaction-engineeri ng-3ed-edition作者-octave-Levenspiel-课后习题答案Corresponding Solutions for Chemical Reaction EngineeringCHAPTER 1 OVERVIEW OF CHEMICAL REACTION ENGINEERING (1)CHAPTER 2 KINETICS OF HOMOGENEOUS REACTIONS (3)CHAPTER 3 INTERPRETATION OF BATCH REACTOR DATA (7)CHAPTER 4 INTRODUCTION TO REACTOR DESIGN (20)CHAPTER 5 IDEAL REACTOR FOR A SINGLE REACTOR (23)CHAPTER 6 DESIGN FOR SINGLE REACTIONS (27)CHAPTER 10 CHOOSING THE RIGHT KIND OF REACTOR (34)CHAPTER 11 BASICS OF NON-IDEAL FLOW (36)CHAPTER 18 SOLID CATALYZED REACTIONS (45)Chapter 1 Overview of Chemical Reaction Engineering1.1 Municipal waste water treatment plant. Consider a municipal water treatment plant for a small community (Fig.P1.1). Waste water, 32000 m 3/day, flows through the treatment plant with a mean residence time of 8 hr, air is bubbled through the tanks, and microbes in the tank attack and break down the organic material (organic waste) +O 2 −−−→−microbes CO 2 + H 2OA typical entering feed has a BOD (biological oxygen demand) of 200 mg O 2/liter, while the effluent has a megligible BOD. Find the rate of reaction, or decrease in BOD in the treatment tanks.Figure P1.1Solution:)/(1017.2)/(75.183132/100010001)0200()(313200031320001343333s m mol day m mol day molgm L mg g L mg day day m dayday m VdtdN r A A ⋅⨯=⋅=-⨯⨯⨯-⨯-=-=--1.2 Coal burning electrical power station. Large central power stations (about 1000 MW electrical) using fluiding bed combustors may be built some day (see Fig.P1.2). These giants would be fed 240 tons of coal/hr (90% C, 10%H 2), 50% of which would burn within the battery of primary fluidized beds, the other 50% elsewhere in the system. One suggested design would use a battery of 10 fluidized beds, each 20 m long, 4 m wide, and containing solids to a depth of 1 m. Find the rate of reaction within theWaste Waste Clean200 mgMean residenZerobeds, based on the oxygen used.Solution:380010)1420(m V =⨯⨯⨯=)/(9000101089.05.01024033hr bed molc hrkgckgcoal kgc hr coal t N c ⋅-=⨯-=⨯⨯⨯-=∆∆ )/(25.111900080011322hr m kmolO t N V r r c c O ⋅=-⨯-=∆∆-=-=)/(12000412000190002hr bed mol dt dO ⋅=+⨯= )/(17.4800)/(105.113422s m mol hr bed mol dt dO V r O ⋅=⋅⨯==-Chapter 2 Kinetics of Homogeneous Reactions2.1 A reaction has the stoichiometric equation A + B =2R . What is the order of reaction?Solution: Because we don’t know whether it is an elementary reaction or not, we can’t tell the index of the reaction.2.2 Given the reaction 2NO 2 + 1/2 O 2 = N 2O 5 , what is the relation between the ratesof formation and disappearance of the three reaction components? Solution: 522224O N O NO r r r =-=-2.3 A reaction with stoichiometric equation 0.5 A + B = R +0.5 S has the following rateexpression-r A = 2 C 0.5 A C BWhat is the rate expression for this reaction if the stoichiometric equation is written asA + 2B = 2R + SSolution: No change. The stoichiometric equation can’t effect the rate equation, so it doesn’t change.2.4 For the enzyme-substrate reaction of Example 2, the rate of disappearance ofsubstrate is given by-r A =A06]][[1760C E A + , mol/m 3·sWhat are the units of the two constants? Solution: ][]6[]][][[][03A A C E A k s m mol r +=⋅=- 3/][]6[m mol C A ==∴sm mol m mol m mol s m mol k 1)/)(/(/][3333=⋅⋅=2.5 For the complex reaction with stoichiometry A + 3B → 2R + S and withsecond-order rate expression-r A = k 1[A][B]are the reaction rates related as follows: r A = r B = r R ? If the rates are not so related, then how are they related? Please account for the sings , + or - .Solution: R B A r r r 2131=-=-2.6 A certain reaction has a rate given by-r A = 0.005 C 2 A , mol/cm 3·min If the concentration is to be expressed in mol/liter and time in hours, what wouldbe the value and units of the rate constant?Solution:min)()(3'⋅⨯-=⋅⨯-cm molr hr L mol r A A 22443'300005.0106610)(minAA A A A C C r r cm mol mol hr L r =⨯⨯=⋅⨯=-⋅⋅⋅=-∴ AA A A A C C cmmol mol L C cmmolC L mol C 33'3'10)()(=⋅⋅=∴⨯=⨯2'42'32'103)10(300300)(AA A A C C C r --⨯=⨯==-∴ 4'103-⨯=∴k2.7 For a gas reaction at 400 K the rate is reported as -dtdp A= 3.66 p 2 A , atm/hr (a) What are the units of the rate constant?(b) What is the value of the rate constant for this reaction if the rate equation isexpressed as-r A = - dtdN V A1 = k C2 A , mol/m 3·s Solution:(a) The unit of the rate constant is ]/1[hr atm ⋅ (b) dtdN V r AA 1-=-Because it’s a gas reaction occuring at the fined terperatuse, so V=constant, and T=constant, so the equation can be reduced to22)(66.366.3)(1RT C RTP RT dt dP RT dt dP VRT V r A A A A A ==-=-=-22)66.3(A A kC C RT ==So we can get that the value of1.12040008205.066.366.3=⨯⨯==RT k2.9 The pyrolysis of ethane proceeds with an activation energy of about 300 kJ/mol.How much faster the decomposition at 650℃ than at 500℃?Solution:586.7)92311731()10/(314.8/300)11(3211212=-⋅⋅=-==KK K mol kJ mol kJ T T R E k k Ln r r Ln7.197012=∴r r2.11 In the mid-nineteenth century the entomologist Henri Fabre noted that French ants (garden variety) busily bustled about their business on hot days but were rather sluggish on cool days. Checking his results with Oregon ants, I findRunning speed, m/hr150160230295370Temperatu re, ℃13 16 22 24 28 What activation energy represents this change in bustliness? Solution:RTE RTE RTE ek eak t cons ion concentrat f let ion concentrat f ek r ---=⋅⋅=⋅='00tan )()(RET Lnk Lnr A 1'-=∴ Suppose Tx Lnr y A 1,==, so ,REslope -= intercept 'Lnk =)/(1-⋅h m r A 150 160 230 295 370 A Lnr-3.1780 -3.1135 -2.7506 -2.5017 -2.2752CT o / 13 16 22 24 28 3101-⨯T3.4947 3.4584 3.3881 3.3653 3.3206-y = 5417.9x - 15.686R2 = 0.9712340.00330.003350.00340.003450.00351/T-L n r-y = -5147.9 x + 15.686Also K REslope 9.5147-=-=, intercept 'Lnk == 15.686 , mol kJ K mol J K E /80.42)/(3145.89.5147=⋅⨯-=Chapter 3 Interpretation of Batch Reactor Data3.1 If -r A = - (dC A /dt) =0.2 mol/liter·sec when C A = 1 mol/liter, what is the rate ofreaction when C A = 10 mol/liter? Note: the order of reaction is not known.Solution: Information is not enough, so we can’t answer this kind of question.3.2 Liquid a sedomposes by first-order kinetics, and in a batch reactor 50% of A isconverted in a 5-minute run. How much longer would it take to reach 75% conversion?Solution: Because the decomposition of A is a 1st -order reaction, so we can express the rate equation as:A A kC r =-We know that for 1st -order reaction, kt C C LnAAo=, 11kt C C LnA Ao =, 22kt C CLn A Ao = Ao A C C 5.01=, Ao A C C 25.02=So 21)24(1)(11212Ln kLn Ln k C C Ln C C Ln k t t A Ao A Ao =-=-=- equ(1) min 521)(111===Ln kC C Ln k t A Ao equ(2) So m in 5112==-t t t3.3 Repeat the previous problem for second-order kinetics. Solution: We know that for 2nd -order reaction, kt C C A A =-011, So we have two equations as follow:min 511211101k kt C C C C C AoAo Ao A A ===-=-, equ(1)2123)1(31411kt kt C C C C C AoAo Ao Ao A ===-=-, equ(2) So m in 15312==t t , m in 1012=-t t3.4 A 10-minute experimental run shows that 75% of liquid reactant is converted to product by a 21-order rate. What would be the fraction converted in a half-hour run?Solution: In a-21order reaction: 5.0AA A kC dt dC r =-=-, After integration, we can get:5.015.02A Ao C C kt -=, So we have two equations as follow:min)10(5.0)41(15.05.05.05.015.0k kt C C C C C Ao Ao AoA Ao ===-=-, equ(1) min)30(25.025.0k kt C C A Ao ==-, equ(2)Combining these two equations, we can get:25.05.1kt C Ao =, but this means 05.02<A C , whichis impossible, so we can conclude that less than half hours, all the reactant is consumed up. So the fraction converted 1=A X .3.5 In a hmogeneous isothermal liquid polymerization, 20% of the monomer disappears in 34 minutes for initial monomer concentration of 0.04 and also for 0.8 mol/liter. What rate equation represents the disappearance of the monomer?Solution: The rate of reactant is independent of the initial concentration of monomers, so we know the order of reaction is first-order,monomer monomer kC r =-And k C C Lnoomin)34(8.0= 1min 00657.0-=kmonomer monomer C r )min 00657.0(1-=-3.6 After 8 minutes in a batch reactor, reactant (C A0 = 1 mol/liter) is 80% converted; after 18 minutes, conversion is 90%. Find a rate equation to represent this reaction. Solution:In 1st order reaction, 43.1511111111212==--=Ln Ln X Lnk X Ln k t t A A , dissatisfied.In 2nd order reaction, 49/4/912.0111.01)11(1)11(11212==--=--=Ao Ao Ao Ao Ao Ao Ao A Ao A C C C C C C C C k C C k t t, satisfied.According to the information, the reaction is a 2nd -order reaction.3.7 nake-Eyes Magoo is a man of habit. For instance, his Friday evenings are all alike —into the joint with his week’s salary of ＄180, steady gambling at “2-up” for two hours, then home to his family leaving ＄45 behind. Snake Eyes’s betting pattern is predictable. He always bets in amounts proportional to his cash at hand, and his losses are also predictable —at a rate proportional to his cash at hand. This week Snake-Eyes received a raise in salary, so he played for three hours, but as usual went home with ＄135. How much was his raise? Solution:180=Ao n , 13=A n , h t 2=,135'=A n , h t 3;=, A A kn r α-So we obtain kt n n LnAAo=, ''')()(tn n Ln t n n Ln AAo A Ao= 3135213180'Ao n Ln Ln =, 28'=An3.9 The first-order reversible liquid reactionA ↔ R , C A0 = 0.5 mol/liter, C R0=0takes place in a batch reactor. After 8 minutes, conversion of A is 33.3% while equilibrium conversion is 66.7%. Find the equation for the this reaction. Solution: Liquid reaction, which belongs to constant volume system,1st order reversible reaction, according to page56 eq. 53b, we obtain121112102110)(1)(-+-+=+-==⎰⎰AX A A tX k k k k Lnk k X k k k dX dt t Amin 8sec 480==t , 33.0=A X , so we obtain eq(1)33.0)(1min8sec 480211121k k k k Ln k k +-+= eq(1) Ae AeAe c X X M C C k k K -+===1Re 21, 0==AoRo C C M , so we obtain eq(2) 232132121=-=-==AeAe c X X k k K ,212k k =∴ eq(2)Combining eq(1) and eq(2), we obtain1412sec 108.4m in 02888.0---⨯==k 14121sec 1063.9m in 05776.02---⨯===k kSo the rate equation is )(21A Ao A AA C C k C k dtdC r --=-=- )(sec 1063.9sec 108.401414A A A C C C -⨯-⨯=----3.10 Aqueous A reacts to form R (A→R) and in the first minute in a batch reactor itsconcentration drops from C A0 = 2.03 mol/liter to C Af = 1.97 mol/liter. Find the rate equation from the reaction if the kinetics are second-order with respect to A.Solution: It’s a irreversible second -order reaction system, according to page44 eq 12, we obtainmin 103.2197.111⋅=-k , so min015.01⋅=mol Lkso the rate equation is 21)min 015.0(A A C r -=-3.15 At room temperature sucrose is hydrolyzed by the catalytic action of the enzymesucrase as follows:Aucrose −−→−sucraseproductsStarting with a sucrose concentration C A0 = 1.0 millimol/liter and an enzyme concentrationC E0= 0.01 millimol/liter, the following kinetic data are obtained in a batch reactor (concentrations calculated from optical rotation measurements):Determine whether these data can be reasonably fitted by a knietic equation of the Michaelis-Menten type, or-r A =MA E A C C C C k +03 where C M = Michaelis constantIf the fit is reasonable, evaluate the constants k 3 and C M . Solve by the integral method.Solution: Solve the question by the integral method:AA M A A Eo A A C k Ck C C C C k dt dC r 5431+=+=-=-, M Eo C C k k 34=, MC k 15= AAo A Ao A Ao C C C C Lnk k k C C t -⋅+=-4451hrt ,AC ,mmol /L A Ao AAo C C C C Ln-AAo C C t -1 0.84 1.0897 6.25 20.681.20526.25C A , millimol /liter0.84 0.68 0.53 0.38 0.27 0.16 0.09 0.04 0.018 0.006 0.0025t,hr 1 2 3 4 5 6 7 8 9 10 113 0.53 1.3508 6.38304 0.38 1.5606 6.45165 0.27 1.7936 6.8493 6 0.16 2.1816 7.14287 0.09 2.6461 7.69238 0.04 3.3530 8.33339 0.018 4.0910 9.1650 10 0.006 5.1469 10.0604 110.00256.006511.0276Suppose y=A Ao C C t-, x=AAo A Ao C C C C Ln-, thus we obtain such straight line graphy = 0.9879x + 5.0497R 2 = 0.99802468101201234567Ln(Cao/Ca)/(Cao-Ca)t /(C a o -C a )9879.0134===Eo M C k C k Slope , intercept=0497.545=k k So )/(1956.00497.59879.015L mmol k C M ===, 14380.1901.09879.01956.0-=⨯==hr C C k k Eo M3.18 Enzyme E catalyzes the transformation of reactant A to product R as follows: A −−→−enzyme R, -r A =min22000⋅+liter molC C C A E AIf we introduce enzyme (C E0 = 0.001 mol/liter) and reactant (C A0 = 10mol/liter) into a batch rector and let the reaction proceed, find the time needed for the concentration of reactant to drop to 0.025 mol/liter. Note that the concentration of enzyme remains unchanged during the reaction.. Solution:510001.020021+=⨯+=-=-AA A A A C C C dC dt r Rearranging and integrating, we obtain:10025.0025.0100)(510)510(⎥⎦⎤⎢⎣⎡-+=+-==⎰⎰A Ao A Ao A A tC C C C Ln dC C dt t min 79.109)(5025.01010=-+=A Ao C C Ln3.20 M.Hellin and J.C. Jungers, Bull. soc. chim. France, 386(1957), present the data in Table P3.20 on thereaction of sulfuric acid with diethylsulfate in a aqueous solution at22.9℃:H 2SO 4 + (C 2H 5)2SO 4 → 2C 2H 5SO 4HInitial concentrations of H 2SO 4 and (C 2H 5)2SO 4 are each 5.5 mol/liter. Find a rate equation for this reaction.Table P3.20 t, minC 2H 5SO 4H , mol/li ter t, minC 2H 5SO 4H , mol/li ter1804.1141 1.18 194 4.31 48 1.38 212 4.45 55 1.63 267 4.86 75 2.24 318 5.15 96 2.75 368 5.32 127 3.31 379 5.35 146 3.76 410 5.42 1623.81∞(5.80)Solution: It’s a constant -volume system, so we can use X A solving the problem: i) We postulate it is a 2nd order reversible reaction system R B A 2⇔+ The rate equation is: 221R B A A A C k C C k dtdC r -=-=- L mol C C Bo Ao /5.5==, )1(A Ao A X C C -=, A A Ao Bo B C X C C C =-=, A Ao R X C C 2=When ∞=t , L mol X C C Ae Ao /8.52Re == So 5273.05.528.5=⨯=Ae X , L mol X C C C Ae Ao Be Ae /6.2)5273.01(5.5)1(=-⨯=-== After integrating, we obtaint C X k X X X X X LnAo AeA Ae A Ae Ae )11(2)12(1-=--- eq (1)The calculating result is presented in following Table.t,mi nLmol C R /,Lmol C A /,AXAAe AAe Ae X X X X X Ln---)12()1(AeAX X Ln -0 0 5.5 0 0 041 1.18 4.91 0.10730.2163 -0.227548 1.38 4.81 0.12540.2587 -0.271755 1.63 4.685 0.14820.3145 -0.329975 2.24 4.38 0.20360.4668 -0.488196 2.75 4.125 0.25 0.6165 -0.642712 7 3.31 3.8450.30090.8140 -0.845614 6 3.76 3.620.34181.0089 -1.044916 2 3.81 3.5950.34641.0332 -1.069718 0 4.11 3.4450.37361.1937 -1.233119 4 4.31 3.3450.39181.3177 -1.359121 2 4.45 3.2750.40451.4150 -1.4578267 4.86 3.07 0.4418 1.7730 -1.8197 318 5.15 2.925 0.4682 2.1390 -2.1886 368 5.32 2.84 0.4836 2.4405 -2.4918 379 5.35 2.825 0.4864 2.5047 -2.5564 4105.42 2.79 0.4927 2.6731 -2.7254 ∞5.82.60.5273——Draw AAe AAe Ae X X X X X Ln---)12(~ t plot, we obtain a straight line:y = 0.0067x - 0.0276R 2= 0.998800.511.522.530100200300400500tL n0067.0)11(21=-=Ao AeC X k Slope ,min)/(10794.65.5)15273.01(20067.041⋅⨯=⨯-=∴-mol L kWhen approach to equilibrium, BeAe c C C C k k K 2Re 21==, so min)/(10364.18.56.210794.642242Re 12⋅⨯=⨯⨯==--mol L C C C k k Be Ae So the rate equation ism in)/()10364.110794.6(244⋅⨯-⨯=---L mol C C C r R B A Aii) We postulate it is a 1st order reversible reaction system, so the rate equation isR A AA C k C k dtdC r 21-=-=- After rearranging and integrating, we obtaint k X X X Ln AeAe A '11)1(=-eq (2) Draw )1(AeAX X Ln -~ t plot, we obtain another straight line: -y = 0.0068x - 0.0156R 2 = 0.998600.511.522.530100200300400500x-L n0068.0'1-==AeX k Slope ,So 13'1m in 10586.35273.00068.0--⨯-=⨯-=k133Re '1'2min 10607.18.56.210586.3---⨯-=⨯⨯-==C C k k AeSo the rate equation ism in)/()10607.110586.3(33⋅⨯+⨯-=---L mol C C r R A AWe find that this reaction corresponds to both a 1st and 2nd order reversible reaction system, by comparing eq.(1) and eq.(2), especially when X Ae =0.5 , the two equations are identical. This means these two equations would have almost the same fitness of data when the experiment data of the reaction show that X Ae =0.5.(The data that we use just have X Ae =0.5273 approached to 0.5, so it causes to this.)3.24 In the presence of a homogeneous catalyst of given concentration, aqueous reactant A is converted to product at the following rates, and C A alone determines this rate:C A ,mol/liter1 2 4 6 7 9 12-r A , mol/liter·hr0.06 0.1 0.25 1.0 2.0 1.0 0.5We plan to run this reaction in a batch reactor at the same catelyst concentration as used in getting the above data. Find the time needed to lower the concentration of A from C A0 = 10 mol/liter to C Af = 2 mol/liter.Solution: By using graphical integration method, we obtain that the shaped area is 50 hr.04812162002 4 68 10 12 14Ca-1/Ra3.31 The thermal decomposition of hydrogen iodide 2HI → H 2 + I 2is reported by M.Bodenstein [Z.phys.chem.,29,295(1899)] as follows:T,℃ 508427 393 356 283k,cm 3/mol·s0.10590.003100.00058880.9×10-60.942×10-6Find the complete rate equation for this reaction. Use units of joules, moles, cm 3,and seconds.According to Arrhenius’ Law,k = k 0e -E/R Ttransform it,- In(k) = E/R·(1/T) －In(k 0)Drawing the figure of the relationship between k and T as follows:y = 7319.1x - 11.567R 2= 0.987904812160.0010.0020.0030.0041/T-L n (k )From the figure, we getslope = E/R = 7319.1 intercept = - In(k 0) = -11.567E = 60851 J/mol k 0 = 105556 cm 3/mol·sFrom the unit [k] we obtain the thermal decomposition is second-order reaction, so the rate expression is- r A = 105556e -60851/R T ·C A 2Chapter 4 Introduction to Reactor Design4.1 Given a gaseous feed, C A0 = 100, C B0 = 200, A +B→ R + S, X A = 0.8. Find X B ,C A ,C B . Solution: Given a gaseous feed, 100=Ao C , 200=Bo C , S R B A +→+0=A X , find B X , A C , B C0==B A εε, 202.0100)1(=⨯=-=A Ao A X C C4.02008.01001=⨯⨯==Bo A Ao B C X bC X 1206.0200)1(=⨯=-=B Bo B X C C4.2 Given a dilute aqueous feed, C A0 = C B0 =100, A +2B→ R + S, C A = 20. Find X A , X B , C B .Solution: Given a dilute aqueous feed, 100==Bo Ao C C ,S R B A +→+2, 20=A C , find A X , B X , B CAqueous reaction system, so 0==B A εε When 0=A X , 200=V When 1=A X , 100=VSo 21-=A ε, 41-==Ao Bo A B bC C εε8.01002011=-=-=Ao A A C C X , 16.11008.010012>=⨯⨯=⋅=Bo A Ao B C X C a b X , which is impossible. So 1=B X , 100==Bo B C C4.3 Given a gaseous feed, C A0 =200, C B0 =100, A +B→ R, C A = 50. Find X A , X B , C B . Solution: Given a gaseous feed, 200=Ao C , 100=Bo C ,R B A →+, 50=A C .find A X , B X , B C75.02005011=-=-=Ao A A C C X , 15.1>==BoAAo B C X bC X , which is impossible. So 100==Bo B C C4.4 Given a gaseous feed, C A0 = C B0 =100, A +2B→ R, C B = 20. Find X A , X B , C A . Solution: Given a gaseous feed, 100=+Bo Ao C C ，R B A →+2, 20=Bo C , Find A X , B X , A C0=B X , 200100100=+=B A V ,1=B X 15010050=+=R A V25.0200200150-=-=B ε, 5.01002110025.0-=⨯⨯-=-A ε842.02025.010020100=⨯--=B X , 421.0100842.010021=⨯⨯=A X34.73421.05.01421.0110011=⨯--⨯=+-=A A A AoA X X C C ε4.6 Given a gaseous feed, T 0 =1000 K, π0＝5atm, C A0=100, C B0=200, A +B→5R,T =400K, π=4atm, C A =20. Find X A , X B , C B .Solution: Given a gaseous feed, K T o 1000=, atm 50=π, 100=Ao C , 200=Bo CR B A 5→+, K T 400=, atm 4=π, 20=A C , find A X , B X , B C .1300300600=-=A ε, 2==Ao Bo AB bC C a εε,5.0410********=⨯⨯=ππT T According to eq page 87,818.05.010020115.0100201110000=⨯⨯+⨯-=+-=ππεππT T C C T T C C X Ao A AAo A A409.0200818.0100=⨯==Bo A Ao B aC X bC X130818.011200)818.0100200(1)(0=⨯+⨯-=+-=A A Ao A Ao Bo B X C T T X a b C C C εππ4.7 A Commercial Popcorn Popping Popcorn Popper. We are constructing a 1-literpopcorn to be operatedin steady flow. First tests in this unit show that 1 liter/min of raw corn feed stream produces 28 liter/minof mixed exit stream. Independent tests show that when raw corn pops its volumegoes from 1 to 31.With this information determine what fraction of raw corn is popped in the unit.Solution: 301131=-=A ε, ..1u a C Ao =, ..281281u a C C Ao A ==%5.462813012811=⨯+-=+-=∴AA Ao A Ao A C C C C X εChapter 5 Ideal Reactor for a single Reactor5.1 Consider a gas-phase reaction 2A → R + 2S with unknown kinetics. If a spacevelocity of 1/min is needed for 90% conversion of A in a plug flow reactor, find the corresponding space-time and mean residence time or holding time of fluid in the plug flow reactor.Solution: min 11==sτ,Varying volume system, so t can’t be found.5.2 In an isothermal batch reactor 70% of a liquid reactant is converted in 13 min.What space-time and space-velocity are needed to effect this conversion in a plug flow reactor and in a mixed flow reactor? Solution: Liquid reaction system, so 0=A ε According to eq.4 on page 92, min 130=-=⎰AX AAAo r dC C t Eq.13, AAAo A A Ao R F M r X C r C C -=--=..τ, R F M ..τ can’t be cert ain. Eq.17, ⎰-=AX AAAo R F P r dX C 0..τ, so m in 13...==R B R F P t τ5.4 We plan to replace our present mixed flow reactor with one having double thebolume. For the same aqueous feed (10 mol A/liter) and the same feed rate find the new conversion. The reaction are represented byA → R, -r A = kC 1.5 ASolution: Liquid reaction system, so 0=A εA A Ao Ao r X C F V -==τ, 5.1)]1([)(A Ao A A Ao A Ao X C k X r C C C -=-- Now we know: V V 2=', Ao Ao F F =', Ao Ao C C =', 7.0=A X So we obtain5.15.15.15.1)1()2)1(2A Ao A A Ao A Ao Ao X kC X X kC X F VF V -='-'==''52.8)7.01(7.02)1(5.15.1=-⨯='-'∴A AX X794.0='A X5.5 An aqueous feed of A and B (400liter/min, 100 mmol A/liter, 200 mmol B/liter) isto be converted to product in a plug flow reactor. The kinetics of the reaction is represented byA +B→ R, -r A = 200C A C Bmin⋅liter molFind the volume of reactor needed for 99.9% conversion of A to product.Solution: Aqueous reaction system, so 0=A εAccording to page 102 eq.19,⎰⎰-=-==Af AfX AA X A A AoAo Ao r dX r dC C C t F V 001⎰-==AfX AAAo or dX C Vντ, m in /400liter o =ν, L r dX r dX C V AAX A A o Ao Af3.1244001.0999.000=-⨯=-=∴⎰⎰ν5.9 A specific enzyme acts as catalyst in the fermentation of reactant A. At a givenenzyme concentration in the aqueous feed stream (25 liter/min) find the volume of plug flow reactor needed for 95% conversion of reactant A (C A0 =2 mol/liter ). The kinetics of the fermentation at this enzyme concentration is given byA −−→−enzymeR , -r A = litermolC C A A ⋅+min 5.011.0Solution: P.F.R, according to page 102 eq.18, aqueous reaction, 0=ε⎰-=A X AA Ao r dX F V 0 )11(21251.05.010A AX A A A Ao X X Ln dX C C F V A+-⨯=+=∴⎰\L Ln4.986)95.005.01(125=+=5.11 Enzyme E catalyses the fermentation of substrate A (the reactant) to product R.Find the size of mixed flow reactor needed for 95% conversion of reactant in a feed stream (25 liter/min ) of reactant (2 mol/liter) and enzyme. The kinetics of the fermentation at this enzyme concentration are given byA −−→−enzyme R , -r A =litermolC C A A ⋅+min 5.011.0Solution: min /25L o =ν, L mol C Ao /2=, m in /50mol F Ao =, 95.0=A X Constant volume system, M.F.R., so we obtainmin 5.199205.05.01205.01.095.02=⨯⨯+⨯⨯⨯=-==AAAo or X C Vντ,39875.4min /25min 5.199m L V o =⨯==τν5.14 A stream of pure gaseous reactant A (C A0 = 660 mmol/liter) enters a plug flowreactor at a flow rate of F A0 = 540 mmol/min and polymerizes the as follows3A → R, -r A = 54min⋅liter mmolHow large a reactor is needed to lower the concentration of A in the exitstream to C Af = 330 mmol/liter?Solution: 321131-=-=A ε, 75.0660330321660330111=⨯--=+-=Ao A A Ao A A C C C C X ε 0-order homogeneous reaction, according to page 103 eq.20A Ao AoAooX C F VC kVkk ===ντ So we obtainL X k C C F V A Ao Ao Ao 5.75475.05401=⨯==5.16 Gaseous reactant A decomposes as follows:A → 3 R, -r A = (0.6min -1)C AFind the conversion of A in a 50% A – 50% inert feed (υ0 = 180 liter/min, C A0 =300 mmol/liter) to a 1 m 3 mixed flow reactor.Solution: 31m V =, M.F.R. 1224=-=A εAccording to page 91 eq.11, AAAoAAo AAAo oX X C X C r X C V+-=-==116.0ντmin/1801000)1(6.0)1(L LX X X A A A =-+=So we obtain 667.0=A XChapter 6 Design for Single Reactions6.1 A liquid reactant stream (1 mol/liter) passes through two mixed flow reactors in aseries. The concentration of A in the exit of the first reactor is 0.5 mol/liter. Find the concentration in the exit stream of the second reactor. The reaction is second-order with respect to A and V 2/V 1 =2.Solution:V 2/V 1 = 2, τ1 =011υV =A A A r C C --10 , 2τ = 022υV = 221A A A r C C --C A0=1mol/l , C A1=0.5mol/l , 0201υυ=, -r A1=kC 2 A1 ,-r A2=kC 2 A2 (2nd-order) , 2×2110A A A kC C C -=2221A A A kC C C - So we obtain 2×(1-0.5)/(k0.52)=(0.5-C A2)/(kC A22)C A2= 0.25 mol/l6.2 Water containing a short-lived radioactive species flows continuously through awell-mixed holdup tank. This gives time for the radioactive material to decay into harmless waste. As it now operates, the activity of the exit stream is 1/7 of the feed stream. This is not bad, but we’d like to lower it still more.One of our office secretaries suggests that we insert a baffle down the middle ofthe tank so that the holdup tank acts as two well-mixed tanks in series. Do you think this would help? If not, tell why; if so calculate the expected activity of the exit stream compared to the entering stream.Solution: 1st-order reaction, constant volume system. From the information offeredabout the first reaction,we obtain1τ=01100117171A A A A A A C k C C kC C C V ⋅-=-=υ If a baffle is added,022220212122212υυτττV V +=+==011υV =2222221210A A A A A A kC C C kC C C -+-=007176A A kC C =6/k …… ①。
Update .NET Framework 3.5 on Windows 8, Windows Server 2012 R2
Update for the .NET Framework 3.5 on Windows 8, Windows 8.1, Windows Server 2012, and Windows Server 2012 R2INTRODUCTIONThis update resolves an issue that prevents the optional Microsoft .NET Framework 3.5 feature from being enabled after you install security update 2966827 or 2966828 (described in Microsoft Security Bulletin MS14-046) for the Microsoft .NET Framework 3.5.This update applies to Windows 8, Windows Server 2012, Windows 8.1, and Windows Server 2012 R2.SummaryMicrosoft has released update 3005628 for the .NET Framework 3.5 on Windows 8, Windows Server 2012, Windows 8.1, and Windows Server 2012 R2.∙Individual, small-business, and organizational users should use the Windowsautomatic-updating feature to install the fixes from Microsoft Update. To do this, see Get security updates automatically .∙ IT professionals should refer to the "Download information" and "Command-line switchesfor this update" sections. SymptomsAfter you install security update 2966827 or 2966828 (described in Microsoft Security Bulletin MS14-046) for the Microsoft .NET Framework 3.5 and then try to enable the optional Microsoft .NET Framework 3.5 feature in Windows Features for the first time, the feature may not be enabled.When this problem occurs, you may receive an error message that resembles one of the following.Error code Error messages0x800F0906 The source files could not be downloaded.Use the "source" option to specify the location of the files that are requiredto restore the feature. For more information on specifying a source location,see /fwlink/?LinkId=243077.The DISM log file can be found at C:\Windows\Logs\DISM\dism.logWindows couldn't complete the requested changes.Windows couldn't connect to the Internet to download necessary files. Make surethat you're connected to the Internet, and click "Retry" to try again.Installation of one or more roles, role services, or features failed.The source files could not be found. Try installing the roles, role services,or features again in a new Add Roles and Features Wizard session, and on theConfirmation page of the wizard, click "Specify an alternate source path" tospecify a valid location of the source files that are required for theinstallation. The location must be accessible by the computer account of thedestination server.0x800F0906 - CBS_E_DOWNLOAD_FAILURE0x800F081F The source files could not be found.Use the "Source" option to specify the location of the files that are requiredto restore the feature. For more information on specifying a source location,see /fwlink/?LinkId=243077. The DISM log file can be found at C:\Windows\Logs\DISM\dism.log0x800F081F - CBS_E_SOURCE_MISSINGError code: 0x800F081FError: 0x800F081FCauseSecurity updates 2966827 and 2966828 (described in Microsoft Security Bulletin MS14-046) for the .NET Framework 3.5 require metadata that is added to the system only when the Microsoft .NET Framework 3.5 feature is enabled. Updates that apply to the .NET Framework3.5 are meant to be installed only after the .NET Framework 3.5 feature is installed. This issue is also described in the following Microsoft Knowledge Base: 3002547 Enabling the Microsoft .NET Framework 3.5 optional Windows feature in Windows 8, Windows Server 2012, Windows 8.1, or Windows Server 2012 R2 may fail after you install security update 2966827 or 2966828ResolutionTo resolve this issue, install update 3005628.Update 3005628 will remove security update 2966827 or 2966828 from any system that does not have the .NET Framework 3.5 feature content installed and present in Windows 8, Windows Server 2012, Windows 8.1, and Windows Server 2012 R2. Going forward, security updates 2966827 and 2966828 will be offered only to applicable systems that have the .NET Framework 3.5 feature enabled.Customers who have security update 2966828 predeployed to their system by their device manufacturer or who have the security update predeployed by their system administrator by using "/StartComponentCleanup" and "/ResetBase" option of the Deployment Image Servicing and Management (DISM) tool should install update 3005628. See the "Download information" section for links to the appropriate updates.Customers who receive and install update 3005628 from Microsoft Update channels do not have to take any additional action. Individual customers or IT professionals who decide to run update 3005628 manually or within their managed environments should refer to the "Download information" and "Command-line switches for this update" sections for more information. Download informationThe following files are available for download from the Microsoft Download Center.For all supported x86-based systemsDownload the NDPFixit-KB3005628-X86.exe package now.For all supported x64-based systemsDownload the NDPFixit-KB3005628-X64.exe package now.Customers who are aware that the DISM "/StartComponentCleanup" and "/ResetBase"option was used on their system should download and install the following update:For all supported x86-based systemsDownload the Windows8.1-KB2966828-x86.msu package now.For all supported x64-based systemsDownload the Windows8.1-KB2966828-x64.msu package now.Command-line switches for this updateThere are no command line switches that are applicable to this update. The update executable must be run by using elevated administrative credentials.Restart informationThis update does not require a system restart after you apply it unless files that are being updated are locked or are being used.More informationThe information in this article applies to the following:∙The Microsoft .NET Framework 3.5 on Windows 8, Windows Server 2012, Windows 8.1, and Windows Server 2012 R2.Security update patch bulletin .NET Framework for MS14-06 KB2966827 KB2966828 KB3002547 KB3005628 enabling .NET Framework 3.5PropertiesArticle ID: 3005628 - Last Review: 04/23/2015 08:27:00 - Revision: 2.2Applies to∙Microsoft .NET Framework 3.5Keywords:kbexpertiseinter kbinfo kbsecadvisory kbsecurity kbsecvulnerability kbmsifixme kbfixme KB3005628Can’t Install Microsoft .NET Framework 3.5 in Windows 8 and LaterBut there are a few people who have faced problems while installing .NET Framework 3.5. When they try to install it, they receive error messages such as the source files could not be found or downloaded, etc. These error messages contain different error codes such as 0x800F081F or 0x800F0906.To fix these problems, we posted a troubleshooting article which allows you to install .NET Framework 3.5 in Windows 8 and later without receiving any error messages:The above mentioned guide helped a lot of people and fixed the error messages but still there were a few people who were not able to install .NET Framework 3.5 in Windows 8 and later. Then an AskVG reader Patrik posted a different solution which worked for those people.Patrik: I have three Surface Pro 3 with Windows Pro 8.1 that have the same issue. .NET Framework 3.5 will not install. I've tried both via DISM, Programs and Features and via full install setup. I have tried to uninstall every single Windows update on the machine and then install via DISM, but that didn't help either. I even tried to install all three KB-updates (KB2966826-8) to see if I can uninstall them to get rid of potential dll-files that could be interfere with the .Net 3.5 setup. So I then decided to wipe it clean via Windows own restore system guide. But it didn't even work after I have done a clean Windows install. Immediately after Windows installed I tried to enable the .NET 3.5 feature, and of course DISM didn't work either. I have called MS support to get some assistance, but I guess they will call back in a few days or so. Finally, I have solved the issue after many hours. I found a setting in the local computer policy.Today in this article, we are going to share the solution given by our reader Patrik so that all Windows 8 and later users can install .NET Framework 3.5 without any problem.This solution should work for all devices whether its a Desktop PC, laptop or Surface tablet running on Windows 8, Windows 8.1 or Windows 10 operating systems.If you are also facing problems while installing .NET Framework 3.5 in Windows 8 or later, check out following method to fix the issue:1.Type gpedit.msc in RUN or Start search box and press Enter. It'll open Group Policy Editor.2. Now go to:Computer Configuration -> Administrative Templates -> System3. In right-side pane, look for "Specify settings for optional component installation and component repair" option.4.The option would be set to Not Configured. Double-click on it, set it to Enabled and enable "Contact Windows Update directly to download repair content instead of Windows Server Update Services (WSUS)" option.That's it. Now try to install .NET Framework 3.5 again using Control Panel -> Programs and Features or by using DISM command and now you'll not face any problem.Attempting to Install .NET Framework 3.5 on Windows Server 2012 R2 Fails with Error Code 0x800F0906 or “the source files could not be downloaded”, even when supplying sourceCharity ShelbourneCharity ShelbourneMSFT11,886 Points 4 2 1Recent AchievementsBlog Commentator II Blog Party Starter Blog Conversation StarterView Profile29 Sep 2014 6:50 AM∙Comments 52∙LikesIn one of my prior posts, I mentioned a step required when installing .Net Framework 3.5 on Windows Server 2012 or later operating systems. Specifically, I mentioned the need to supply the source for the .Net files because this is one of the few components we do not stage to the component store on Windows Server 2012 and later. One of the bullets in the things to keep in mind section at the bottom of this blog reads:If you are attempting to install .Net Framework 3.5 on Windows Server 2012, instead of specifying the install.wim, you need to specify the Sources\SxS directory on the DVD or if providing the source, the files in the SXS directory copied from the DVD are just for .Net Framework 3.5. You can host them on a share and supply them through the GUI or through Group Policy just like any other feature.Recently one of my customers ran into an issue when attempting to install .Net Framework 3.5 on Windows Server 2012 R2 using the following command:Install-WindowsFeature –name NET-Framework-Core –source F:\sources\sxsF:\ is the mounted Windows Server 2012 R2 ISO.Yet it still failed with the following error:The request to add or remove features on the specified server failed. The source files could not be downloaded. Use the "source" option to specify the location of the files that are required to restore the feature. Error: 0x800F0906Or if you doing this via the GUI, it will look something like the following:Looking at the cbs.log file from the C:/Windows/logs/cbs directory, we see the following:2014-09-22 14:27:18, Info CBS Not able to find package:Package_78_for_KB2966828~31bf3856ad364e35~amd64~~6.3.1.4 from the cached windows update index. [HRESULT = 0x800f090e - CBS_E_EMPTY_PACKAGE_MAPPING_INDEX] 2014-09-22 14:27:18, Info CBS Failed to find package:Package_78_for_KB2966828~31bf3856ad364e35~amd64~~6.3.1.4 from the index mapping [HRESULT = 0x800f090e - CBS_E_EMPTY_PACKAGE_MAPPING_INDEX]2014-09-22 14:27:18, Info CBS Failed to get WU category/updateID for package: Package_78_for_KB2966828~31bf3856ad364e35~amd64~~6.3.1.4 [HRESULT = 0x800f090e - CBS_E_EMPTY_PACKAGE_MAPPING_INDEX]2014-09-22 14:27:18, Info CBS Failed to get the category and update ID for package: Package_78_for_KB2966828~31bf3856ad364e35~amd64~~6.3.1.4, component:amd64_microsoft-windows-wpfcorecomp.resources_31bf3856ad364e35_6.3.9600.20708_en-us _a3b67b207d2057ab, file: PresentationHostDLL.dll.mui [HRESULT = 0x800f090e -CBS_E_EMPTY_PACKAGE_MAPPING_INDEX]2014-09-22 14:27:18, Info CBS Exec: Missing files are found during staging attempt but failed to download content from WU. [HRESULT = 0x800f090e -CBS_E_EMPTY_PACKAGE_MAPPING_INDEX]2014-09-22 14:27:18, Info CBS CommitPackagesState: Started persisting state of packages2014-09-22 14:27:18, Info CBS CommitPackagesState: Completed persisting state of packages2014-09-22 14:27:18, Info CSI 00000010@2014/9/22:19:27:18.968 CSI Transaction @0x69943b49e0 destroyed2014-09-22 14:27:18, Info CBS Perf: Stage chain complete.2014-09-22 14:27:18, Info CBS Failed to stage execution chain. [HRESULT =0x800f0906 - CBS_E_DOWNLOAD_FAILURE]2014-09-22 14:27:18, Error CBS Failed to process single phase execution. [HRESULT = 0x800f0906 - CBS_E_DOWNLOAD_FAILURE]So why are we getting these errors despite supplying the source?We released a security update in August 2014 that updates .Net components. The security updates are as follows:KB2966828: MS14-046: Description of the security update for the .NET Framework 3.5 on Windows 8.1 and Windows Server 2012 R2: August 12, 2014KB2966827: MS14-046: MS14-046: Description of the security update for the .NET Framework 3.5 on Windows 8 and Windows Server 2012: August 12, 2014If either of these updates are installed, you will run into the above issue if your server does not have access to the Internet to pull the updated components.How do we resolve this?Since this customer’s servers do not have internet access, in their case, they did the following:1) Uninstalled the security update2) Installed .Net Framework 3.5 (which installed without error)3) Reinstalled the update==================================================================== ==============**UPDATE**A fixit package just released for this issue. Please install the update in the following article to correct this issue:/kb/3005628==================================================================== ==============Take a look at the following TechNet article for .Net Framework 3.5 deployment considerations:/en-us/library/dn482066.aspxHere’s a link to the official Knowledge Base article on this issue:https:///kb/3002547How do you keep from running into this in the future?I personally recommend that you proactively enable .Net Framework 3.5 on the server images and templates in your environment to prevent having to troubleshoot or take additional steps such as this going forward.~ Charity .net Shelbourne。
3GPP 5G基站(BS)R16版本一致性测试英文原版(3GPP TS 38.141-1)
4.2.2
BS type 1-H.................................................................................................................................................. 26
4.3
Base station classes............................................................................................................................................27
1 Scope.......................................................................................................................................................13
All rights reserved. UMTS™ is a Trade Mark of ETSI registered for the benefit of its members 3GPP™ is a Trade Mark of ETSI registered for the benefit of its Members and of the 3GPP Organizational Partners LTE™ is a Trade Mark of ETSI registered for the benefit of its Members and of the 3GPP Organizational Partners GSM® and the GSM logo are registered and owned by the GSM Association
思科交换机N5K N2K-网络设备技术指标要求-公开版
*安全性 *访问控制列表 *Private VLAN 支持 *管理性 *网管协议 *端口镜像 *智能诊断功能
实配支持 SNMP v1、v2 和 v3 支持在物理端口、PortChannel、VLAN 和 FC 光纤通道端口 上的镜像； 支持主动对硬件和内部数据路径进行检测功能，支持引导 时间诊断、持续监控，以及按需和定期测试等
2 数据中心柜顶接入交换机
设备名称：柜顶接入交换机 指标项 每台配置要求： 千兆以太网电口≥48，万兆以太网光纤接口≥4 指标项 *性能要求 *线速转发端口数 *转发延迟 *功能要求 *虚拟服务器支持 *可靠性 *电源冗余 *风扇冗余 指标要求 千兆以太网端口≥40，万兆以太网端口≥4 端口到端口转发延迟≤3.2us 支持硬件化 VN-Link 协议，实现与 VMware 虚拟机的策略定 位 实配冗余电源 实配冗余可在线更换风扇模块 数量: ≥
华为宽带MA5200 业务应用命令
目录第7章业务应用命令..............................................................................................................7-17.1 端口VLAN配置命令..........................................................................................................7-17.1.1 access-limit..............................................................................................................7-17.1.2 access-type.............................................................................................................7-27.1.3 accounting-copy......................................................................................................7-37.1.4 arp-proxy.................................................................................................................7-47.1.5 authentication-method.............................................................................................7-47.1.6 client-option82.........................................................................................................7-57.1.7 default-domain.........................................................................................................7-67.1.8 dhcp-shortlease.......................................................................................................7-77.1.9 dhcp-broadcast........................................................................................................7-87.1.10 display ephone-user..............................................................................................7-97.1.11 display { vlan-leased-line | proxy-leased-line | relay-leased-line | pppoe-leased-line }................................................................................................................................7-107.1.12 display layer3-subscriber....................................................................................7-117.1.13 display portvlan...................................................................................................7-127.1.14 display static-user...............................................................................................7-137.1.15 ephone-user........................................................................................................7-157.1.16 ip trigger..............................................................................................................7-157.1.17 layer3-subscriber.................................................................................................7-167.1.18 multicast-control..................................................................................................7-177.1.19 nas-port-type.......................................................................................................7-187.1.20 permit-domain.....................................................................................................7-197.1.21 permit-ip..............................................................................................................7-207.1.22 pnp......................................................................................................................7-217.1.23 portvlan-name.....................................................................................................7-217.1.24 roam-domain.......................................................................................................7-227.1.25 static-user............................................................................................................7-237.1.26 vbas.....................................................................................................................7-247.1.27 vlan-host-car........................................................................................................7-257.1.28 wlan.....................................................................................................................7-267.2 PPPoE配置命令..............................................................................................................7-277.2.1 access-threshold pppoe-server.............................................................................7-277.2.2 ppp connection chasten........................................................................................7-277.2.3 pppoa enable.........................................................................................................7-287.2.4 pppoe invalid-server-detecting..............................................................................7-297.2.5 pppoe-server bind.................................................................................................7-297.2.6 pppoe-server bind virtual-template........................................................................7-307.2.7 pppoe-server service-name..................................................................................7-317.3 L2TP业务配置命令..........................................................................................................7-337.3.1 display l2tp session...............................................................................................7-337.3.2 display l2tp tunnel.................................................................................................7-347.3.3 display l2tp-group..................................................................................................7-357.3.4 l2tp aging...............................................................................................................7-367.3.5 l2tp enable.............................................................................................................7-367.3.6 l2tp-group..............................................................................................................7-377.3.7 reset l2tp tunnel.....................................................................................................7-387.3.8 start l2tp.................................................................................................................7-387.3.9 tunnel.....................................................................................................................7-397.3.10 tunnel authentication...........................................................................................7-407.3.11 tunnel avp-hidden................................................................................................7-407.3.12 tunnel load-sharing..............................................................................................7-417.3.13 tunnel name.........................................................................................................7-427.3.14 tunnel password..................................................................................................7-427.3.15 tunnel radius-force..............................................................................................7-437.3.16 tunnel source.......................................................................................................7-447.3.17 tunnel timer hello.................................................................................................7-45 7.4 802.1X配置命令..............................................................................................................7-457.4.1 authentication timeout...........................................................................................7-457.4.2 display dot1x template..........................................................................................7-467.4.3 dot1x authentication trigger...................................................................................7-467.4.4 dot1x-template.......................................................................................................7-477.4.5 keepalive retransmit..............................................................................................7-487.4.6 reauthentication interval........................................................................................7-497.4.7 request..................................................................................................................7-497.4.8 reset......................................................................................................................7-50 7.5 NAT配置命令...................................................................................................................7-517.5.1 display nat.............................................................................................................7-517.5.2 display nat statistics..............................................................................................7-527.5.3 display proxy-server..............................................................................................7-527.5.4 nat acl....................................................................................................................7-537.5.5 nat address-group.................................................................................................7-537.5.6 nat aging-time........................................................................................................7-547.5.7 nat connection-limit...............................................................................................7-557.5.8 nat outbound.........................................................................................................7-557.5.9 nat outside.............................................................................................................7-567.5.10 nat server............................................................................................................7-577.5.11 proxy-server........................................................................................................7-58 7.6 多播配置命令...................................................................................................................7-587.6.1 display igmp configuration.....................................................................................7-587.6.2 display igmp group................................................................................................7-597.6.3 display igmp interface...........................................................................................7-607.6.4 display igmp user..................................................................................................7-607.6.5 display multicast configuration..............................................................................7-617.6.6 display multicast member......................................................................................7-617.6.7 display multicast routing-table...............................................................................7-62 7.6.8 display multicast source........................................................................................7-63 7.6.9 display multicast vif...............................................................................................7-63 7.6.10 igmp designated-router.......................................................................................7-64 7.6.11 igmp group-policy................................................................................................7-64 7.6.12 igmp max-response-time.....................................................................................7-65 7.6.13 igmp proxy...........................................................................................................7-66 7.6.14 igmp timer other-querier-present.........................................................................7-67 7.6.15 igmp timer query..................................................................................................7-67 7.6.16 igmp version........................................................................................................7-68 7.6.17 multicast free-group............................................................................................7-69 7.6.18 multicast hmcp....................................................................................................7-70 7.6.19 multicast hmcp keep-alive interval......................................................................7-70 7.6.20 multicast hmcp timeout.......................................................................................7-71 7.6.21 multicast routing-enable......................................................................................7-72 7.6.22 multicast source car............................................................................................7-72 7.6.23 multicast traffic-collection interval.......................................................................7-73 7.6.24 multicast-vlan......................................................................................................7-74第7章业务应用命令7.1 端口VLAN配置命令7.1.1 access-limit【命令】access-limit user-numberundo access-limit【视图】端口VLAN视图【参数】user-number：端口VLAN允许接入的最大用户数。
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