2008年8月20日星期三

电动机负荷的负载系数、使用系数、不同时系数

www.ise.ufl.edu/capehart/papers/diversity.doc
http://www.ise.ufl.edu/capehart/papers/diversity.html
http://www.ise.ufl.edu/capehart/papers/ebaltemp.xls

这些东西是一个叫Barney L. Capehart教授的资料里面找到的,主要是用于计算用电费用。他的Use (or utilization) factor是用于计算用电量的,不是一般所说的利用系数,这里暂时称为使用系数,使用系数x年使用小时基数=年使用小时数。利用系数和这里Load factor 比较接近,这里暂时称为负载系数。Diversity factor 就是不同时系数。

Definitions:
  1. Load factor - the ratio of the load that a piece of equipment actually draws when it is in operation to the load it could draw (which we call full load).
     For example, an oversized motor - 20 hp - drives a constant 15 hp load whenever it is on. The motor load factor is then 15/20 = 75%.  

  2. Use (or utilization) factor - the ratio of the time that a piece of equipment is in use to the total time that it could be in use.
     For example, the motor above may only be used for eight hours a day, 50 weeks a year. The hours of operation would then be 2000 hours, and the motor use factor for a base of 8760 hours per year would be 2000/8760 = 22.83%. With a base of 2000 hours per year, the motor use factor would be 100%. The bottom line is that the use factor is applied to get the correct number of hours that the motor is in use.  

  3. Diversity factor - the probability that a particular piece of equipment will come on at the time of the facility's peak load.
     The diversity factor is the most complicated of these factors. For example, we might have ten air conditioning units that are 20 tons each at a facility. In Florida we typically assume that the average full load equivalent operating hours for the units are 2000 hours per year. However, since the units are each thermostatically controlled, we do not know exactly when each unit turns on. If the ten units are substantially bigger than the facility's actual peak A/C load, then fewer than all ten units will likely come on at once. Thus, even though each unit runs a total of 2000 hours a year, they do not all come on at the same time to affect the facility's peak load. The diversity factor gives us a correction factor to use, which results in a lower total kW load for the ten A/C units. If the energy balance we do for this facility comes out within reason, but the demand balance shows far too many kW for the peak load, then we can use the diversity factor to bring the kW into line with the facility's true peak load. The diversity factor does not affect the kWh; it only affects the kW.


Motor load factors in many facilities are more in the range of 40% - 50%, than in the range of 80% that had been a standard assumption for many years of doing audits. Rarely do you find a motor running at 100% load factor.

However, not all motors at a facility are running at the same load factors. Ventilating fans that come from a supplier as a packaged unit with a fan and a motor are most often assumed to be operating at near full load. You should probably use a load factor of 80% here, since the manufacturer of the ventilating fans should have reasonably matched these loads. Other motors may also be in this category - some engineering judgment and common sense are required to determine which other motors these are.  

Motors with variable loads are going to have the lowest load factors in general. A dust collector fan motor will normally have quite a variable load, and would often be expected to have a low load factor. Other examples are saws, presses, milling machines, sanders and grinders, waste grinders, water pumps, hydraulic pumps, etc. 

If a group of motors do not all operate together all of the time, then using a diversity factor is appropriate. This is the case with a number of separate air conditioning units (considering the motors for the compressors) that are individually thermostatically controlled. It could also be the case for a group of production motors if some of the motors are not in use all of the time. You should use a diversity factor in your motor calculations, since it is not often the case that a facility has all of the motors on at the same time.

Reconciling the energy balance: When you perform an energy balance, do not use the motor load factor as the first and only adjustment made to reconcile the estimated energy use (energy balance) with the energy bills. Making this adjustment too quickly results in failure to pick up other things that have been overlooked.

For example, if the energy use does not balance with the energy bills, the first step is to check to see that all of the equipment and uses have been accounted for.

Do the items on the energy balance spreadsheet match your recollection of the equipment you saw in the facility?
Does anything appear to be missing?
Are the utility bills for total energy use and peak kW recorded correctly?

The next step is to check the hours of use for lights and other equipment to see if it matches your knowledge of the facility's operation. Remember that each motor - as well as each other piece of equipment - does not necessarily operate the same number of hours each day or year. Finally, if some of the equipment does not come on at the same time as the facility peaks in kW use, then utilize the diversity factor to account for this.

Adjusting the motor load factors should probably be the last thing you do to reconcile the energy and demand balances. Now, if all other information and all other factors are correct to the best of your knowledge, then adjust the load factors. While motor load factors are not often in the range of 80-100%, you should be equally suspect of very low motor load factors. If you get motor load factors in the range of 20-30%, it is more likely that you have the hours of use wrong than that you have a facility which is using motors that are an average of four times too big for the job they are doing. Lumber mills and wood products facilities using lots of saws may have these low load factors. Most other places should have motors with a higher load factor.

Basic motor load measurements should be taken at the plant visit. The electrical person at the facility is generally willing to measure the current being drawn by a motor of interest. Air compressors are ones that are usually easy to do, and you should ask the plant personnel to do this for you. Let them open the motor controller or switch box and connect a clamp-on ammeter to see what the current for the motor is. You then need to know the full load current from the nameplate of the motor. The ratio of the actual current to the full load current is the approximate load factor on the motor at that time. This procedure works as long as the current is greater than or equal to about 50% of the full load current. Try to take this measurement for each of the large motors in the facility - i.e. motors of 50 hp and above; or even 20 hp or above if the facility does not have a lot of big motors. If you have not received formal electrical safety training, you should not make these electrical measurements yourself. If the facility electrician does not want to make these for you, then let it go at that.

Air handlers―use factor: Air handlers use motors and are subject to all of the comments made in the motor section. In addition, you may be able to get a better handle on the hours of use for the air handlers by knowing how the A/C system works. Ask if the air handlers run constantly when the facility is occupied. They might if the facility wants the ventilation, even though the compressors might not come on except to periodically provide some temperature reduction or moisture removal. If this is the case, then the use factor for these air handler motors should reflect an hours-of -use that matches the offices or other area that the air handlers supply. In addition, the hours-of-use must also consider the compressor run hours. Thus the total hours for the air handlers must be at least the same as the compressor hours, and may be higher if the A/C unit is left on during periods that the facility is not occupied, or if ventilation is provided.

If the air handlers only come on when the thermostat orders cooling, then the hours-of -use must be the same as the hours-of-use of the compressors.

It is important to get adequate information on the operation of the air conditioning system. To get complete data on the air handler motors for an air-conditioned facility, you will need all of the standard information - size, maker, single or three phase, etc - together with the operating basis for the air handlers discussed above.

You should also collect data on the drive belt system for air handlers. Record the number of belts, the lengths, and the types of belts. Ask about motor and drive lubrication and cleaning. Also check the A/C filters to see if they are reasonably clean.

Sometimes a visual inspection will show some real problems. Ask the maintenance person to open up one of the air handlers - or just look into it (SAFELY) if it is accessible - and see if the belt is tight, slack, or really loose. Do not stick your hand into an air handler that is off at the moment, and may come back on when the thermostat kicks in. Have the maintenance person turn the air handler motor off with the circuit breaker or control box. Do not put your finger on a moving drive belt.

Is the belt in good shape?
Is it frayed, cracked or coming apart?
Does it look like the pulleys for the motor and the fan are lined up?

Ask the electrician to measure the current that the air handler motor is drawing to see what its load factor is while driving the fan. It should be very near full load - but you never know. Maybe the original motor burned out and was replaced with a bigger one to "make sure it did not burn out again." Remember to take the full load current off the nameplate to find the load factor.

2008年8月19日星期二

只剩下一个月

离注册考试只有一个月了,上个星期因为种种原因没有静下心来。

首先是奥运,这种体育盛会随便挑出一个时段来都会有好看的节目,只有你看不完。

然后是熊猫猪出现了,还是一头病的熊猫猪.虽然被封为小熊猫,还是要不时打打杂。

最后就是最浪费时间的事--看网络小说.最近看的是<魔法炒手>,感觉不错。

最大的感受是只要你有理论、实践、创新,到哪里都可以发挥作用。东西还是要好好学,用心体会。

希望能再次回到轨道上来。

2008年8月12日星期二

开式系统和闭式系统水泵变频的经典分析

http://www.ehvacr.com/lw/HTML/4448.html

开式系统中变速泵的节能分析
狄洪发,男,1945年1月生,大学,教授,北京清华大学建筑技术科学系

看了一些书和资料,觉得这篇文章介绍得最好了,可以说是经典。很少书对相似定律讲得比较清楚,这篇文章清楚地介绍了相似工况的定义、成立的条件,并举了两个例子说明这个问题,还进行了详细的变频节能计算分析。清华大学的老教授,是有真才实学,讲得清晰易懂。

当然,文章论述的功率,是指水泵的轴功率,和电机消耗的电功率还有点不同。

开式系统中变速泵的节能分析

摘要:变速泵在开式和闭式管路中的节能效果不同,引起这种差别的原因是开式系统受水的静压作用改变了泵的效率。分析了开式管路系统中变速泵的节能效果并给出了计算实例。
关键词:变速泵 节能 开式系统

0 引言

  变速泵在空调供暖系统中的应用,对于改善系统调节性能、节省运行费用发挥了很大作用。在闭式循环系统中,其节能效果非常明显,因此得到广泛的应 用。目前在不少开工系统中,也使用了变速泵,例如在供暖补水泵定压系统中,补水泵也采用了变速调节。由于水的重力作用,在闭式系统与开式系统中,采用变速 泵的节能效果显然不同。随着开式系统水提升高度的增加,变速泵的节能效果进下降低,在流量变化大的闭式系统中应用变速泵肯定有明显的节能效果,但在开式系 统中采用变速泵是否节能,要根据管路的实际情况从投资和运行费用角度综合分析。

  1 基本方程

  1.1 泵的相似工况

  由泵的相似定律       (1)

  即

  有     (2)

  式中p1,p2,分别为转速n1,n2下泵的扬程,G1,G2分别为转速n1,n2下泵的流量,N1,N2分别为转速n1,n2下泵的功率,C为常数。

  式(2)表明,只要泵在转速n1下的工作点A和转速n2下的工作点B(简称工况A,B)相似,则泵扬程与流量平方的比值是一个常数;同时可以得知,如果两个工相似,则这两个工况下泵的效率必然相同。实际上,式(2)也表示了泵的等效率曲线式

          (3)

  在一定转速下,泵的工作点管路阻力的变化而变化。对于某一确定的管路,常数C可用式(2)求得。式(2)还表明,即使泵的工作点随着运行情况变化,但只工作点在该曲线上,则泵的效率就不会发生变化。

  1.2 变速泵节能分析

  泵的转速泵从n1从调整到n2后泵的轴功率从N1,变到N1,由轴功率 知,若把水看成不可压缩流体,则在工况A,B下功率之比为

           (4)

  如果工况A,B在等效率曲线上,则ηab,且 。也就是说,式(1)成立的条件必须是泵的工作点A,B均在等效率曲线上。

  1.3 管路水力特征

  管路系统A,B两点间流动的压力总损失pab与流量G存在着上述特征关系

          pab= po+SG2 (5)

  式中po为A与B两点间水的静压。闭式管路系统的终点和起点生命,因此po=0;开式管路系统的终点和起点不重合,且一般情况下A,B两点之间的压差不0。S为管路阻力系数,在水管路系统中可以认为S值的大小与流量无关而仅与管路结构有关,因此对于一个具体的管路系统而言,S是一个不随流量变化的常数值。记SG2=Δp,它表示了流体磨擦阻力的大小与流量的平方成正比;当知道了管路结构后就可以求得po和S值,从而由式(5)得到管路的特征曲线。

  2 闭式系统中变速泵的节能分析

  在闭式系统中,泵作功所输出的能量完全消耗在克服水在管路中流动时的磨擦阻力,也就是说泵的扬程p等于整个管路的压降。这样,当流量发生变化后,泵的能耗自然也成比例地变化。

  2.1 泵与管路联合工作点

  泵的工作点为泵的特征曲线与管路特征曲线的交点,如图1所示的闭式管路系统中的A点,其对应的流量、压力及效率分别为Ga,pa及ηa。 图中效率曲线η是针对转速n1而言的。

图1 闭式系统性能曲线

  闭式管路中流动阻力损失等于泵的扬程,即Δp=p,且po=0,则由式(5)有

      (6)

  对比式(2)与式(6)不难发现C=S,这表明在闭式管路系统中泵的等效率曲线与管路特征收与管路特曲线完全重合。把该管路的阻力系统记作 ,则C=Sa,因此泵的等效率曲线方程和管路特征曲线均为p=SaG2。当泵转速从n1变到n2时,泵的工作点从A点变到B点,B点仍在管路特征曲线上。由于泵的等效率曲线与管路特征曲线重合,则泵在A,B两点的效率ηab,这是闭式管路系统所具有的特点。

  2.2 变速泵的节能

  当泵转速从n1调整到n2后,泵的特征曲线变化如图1所示。在ηab的条件下,由式(4)知A,B工况功率之比为

               (7)

  也就是说,泵的功率与转速的三次方成正比。因此在闭式水系统中变速泵的节能效果是非常明显的。

  3 开式管路系统中变速泵的节能分析

  一供热系统如图2所示,使用高位膨胀水箱定压。该供热系统中,补水泵在开式管路系统中运行,而循环泵则在闭式管路系统中运行。当用户热负荷发生 变化后,如使用中央量调、循环泵为变速泵,如上分析则循环泵所消耗的功率与转速的三次方成比。但补水泵在开式管路系统内,当补水量发生变化且补水泵亦为变 速泵时,泵转速变化后泵所消耗的功率并非与转速的三次方成正比,因为受到了开式管路系统进出口高差H的影响。

  3.1 开式管路系统水力特性

  开式管路中泵的提供的扬程一方面消耗在克服流动阻力上,另一方面还消耗在提升水的高度上。如图2,式(5)所示压力总损失p=po+SG2中的po即时H=20m带来的静压。知道管路结构后即可求得管路的阻力系数S值。因此,知道po和S后就可以画出管路特征曲线,如图3所示。由泵的特征曲线和管路特征曲线的交点得到泵在转速n1下的工作点A,从而得到A点的流量Ga及泵的扬程pa,此处pa即为式(5)中的pab。为表示明确,管路的阻力系数也可写为:

            (8)

图2 系统示意图

  3.2 开式管路中变速泵的等效率曲线

  为分析比较方便,设开式管系统中变速泵在转速n1的工作点A与闭式管路系统相同,如图3。当补水量发生变化后泵的转速从n1变到n2,此时泵的工作点沿着管路特征曲线从A变到B′点,B′点所对应的流量和压力为Gb′,pb′。泵在B′点效率η′b并不一定等于A点的效率ηa,因为A,B′两个工况并不相似。

  如上分析在闭式管路中过A点的等效率曲线方程中C=Sa,作p=SaG2曲线与泵在n1下的性能曲线之交点得到对应的效率ηa。在开式管路系统中等效率曲线式(3)中的p此时对应p ab,同样把式(3)中的常数C记作C a,则 。但开式管路系统的阻力系统数 ,因为p0≠0,所以C a ≠S0。也就是说在p0≠0的开式管路系统中泵的等效率曲线与管路特征曲线不重合。把A,B′点的等效率曲线常数:

               (9)

            (10)

  已知Gb′后可以由等效率曲线方程p= Gb′G2得到过程B′点的等效率曲线,该等效率曲线与转速n1下的泵性能曲线交点为B′′,由B′′得到此时泵在转速n1下的效率η′b,如图3。

                     图3 开式系统性能曲线1

  3.3 节能分析

  在泵转速从n1调整到n2后,把A,B′工况对应的Ga,pa,ηa及Gb′,pb′,η′b值代入式(4),即得到节能比。需要注意,即使假设此时ηab,转速变化后的节能比也只能由 求得,而不等于 。因为在开式管路系统中po≠0,A,B′点不在等效率曲线上,因此不能直接使用式(1)。

  4 算例

  可以用解析法或作图法来求解。假设泵在n1下的性能曲线方程为p=α01G+α2G2,则在n2下的方程为p=α0′+α1′G+α2′G2,其中α0,α1,α2为已知系数α0′=α0k2,α1′=α1 k,α2′=α2,转速比。根据上两式及式(5)、(6)、(7)、(9)及(4)联立即可求得相应的数值,本文利用作图法求解,以便于理解。

  4.1 闭式系统

  如图2所示闭式管路系统中循环泵,设转速n1=1100r/min调整到n2=860 r/min,工作点分别为A,B。由图1得A点G a =12.85m3/h,p a =330.2kpa,η a =0.75,B点G b =10.05m3/h,p b =201.7kpa,则由式(3)得功率比

  4.2 开式系统中保持转速n2 与闭式系统的转速相同

  如图2,补水泵在开式管路系统中,同样,泵转速从1100r/min调整到n2=860 r/min,即n2相同,工作点由A变到B′,由图3知B′点的Gb′=5.95m3/h,p′ b =227.8kPa.由式(10)得到过B′点的等效率曲线系数Gb′=0.643,等效率曲线p=0.643G2与转速n1下的泵性能曲线的交点为B′′,B′′对应的效率ηb′=60%。因此在A,B′两工况下的功率。泵在A,B′两点效率不同对节能的影响不可忽视,如果不计泵效率的变化,则。与实际值的相对偏差为20%,而这一点往往容易被忽略。

  4.3 开式系统中,保持流量Gb′与闭式系统的流量相同

  当n2相同时,开式系统B′点的流量Gb′肯定小于闭式系统B点的流量Gb。假若仍希望Gb′= Gb=10.05 m3/h,则闭式、开式系统的n2肯定不相同。如图4,当开式系统的转速调整到n2′=998r/min时,则Gb′=10.05 m3/h,此时闭式系统n2=860r/min下的流量相同,闭式系统的等效率曲线为过A,B的曲线,而开式系统的等效率曲线为过B′,B′′的曲线。从图4可以求得,闭式系统在转速n2=860r/min下所对应的效率为0.75,开式系统在转速n2′=998r/min时所对应的效率为0.68。据式(4)知, 。因此,对于开式、闭式系统当流量均从12.85 m3/h变到10.05 m3/h时,开式系统节能效果只有闭式系统的73%。

                     图4 开式系统性能曲线2

  5 结论

  5.1 在开式管路系统中,泵变速前后工作点A,B′并不在同一等效率曲线上,因此A,B′点所对应的效率不相等。在分析变速泵的节能效果时,应考虑效率不同所带来的影响。

  5.2 在闭式和开式管路系统中,泵转速虽然都是从n1变到n2,但两者的节能效果并不相同。在本文的算例中,闭式系统中泵的节能效率47.8%,而开式系统中节能效率为39.9%。这主要是由于开式系统受静水压p0的影响。如果p0=0,则节能效率相同;p0值越大,两者的节能效果差别也就越大。

  5.3 在闭式和开式管路系统中,如果保持变速前后的流量都相同,则开式系统的节能效果要小于闭式系统,p0越大,开式系统节能效果越差。

  参考文献

  1 钱以明,高层建筑空调与节能,上海:同济大学出版社,1990

  2 江亿,用变速泵和变速风机代替调节用风阀水阀,暖通空调,1997,27(2):66~71

  3 李向东,李百萍,牟灵泉,变频调速装置在空调水系统中的应用,暖通空调,1997,27(4):62~65

  作者简介:

  1.狄洪发,男,1945年1月生,大学,教授,100084 北京清华大学建筑技术科学系,(010)62779996。

 



灭魔还是灭神

《灭魔志》两本终于看完了。

不知道为什么会用灭魔志这个书名?

开始的魔可以升天为神,开始的神会沦为阶下囚。

在网络小说里面,人鬼神三者中,一般人是正、神是邪、鬼比较中立。《无赖邪佛》也是这样。

是现在的作家要挑战权贵,还是现在的权贵让人不屑?

人类中也不乏顶尖的高手,在和游手好闲的神仙搏斗时,也不会落于下风。

高高在上的神仙们啊,如果做出伤害人类的事情,迟早会被灭掉的。

神仙映射的是什么?不言而喻了。

《灭魔志》写得挺好的,除了开始的几节比较无厘头。

《无赖邪佛》从开始就写得不错,可惜后面更新太慢了。

2008年8月11日星期一

电脑功率

http://diybbs.pconline.com.cn/topic.jsp?tid=8736182

【我们如何测试?】

本次测试的平台至少都是双核处理器,而高端平台更是选择了Intel的4核处理器。在测试中,我们将就每一种平台在各方面的实际应用环境进行功耗测试,同时考虑到用户的升级需求,我们在每一种平台上都采用了两款显卡进行测试,其中一款比原有平台显卡性能更高,这是因为升级显卡是目前最有效的提升性能手段。

在测试仪器上,我们选择了专业的功率测试仪,能有效地测试出每个平台在不同环境下的即时功率。此外,在这里我们测试的系统整体平台功耗包括了20英寸LCD显示器。


【低端平台实际功耗测试】

测试平台:处理器:Athlon64 X2 4000+ 主板:NV C68 显卡:集成/8600GT
内存:金士顿 DDR2 800 1G×2 硬盘:希捷7200.10 250G 显示器:华硕MW201U
电源:长城BTX- 500SD

集成显卡环境下功耗测试: 待机116.6W 最高160.6W
8600GT显卡环境下功耗测试: 待机125.4W 最高178.2W


【中端平台实际功耗测试】

测试平台:处理器:Intel Core2 Duo E4300 主板:Intel P31 显卡:
7600GT/8800GT 内存:金士顿DDR2 800 1G×2 硬盘:希捷7200.10 250G
显示器:华硕MW201U 电源:长城BTX-500SD

7600GT显卡环境下功耗测试: 待机121W 最高158.4W
8800GT显卡环境下功耗测试: 待机121W 最高200.2W


【高端平台实际功耗测试】

测试平台:处理器:Intel Core2 Quad Q6600 主板:Intel X38 显卡:
8800GT/8800Ultra 内存:金士顿DDR2 800 1G×2 硬盘:希捷7200.10 250G
显示器:华硕MW201U 电源:长城BTX-500SD

8800GT显卡环境下功耗测试: 待机169.4W 最高250.8W
8800Ultra显卡环境下功耗测试: 待机228.8W 最高341W


随着电脑速度越来越快,配电时按400W,0.8系数也不过分了。

2008年8月8日星期五

奥运开幕式看点不多

从8点到12点,看了整整4个小时。

最有印象的是29个烟花脚印,很震撼;小朋友们唱国歌和抬国旗,很可爱;中国队出场,姚明和一个映秀的小英雄,很感动。

开始像打鼓的那一幕也算还可以,不过时间长了点,翻来覆去的;最后李宁点火炬有点夸张,估计要把他给累坏了,而且场面上感觉也很一般。

其他什么中国画、京剧、活字印刷之类的简直不知道想干什么,色彩也不好看,比较失败。

观众散场要到家估计凌晨2、3点左右吧;运动员可能优先一点,但明天如果有比赛也会影响竞技状态的。

张艺谋就不能少耗点时间,放大家一马。

Firefox 原来指的是 小熊猫

zh.wikipedia.org/wiki/小熊�

小熊猫又名红熊猫、红猫熊、小猫熊,有时在中文中也称火狐,英文中亦有"Fire
Fox"既是对其的直接译名,是一种哺乳类动物,分布在中国南方到喜马拉雅山麓。属于食肉目、小熊猫科。它究竟应该列在熊科或浣熊科是多年来一直被争论的问题,最近经过基因分析,认为与美洲大陆的浣熊亲缘关系最接近,应该单独列为小熊猫科。

小熊猫长约50至60厘米,体重约3至5公斤,身体躯干毛为栗色,四肢和腹有些黑色毛,面部白色花纹不像大熊猫那样明显,但前爪根骨的一部分突起成手指状,可作为第6手指抓握物品,和大熊猫相同。

"熊猫"这个名称其实是小熊猫先取得的,但是后来的大熊猫更广为人知,所以单称"熊猫"的时候多指的是大熊猫。

法国的博物学者乔治・居维叶(Georges Cuvier)的弟弟动物学者弗列德利克・居维叶(Frédéric
Cuvier)看到小熊猫的标本相当感动,因此以希腊文中的"火焰色的猫(Ailurus fulgens)"作为其学名。

英文中则有"Red Panda"、"Wah"、"Firefox"等名称,最后的Firefox即是有名的开放源代码浏览器Mozilla
Firefox的来源,但是该浏览器的图案画的却比较接近有着火尾巴的狐狸。