理解反渗透产水率与浓缩系数间的重要关系

UnderstandingThe Critical Relationship Between Reverse Osmosis Recovery Rates And ConcentrationFactors 理解反渗透产水率与浓缩系数间的重要关系

By Harold G. Fravel, Jr., ExecutiveDirector, American Membrane Technology Association (AMTA)

Every aspect of a reverse osmosis (RO) membrane operation requires somedegree of process math. Changes in operating conditions that may be consideredminor can actually have significant and detrimental effects on systemperformance. This is certainly the case when calculating system recoveryrates, and the subsequent changes in concentrationfactors, in RO plant operation.

反渗透膜操作的各个方面都需要一定程度的数学知识。事实上，略微的改变运行条件会很大程度地影响系统性能，在反渗透的运行中正是这种情况，尤其是在计算系统产水率和浓缩系数间的关系。

The function of both RO and nanofiltration (NF) membranes is to rejectdissolved ions (salts). RO membranes reject all ions at some percentagewhile NF rejects primarily divalent ions such as calcium and sulfate. Theindustry commonly refers to NF as a “softening membrane” meaning that it morereadily rejects calcium and magnesium. NF does not require the higher netdriving pressures necessary for RO operation, making NF a more economicalternative when primarily divalent ion removal is required. 反渗透膜和纳滤膜的作用是除盐。从某种程度而言，反渗透膜能够去除所有离子，纳滤膜主要去除像钙、硫这样的二价离子。工业上通常把纳滤膜

称为柔和膜,就是说它更容易去除钙镁离子。纳滤不需要高压，因此在主要去除二价离子的时候，纳滤是个更经济的选择。

Both RO and NF processes involve the separation of the incoming raw waterinto two streams, permeate and concentrate. As the feedwater passes through themembranes, pressure drives a relatively high percentage of water moleculesthrough the membrane barrier to produce a low-salinity permeate stream whileleaving a lower percentage of water behind in a higher-salinity concentratestream. A process challenge in successful membrane system operation ismaintaining a balance between the desired permeate volume and a sufficientconcentrate volume to keep sparingly soluble salts in solution. 反渗透和纳滤都会把进来的原水分成两股，一股是纯水，一股是浓缩水。当原水透过膜，由于水压从而使得大量水分子透过膜于是制出低盐度的纯水，同时一小部分水未能透过膜继而成为高浓度水。膜运行系统的成功与否在于如何保持纯水和浓水之间的平衡，因为需要一定量的浓水溶解可溶性盐类。

Published rejection rates of today’s RO membranes exceed 99 percent, whichmeans that the majority of the feedwater salts are rejected by the membrane andpass into the concentrate stream. As operators increase system recoveryto produce more permeate, they effectively remove water from the feed stream,leaving less water to dilute the increasingly higher concentration ofsalts. When the solubility limits are exceeded, the salts fall out ofsolution and combine to form scale foulants on the membrane surface. Common scale formers in membrane applications include calcium carbonate,calcium sulfate, barium sulfate, and strontium sulfate.

据文献显示，如今反渗透膜的除盐率已经高于99%，也就是说原水中的绝大部分盐类被膜分离，从而进入浓水段。操作员提高系统的产水率，于是

透过膜的水量增加，剩下较少的水来溶解高浓度盐类。当溶解度超过了溶解极限值，盐类就不能再溶解从而变成污垢贴附到膜表面。通常在膜表面生成的污垢有碳酸钙，硫酸钙，硫酸钡和硫酸锶。

Finding the proper balance between acceptable permeate production andreduced membrane fouling is crucial, and why the process math is soimportant. RO and NF system recovery rates are calculated based on theamount of permeate produced divided by the total amount of raw water fed to thetrain. This calculation is illustrated as Qp÷Qf. anddescribed below: Qf. = feed flow Qp = permeate flowand Qc = concentrate flow

确定纯水和浓水的比例以及减少膜污染是十分关键的，上述过程中数学运算很重要。反渗透和纳滤系统的产水率计算如下：所产纯水总量除以原水量，可以用下式Qp÷Qf表示，其中：

Qf. = feed flow 原水 Qp = permeate flow纯水（渗透液） Qc = concentrate flow 浓水

If a system feed flow (Qf) is 1,000,000 gallons per day (GPD)and the permeate flow is 750,000 GPD, then the calculated recovery rate wouldbe 0.75 or 75 percent.

750,000 permeate GPD ÷ 1,000,000 feed GPD = 0.75 or 75percent

如果一个系统的原水是1000000加仑每天，纯水每天生产750000加仑每天，那么产水率就是0.75或75%：

750,000 加仑纯水 ÷ 1,000,000加仑原水= 0.75 or 75 %