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现有一台5级变频卧式油泵，有两种工况1,2，对应的2种不同的介质，流量相同，扬程不同，工况一大于工况二，通过变频降速可以达到工况2，请问这时的效率怎么计算，有没有哪个标准或者书籍提到这个计算公式的
我们遇到这样的情形一把考虑效率基本不变，当然，肯定有出入，应该不大；看看其他海友怎么说:lol
本帖最后由 3983596_FPPZ 于 2019-1-24 13:24 编辑 如果是纯理论上的计算，那么需要的已知条件很多，，计算过程过于繁琐，你可以在关醒凡老师的《现代泵设计手册》中的第三章第五节中：相似抛物线及其应用 章节中找到这方面的理论依据（P45）。 但是实际使用过程中，如扬程转速相差很小的话，可以视为效率基本不变。
根据流量扬程介质密度算出来有效功率，再根据电机的电流、效率算出来电机的输出功率，联轴器的传动效率可以忽略了，反推一下效率吧
两种介质粘度相差多少？
计算过于理论和复杂，最好的方法就是实测一下工况点运行的电流电压，扬程流量，计算效率
实际上是三种工况，前两种流量都是2000m3/h，粘度都不到40cP，可以忽略粘度。第三种流量1200m3/h,粘度就是200多cP了，扬程有影响. 要同时考虑三种工况，计算效率和功率，没什么头绪。
现在还处于设计阶段，没有实际产品
要给客户提交文件，必须体现出差别的
根据离心泵相似性原则： 1. 离心泵效率相似关系：η2=η1 2. 离心泵流量相似关系：Q2=(D2/D1)3*(n2/n1)*Q1 3. 离心泵扬程相似关系：H2=(D2/D1)2*(n2/n1)2*H1 4. 离心泵功率相似关系：P2=(D2/D1)5*(n2/n1)3*P1
没有理论值，只有检测经验值。当频率降到45.HZ时泵效率下降约为3%.降到40HZ时效率下降约7%.当35HZ时效率下降约11%,当30HZ时，泵效率下降约17%.这个效率是与工频相比较。如果你工况稳定，建议你选一大一小两台泵工频运行。变频器耗电约为5%.