IEC 60335-1 clause 5.13怎么理解

kuiyin01 · 发表于 2010-5-27 11:30

5.13 The tests for appliances with PTC heating elements are carried out at a voltage corresponding to the specified power input.

When a power input greater than the rated power input is specified, the factor for multiplying the voltage is equal to the square root of the factor for multiplying the power input.

这段话怎么理解？第一段意思应该是以功率来调电压。第二段好像又是测试按电压乘于一个系数来进行。到底是测试是以功率还是电压为基础？

杨柳风 · 发表于 2010-5-27 11:52

第一句就是说在额定电压下测试了.

第二句的意思是这样的:大致的意思是如要求输入一个大于额定功率的功率时,可按换算成电压来测试的.

如11.8的温生测试要求在1.15倍的额定功率下测试, 但PTC的功率不会随电压的升高而升高,无法达到1.15倍的功率.只能根据P=U2/R换算成电压, 即提高1.07倍电压即可.

kuiyin01 · 发表于 2010-5-27 12:02

第一段还是有些疑问carried out at a voltage corresponding to the specified power input. 这句说的应该是电压调到功率下。如温升是1.15倍功率。这句的意思已经就是把电压调器具是1.15倍功率下。

kuiyin01 · 发表于 2010-5-27 12:27

现在有了点认识，请指教
假如测试功率小于额定功率的时候（如19.2章节0.85倍功率）直接按功率来测试，即0.85倍额定功率下测试。
当测试功率大于额定功率时，如11章的1.15倍，19.3章的1.24倍时按电压来测试即额定电压X1.15的平方根。

hexiling · 发表于 2010-5-27 13:00

第一句话是在额定功率所对应的电压下工作，就是使测试样品达到额定功率

hexiling · 发表于 2010-5-27 13:05

第二句话是：如果样机的实际输入功率大于额定电压，那么就要调整电压，给电压乘以一个系数，由于电压和功率间的换算是平方的关系，所以电压乘的系数为功率乘的系数的开平方

杨柳风 · 发表于 2010-5-27 13:07

引用第3楼kuiyin01于2010-05-27 12:27发表的 : _1 x% R- s. _, ~% D1 a
现在有了点认识，请指教4 c; H0 ^" v1 l( u" f, u( `" V
假如测试功率小于额定功率的时候（如19.2章节0.85倍功率）直接按功率来测试，即0.85倍额定功率下测试。
" H- ^4 q+ p E! N; C$ e当测试功率大于额定功率时，如11章的1.15倍，19.3章的1.24倍时按电压来测试即额定电压X1.15的平方根。

无法调功率,都按调电压如0.85, 1.15,1.24的开平方根.

kuiyin01 · 发表于 2010-5-27 14:04

楼上同志那么这句怎么理解呢
5.13 The tests for appliances with PTC heating elements are carried out at a voltage corresponding to the specified power input.

kamleung · 发表于 2010-5-31 17:52

To my understanding, this clause is taking about as below:

1) The first sentence:
At the very beginning,  the test voltage is selected as the "rated voltage" given on the rating plate.

2) Second sentence:
With the test volatge setting as the rated voltage, the actual power is checked against with the "rated power" on the rating plate. If the "actual power" is greater than the "rated power", then the test voltage will be adjusted with the square root of the factor for multiplying the power input. Perhaps, an example may help to show my point as below:
- the rating plate is marked with "rated voltage 240Vac" and "rated power 240W"
- according to the 1st sentence, the test volatge is set as 240Vac
- according to the 2nd sentence, the power is measured.  Say for example, the power is measured as 250W. Then the test volatge will be then adjusted with the below factor:
                     factor = square root (240W/250W) = 0.98
- the final test votage is thus given as 240V x 0.98 = 235.2V
With the test voltageas 235Vac, the product is therefore tested at its rated power.

Please feel free to correct me if you get any mistake in my undertanding.

kuiyin01 · 发表于 2010-5-31 17:57

呵呵答案真是丰富呀
期待各位各抒己见

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