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在设计微型阀时,关键点之一是对响应时间进行估算。设计一种热气致动双稳态微型阀,应用热电类比方法建立简化模型,对开阀响应时间进行估算。在流量测量的基础上,建立气流升压模型,计算了流经此微型阀的气体填充一定空间所需要的响应时间。分析结果显示:当加热膜片的厚度分别为8μm和25μm时,开阀响应时间分别为11.4ms和20.2ms。在200kPa压强差下,气流在55.5μL空间内的升压响应时间为7.6ms。对微型阀的性能进行测试,证明了热电类比简化模型能够很好的模拟开阀响应时间,气流升压模型能够对微型阀开阀后性能进行很好的估计。
One of the key points in designing a microvalve is to estimate the response time. A hot gas actuated bistable microvalve was designed. A simplified model was established by using the thermoelectric analogy method, and the valve response time was estimated. On the basis of the flow measurement, an air flow boost model is established, and the response time required to fill a certain space with the gas flowing through the micro valve is calculated. The analysis results show that when the thickness of heating diaphragm is 8μm and 25μm, the valve opening response time is 11.4ms and 20.2ms respectively. At a pressure differential of 200 kPa, the boost response time of the airflow in 55.5 μL space is 7.6 ms. Testing the performance of the microvalve proves that the thermoelectric analogy simplified model can well simulate the valve opening response time, and the airlift model can estimate the performance of the microvalve after the valve is opened.