Question

用变压变送器测流量,差压为25kpa,二次表量程为0-200T\h,差压变送器输入电流为4-20mA,试求电流为6.56

Answer 1

输入的电压是4~20mA,所以4mA对应的差压是0,流量也是0；20mA对应的是满量程值.则(6.56-4)/(20-4)=2.56/16=0.16 .6.56mA时对应的差压为25*0.16=4kpa；流量为200*0.16=32T/h;

同理(8-4)/(20-4)=0.25.8mA时对应的差压为25*0.25=6.25kpa；流量为200*0.25=50T/h;