以下是红外热像仪应用中套管电压致热缺陷的诊断根据：套管外壁是一层导热性能较差的绝缘陶瓷，通过测量表面温度，我们很难得到设备内部的真实温度。一般来说，我们建议采用三相对比的方式来衡量套管是否异常发热。对于套管缺油的情况，是一种电压致热故障。这种故障对于电力部门的安全生产是一种较大隐患。在实际运行中检测人员应当注意观察，以免造成重大损失。以下是套管缺油的诊断判据：变压器套管热缺陷处理建议如果异常发热是由于将军帽与外部接线板或内部导电杆接触不良所造成的。
PV023R1K1T1NHCC
PV023R1K1T1VMMC
PV023R1K1T1WMRC
PV023R9K1T1NFWS
PV023R1K1JHNMMC
PV023R1K4T1NFHS
PV023R1L1T1NMMC
PV023R1K8T1VMMC
PV023R1K1T1NFRZ
PV023R9K1T1NMMC
PV023R1K4T1NMR1
PV023R1K1T1VFDS
PV023R1K8T1NMMC
PV023R1K8T1NFWS
PV023R1K1T1WMR1
PV023R1K8T1N001
PV023L1K1T1NMMC
PV023R1K4T1NMMC
PV023R1K1T1NKLC
PV023R1K4T1NFR1
PV023R1K1S1NFWS
PV023R1K1T1NFFC
PV023L1K1T1NFWS
PV023R1K1T1WFDS
PV023R1K1T1NFFP
PV023R1L1T1NF
PV023R1K1T1WMM1
PV023R1K1T1NHLC
PV023R1K1T1NMRZ
PV023R1K1T1NMRK
PV023R1K1T1WMMC
PV023R1K1T1NFF1
PV028R1K1T1N001
PV028R1K1T1N100
PV028R1K1T1NFDS
PV028R1K1T1NFR1
PV028R1K1T1NFHS
PV028R1K1T1NMMC
PV028R1K1T1NMM1
PV028R1K1T1NMRC
PV028R1K1T1NFWS
PV028R1K1T1NFRC
PV028R1K1T1NFF1
PV028R1K1T1WMM1
PV028R1K1T1WFR1
PV028R1K8T1NFWS
PV028R1K4T1NFR1
PV028R1K1T1VMMC
PV028R1K1T1NHCC
PV028R1K4T1NMMC
PV028R1K1T1NELC
PV028R1K1T1NHLC
PV028R1K8T1N001
PV028R1K1T1NF
PV028R9K1T1NFWS
PV028R9K1T1NMMC
PV028R1K1T1VFDS
PV028R1K1AYNMRZ
PV028R1K8T1NMMC
PV028R1K1T1NMRK
PV028R1K1T1NFFP
PV028L1K1T1NMMC
PV028R1K1T1NFRZ
PV028R1K1S1NFWS
PV028R1K1T1NMMZ
PV028R1K1T1NMR1
PV028R1K1T1NMFC
PV028R1K1T1WFDS
PV028L1K1T1NFWS
PV028R1K1JHNMMC
PV028R1K1T1NMRZ
PV028R1K4T1NFHS
PV028R1K1T1NMF1
PV028R1K1T1NGLC
PV028R1K1T1WMRC
PV028R1L1T1NMMC
PV028R1K1T1WMMC
PV028R1K1T1WMR1
PV028R1K8T1VMMC
PV028R1K1T1NFFC
PV028R1K1T1NMMK
PV028R1K4T1NMR1
PV032R1K1T1N001
PV032R1K1T1N100
PV032R1K1T1NFDS
PV032R1K1T1NFR1
PV032R1K1T1NFHS
PV032R1K1T1NMMC
PV032R1K1T1NMM1
PV032R1K1T1NMRC
PV032R1K1T1NFWS
PV032R1K1T1NFRC
PV032R1K1T1NFF1
PV032R1K1AYNMTZ
PV032R1K1T1NMFC
PV032R1K1T1NMR1
PV032R1K1T1NMF1
PV032R1K1T1NMRK
PV032R1K1T1NFFP
PV032R1K1T1WMMC
PV032R1K4T1NMMC
PV032R1K8T1VMMC
PV032R1K1T1WFR1
PV032R1K1T1NFFC
PV032R1K1JHNMMC
PV032R1K1T1WMRC
PV032R9K1T1NFWS
每一种测量模式是怎么计算的，如何应用，本文将进行详细说明。RMS（真有效值）真有效值简单而言即代表一交流电相当于多大数值的直流电在单位时间内所做的功，真有效值为10V的交流电与10V的直流电对相同的负载在相同的时间下所做的功相同。举个例子来说有一组100伏的电池组，每次供电10分钟之后停10分钟（模拟出交流信号），如果这组电池带动的是10Ω电阻，供电的10分钟内，产生的电流I=U/R=10A，功率P=U*I=1000W的功率，停电时电流和功率为零,那么在20分钟的其平均功率为500W。
