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문제설명

문제설명. T water = 20 ℃ (냉각수 온도) h = 500 W/m 2  K ( 다음페이지 또는 열전달 책 참고 ) SFL 명령어로 가해주면 됩니다. T air = 25 ℃ (외부 공기온도) h = 5 W/m 2  K ( 다음페이지 참고, 최대20이하 … ) SFL 명령어로 가해주면 됩니다. I²× R 로 heat generation (W) 계산하신후 전체 볼륨으로 나누면 ( W/m³) 이 계산됩니다. 이 값( W/m³) 을 BFA 명령어로 가해주면 됩니다.

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문제설명

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  1. 문제설명 Twater = 20 ℃ (냉각수 온도) h = 500 W/m2K (다음페이지 또는 열전달 책참고) SFL 명령어로 가해주면 됩니다. Tair = 25 ℃ (외부 공기온도) h = 5 W/m2K (다음페이지 참고, 최대20이하…) SFL 명령어로 가해주면 됩니다. I²× R 로 heat generation (W) 계산하신후전체 볼륨으로 나누면 (W/m³) 이 계산됩니다.이 값(W/m³)을 BFA 명령어로 가해주면 됩니다.

  2. Convective Heat Transfer Coefficients

  3. 간단한 해석모델 Twater = 20 ℃ h = 500 W/m2K Tair = 25 ℃ h = 5 W/m2K Motor coil Heat generation ( 임의의 값 ) = 100 W/m3

  4. Input file /solu SFL,1,CONV,5, ,25, SFL,2,CONV,5, ,25, SFL,3,CONV,5, ,25, SFL,4,CONV,5, ,25, SFL,9,CONV,500, ,20, SFL,10,CONV,500, ,20, SFL,11,CONV,500, ,20, SFL,12,CONV,500, ,20, BFA,2,HGEN,100 solve /POST1 PLNSOL,temp, /Prep7 Et,1,77 Mp,kxx,1,0.78 ! thermal conductivity of the motor coil Mp,kxx,2,0.78 ! thermal conductivity of the S18 rect,,1,,1 rect,0.3,0.7,0.1,0.5 wpoff,0.5,0.75 PCIRC,0.1, ,0,360, AOVLAP,all ADELE, 3, , ,1 esize,0.05 mat,1 amesh,2 mat,2 amesh,4

  5. Result (Temp)

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