10-12;查
表
关于同志近三年现实表现材料材料类招标技术评分表图表与交易pdf视力表打印pdf用图表说话 pdf
可得,λ1=0.8+0.00058t,λ2=0.4+0.00026t假设,tw2=tw1+tw22=1200+802=6400C则λ1=0.8+0.00058×1200+6402=1.334λ2=0.4+0.00026×640+802=0.494q=tw1-tw2δ1λ1+δ2λ2=1200-800.3481.334+0.1160.494=2259tw2=tw1-δ1λ1q=1200-0.3481.334*2250=6100C假设与检验相差不大于4%,设tw2=6100Cλ1=1.325wm.0C,λ2=0.490wm.0C,q=2243Wm2,tw2=6610Ctw2与假设相差不大则
说明
关于失联党员情况说明岗位说明总经理岗位说明书会计岗位说明书行政主管岗位说明书
结果可信Q=qAτ=2243*1*3600=8075KJ当取λ2=0.116W(m.0C)代替外层材料时由q=tw1-tw2δ1λ1+δ2λ2⟹2243=1200-800.3481.325+820.116δ2=27mm10-13,R1=12πλ1lnd2d1=12π58ln170160=1.66*10-4(m.0C)w蒸汽管(R1)R2=12πλ2lnd3d2=12π0.17ln230170=0.283(m.0C)w第一层保温材料(R2)R3=12πλ3lnd4d3=12π0.093ln330230=0.618(m.0C)w第二层保温材料(R3)(2)qL=t1-t4i=13Ri=300-501.66*10-4+0.293+0.618=277.46wm(3)t2=t1-qL*R1=300-277.42*1.66*10-4=299.950Ct2=t2-qL*R2=299.95-277.42*0.283=221.440C10-19,经过10h后F0=aτx2=0.1*1042=0.0625x2aτ=12F0=120.0625=2.0erfx2aτ=0.995332又可知的是,tw=8000C,t0=200C,tw-ttw-t0=0.995332t=tw-0.995332*tw-t0=800-0.995332*800-20=23.650C10-21(1)Bi=hδλ=2000*0.00925=0.720由1Bi=1.39,可查图ΘWΘm~1Bi,ΘWΘm=0.717ΘWΘ0=tf-twt0-tf=800-175030-1750=0.552ΘmΘ0=ΘWΘ0ΘmΘw=0.552*10.717=0.725查图ΘmΘ0~1Bi,F0=0.6τ=F0*σ2a=0.6*0.009225/(560*8400)=9.14s(2)Θm=0.725Θ0,tm=0.717Θ0+tf,又,ΘmΘ0=tm-tft0-tftm=0.725*30-1750+1750=5030C则有,Δt=800-503=2970C(3)gardt平均=ΔtmaxΔx=2970.09=330000C/m