μ=1.8mPa·s,Cp=2070J/kg·℃.解:Re=duρ/μ,Pr=Cpμ/λα1=(0.023λ/d)(Re的0.8次方)(Pr的0.4次方)传热系数1/K=(1/α2)×(d2/d1)+(d2/λ)ln(d2/d1)+1/α1∵管壁较薄热阻忽略∴1/K=1/α1+1/α2K=α1α2/(α1+α2)=[(0.023λ/d)(Re的0.8次方)(Pr的0.4次方)α2]/[(0.023λ/d)(Re的0.8次方)(Pr的0.4次方)+α2]Δtm=[(T1-t2)-(T2-t1)]/ln[(T1-t2)(T2-t1)]=[(75-45)-(65-30)]/ln[(75-45)(65-30)]=32.44℃Q=KAΔtm=QmCp(T1-T2)=uAρCp(T1-T2)....(终式)
将所有式子带入终式
可以解方程得到流速u,带入Qm=uAρ,解得Qm=1.24㎏/s