(rho )_(2)=(G)_(2)((l)_(i)-(t)_(i))=rho dfrac (pi {l)^2}(4),(k(g)^n-(t)_(i))=313318W-|||-^n=(100)^circ C-|||-Delta t=110-50=(60)^circ C-|||-_(总)=dfrac ({W)^2-(A)^2}(ln dfrac {N')(N)}=dfrac (10-60)(h)dfrac (10)(60)=(279)^circ C-|||-_(r)=(t)_(n)-(Delta )_(n)=110-279=(82.1)^circ C-|||-_(总)=dfrac ({W)^2-(A)^2}(ln dfrac {N')(N)}=dfrac (10-60)(h)dfrac (10)(60)=(279)^circ C-|||-_(r)=(t)_(n)-(Delta )_(n)=110-279=(82.1)^circ C-|||-ρ=970.4 kg/,c,=4.197kJ/ kg-C λ=0.675W/-|||-m-c-|||-=0.357times (10)^-6(m)^2/(H)_(3)=347times (10)^-6N... (g)^2(m)^2-|||-_(n)=259.0times (10)^-6N-8(m)^2PH=2.155-|||-_(c)=dfrac ({U)_(a)d}(V)=dfrac (13times 0.014)(0.357times {10)^-6}=509880.4-|||-_(r)=0.027R(c)_(2)(r)_(1)(R)_(1)(r)_(r)((dfrac {{H)_(3)}({R)_(6)})}^014=211.94-|||-alpha =(N)_(2)dfrac (lambda )(d)=10218.7-|||-_(r)=0.027R(c)_(2)(r)_(1)(R)_(1)(r)_(r)((dfrac {{H)_(3)}({R)_(6)})}^014=211.94-|||-alpha =(N)_(2)dfrac (lambda )(d)=10218.7-|||-_(1)=a(F)_(N)(I)_(2)=313389N-|||-(rho )_(2)=(G)_(n)((xi )_(i)(t)_(i))=rho dfrac (pi {l)^2}(4),(k(g)^n-4)=313318W山东大学热工基础课后习题解答可见 Q1 和 Q2 相差很小,所以10-6 用内径为 0.016 m、长为 2.5 m 的不锈钢管进行管内对流换热实验,实验时直接对不锈钢管通以直流电加热,电压为 5V,电流为 900A,水的进口温度为 20 C ,流速为 0.5 m/s,管外用保温材料保温,忽略热损失。试求管内对流换热的表面传热系数及换热温差。