题目
(25分) 为了得到热水,0.361 MPa (ts=140℃) 的水蒸气在管外凝结(如图3所示),其表面传热系数_(0)=9500W/((m)^2cdot R)。冷却水在盘管内流动,流速为_(0)=9500W/((m)^2cdot R),黄铜管外径为_(0)=9500W/((m)^2cdot R),壁厚为_(0)=9500W/((m)^2cdot R),导热系数为_(0)=9500W/((m)^2cdot R),盘管的弯曲半径为_(0)=9500W/((m)^2cdot R)。冷水进换热器时的温度为_(0)=9500W/((m)^2cdot R),加热到_(0)=9500W/((m)^2cdot R)。试求所需的换热面积及盘管长度。不计管内入口效应修正及温差修正。附注:(1) 管内湍流强制对流换热实验关联式为:_(0)=9500W/((m)^2cdot R) (流体被加热n=0.4;流体被冷却n=0.3)(2) 60oC时水的物性:ρ=983.1 kg/m3, cp=4.179 kJ/(kg K),λ=65.9×10-2 W/(m K),ν=0.478×10-6 m2/s, Pr=2.99;(3) 弯管修正系数:_(0)=9500W/((m)^2cdot R)
(25分) 为了得到热水,0.361 MPa (ts=140℃) 的水蒸气在管外凝结(如图3所示),其表面传热系数
。冷却水在盘管内流动,流速为
,黄铜管外径为
,壁厚为
,导热系数为
,盘管的弯曲半径为
。冷水进换热器时的温度为
,加热到
。试求所需的换热面积及盘管长度。不计管内入口效应修正及温差修正。
附注:
(1) 管内湍流强制对流换热实验关联式为:
(流体被加热n=0.4;流体被冷却n=0.3)
(2) 60oC时水的物性:ρ=983.1 kg/m3, cp=4.179 kJ/(kg K),λ=65.9×10-2 W/(m K),
ν=0.478×10-6 m2/s, Pr=2.99;
(3) 弯管修正系数:
题目解答
答案
解:
(1) 求管内表面传热系数
定性温度: 
查物性:ρ=983.1kg/m3, cp=4.179kJ/(kg·K),λ=0.659W/(m·K),
ν=0.478×10-6 m2/s,Pr=2.99
∴ 
……………………..(8分)
(2) 传热量
…………………………….(3分)
(3) 对数平均温差
………………………(4分)
(4)传热系数
………...(6分)
(5)换热面积及管长
………………………………...(4分)
解析
步骤 1：计算管内表面传热系数
定性温度：${t}_{r}=\dfrac {1}{2}({t}_{1}+{t}_{1})=\dfrac {1}{2}(25+95)=60({}^{\circ }C)$
查物性：ρ=983.1kg/m3, cp=4.179kJ/(kg·K),λ=0.659W/(m·K),
ν=0.478×10-6 m2/s,Pr=2.99
$Re=\dfrac {Ud,}{V}=\dfrac {0.8\times 0.015}{0.478\times {10}^{-2}}=2.51\times {10}^{4}\gt {10}^{4}$ ∴ 湍流
${c}_{t}=1+10.3{(\dfrac {d}{R})}^{3}=1+10.3{(\dfrac {15}{90})}^{3}=1.05$
${h}_{1}=\dfrac {\lambda }{d},\mu u=\dfrac {65.9\times {10}^{-2}}{0.015}\times 124=5.45\times {10}^{5}({m}^{2},\erasure )$
步骤 2：计算传热量
$\bigcirc {1}=pu\cdot \dfrac {\pi }{4}{d}^{2}$ ·cp $(i-i)=968.1\times 0.8\times \dfrac {\pi }{4}\times 0.1115\times 4179\times 19.5-25j=4.07\times {10}^{4}(N)$
步骤 3：计算对数平均温差
$\Delta t=\dfrac {(140-25)-(140-95)}{\ln \dfrac {140-25}{140-25}}=\dfrac {70}{\ln \dfrac {115}{45}}=74.6({}^{\circ }C)$
步骤 4：计算传热系数
$k0= \dfrac {1}{{a}_{1}}-\dfrac {{d}_{2}}{{d}_{2}}-\dfrac {{d}_{2}}{22}\ln \dfrac {{d}_{1}}{{a}_{1}}+\dfrac {1}{{b}_{2}}=\dfrac {1}{5.45\times {10}^{3$ __ 2.20× $\dfrac {1}{{10}^{-4}+1.24\times {10}^{-5}+1.05\times {10}^{-4}}=2.96\times {10}^{3}({m}^{2}\cdot k)$
步骤 5：计算换热面积及管长
$A=\dfrac {\phi }{{k}_{0}\Delta {c}_{m}}=\dfrac {4.07\times {10}^{4}}{2.96\times {10}^{3}\times 74.6}=0.184({m}^{2})$ $I=\dfrac {A}{\pi {d}_{0}}=\dfrac {0.184}{\pi \times 0.018}=3.25({m}^{2})$
定性温度：${t}_{r}=\dfrac {1}{2}({t}_{1}+{t}_{1})=\dfrac {1}{2}(25+95)=60({}^{\circ }C)$
查物性：ρ=983.1kg/m3, cp=4.179kJ/(kg·K),λ=0.659W/(m·K),
ν=0.478×10-6 m2/s,Pr=2.99
$Re=\dfrac {Ud,}{V}=\dfrac {0.8\times 0.015}{0.478\times {10}^{-2}}=2.51\times {10}^{4}\gt {10}^{4}$ ∴ 湍流
${c}_{t}=1+10.3{(\dfrac {d}{R})}^{3}=1+10.3{(\dfrac {15}{90})}^{3}=1.05$
${h}_{1}=\dfrac {\lambda }{d},\mu u=\dfrac {65.9\times {10}^{-2}}{0.015}\times 124=5.45\times {10}^{5}({m}^{2},\erasure )$
步骤 2：计算传热量
$\bigcirc {1}=pu\cdot \dfrac {\pi }{4}{d}^{2}$ ·cp $(i-i)=968.1\times 0.8\times \dfrac {\pi }{4}\times 0.1115\times 4179\times 19.5-25j=4.07\times {10}^{4}(N)$
步骤 3：计算对数平均温差
$\Delta t=\dfrac {(140-25)-(140-95)}{\ln \dfrac {140-25}{140-25}}=\dfrac {70}{\ln \dfrac {115}{45}}=74.6({}^{\circ }C)$
步骤 4：计算传热系数
$k0= \dfrac {1}{{a}_{1}}-\dfrac {{d}_{2}}{{d}_{2}}-\dfrac {{d}_{2}}{22}\ln \dfrac {{d}_{1}}{{a}_{1}}+\dfrac {1}{{b}_{2}}=\dfrac {1}{5.45\times {10}^{3$ __ 2.20× $\dfrac {1}{{10}^{-4}+1.24\times {10}^{-5}+1.05\times {10}^{-4}}=2.96\times {10}^{3}({m}^{2}\cdot k)$
步骤 5：计算换热面积及管长
$A=\dfrac {\phi }{{k}_{0}\Delta {c}_{m}}=\dfrac {4.07\times {10}^{4}}{2.96\times {10}^{3}\times 74.6}=0.184({m}^{2})$ $I=\dfrac {A}{\pi {d}_{0}}=\dfrac {0.184}{\pi \times 0.018}=3.25({m}^{2})$