热传导
- 傅里叶定律
- Q̇ = kAΔT/L
- 导热热阻
- R = L/(kA)
- 对流热阻
- R = 1/(hA)
- 串联热阻
- Q̇ = ΔTtotal/ΣR
- U 值
- U = 1/(ΣR·A)
传热学：速成包
公式表、每个专题的关键方法步骤和 4 道练习题，完整解答附在最后。
1公式与方法
换热器 (LMTD)
- 热负荷
- Q = U·A·ΔTlm
- 对数平均温差
- ΔTlm = (ΔT1 − ΔT2)/ln(ΔT1/ΔT2)
- 物流热量衡算
- Q = ṁcp(Tin − Tout)
2练习题
1.热传导中等
一面墙由 150 mm 的石膏板和 150 mm 的木材组成，面积为 6 m²。外表面温度为 23 °C 和 9 °C。 求两层交界面处的温度 Ti。
L1 = 150 mm, k1 = 0.17 W/(m·K)L2 = 150 mm, k2 = 0.17 W/(m·K)A = 6 m²T1 = 23 °CT3 = 9 °C
Ti = °C
2.热传导困难
由 250 mm 混凝土和 25 mm 发泡聚苯乙烯组成的墙，将 21 °C 的房间与 -10 °C 的室外空气隔开。对流换热系数室内为 7.7 W/(m²·K)，室外为 30 W/(m²·K)，墙面积为 8 m²。 求通过墙体的热损失 Q̇。
L1 = 250 mm, k1 = 1.4 W/(m·K)L2 = 25 mm, k2 = 0.035 W/(m·K)A = 8 m²hi = 7.7 W/(m²·K)ho = 30 W/(m²·K)Troom = 21 °CTout = -10 °C
Q̇ = W
3.换热器 (LMTD)中等
一台并流换热器传递 220 kW。 其传热面积为 5 m²。热流体从 115 °C 冷却到 90 °C，冷流体从 30 °C 加热到 70 °C。求总传热系数 U。
Q̇ = 220 kWA = 5 m²Th: 115 → 90 °CTc: 30 → 70 °C
U = W/(m²·K)
4.换热器 (LMTD)困难
在逆流套管式换热器中，热油（0.2 kg/s，cp = 2.1 kJ/(kg·K)）在内管中流动，从 120 °C 冷却到 65 °C。 环隙中的水（0.3 kg/s，cp = 4.18 kJ/(kg·K)）以 15 °C 进入。内管直径为 50 mm，U = 150 W/(m²·K) 以该直径为基准。求所需的管长 L。
ṁh = 0.2 kg/scp,h = 2.1 kJ/(kg·K)Th: 120 → 65 °Cṁc = 0.3 kg/sTc,in = 15 °CD = 50 mmU = 150 W/(m²·K)
L = m
3溶液
1.热传导
答案: Ti = 16 °C
热阻：通过串联各层的稳态热流类似于通过串联电阻的电流：Q̇ = ΔT/ΣR，导热层 R = L/(kA)，对流表面 R = 1/(hA)。各层的温降与其热阻成正比。
各层热阻
R1 = L1/(k1A) = (150/1000)/(0.17·6) = 0.1471 K/W
R2 = L2/(k2A) = (150/1000)/(0.17·6) = 0.1471 K/W
热流量
Q̇ = (T1 − T3)/(R1 + R2) = 14/0.2941 = 47.6 W
交界面温度
第一层中温度下降 Q̇·R₁。
Ti = T1 − Q̇·R1 = 23 − 47.6·0.1471 = 16 °C
答案
Ti = 16 °C
2.热传导
答案: Q̇ = 234.8 W
热阻：通过串联各层的稳态热流类似于通过串联电阻的电流：Q̇ = ΔT/ΣR，导热层 R = L/(kA)，对流表面 R = 1/(hA)。各层的温降与其热阻成正比。
串联热阻
Ri = 1/(hiA) = 1/(7.7·8) = 0.01623 K/W
R1 = L1/(k1A) = (250/1000)/(1.4·8) = 0.02232 K/W
R2 = L2/(k2A) = (25/1000)/(0.035·8) = 0.08929 K/W
Ro = 1/(hoA) = 1/(30·8) = 0.004167 K/W
ΣR = 0.132 K/W
热损失
Q̇ = (21 − (-10))/0.132 = 234.8 W
答案
Q̇ = 234.8 W
3.换热器 (LMTD)
答案: U = 979.5 W/(m²·K)
LMTD method: the temperature difference between the streams changes along the exchanger, so the rate equation uses its logarithmic mean: Q̇ = U·A·ΔTlm, with ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂), where ΔT₁ and ΔT₂ are the differences at the two ends. Counter-flow gives a larger ΔTlm than parallel flow for the same inlet and outlet temperatures.
两端温差
并流时，两个入口在同一端，两个出口在另一端。
ΔT₁ = Th,in − Tc,in = 115 − 30 = 85 K
ΔT₂ = Th,out − Tc,out = 90 − 70 = 20 K
对数平均温差
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (85 − 20)/ln(85/20) = 44.92 K
总传热系数
U = Q̇/(A·ΔTlm) = 220000/(5·44.92) = 979.5 W/(m²·K)
答案
U = 979.5 W/(m²·K)
4.换热器 (LMTD)
答案: L = 14.72 m
LMTD method: the temperature difference between the streams changes along the exchanger, so the rate equation uses its logarithmic mean: Q̇ = U·A·ΔTlm, with ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂), where ΔT₁ and ΔT₂ are the differences at the two ends. Counter-flow gives a larger ΔTlm than parallel flow for the same inlet and outlet temperatures.
热负荷和水的出口温度
Q̇ = 0.2·2.1·(120 − 65) = 23.1 kW
Tc,out = 15 + 23.1/(0.3·4.18) = 33.42 °C
两端温差
逆流时，热流体入口对着冷流体出口，热流体出口对着冷流体入口。
ΔT₁ = Th,in − Tc,out = 120 − 33.42 = 86.58 K
ΔT₂ = Th,out − Tc,in = 65 − 15 = 50 K
对数平均温差
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (86.58 − 50)/ln(86.58/50) = 66.62 K
面积和长度
A = 23100/(150·66.62) = 2.311 m²
L = A/(πD) = 2.311/(π·0.05) = 14.72 m
答案
L = 14.72 m
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