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 的砖和 50 mm 的发泡聚苯乙烯组成，面积为 20 m²。外表面温度为 20 °C 和 -8 °C。 求两层交界面处的温度 Ti。
L1 = 150 mm, k1 = 0.72 W/(m·K)L2 = 50 mm, k2 = 0.035 W/(m·K)A = 20 m²T1 = 20 °CT3 = -8 °C
Ti = °C
2.热传导困难
由 50 mm 木材和 150 mm 玻璃组成的墙，将 21 °C 的房间与 -1 °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 20 W/(m²·K)，墙面积为 15 m²。 求墙内表面的温度 Ts。
L1 = 50 mm, k1 = 0.17 W/(m·K)L2 = 150 mm, k2 = 0.96 W/(m·K)A = 15 m²hi = 7.7 W/(m²·K)ho = 20 W/(m²·K)Troom = 21 °CTout = -1 °C
Ts = °C
3.换热器 (LMTD)中等
一台并流换热器传递 240 kW。 其传热面积为 25 m²。热流体从 175 °C 冷却到 80 °C，冷流体从 20 °C 加热到 45 °C。求总传热系数 U。
Q̇ = 240 kWA = 25 m²Th: 175 → 80 °CTc: 20 → 45 °C
U = W/(m²·K)
4.换热器 (LMTD)困难
在并流套管式换热器中，热油（0.5 kg/s，cp = 2.5 kJ/(kg·K)）在内管中流动，从 90 °C 冷却到 50 °C。 环隙中的水（1.5 kg/s，cp = 4.18 kJ/(kg·K)）以 20 °C 进入。内管直径为 40 mm，U = 200 W/(m²·K) 以该直径为基准。求所需的管长 L。
ṁh = 0.5 kg/scp,h = 2.5 kJ/(kg·K)Th: 90 → 50 °Cṁc = 1.5 kg/sTc,in = 20 °CD = 40 mmU = 200 W/(m²·K)
L = m
3溶液
1.热传导
答案: Ti = 16.44 °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)/(0.72·20) = 0.01042 K/W
R2 = L2/(k2A) = (50/1000)/(0.035·20) = 0.07143 K/W
热流量
Q̇ = (T1 − T3)/(R1 + R2) = 28/0.08185 = 342.1 W
交界面温度
第一层中温度下降 Q̇·R₁。
Ti = T1 − Q̇·R1 = 20 − 342.1·0.01042 = 16.44 °C
答案
Ti = 16.44 °C
2.热传导
答案: Ts = 16.47 °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。
串联热阻
Ri = 1/(hiA) = 1/(7.7·15) = 0.008658 K/W
R1 = L1/(k1A) = (50/1000)/(0.17·15) = 0.01961 K/W
R2 = L2/(k2A) = (150/1000)/(0.96·15) = 0.01042 K/W
Ro = 1/(hoA) = 1/(20·15) = 0.003333 K/W
ΣR = 0.04202 K/W
热损失
Q̇ = (21 − (-1))/0.04202 = 523.6 W
Ts = Troom − Q̇·Ri = 21 − 523.6·0.008658 = 16.47 °C
答案
Ts = 16.47 °C
3.换热器 (LMTD)
答案: U = 119 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 = 175 − 20 = 155 K
ΔT₂ = Th,out − Tc,out = 80 − 45 = 35 K
对数平均温差
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (155 − 35)/ln(155/35) = 80.64 K
总传热系数
U = Q̇/(A·ΔTlm) = 240000/(25·80.64) = 119 W/(m²·K)
答案
U = 119 W/(m²·K)
4.换热器 (LMTD)
答案: L = 47.95 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 = 0.5·2.5·(90 − 50) = 50 kW
Tc,out = 20 + 50/(1.5·4.18) = 27.97 °C
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 = 90 − 20 = 70 K
ΔT₂ = Th,out − Tc,out = 50 − 27.97 = 22.03 K
对数平均温差
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (70 − 22.03)/ln(70/22.03) = 41.49 K
面积和长度
A = 50000/(200·41.49) = 6.026 m²
L = A/(πD) = 6.026/(π·0.04) = 47.95 m
答案
L = 47.95 m
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