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 50 mm 的混凝土和 25 mm 的木材组成，面积为 20 m²。外表面温度为 22 °C 和 -14 °C。 求通过墙体的热流量 Q̇。
L1 = 50 mm, k1 = 1.4 W/(m·K)L2 = 25 mm, k2 = 0.17 W/(m·K)A = 20 m²T1 = 22 °CT3 = -14 °C
Q̇ = W
2.热传导困难
由 150 mm 玻璃和 100 mm 木材组成的墙，将 21 °C 的房间与 -15 °C 的室外空气隔开。对流换热系数室内为 6 W/(m²·K)，室外为 30 W/(m²·K)，墙面积为 20 m²。 求通过墙体的热损失 Q̇。
L1 = 150 mm, k1 = 0.96 W/(m·K)L2 = 100 mm, k2 = 0.17 W/(m·K)A = 20 m²hi = 6 W/(m²·K)ho = 30 W/(m²·K)Troom = 21 °CTout = -15 °C
Q̇ = W
3.换热器 (LMTD)中等
一台并流换热器传递 370 kW。 其传热面积为 12 m²。热流体从 100 °C 冷却到 80 °C，冷流体从 20 °C 加热到 60 °C。求总传热系数 U。
Q̇ = 370 kWA = 12 m²Th: 100 → 80 °CTc: 20 → 60 °C
U = W/(m²·K)
4.换热器 (LMTD)困难
在逆流套管式换热器中，热油（0.5 kg/s，cp = 2.1 kJ/(kg·K)）在内管中流动，从 90 °C 冷却到 50 °C。 环隙中的水（0.8 kg/s，cp = 4.18 kJ/(kg·K)）以 20 °C 进入。内管直径为 50 mm，U = 300 W/(m²·K) 以该直径为基准。求所需的管长 L。
ṁh = 0.5 kg/scp,h = 2.1 kJ/(kg·K)Th: 90 → 50 °Cṁc = 0.8 kg/sTc,in = 20 °CD = 50 mmU = 300 W/(m²·K)
L = m
3溶液
1.热传导
答案: Q̇ = 3939 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。
各层热阻
R1 = L1/(k1A) = (50/1000)/(1.4·20) = 0.001786 K/W
R2 = L2/(k2A) = (25/1000)/(0.17·20) = 0.007353 K/W
热流量
Q̇ = (T1 − T3)/(R1 + R2) = 36/0.009139 = 3939 W
答案
Q̇ = 3939 W
2.热传导
答案: Q̇ = 762.3 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/(6·20) = 0.008333 K/W
R1 = L1/(k1A) = (150/1000)/(0.96·20) = 0.007813 K/W
R2 = L2/(k2A) = (100/1000)/(0.17·20) = 0.02941 K/W
Ro = 1/(hoA) = 1/(30·20) = 0.001667 K/W
ΣR = 0.04722 K/W
热损失
Q̇ = (21 − (-15))/0.04722 = 762.3 W
答案
Q̇ = 762.3 W
3.换热器 (LMTD)
答案: U = 712.4 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 = 100 − 20 = 80 K
ΔT₂ = Th,out − Tc,out = 80 − 60 = 20 K
对数平均温差
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (80 − 20)/ln(80/20) = 43.28 K
总传热系数
U = Q̇/(A·ΔTlm) = 370000/(12·43.28) = 712.4 W/(m²·K)
答案
U = 712.4 W/(m²·K)
4.换热器 (LMTD)
答案: L = 21.1 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 = 0.5·2.1·(90 − 50) = 42 kW
Tc,out = 20 + 42/(0.8·4.18) = 32.56 °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 = 90 − 32.56 = 57.44 K
ΔT₂ = Th,out − Tc,in = 50 − 20 = 30 K
对数平均温差
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (57.44 − 30)/ln(57.44/30) = 42.25 K
面积和长度
A = 42000/(300·42.25) = 3.314 m²
L = A/(πD) = 3.314/(π·0.05) = 21.1 m
答案
L = 21.1 m
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