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 250 mm 的矿棉和 150 mm 的砖组成，面积为 15 m²。外表面温度为 19 °C 和 -2 °C。 求通过墙体的热流量 Q̇。
L1 = 250 mm, k1 = 0.04 W/(m·K)L2 = 150 mm, k2 = 0.72 W/(m·K)A = 15 m²T1 = 19 °CT3 = -2 °C
Q̇ = W
2.热传导困难
由 200 mm 发泡聚苯乙烯和 50 mm 玻璃组成的墙，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 8 W/(m²·K)，室外为 25 W/(m²·K)，墙面积为 20 m²。 求墙体的总传热系数 U。
L1 = 200 mm, k1 = 0.035 W/(m·K)L2 = 50 mm, k2 = 0.96 W/(m·K)A = 20 m²hi = 8 W/(m²·K)ho = 25 W/(m²·K)Troom = 21 °CTout = -15 °C
U = W/(m²·K)
3.换热器 (LMTD)中等
一台管壳式换热器（单壳程、双管程）须传递 340 kW。热流体从 115 °C 冷却到 60 °C，冷流体从 25 °C 加热到 85 °C。该布置的 LMTD 修正系数为 F = 0.85，U = 300 W/(m²·K)。求所需的传热面积 A。
Q̇ = 340 kWF = 0.85U = 300 W/(m²·K)Th: 115 → 60 °CTc: 25 → 85 °C
A = m²
4.换热器 (LMTD)困难
在一台逆流换热器中，两股流体的热容量流率相同，ṁ·cp = 2 kW/K。热流体以 170 °C 进入，冷流体以 15 °C 进入。换热器的 A = 2 m²，U = 500 W/(m²·K)。求热流体出口温度 Th,out。
Ch = Cc = 2 kW/KTh,in = 170 °CTc,in = 15 °CA = 2 m²U = 500 W/(m²·K)
Th,out = °C
3溶液
1.热传导
答案: Q̇ = 48.77 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) = (250/1000)/(0.04·15) = 0.4167 K/W
R2 = L2/(k2A) = (150/1000)/(0.72·15) = 0.01389 K/W
热流量
Q̇ = (T1 − T3)/(R1 + R2) = 21/0.4306 = 48.77 W
答案
Q̇ = 48.77 W
2.热传导
答案: U = 0.1686 W/(m²·K)
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/(8·20) = 0.00625 K/W
R1 = L1/(k1A) = (200/1000)/(0.035·20) = 0.2857 K/W
R2 = L2/(k2A) = (50/1000)/(0.96·20) = 0.002604 K/W
Ro = 1/(hoA) = 1/(25·20) = 0.002 K/W
ΣR = 0.2966 K/W
总传热系数
U = 1/(A·ΣR) = 1/(20·0.2966) = 0.1686 W/(m²·K)
答案
U = 0.1686 W/(m²·K)
3.换热器 (LMTD)
答案: A = 41.11 m²
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（逆流）
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 = 115 − 85 = 30 K
ΔT₂ = Th,out − Tc,in = 60 − 25 = 35 K
对数平均温差
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (30 − 35)/ln(30/35) = 32.44 K
面积
A = Q̇/(U·F·ΔTlm,cf) = 340000/(300·0.85·32.44) = 41.11 m²
答案
A = 41.11 m²
4.换热器 (LMTD)
答案: Th,out = 118.3 °C
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.
恒定温差
冷流体出口对着热流体入口：Tc,out = Tc,in + (Th,in − Th,out)，所以 ΔT₁ = Th,in − Tc,out = Th,out − Tc,in = ΔT₂。
ΔTlm = Th,out − Tc,in
能量平衡等于传热方程
2·(170 − Th,out) = 1·(Th,out − 15)
Th,out = (2·170 + 1·15)/(2 + 1) = 118.3 °C
答案
Th,out = 118.3 °C
传递的热量：Q̇ = 103.3 kW；冷流体以 66.67 °C 离开。
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