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 的玻璃和 25 mm 的混凝土组成，面积为 12 m²。外表面温度为 24 °C 和 9 °C。 求两层交界面处的温度 Ti。
L1 = 250 mm, k1 = 0.96 W/(m·K)L2 = 25 mm, k2 = 1.4 W/(m·K)A = 12 m²T1 = 24 °CT3 = 9 °C
Ti = °C
2.热传导困难
由 50 mm 发泡聚苯乙烯和 25 mm 玻璃组成的墙，将 20 °C 的房间与 -2 °C 的室外空气隔开。对流换热系数室内为 10 W/(m²·K)，室外为 25 W/(m²·K)，墙面积为 6 m²。 求墙内表面的温度 Ts。
L1 = 50 mm, k1 = 0.035 W/(m·K)L2 = 25 mm, k2 = 0.96 W/(m·K)A = 6 m²hi = 10 W/(m²·K)ho = 25 W/(m²·K)Troom = 20 °CTout = -2 °C
Ts = °C
3.换热器 (LMTD)中等
在冷凝器中，蒸汽在管外以恒定的 120 °C 冷凝。冷却水（2 kg/s，cp = 4.18 kJ/(kg·K)）在管内流动，从 20 °C 加热到 55 °C。总传热系数为 3000 W/(m²·K)。求所需的传热面积 A。
Ts = 120 °Cṁ = 2 kg/scp = 4.18 kJ/(kg·K)Tc: 20 → 55 °CU = 3000 W/(m²·K)
A = m²
4.换热器 (LMTD)困难
在一台传热面积为 1.5 m²、U = 2000 W/(m²·K) 的冷凝器中，蒸汽在管外以恒定的 130 °C 冷凝。冷却水（0.5 kg/s，cp = 4.18 kJ/(kg·K)）以 30 °C 进入管内。求水的出口温度 Tc,out。
Ts = 130 °CA = 1.5 m²U = 2000 W/(m²·K)ṁ = 0.5 kg/scp = 4.18 kJ/(kg·K)Tc,in = 30 °C
Tc,out = °C
3溶液
1.热传导
答案: Ti = 9.963 °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) = (250/1000)/(0.96·12) = 0.0217 K/W
R2 = L2/(k2A) = (25/1000)/(1.4·12) = 0.001488 K/W
热流量
Q̇ = (T1 − T3)/(R1 + R2) = 15/0.02319 = 646.8 W
交界面温度
第一层中温度下降 Q̇·R₁。
Ti = T1 − Q̇·R1 = 24 − 646.8·0.0217 = 9.963 °C
答案
Ti = 9.963 °C
2.热传导
答案: Ts = 18.62 °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/(10·6) = 0.01667 K/W
R1 = L1/(k1A) = (50/1000)/(0.035·6) = 0.2381 K/W
R2 = L2/(k2A) = (25/1000)/(0.96·6) = 0.00434 K/W
Ro = 1/(hoA) = 1/(25·6) = 0.006667 K/W
ΣR = 0.2658 K/W
热损失
Q̇ = (20 − (-2))/0.2658 = 82.78 W
Ts = Troom − Q̇·Ri = 20 − 82.78·0.01667 = 18.62 °C
答案
Ts = 18.62 °C
3.换热器 (LMTD)
答案: A = 1.2 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.
由水侧求热负荷
Q̇ = 2·4.18·(55 − 20) = 292.6 kW
两端温差
ΔT₁ = 120 − 20 = 100 K
ΔT₂ = 120 − 55 = 65 K
对数平均温差
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (100 − 65)/ln(100/65) = 81.25 K
面积
A = Q̇/(U·ΔTlm) = 292600/(3000·81.25) = 1.2 m²
答案
A = 1.2 m²
4.换热器 (LMTD)
答案: Tc,out = 106.2 °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.
联立能量平衡与传热方程
取 ΔT₁ = Ts − Tc,in，ΔT₂ = Ts − Tc,out，则 ΔT₁ − ΔT₂ 等于水的温升，所以温升被约去。
ṁcp(ΔT₁ − ΔT₂) = U·A·(ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂)
ln(ΔT₁/ΔT₂) = U·A/(ṁcp) = 2000·1.5/(0.5·4180) = 1.435
出口温度
ΔT₂ = ΔT₁·e−1.435 = 100·0.238 = 23.8 K
Tc,out = 130 − 23.8 = 106.2 °C
答案
Tc,out = 106.2 °C
传递的热量：Q̇ = 159.3 kW。
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