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 100 mm 的矿棉和 75 mm 的发泡聚苯乙烯组成，面积为 15 m²。外表面温度为 23 °C 和 5 °C。 求通过墙体的热流量 Q̇。
L1 = 100 mm, k1 = 0.04 W/(m·K)L2 = 75 mm, k2 = 0.035 W/(m·K)A = 15 m²T1 = 23 °CT3 = 5 °C
Q̇ = W
2.热传导困难
由 50 mm 矿棉和 150 mm 混凝土组成的墙，将 20 °C 的房间与 3 °C 的室外空气隔开。对流换热系数室内为 8 W/(m²·K)，室外为 30 W/(m²·K)，墙面积为 20 m²。 求墙内表面的温度 Ts。
L1 = 50 mm, k1 = 0.04 W/(m·K)L2 = 150 mm, k2 = 1.4 W/(m·K)A = 20 m²hi = 8 W/(m²·K)ho = 30 W/(m²·K)Troom = 20 °CTout = 3 °C
Ts = °C
3.换热器 (LMTD)中等
一台逆流换热器传递 500 kW。 其传热面积为 8 m²。热流体从 130 °C 冷却到 85 °C，冷流体从 15 °C 加热到 85 °C。求总传热系数 U。
Q̇ = 500 kWA = 8 m²Th: 130 → 85 °CTc: 15 → 85 °C
U = W/(m²·K)
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 120 °C 冷凝。冷却水（0.5 kg/s，cp = 4.18 kJ/(kg·K)）以 10 °C 进入管内。求水的出口温度 Tc,out。
Ts = 120 °CA = 1.5 m²U = 2000 W/(m²·K)ṁ = 0.5 kg/scp = 4.18 kJ/(kg·K)Tc,in = 10 °C
Tc,out = °C
3溶液
1.热传导
答案: Q̇ = 58.15 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) = (100/1000)/(0.04·15) = 0.1667 K/W
R2 = L2/(k2A) = (75/1000)/(0.035·15) = 0.1429 K/W
热流量
Q̇ = (T1 − T3)/(R1 + R2) = 18/0.3095 = 58.15 W
答案
Q̇ = 58.15 W
2.热传导
答案: Ts = 18.6 °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/(8·20) = 0.00625 K/W
R1 = L1/(k1A) = (50/1000)/(0.04·20) = 0.0625 K/W
R2 = L2/(k2A) = (150/1000)/(1.4·20) = 0.005357 K/W
Ro = 1/(hoA) = 1/(30·20) = 0.001667 K/W
ΣR = 0.07577 K/W
热损失
Q̇ = (20 − (3))/0.07577 = 224.4 W
Ts = Troom − Q̇·Ri = 20 − 224.4·0.00625 = 18.6 °C
答案
Ts = 18.6 °C
3.换热器 (LMTD)
答案: U = 1105 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
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 = 130 − 85 = 45 K
ΔT₂ = Th,out − Tc,in = 85 − 15 = 70 K
对数平均温差
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (45 − 70)/ln(45/70) = 56.58 K
总传热系数
U = Q̇/(A·ΔTlm) = 500000/(8·56.58) = 1105 W/(m²·K)
答案
U = 1105 W/(m²·K)
4.换热器 (LMTD)
答案: Tc,out = 93.82 °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 = 110·0.238 = 26.18 K
Tc,out = 120 − 26.18 = 93.82 °C
答案
Tc,out = 93.82 °C
传递的热量：Q̇ = 175.2 kW。
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