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 的矿棉和 100 mm 的砖组成，面积为 6 m²。外表面温度为 15 °C 和 -11 °C。 求两层交界面处的温度 Ti。
L1 = 150 mm, k1 = 0.04 W/(m·K)L2 = 100 mm, k2 = 0.72 W/(m·K)A = 6 m²T1 = 15 °CT3 = -11 °C
Ti = °C
2.热传导困难
由 100 mm 石膏板和 25 mm 矿棉组成的墙，将 22 °C 的房间与 -4 °C 的室外空气隔开。对流换热系数室内为 8 W/(m²·K)，室外为 20 W/(m²·K)，墙面积为 6 m²。 求墙内表面的温度 Ts。
L1 = 100 mm, k1 = 0.17 W/(m·K)L2 = 25 mm, k2 = 0.04 W/(m·K)A = 6 m²hi = 8 W/(m²·K)ho = 20 W/(m²·K)Troom = 22 °CTout = -4 °C
Ts = °C
3.换热器 (LMTD)中等
一台管壳式换热器（单壳程、双管程）须传递 370 kW。热流体从 150 °C 冷却到 130 °C，冷流体从 15 °C 加热到 70 °C。该布置的 LMTD 修正系数为 F = 0.88，U = 200 W/(m²·K)。求所需的传热面积 A。
Q̇ = 370 kWF = 0.88U = 200 W/(m²·K)Th: 150 → 130 °CTc: 15 → 70 °C
A = m²
4.换热器 (LMTD)困难
在一台传热面积为 4 m²、U = 2500 W/(m²·K) 的冷凝器中，蒸汽在管外以恒定的 110 °C 冷凝。冷却水（3 kg/s，cp = 4.18 kJ/(kg·K)）以 20 °C 进入管内。求水的出口温度 Tc,out。
Ts = 110 °CA = 4 m²U = 2500 W/(m²·K)ṁ = 3 kg/scp = 4.18 kJ/(kg·K)Tc,in = 20 °C
Tc,out = °C
3溶液
1.热传导
答案: Ti = −10.07 °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)/(0.04·6) = 0.625 K/W
R2 = L2/(k2A) = (100/1000)/(0.72·6) = 0.02315 K/W
热流量
Q̇ = (T1 − T3)/(R1 + R2) = 26/0.6481 = 40.11 W
交界面温度
第一层中温度下降 Q̇·R₁。
Ti = T1 − Q̇·R1 = 15 − 40.11·0.625 = −10.07 °C
答案
Ti = −10.07 °C
2.热传导
答案: Ts = 19.66 °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/(8·6) = 0.02083 K/W
R1 = L1/(k1A) = (100/1000)/(0.17·6) = 0.09804 K/W
R2 = L2/(k2A) = (25/1000)/(0.04·6) = 0.1042 K/W
Ro = 1/(hoA) = 1/(20·6) = 0.008333 K/W
ΣR = 0.2314 K/W
热损失
Q̇ = (22 − (-4))/0.2314 = 112.4 W
Ts = Troom − Q̇·Ri = 22 − 112.4·0.02083 = 19.66 °C
答案
Ts = 19.66 °C
3.换热器 (LMTD)
答案: A = 21.8 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 = 150 − 70 = 80 K
ΔT₂ = Th,out − Tc,in = 130 − 15 = 115 K
对数平均温差
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (80 − 115)/ln(80/115) = 96.44 K
面积
A = Q̇/(U·F·ΔTlm,cf) = 370000/(200·0.88·96.44) = 21.8 m²
答案
A = 21.8 m²
4.换热器 (LMTD)
答案: Tc,out = 69.46 °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) = 2500·4/(3·4180) = 0.7974
出口温度
ΔT₂ = ΔT₁·e−0.7974 = 90·0.4505 = 40.54 K
Tc,out = 110 − 40.54 = 69.46 °C
答案
Tc,out = 69.46 °C
传递的热量：Q̇ = 620.2 kW。
ExamProblems · examproblems.com/courses/heat-transfer