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 的木材和 25 mm 的混凝土组成，面积为 4 m²。外表面温度为 16 °C 和 -5 °C。 求两层交界面处的温度 Ti。
L1 = 100 mm, k1 = 0.17 W/(m·K)L2 = 25 mm, k2 = 1.4 W/(m·K)A = 4 m²T1 = 16 °CT3 = -5 °C
Ti = °C
2.热传导困难
由 200 mm 砖和 150 mm 发泡聚苯乙烯组成的墙，将 21 °C 的房间与 4 °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。 求墙内表面的温度 Ts。
L1 = 200 mm, k1 = 0.72 W/(m·K)L2 = 150 mm, k2 = 0.035 W/(m·K)A = 20 m²hi = 6 W/(m²·K)ho = 30 W/(m²·K)Troom = 21 °CTout = 4 °C
Ts = °C
3.换热器 (LMTD)中等
在冷凝器中，蒸汽在管外以恒定的 130 °C 冷凝。冷却水（4 kg/s，cp = 4.18 kJ/(kg·K)）在管内流动，从 25 °C 加热到 105 °C。总传热系数为 2500 W/(m²·K)。求所需的传热面积 A。
Ts = 130 °Cṁ = 4 kg/scp = 4.18 kJ/(kg·K)Tc: 25 → 105 °CU = 2500 W/(m²·K)
A = m²
4.换热器 (LMTD)困难
在一台传热面积为 2 m²、U = 2500 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 = 2 m²U = 2500 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 = −4.381 °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) = (100/1000)/(0.17·4) = 0.1471 K/W
R2 = L2/(k2A) = (25/1000)/(1.4·4) = 0.004464 K/W
热流量
Q̇ = (T1 − T3)/(R1 + R2) = 21/0.1515 = 138.6 W
交界面温度
第一层中温度下降 Q̇·R₁。
Ti = T1 − Q̇·R1 = 16 − 138.6·0.1471 = −4.381 °C
答案
Ti = −4.381 °C
2.热传导
答案: Ts = 20.41 °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/(6·20) = 0.008333 K/W
R1 = L1/(k1A) = (200/1000)/(0.72·20) = 0.01389 K/W
R2 = L2/(k2A) = (150/1000)/(0.035·20) = 0.2143 K/W
Ro = 1/(hoA) = 1/(30·20) = 0.001667 K/W
ΣR = 0.2382 K/W
热损失
Q̇ = (21 − (4))/0.2382 = 71.38 W
Ts = Troom − Q̇·Ri = 21 − 71.38·0.008333 = 20.41 °C
答案
Ts = 20.41 °C
3.换热器 (LMTD)
答案: A = 9.598 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 = 4·4.18·(105 − 25) = 1338 kW
两端温差
ΔT₁ = 130 − 25 = 105 K
ΔT₂ = 130 − 105 = 25 K
对数平均温差
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (105 − 25)/ln(105/25) = 55.75 K
面积
A = Q̇/(U·ΔTlm) = 1.338 × 10⁶/(2500·55.75) = 9.598 m²
答案
A = 9.598 m²
4.换热器 (LMTD)
答案: Tc,out = 120.9 °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) = 2500·2/(0.5·4180) = 2.392
出口温度
ΔT₂ = ΔT₁·e−2.392 = 100·0.09142 = 9.142 K
Tc,out = 130 − 9.142 = 120.9 °C
答案
Tc,out = 120.9 °C
传递的热量：Q̇ = 189.9 kW。
ExamProblems · examproblems.com/courses/heat-transfer