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 200 mm 的石膏板和 25 mm 的木材组成，面积为 4 m²。外表面温度为 20 °C 和 -3 °C。 求通过墙体的热流量 Q̇。
L1 = 200 mm, k1 = 0.17 W/(m·K)L2 = 25 mm, k2 = 0.17 W/(m·K)A = 4 m²T1 = 20 °CT3 = -3 °C
Q̇ = W
2.热传导困难
由 100 mm 砖和 75 mm 矿棉组成的墙，将 20 °C 的房间与 -9 °C 的室外空气隔开。对流换热系数室内为 6 W/(m²·K)，室外为 25 W/(m²·K)，墙面积为 6 m²。 求墙体的总传热系数 U。
L1 = 100 mm, k1 = 0.72 W/(m·K)L2 = 75 mm, k2 = 0.04 W/(m·K)A = 6 m²hi = 6 W/(m²·K)ho = 25 W/(m²·K)Troom = 20 °CTout = -9 °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 260 kW。热流体从 170 °C 冷却到 85 °C，冷流体从 25 °C 加热到 55 °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 = 260 kWF = 0.85U = 300 W/(m²·K)Th: 170 → 85 °CTc: 25 → 55 °C
A = m²
4.换热器 (LMTD)困难
在逆流套管式换热器中，热油（1 kg/s，cp = 2.5 kJ/(kg·K)）在内管中流动，从 120 °C 冷却到 50 °C。 环隙中的水（1 kg/s，cp = 4.18 kJ/(kg·K)）以 20 °C 进入。内管直径为 25 mm，U = 150 W/(m²·K) 以该直径为基准。求所需的管长 L。
ṁh = 1 kg/scp,h = 2.5 kJ/(kg·K)Th: 120 → 50 °Cṁc = 1 kg/sTc,in = 20 °CD = 25 mmU = 150 W/(m²·K)
L = m
3溶液
1.热传导
答案: Q̇ = 69.51 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) = (200/1000)/(0.17·4) = 0.2941 K/W
R2 = L2/(k2A) = (25/1000)/(0.17·4) = 0.03676 K/W
热流量
Q̇ = (T1 − T3)/(R1 + R2) = 23/0.3309 = 69.51 W
答案
Q̇ = 69.51 W
2.热传导
答案: U = 0.4503 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/(6·6) = 0.02778 K/W
R1 = L1/(k1A) = (100/1000)/(0.72·6) = 0.02315 K/W
R2 = L2/(k2A) = (75/1000)/(0.04·6) = 0.3125 K/W
Ro = 1/(hoA) = 1/(25·6) = 0.006667 K/W
ΣR = 0.3701 K/W
总传热系数
U = 1/(A·ΣR) = 1/(6·0.3701) = 0.4503 W/(m²·K)
答案
U = 0.4503 W/(m²·K)
3.换热器 (LMTD)
答案: A = 12.06 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 = 170 − 55 = 115 K
ΔT₂ = Th,out − Tc,in = 85 − 25 = 60 K
对数平均温差
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (115 − 60)/ln(115/60) = 84.54 K
面积
A = Q̇/(U·F·ΔTlm,cf) = 260000/(300·0.85·84.54) = 12.06 m²
答案
A = 12.06 m²
4.换热器 (LMTD)
答案: L = 349.3 m
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.
热负荷和水的出口温度
Q̇ = 1·2.5·(120 − 50) = 175 kW
Tc,out = 20 + 175/(1·4.18) = 61.87 °C
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 = 120 − 61.87 = 58.13 K
ΔT₂ = Th,out − Tc,in = 50 − 20 = 30 K
对数平均温差
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (58.13 − 30)/ln(58.13/30) = 42.53 K
面积和长度
A = 175000/(150·42.53) = 27.43 m²
L = A/(πD) = 27.43/(π·0.025) = 349.3 m
答案
L = 349.3 m
ExamProblems · examproblems.com/courses/heat-transfer