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 50 mm 的发泡聚苯乙烯和 150 mm 的石膏板组成，面积为 10 m²。外表面温度为 22 °C 和 -5 °C。 求通过墙体的热流量 Q̇。
L1 = 50 mm, k1 = 0.035 W/(m·K)L2 = 150 mm, k2 = 0.17 W/(m·K)A = 10 m²T1 = 22 °CT3 = -5 °C
Q̇ = W
2.热传导困难
由 250 mm 矿棉和 25 mm 玻璃组成的墙，将 22 °C 的房间与 5 °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 6 m²。 求墙内表面的温度 Ts。
L1 = 250 mm, k1 = 0.04 W/(m·K)L2 = 25 mm, k2 = 0.96 W/(m·K)A = 6 m²hi = 8 W/(m²·K)ho = 30 W/(m²·K)Troom = 22 °CTout = 5 °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 640 kW。热流体从 160 °C 冷却到 85 °C，冷流体从 20 °C 加热到 35 °C。该布置的 LMTD 修正系数为 F = 0.8，U = 1000 W/(m²·K)。求所需的传热面积 A。
Q̇ = 640 kWF = 0.8U = 1000 W/(m²·K)Th: 160 → 85 °CTc: 20 → 35 °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 = 2.3 kJ/(kg·K)）在内管中流动，从 115 °C 冷却到 55 °C。 环隙中的水（0.8 kg/s，cp = 4.18 kJ/(kg·K)）以 25 °C 进入。内管直径为 20 mm，U = 400 W/(m²·K) 以该直径为基准。求所需的管长 L。
ṁh = 1 kg/scp,h = 2.3 kJ/(kg·K)Th: 115 → 55 °Cṁc = 0.8 kg/sTc,in = 25 °CD = 20 mmU = 400 W/(m²·K)
L = m
3溶液
1.热传导
答案: Q̇ = 116.8 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) = (50/1000)/(0.035·10) = 0.1429 K/W
R2 = L2/(k2A) = (150/1000)/(0.17·10) = 0.08824 K/W
热流量
Q̇ = (T1 − T3)/(R1 + R2) = 27/0.2311 = 116.8 W
答案
Q̇ = 116.8 W
2.热传导
答案: Ts = 21.67 °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) = (250/1000)/(0.04·6) = 1.042 K/W
R2 = L2/(k2A) = (25/1000)/(0.96·6) = 0.00434 K/W
Ro = 1/(hoA) = 1/(30·6) = 0.005556 K/W
ΣR = 1.072 K/W
热损失
Q̇ = (22 − (5))/1.072 = 15.85 W
Ts = Troom − Q̇·Ri = 22 − 15.85·0.02083 = 21.67 °C
答案
Ts = 21.67 °C
3.换热器 (LMTD)
答案: A = 8.719 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 = 160 − 35 = 125 K
ΔT₂ = Th,out − Tc,in = 85 − 20 = 65 K
对数平均温差
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (125 − 65)/ln(125/65) = 91.75 K
面积
A = Q̇/(U·F·ΔTlm,cf) = 640000/(1000·0.8·91.75) = 8.719 m²
答案
A = 8.719 m²
4.换热器 (LMTD)
答案: L = 142.2 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 = 1·2.3·(115 − 55) = 138 kW
Tc,out = 25 + 138/(0.8·4.18) = 66.27 °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 = 115 − 66.27 = 48.73 K
ΔT₂ = Th,out − Tc,in = 55 − 25 = 30 K
对数平均温差
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (48.73 − 30)/ln(48.73/30) = 38.61 K
面积和长度
A = 138000/(400·38.61) = 8.935 m²
L = A/(πD) = 8.935/(π·0.02) = 142.2 m
答案
L = 142.2 m
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