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 的木材和 150 mm 的石膏板组成，面积为 15 m²。外表面温度为 15 °C 和 7 °C。 求两层交界面处的温度 Ti。
L1 = 200 mm, k1 = 0.17 W/(m·K)L2 = 150 mm, k2 = 0.17 W/(m·K)A = 15 m²T1 = 15 °CT3 = 7 °C
Ti = °C
2.热传导困难
由 50 mm 矿棉和 50 mm 石膏板组成的墙，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 15 m²。 求墙体的总传热系数 U。
L1 = 50 mm, k1 = 0.04 W/(m·K)L2 = 50 mm, k2 = 0.17 W/(m·K)A = 15 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
一台逆流换热器传递 160 kW。 其传热面积为 10 m²。热流体从 100 °C 冷却到 65 °C，冷流体从 20 °C 加热到 85 °C。求总传热系数 U。
Q̇ = 160 kWA = 10 m²Th: 100 → 65 °CTc: 20 → 85 °C
U = W/(m²·K)
4.换热器 (LMTD)困难
在逆流套管式换热器中，热油（0.3 kg/s，cp = 2.5 kJ/(kg·K)）在内管中流动，从 125 °C 冷却到 55 °C。 环隙中的水（1.5 kg/s，cp = 4.18 kJ/(kg·K)）以 15 °C 进入。内管直径为 32 mm，U = 300 W/(m²·K) 以该直径为基准。求所需的管长 L。
ṁh = 0.3 kg/scp,h = 2.5 kJ/(kg·K)Th: 125 → 55 °Cṁc = 1.5 kg/sTc,in = 15 °CD = 32 mmU = 300 W/(m²·K)
L = m
3溶液
1.热传导
答案: Ti = 10.43 °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) = (200/1000)/(0.17·15) = 0.07843 K/W
R2 = L2/(k2A) = (150/1000)/(0.17·15) = 0.05882 K/W
热流量
Q̇ = (T1 − T3)/(R1 + R2) = 8/0.1373 = 58.29 W
交界面温度
第一层中温度下降 Q̇·R₁。
Ti = T1 − Q̇·R1 = 15 − 58.29·0.07843 = 10.43 °C
答案
Ti = 10.43 °C
2.热传导
答案: U = 0.5712 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/(6·15) = 0.01111 K/W
R1 = L1/(k1A) = (50/1000)/(0.04·15) = 0.08333 K/W
R2 = L2/(k2A) = (50/1000)/(0.17·15) = 0.01961 K/W
Ro = 1/(hoA) = 1/(25·15) = 0.002667 K/W
ΣR = 0.1167 K/W
总传热系数
U = 1/(A·ΣR) = 1/(15·0.1167) = 0.5712 W/(m²·K)
答案
U = 0.5712 W/(m²·K)
3.换热器 (LMTD)
答案: U = 585.9 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 = 100 − 85 = 15 K
ΔT₂ = Th,out − Tc,in = 65 − 20 = 45 K
对数平均温差
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (15 − 45)/ln(15/45) = 27.31 K
总传热系数
U = Q̇/(A·ΔTlm) = 160000/(10·27.31) = 585.9 W/(m²·K)
答案
U = 585.9 W/(m²·K)
4.换热器 (LMTD)
答案: L = 26.34 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 = 0.3·2.5·(125 − 55) = 52.5 kW
Tc,out = 15 + 52.5/(1.5·4.18) = 23.37 °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 = 125 − 23.37 = 101.6 K
ΔT₂ = Th,out − Tc,in = 55 − 15 = 40 K
对数平均温差
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (101.6 − 40)/ln(101.6/40) = 66.09 K
面积和长度
A = 52500/(300·66.09) = 2.648 m²
L = A/(πD) = 2.648/(π·0.032) = 26.34 m
答案
L = 26.34 m
ExamProblems · examproblems.com/courses/heat-transfer