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 的玻璃和 150 mm 的混凝土组成，面积为 6 m²。外表面温度为 20 °C 和 -13 °C。 求通过墙体的热流量 Q̇。
L1 = 200 mm, k1 = 0.96 W/(m·K)L2 = 150 mm, k2 = 1.4 W/(m·K)A = 6 m²T1 = 20 °CT3 = -13 °C
Q̇ = W
2.热传导困难
由 150 mm 矿棉和 150 mm 砖组成的墙，将 21 °C 的房间与 -12 °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。 求通过墙体的热损失 Q̇。
L1 = 150 mm, k1 = 0.04 W/(m·K)L2 = 150 mm, k2 = 0.72 W/(m·K)A = 15 m²hi = 6 W/(m²·K)ho = 25 W/(m²·K)Troom = 21 °CTout = -12 °C
Q̇ = W
3.换热器 (LMTD)中等
一台并流换热器传递 290 kW。 其传热面积为 4 m²。热流体从 170 °C 冷却到 50 °C，冷流体从 25 °C 加热到 45 °C。求总传热系数 U。
Q̇ = 290 kWA = 4 m²Th: 170 → 50 °CTc: 25 → 45 °C
U = W/(m²·K)
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 155 °C 冷却到 95 °C。 环隙中的水（0.5 kg/s，cp = 4.18 kJ/(kg·K)）以 25 °C 进入。内管直径为 40 mm，U = 500 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: 155 → 95 °Cṁc = 0.5 kg/sTc,in = 25 °CD = 40 mmU = 500 W/(m²·K)
L = m
3溶液
1.热传导
答案: Q̇ = 627.6 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)/(0.96·6) = 0.03472 K/W
R2 = L2/(k2A) = (150/1000)/(1.4·6) = 0.01786 K/W
热流量
Q̇ = (T1 − T3)/(R1 + R2) = 33/0.05258 = 627.6 W
答案
Q̇ = 627.6 W
2.热传导
答案: Q̇ = 118.8 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。
串联热阻
Ri = 1/(hiA) = 1/(6·15) = 0.01111 K/W
R1 = L1/(k1A) = (150/1000)/(0.04·15) = 0.25 K/W
R2 = L2/(k2A) = (150/1000)/(0.72·15) = 0.01389 K/W
Ro = 1/(hoA) = 1/(25·15) = 0.002667 K/W
ΣR = 0.2777 K/W
热损失
Q̇ = (21 − (-12))/0.2777 = 118.8 W
答案
Q̇ = 118.8 W
3.换热器 (LMTD)
答案: U = 1744 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
并流时，两个入口在同一端，两个出口在另一端。
ΔT₁ = Th,in − Tc,in = 170 − 25 = 145 K
ΔT₂ = Th,out − Tc,out = 50 − 45 = 5 K
对数平均温差
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (145 − 5)/ln(145/5) = 41.58 K
总传热系数
U = Q̇/(A·ΔTlm) = 290000/(4·41.58) = 1744 W/(m²·K)
答案
U = 1744 W/(m²·K)
4.换热器 (LMTD)
答案: L = 32.81 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·(155 − 95) = 138 kW
Tc,out = 25 + 138/(0.5·4.18) = 91.03 °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 = 155 − 91.03 = 63.97 K
ΔT₂ = Th,out − Tc,in = 95 − 25 = 70 K
对数平均温差
ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (63.97 − 70)/ln(63.97/70) = 66.94 K
面积和长度
A = 138000/(500·66.94) = 4.123 m²
L = A/(πD) = 4.123/(π·0.04) = 32.81 m
答案
L = 32.81 m
ExamProblems · examproblems.com/courses/heat-transfer