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。

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1公式与方法

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)

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
砖, k = 0.72100 mm石膏板, k = 0.1725 mm20 °C3 °CQ̇ = ?

一面墙由 100 mm 的砖和 25 mm 的石膏板组成，面积为 6 m²。外表面温度为 20 °C 和 3 °C。 求通过墙体的热流量 Q̇。

L1 = 100 mm, k1 = 0.72 W/(m·K)L2 = 25 mm, k2 = 0.17 W/(m·K)A = 6 m²T1 = 20 °CT3 = 3 °C
Q̇ = W
2.热传导困难
玻璃, k = 0.96100 mm石膏板, k = 0.1750 mm室内 21 °C室外 -3 °CQ̇

由 100 mm 玻璃和 50 mm 石膏板组成的墙，将 21 °C 的房间与 -3 °C 的室外空气隔开。对流换热系数室内为 10 W/(m²·K)，室外为 30 W/(m²·K)，墙面积为 4 m²。 求墙体的总传热系数 U。

L1 = 100 mm, k1 = 0.96 W/(m·K)L2 = 50 mm, k2 = 0.17 W/(m·K)A = 4 m²hi = 10 W/(m²·K)ho = 30 W/(m²·K)Troom = 21 °CTout = -3 °C
U = W/(m²·K)
3.换热器 (LMTD)中等
长度T110 °C10 °C55 °CΔT₁ΔT₂冷凝蒸汽

在冷凝器中，蒸汽在管外以恒定的 110 °C 冷凝。冷却水（5 kg/s，cp = 4.18 kJ/(kg·K)）在管内流动，从 10 °C 加热到 55 °C。总传热系数为 2000 W/(m²·K)。求所需的传热面积 A。

Ts = 110 °Cṁ = 5 kg/scp = 4.18 kJ/(kg·K)Tc: 10 → 55 °CU = 2000 W/(m²·K)
A = m²
4.换热器 (LMTD)困难
长度T90 °C50 °C15 °C?ΔT₁ΔT₂并流

在并流套管式换热器中，热油（0.5 kg/s，cp = 2.1 kJ/(kg·K)）在内管中流动，从 90 °C 冷却到 50 °C。 环隙中的水（0.8 kg/s，cp = 4.18 kJ/(kg·K)）以 15 °C 进入。内管直径为 32 mm，U = 400 W/(m²·K) 以该直径为基准。求所需的管长 L。

ṁh = 0.5 kg/scp,h = 2.1 kJ/(kg·K)Th: 90 → 50 °Cṁc = 0.8 kg/sTc,in = 15 °CD = 32 mmU = 400 W/(m²·K)
L = m

3溶液

1.热传导

答案: Q̇ = 356.7 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) = (100/1000)/(0.72·6) = 0.02315 K/W
R2 = L2/(k2A) = (25/1000)/(0.17·6) = 0.02451 K/W

热流量

Q̇ = (T1 − T3)/(R1 + R2) = 17/0.04766 = 356.7 W

答案

Q̇ = 356.7 W
2.热传导

答案: U = 1.881 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/(10·4) = 0.025 K/W
R1 = L1/(k1A) = (100/1000)/(0.96·4) = 0.02604 K/W
R2 = L2/(k2A) = (50/1000)/(0.17·4) = 0.07353 K/W
Ro = 1/(hoA) = 1/(30·4) = 0.008333 K/W
ΣR = 0.1329 K/W

总传热系数

U = 1/(A·ΣR) = 1/(4·0.1329) = 1.881 W/(m²·K)

答案

U = 1.881 W/(m²·K)
3.换热器 (LMTD)

答案: A = 6.247 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 = 5·4.18·(55 − 10) = 940.5 kW

两端温差

ΔT₁ = 110 − 10 = 100 K
ΔT₂ = 110 − 55 = 55 K

对数平均温差

ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (100 − 55)/ln(100/55) = 75.27 K

面积

A = Q̇/(U·ΔTlm) = 940500/(2000·75.27) = 6.247 m²

答案

A = 6.247 m²
4.换热器 (LMTD)

答案: L = 23.98 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 = 0.5·2.1·(90 − 50) = 42 kW
Tc,out = 15 + 42/(0.8·4.18) = 27.56 °C

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 = 90 − 15 = 75 K
ΔT₂ = Th,out − Tc,out = 50 − 27.56 = 22.44 K

对数平均温差

ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (75 − 22.44)/ln(75/22.44) = 43.56 K

面积和长度

A = 42000/(400·43.56) = 2.411 m²
L = A/(πD) = 2.411/(π·0.032) = 23.98 m

答案

L = 23.98 m

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