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。

新题目返回课程

在打印窗口中选择“另存为 PDF”，即可在设备上保存一份副本。

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
混凝土, k = 1.4200 mm玻璃, k = 0.96100 mm17 °C8 °CQ̇ = ?

一面墙由 200 mm 的混凝土和 100 mm 的玻璃组成，面积为 12 m²。外表面温度为 17 °C 和 8 °C。 求通过墙体的热流量 Q̇。

L1 = 200 mm, k1 = 1.4 W/(m·K)L2 = 100 mm, k2 = 0.96 W/(m·K)A = 12 m²T1 = 17 °CT3 = 8 °C
Q̇ = W
2.热传导困难
砖, k = 0.72200 mm发泡聚苯乙烯, k = 0.035150 mm室内 22 °C室外 -13 °CQ̇

由 200 mm 砖和 150 mm 发泡聚苯乙烯组成的墙，将 22 °C 的房间与 -13 °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。 求墙体的总传热系数 U。

L1 = 200 mm, k1 = 0.72 W/(m·K)L2 = 150 mm, k2 = 0.035 W/(m·K)A = 6 m²hi = 8 W/(m²·K)ho = 30 W/(m²·K)Troom = 22 °CTout = -13 °C
U = W/(m²·K)
3.换热器 (LMTD)中等
长度T110 °C30 °C95 °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（1 kg/s，cp = 4.18 kJ/(kg·K)）在管内流动，从 30 °C 加热到 95 °C。总传热系数为 2500 W/(m²·K)。求所需的传热面积 A。

Ts = 110 °Cṁ = 1 kg/scp = 4.18 kJ/(kg·K)Tc: 30 → 95 °CU = 2500 W/(m²·K)
A = m²
4.换热器 (LMTD)困难
长度T110 °C80 °C10 °C?ΔT₁ΔT₂并流

在并流套管式换热器中，热油（0.5 kg/s，cp = 2.5 kJ/(kg·K)）在内管中流动，从 110 °C 冷却到 80 °C。 环隙中的水（0.5 kg/s，cp = 4.18 kJ/(kg·K)）以 10 °C 进入。内管直径为 20 mm，U = 250 W/(m²·K) 以该直径为基准。求所需的管长 L。

ṁh = 0.5 kg/scp,h = 2.5 kJ/(kg·K)Th: 110 → 80 °Cṁc = 0.5 kg/sTc,in = 10 °CD = 20 mmU = 250 W/(m²·K)
L = m

3溶液

1.热传导

答案: Q̇ = 437.2 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)/(1.4·12) = 0.0119 K/W
R2 = L2/(k2A) = (100/1000)/(0.96·12) = 0.008681 K/W

热流量

Q̇ = (T1 − T3)/(R1 + R2) = 9/0.02059 = 437.2 W

答案

Q̇ = 437.2 W
2.热传导

答案: U = 0.2118 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/(8·6) = 0.02083 K/W
R1 = L1/(k1A) = (200/1000)/(0.72·6) = 0.0463 K/W
R2 = L2/(k2A) = (150/1000)/(0.035·6) = 0.7143 K/W
Ro = 1/(hoA) = 1/(30·6) = 0.005556 K/W
ΣR = 0.787 K/W

总传热系数

U = 1/(A·ΣR) = 1/(6·0.787) = 0.2118 W/(m²·K)

答案

U = 0.2118 W/(m²·K)
3.换热器 (LMTD)

答案: A = 2.799 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 = 1·4.18·(95 − 30) = 271.7 kW

两端温差

ΔT₁ = 110 − 30 = 80 K
ΔT₂ = 110 − 95 = 15 K

对数平均温差

ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (80 − 15)/ln(80/15) = 38.83 K

面积

A = Q̇/(U·ΔTlm) = 271700/(2500·38.83) = 2.799 m²

答案

A = 2.799 m²
4.换热器 (LMTD)

答案: L = 32.51 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 = 0.5·2.5·(110 − 80) = 37.5 kW
Tc,out = 10 + 37.5/(0.5·4.18) = 27.94 °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 = 110 − 10 = 100 K
ΔT₂ = Th,out − Tc,out = 80 − 27.94 = 52.06 K

对数平均温差

ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (100 − 52.06)/ln(100/52.06) = 73.44 K

面积和长度

A = 37500/(250·73.44) = 2.043 m²
L = A/(πD) = 2.043/(π·0.02) = 32.51 m

答案

L = 32.51 m

ExamProblems · examproblems.com/courses/heat-transfer