b). 热交换器内压降
Re 2 = Re1 = 3.018 × 10
.8
200.4
5
Pr = 0.858
k = 567.7 × 10 W 0
mC
?3Cp=5.448×10J0
C kg3
Nu ? k 5235 567.7 × 10 . ×W 0
h = = = 11888?3 mC 25 × 10 d
?3
hFΔt 11888 × π × 25 × 10. × 0
Δt 2 = = = 19.4 C
WC p 15 × 40 ×0.5294 5.448 ×
3 10
0Δt 2 t = t1 ? = 290.3 C
2
Nu = 0.023 Re Pr = 5235.
?3
3 ?6kg 1m ==747.3∴ v2 = 0.001333μ2 = 93.36 × 10 Pa ? s ρ 2 v2 kg m m=144381
3 s =
d 22 × ( 25 × 1)
Wv2 0.5294 ×0.0013381
V2 = 0.
π π ?3 2
4 4
.
热交换器 试验段
Re 2 =
d 2V2 ρ 2
μ 2
25 × 10 × 1443 × 7=
47.33
?3
93.36 × 10
?6
= 2.888 × 10
5
13:14:49 习题讲解 36
1 3? ? ? 0.0015 6 ? 10 ?f2 = 0.0055 ?1 + ? 20000 ×+ 5 ? = 0.0146
? ? 25 3.018 ×10? ?? ?
? μ w
f = f 2 ? ? = 0.0160
?
? μf ?
0.6
ΔPf2 = f2 L2 ρ2V= 752Pa
d2 2 a
ΔPel2 = 0
ΔP 2 = G ( v3 ? v1 ) = ?1751Pa
2
2 2
13:14:49 习题讲解 37
