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换热器的计算---请教

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发表于 2007-11-30 11:19:26 | 显示全部楼层 |阅读模式
最近单位上新的项目,要选择换热器,但是如何计算?板式和管式有啥区别?
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发表于 2007-12-6 21:25:50 | 显示全部楼层
管式的不好,建议选用板式的。
发表于 2008-1-19 13:19:25 | 显示全部楼层
管式的冷却面积没有板式冷却面积大
发表于 2008-2-11 02:54:59 | 显示全部楼层
可以查阅化工版的机械设计手册第4册上面有详细的介绍换热面积的算法
发表于 2008-2-11 20:23:09 | 显示全部楼层
板式的换热效率高,不过最好选择好一些的制造厂家.小的可以用管式的划算.
发表于 2008-2-12 09:13:20 | 显示全部楼层
这得看你需要得散热功率是多大,一般大的用板式散热器
发表于 2008-2-15 14:08:16 | 显示全部楼层
COOLER; ~2 t" Q4 I! |
SELECTION
9 M  V$ a( p6 |  `' s3 qDesignation:
2 Q% b  ]: Q) |) k" k5 h' QPV = Power loss [kW]
2 a, x' m2 S+ ~- n! VP01 = Specific cooling capacity [kW/°C]& G  l8 v3 g; o& {2 \* S+ U
V = Tank contents [l]
- p, Z  n* z, b0 r; L6 N2 \: wρoil = Density of the oil [kg/l]
" ^# @' H% w; z  j7 r+ efor mineral oil: 0.915 kg/l
# J& K: P0 s; {Coil = Specific heat capacity [kJ/kgK]
! d) U6 v0 n' J& ]$ Efor mineral oil 1.88 kJ/kgK
6 O" }/ t, \) w∆T = Temperature increase in the
$ m  R& w9 `$ I) f0 hsystem [°C]  x/ H/ U! D, K! `& e( r
t = Operating time [min]
/ C( i. O0 Y5 \6 N$ a/ l8 A& eT1 = Desired oil temperature [°C]
7 R' I# ?8 O, ]1 J7 L8 ET3 = Ambient temperature [°C]: P/ O$ {8 q1 _; j) W! O2 i4 T; @
Example 1:" r+ Q- |+ ]: D7 h
Measurement of the power loss
+ c9 Q* ~4 V/ i1 S; f% a  }0 kon existing units and machinery.1 V! v1 v9 O6 E0 ]
For this method the temperature, F* {/ T9 |% C! g! G) }
increase of the oil is measured, j- p5 p5 t4 I% s0 }# y- Z
over a certain period. The power
% D! H3 x; b6 E6 x1 I( Xloss can be calculated from the4 f3 G: J6 O- S: r% w
temperature increase.1 Y3 e# Z5 f6 f3 u
Parameters:
  E8 L# K7 A- A7 P7 S3 cThe oil temperature increases, ?; c' {5 r3 e+ P. d
from 20 °C to 45 °C over4 ^- z% S& m  h# i/ h$ M1 e2 p! t
15 minutes.- B" R7 ]0 I; H1 f# f. ]7 J
The tank contains 100 l.
9 ]: R$ {0 N' I7 r+ qHeat to be dissipated:
# I+ [" Q/ W  y1 q. E+ @PV =(∆T × C oil × ρoil × V)/t × 60    [kW]
' q7 B5 @2 \" F! z4 r3 p' ^% t0 ~$ F
PV =25 × 1.88 × 0.915 × 100 / (15 × 60 )
/ y, [9 W  z8 [! J= 4.78 [kW]* X) C$ E3 \2 {& e( G/ _
Cooler selection:
1 e" n# y! a+ k- {– Desired oil temperature: 60 °C
. |3 c  u5 a; n5 M– Ambient temperature air: 30 °C
( M( e% U& ~/ i  t& ~' Q& d
/ J9 ^3 z# B) xP01 =PV    /  (T1 - T3  ) [kW/°C] =0.159! ]. f* a0 A7 w, `5 y0 q% o& x+ X- s
A 10% safety margin is
5 R+ n1 k% Q8 nrecommended to allow for- y# S6 h5 `6 k0 j
element contamination, and, }3 e7 b- J' C! B0 ~
therefore the specific power is:( S# e, |8 F% y( ^# x& @
P01 × 1.1 = 0.175 kW/°C.* q! Z, u7 j' w1 ~/ h
The power loss 0.175 kW/°C must6 |9 F5 c8 z: |/ O
be dissipated by an oil cooler.
) s9 ~3 L" h, C8 E; uSuggestion:" x/ s: ~/ L0 t% g$ z# W$ k. v
–Cooler OK-ELH2 - 3000 rpm,
5 W& u7 L8 \  h+ FP01 = 0.20 kW/°C at 80 l/min* C" _, c  t0 |
. n9 p$ F% y4 {+ p2 g) w- B: t' M

, ?1 D# {+ @% H* z* E我发了一段在上面!
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