COOLER2 ^( m! x/ o2 l) q) I/ V7 }
SELECTION
& e6 t% Y& E& \* z" ~9 D+ ^+ eDesignation:
9 _3 \8 l" n0 M) XPV = Power loss [kW]
! c W# d; z K* D" xP01 = Specific cooling capacity [kW/°C]% l8 y6 y# H- O. G0 D* x A
V = Tank contents [l]
I$ c' l, {" O) sρoil = Density of the oil [kg/l]
! ]3 S6 Z- X( ?3 M9 s- |for mineral oil: 0.915 kg/l5 r) [6 q8 }) N. c! D
Coil = Specific heat capacity [kJ/kgK]3 o, H* h( f$ C' x
for mineral oil 1.88 kJ/kgK
. N8 C+ n: I9 i* ^8 P) `, t# n/ b∆T = Temperature increase in the
3 A! O7 L$ C- c2 ?5 [1 xsystem [°C]
% ` G, C0 p, [0 j9 ]t = Operating time [min]) Z; P# P, y i3 L* P& C% _
T1 = Desired oil temperature [°C]4 |! ]+ D1 Y& K8 g, J" C
T3 = Ambient temperature [°C]: G9 q. }: h" c6 A3 h: K
Example 1:
! t7 L! D0 b+ n0 SMeasurement of the power loss
# g) q Q% @: @% N8 eon existing units and machinery.% Z% a: E; R$ Y" z* V# c
For this method the temperature
) J' C# k) S6 p4 ]8 J* @) cincrease of the oil is measured
2 |0 X! D: x: D1 ?8 a, m$ t7 mover a certain period. The power ~& x6 Z5 `7 G. d
loss can be calculated from the
: ?% x$ a' `2 ]" o2 [4 m4 @temperature increase.
' x; A+ Q" _) t) LParameters:6 B$ a% _- J* Q2 w4 l- G5 Q( x
The oil temperature increases( o4 d) E6 Y* J! T0 a
from 20 °C to 45 °C over- u8 J Z) u6 P+ g. i: @5 J0 L
15 minutes.
' h) }6 t9 a0 E! v- A4 M3 z) B1 V+ GThe tank contains 100 l.
/ D. h0 t" Y+ y8 h7 y" {Heat to be dissipated:3 J7 ` P; T8 I+ Z$ j- }1 V* f
PV =（∆T × C oil × ρoil × V）/t × 60 [kW]
( ?3 \" j: U. D( B; E5 T) H5 f# _% s- `! Y; q% P) [
PV =25 × 1.88 × 0.915 × 100 / (15 × 60 ); R$ i* Y+ y7 G0 ~
= 4.78 [kW]: p6 r: U/ s/ a- o& ~& n; L
Cooler selection:( P6 }$ P8 j0 u
– Desired oil temperature: 60 °C
6 P3 o% Z1 u9 w8 } n% c; M+ ]– Ambient temperature air: 30 °C
4 M. @5 N8 f# d2 V7 f! `
3 {+ e: j3 S9 JP01 =PV / (T1 - T3 ) [kW/°C] =0.159 t1 ]' B+ a# O! G1 T
A 10% safety margin is; f. y9 |7 e2 K4 S$ i
recommended to allow for
$ J1 C8 U9 m# k' n/ b- g# Lelement contamination, and
' j6 H! x" f; r3 S& W7 Gtherefore the specific power is:+ y$ B! ?2 {6 l$ x% D- [
P01 × 1.1 = 0.175 kW/°C.; C$ K3 x- ~0 F
The power loss 0.175 kW/°C must. ?# q5 [- b* V( q6 F
be dissipated by an oil cooler.
2 h; n) @7 K8 S6 F3 I5 @( I6 zSuggestion:
" E# s0 ?3 j5 T/ Y0 v–Cooler OK-ELH2 - 3000 rpm,9 w2 a7 D- [( H$ V7 b4 t) |
P01 = 0.20 kW/°C at 80 l/min
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: l0 _3 F4 d) {0 X$ u! z- i) G* m; k( L4 D& E: r4 z2 S
我发了一段在上面！ |
发表于 2007-11-30 11:19:26
