COOLER( S: _( S: |: q8 Y
SELECTION
9 K- ?6 E# y/ |Designation:
/ w3 e; _' _4 p0 `( D3 FPV = Power loss [kW]
u/ ]' e# l( v5 m4 l3 l1 jP01 = Specific cooling capacity [kW/°C] E6 t( g; w' M" b
V = Tank contents [l]- @' g' b6 R! F0 X% G5 p" R) u5 F
ρoil = Density of the oil [kg/l], M: w1 }- [" f
for mineral oil: 0.915 kg/l/ D! s$ \, Q) s0 }' ?/ Y
Coil = Specific heat capacity [kJ/kgK]
1 E0 W4 ]. C: j1 D$ i' |for mineral oil 1.88 kJ/kgK
. p6 c9 _# k* k( i: H1 U2 |/ O∆T = Temperature increase in the
4 n$ Z2 T, D) w0 D' Q0 E: dsystem [°C]
: Q L5 J& x+ S5 Wt = Operating time [min]
4 | E; `1 ]7 mT1 = Desired oil temperature [°C]% K, p2 i& Q; E3 M) K
T3 = Ambient temperature [°C]3 ]. ?" Q% H0 ]2 _0 W4 E0 c; S
Example 1:
6 r! h5 [( K7 E1 {5 hMeasurement of the power loss! M g% Q. ? H$ Z* C" G- d8 Y! \
on existing units and machinery.
" w% N" r2 Z9 h7 ~3 ?, H* w" ~For this method the temperature7 N- [1 F# g2 j0 L# w# L
increase of the oil is measured( v5 V4 ~3 i# {; i j
over a certain period. The power
) \: m- N% K" O6 R/ \loss can be calculated from the. o6 z5 A' h& f) ]1 R) g8 |4 _
temperature increase.) R5 W- L. ^3 G+ j6 J
Parameters:4 I. Q& G. A9 ^2 A6 E
The oil temperature increases7 E# J0 w" D! c Q& K
from 20 °C to 45 °C over! X0 T' A/ N6 q3 {: ]
15 minutes.5 F9 u# C- X5 u
The tank contains 100 l.
/ k1 R7 |+ c! R" ^0 p- \# j! jHeat to be dissipated:
" `/ H- t3 w7 c& c" G8 uPV =（∆T × C oil × ρoil × V）/t × 60 [kW]
( i9 {" u' v1 e" z$ m3 t
& o6 P/ b2 q- c: l1 OPV =25 × 1.88 × 0.915 × 100 / (15 × 60 ). }1 r) k ]$ r- Q. z! P2 K
= 4.78 [kW]
# \9 G& `8 r% X: ?; F9 Y8 j3 n0 K1 iCooler selection:6 j& W( w9 {" v$ V
– Desired oil temperature: 60 °C9 E' Q8 r8 _* r
– Ambient temperature air: 30 °C
2 s- R6 }# Y$ N1 R" n& k: z' ]( u! P F' d( o F0 d2 S
P01 =PV / (T1 - T3 ) [kW/°C] =0.159) L' z: z; Q! |
A 10% safety margin is# r, P' G: X5 {7 m
recommended to allow for
% S; \2 x3 J1 V9 n' u( Z' belement contamination, and
; g& i5 ]! f* p3 Btherefore the specific power is:/ J$ j' I6 x9 r- ]
P01 × 1.1 = 0.175 kW/°C.
% T( Y: d l% F2 I7 aThe power loss 0.175 kW/°C must
) D5 l, i. d. G6 sbe dissipated by an oil cooler.9 w6 j; R6 Y/ n5 H. v8 v: Y
Suggestion:
- a6 d6 `" Y8 U5 J' F0 t7 x* `–Cooler OK-ELH2 - 3000 rpm,
3 j& \4 z" p1 a9 bP01 = 0.20 kW/°C at 80 l/min
( f$ y5 L* R4 T# \- g A% p6 G3 r; Y9 p" I+ _" j
7 | N' v- N4 U# f! F5 P! w) H6 O我发了一段在上面！ |
发表于 2007-11-30 11:19:26
