COOLER
: N! [* T9 Z( {' ISELECTION
1 F% b8 S, Z" }7 Z( [$ g% vDesignation:
: E2 j: C/ ^0 z1 }: kPV = Power loss [kW]
5 l# f+ r/ O3 H4 j7 _P01 = Specific cooling capacity [kW/°C]/ T% I- P& o# U) {# s" z
V = Tank contents [l]
- V3 {" S2 `5 G! a Qρoil = Density of the oil [kg/l]
/ \0 d5 g8 I2 o" ~for mineral oil: 0.915 kg/l% Z9 f! q* ]' A ^3 C- d
Coil = Specific heat capacity [kJ/kgK]
* g- K8 r. O; Vfor mineral oil 1.88 kJ/kgK8 M Y6 J" P0 [+ s8 B! C/ F+ a+ X
∆T = Temperature increase in the% d S" E7 z: A# k( }
system [°C]
+ i% W# v/ A& H5 ft = Operating time [min]/ W) ^2 h, h: v. R' v* j
T1 = Desired oil temperature [°C]0 [* t7 S2 E5 A7 T$ Z' @' P0 M' e$ Z
T3 = Ambient temperature [°C]
Y/ h' ~0 @8 W- G5 [. ~# y0 kExample 1:
x+ u) g1 c+ W7 Z% u" Q6 ~0 DMeasurement of the power loss
2 J) I, e, J6 R3 qon existing units and machinery.4 o, q# M* u. S! a- P. P
For this method the temperature
1 F6 J- q2 D6 l3 x& g- b* u# Qincrease of the oil is measured
8 ^2 {+ ?$ h; ?1 ]+ I6 I' o9 R) cover a certain period. The power+ R% F/ x) T8 S J2 O3 l- G8 N A) k
loss can be calculated from the8 W7 ^* @; `% k
temperature increase.4 B# X, k, ?8 c: k. x: X
Parameters:
: |& V8 w) Y4 e- w; n4 t) `& d% qThe oil temperature increases
+ x5 y6 ?2 |& lfrom 20 °C to 45 °C over4 c* z) u2 y% j- ]5 Y" {1 ]
15 minutes.
% d! f6 {7 ^6 F4 A# S" S9 GThe tank contains 100 l.) r9 }6 p, v+ y: t4 L9 L8 N* [- J
Heat to be dissipated:
# \; G S" ]: _" `- gPV =（∆T × C oil × ρoil × V）/t × 60 [kW]% M5 j0 z- }$ H1 }' q9 Q7 _( _
- M6 j, W! k4 p0 C6 }2 BPV =25 × 1.88 × 0.915 × 100 / (15 × 60 )
( l$ `. F' v( @3 |1 I- S= 4.78 [kW]6 c4 p& q) V1 P/ T) R# K- p# i
Cooler selection:
' Y) L& W$ r% ?% Y– Desired oil temperature: 60 °C& e, j/ _/ d# _$ J. x* R
– Ambient temperature air: 30 °C
1 j8 d5 g. G$ Y; ^/ E
; ^( L- n4 F+ C0 HP01 =PV / (T1 - T3 ) [kW/°C] =0.159
5 {5 I2 O% [9 VA 10% safety margin is+ k: q |4 v. Z5 c& E
recommended to allow for
8 G' s' F' [* c; W6 l. G9 melement contamination, and
# R7 T3 R$ u* Jtherefore the specific power is:2 C( D2 O1 S' @3 m4 L8 K5 }, ^% h
P01 × 1.1 = 0.175 kW/°C.
& L; k" |( H! q, o' I% }The power loss 0.175 kW/°C must
4 U" C8 k7 \- ^4 ^be dissipated by an oil cooler.
# y }3 z8 y1 j; e& cSuggestion:
# _6 l9 c- q0 _: n! F8 H1 d! i. u3 c–Cooler OK-ELH2 - 3000 rpm,
1 x T+ ]- [, G- x* M$ GP01 = 0.20 kW/°C at 80 l/min
) V+ G9 ] ~7 U$ Z) Y1 j$ w( k% l. i. Q: b
5 y0 k+ l' e- m我发了一段在上面！ |
发表于 2007-11-30 11:19:26
