COOLER
7 G# k; N- r- E- C7 y: ~4 CSELECTION, p/ @5 x: n; H1 S" X0 z4 ^
Designation: l" V7 c% }$ [. ]7 r+ m
PV = Power loss [kW]
0 ^" d4 }# H* f- c0 y% bP01 = Specific cooling capacity [kW/°C]0 i9 ]6 y X" C u! H
V = Tank contents [l]
5 b7 Z% ?. {% p, Aρoil = Density of the oil [kg/l]
9 p4 {# T0 ?' kfor mineral oil: 0.915 kg/l
( r Y$ }- a9 N: ZCoil = Specific heat capacity [kJ/kgK]
5 X) w& y' _5 q( s3 d& c( ]: mfor mineral oil 1.88 kJ/kgK
2 D9 Q9 b' o: p7 k* K: O% D3 d7 S( v∆T = Temperature increase in the" f! U$ G5 l, N' h
system [°C]
* a: F! n" a* y: N5 Xt = Operating time [min]1 @- A! z# K% O( r/ G- X) |7 J
T1 = Desired oil temperature [°C]3 M7 _$ h! c, O6 I
T3 = Ambient temperature [°C]8 l+ J- D5 h" u6 x2 l
Example 1:( d- j# O4 `* q C/ s* J
Measurement of the power loss5 G& W8 J4 j+ o; C4 b5 c" D0 T
on existing units and machinery.5 H% A( V- b* } V
For this method the temperature( R+ E4 K6 T: n& D4 n8 k
increase of the oil is measured! O, Q" |$ _0 n) ?
over a certain period. The power
' G, z6 Q3 p: z: ~loss can be calculated from the
8 t; s; L) N. D4 {+ _4 rtemperature increase.2 X5 k: |2 f7 ~3 l2 H
Parameters:
: }9 I1 Y: Q5 Z5 {2 i0 L dThe oil temperature increases- _7 W) Y& X; g3 r1 `1 q' h
from 20 °C to 45 °C over# U( [! t* ]- P2 V/ L' F
15 minutes.
( \# T. C9 _6 |+ d5 ~3 Y. q! _The tank contains 100 l.
- {8 e, V F8 R% m- DHeat to be dissipated:, ]- U7 |4 o1 F0 N# b+ W
PV =（∆T × C oil × ρoil × V）/t × 60 [kW]
* y/ }/ m3 U0 E8 L; n) i1 H$ c$ g& `
PV =25 × 1.88 × 0.915 × 100 / (15 × 60 )% e( U6 b9 `, v# H7 L4 ^8 \+ `
= 4.78 [kW]3 g6 g6 K3 E" o) B$ c# l
Cooler selection:
1 M% v) B/ w( k– Desired oil temperature: 60 °C
, f$ V/ ^7 ~; j8 e& A$ o' ^3 e– Ambient temperature air: 30 °C" h6 B& `* q, e" `
2 }5 S' i# ?+ N, h% `+ N6 N
P01 =PV / (T1 - T3 ) [kW/°C] =0.159
) ~# W/ n6 ?* H* [" i; Y- l* x; t& [/ [A 10% safety margin is
* d) Z. I* q+ M9 _/ Krecommended to allow for
' Y2 r. v9 h- A, t0 V. s" eelement contamination, and
& \# v1 {4 N2 \8 }therefore the specific power is:
' t) t4 l' B1 y9 O4 }6 WP01 × 1.1 = 0.175 kW/°C.* T% l* \! x; n; K% r
The power loss 0.175 kW/°C must
, S$ G( D3 } @/ r* m8 S' Mbe dissipated by an oil cooler.
) l) j6 c; O$ R. W. CSuggestion:
- }# x/ L* H( \8 h0 I–Cooler OK-ELH2 - 3000 rpm,
1 Y1 ]/ u& d" P0 t! [P01 = 0.20 kW/°C at 80 l/min P$ i, ^% M* b: v. M6 O$ r0 r
9 `9 D2 C8 H/ g% r: m% w
3 I- I$ L6 \" w- [9 v" N我发了一段在上面！ |
发表于 2007-11-30 11:19:26
