COOLER+ U' J3 Z3 p: t5 X( Q% L' x
SELECTION
, d" X9 M- F+ c2 w0 ODesignation:/ P; y% p) h, C, g9 f4 [
PV = Power loss [kW]$ }4 `3 R7 p+ w! v9 D( ?# t, e
P01 = Specific cooling capacity [kW/°C]
0 E& ^. [, a7 B$ \! l4 }V = Tank contents [l]8 X `! {2 {8 w9 \
ρoil = Density of the oil [kg/l]
! ?5 p% ?1 K8 W0 zfor mineral oil: 0.915 kg/l
3 E3 g+ |! y. c' O( G7 @Coil = Specific heat capacity [kJ/kgK]/ X% z" c9 W0 c, e1 t* V
for mineral oil 1.88 kJ/kgK+ t% A4 u% | W, \! I
∆T = Temperature increase in the+ c8 D! j! ]& k
system [°C]9 u8 O3 d! ?6 D( O, P
t = Operating time [min]
x4 o( {: X* a c$ S; H dT1 = Desired oil temperature [°C]
4 @. l r+ I7 w. f: c; N1 lT3 = Ambient temperature [°C]: h1 ~0 W' h+ f6 [- c! H
Example 1:
- T& F9 v) t6 t# }Measurement of the power loss$ ]. P+ B$ p4 m; X) `, j
on existing units and machinery.& v" r) [# ^; h8 d- D
For this method the temperature$ N# V; Q+ \1 i( c! `4 n7 y* q/ c
increase of the oil is measured Q: d9 P& _# p9 _/ G4 H5 r- z
over a certain period. The power
i' g( {7 p. Bloss can be calculated from the0 q8 T# H7 [4 r) k% c
temperature increase.
! a- u/ g- F. k1 t( P1 n7 PParameters:
1 r7 P8 E! ]* }5 F8 DThe oil temperature increases
& a) c) E1 ^' A- z6 i+ u: Lfrom 20 °C to 45 °C over
6 H0 r0 @& ^, M: E15 minutes.$ y9 L4 B2 X8 F/ q
The tank contains 100 l.8 ?( ]8 R5 h& v# N
Heat to be dissipated:4 r; W! \ i5 {, s8 m
PV =（∆T × C oil × ρoil × V）/t × 60 [kW]
$ F( } g; R/ M! R% ?7 n) i
* l% x7 z, U9 A3 |. V3 pPV =25 × 1.88 × 0.915 × 100 / (15 × 60 )
0 S! d* E. J+ k! q) s= 4.78 [kW]* P$ G5 \* L4 w$ Q0 L& R2 ^5 H
Cooler selection:
( z' n& W5 T; c; b! F– Desired oil temperature: 60 °C
% S2 G S* ~$ n4 P– Ambient temperature air: 30 °C" j' y: K, I% B2 x
3 i- ]8 n# L# }9 T2 P/ x m
P01 =PV / (T1 - T3 ) [kW/°C] =0.159$ H p% C! Z$ O, C
A 10% safety margin is
; ]& }' L# G Q" `( Zrecommended to allow for
4 |& v. N3 F, v3 k% Melement contamination, and | C: e0 @7 |) F
therefore the specific power is:1 D- |2 A" p g4 |; H! D' q( d% X
P01 × 1.1 = 0.175 kW/°C.
, P1 Y0 T$ f+ ?' k& tThe power loss 0.175 kW/°C must
' ?" e9 ^, w; W4 vbe dissipated by an oil cooler.
9 u. |7 e4 `- y) H% i. y# D: sSuggestion:
7 P4 W5 g4 K+ l" h1 E6 ~–Cooler OK-ELH2 - 3000 rpm," K0 i. ^3 j- q2 q
P01 = 0.20 kW/°C at 80 l/min
4 k6 H' ?. s5 M) l
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' q' p# `% e$ p( `9 ]我发了一段在上面！ |
发表于 2007-11-30 11:19:26
