COOLER2 r+ d0 E+ [1 m1 U" i0 Q
SELECTION
; M5 k) {( y' I7 UDesignation:& S) ?# ]6 G- S5 u# @
PV = Power loss [kW]
- q0 a Q* X. U2 iP01 = Specific cooling capacity [kW/°C]6 y! S# O8 O( \* A( l5 }; |
V = Tank contents [l]
7 N1 k/ X8 i6 J0 T- C& i2 }; ?ρoil = Density of the oil [kg/l]
) s" E; f; }; `) L4 b' vfor mineral oil: 0.915 kg/l
" R; f7 Q6 Z# L) M3 I) QCoil = Specific heat capacity [kJ/kgK]0 W7 Z. u9 G# N! `8 P6 G
for mineral oil 1.88 kJ/kgK
' u7 M; @" a& a: r& H0 U∆T = Temperature increase in the
& a) b0 T/ j; n7 Bsystem [°C]
. b2 V9 }8 X% G- |4 e4 gt = Operating time [min]( }. E+ w9 a+ ?
T1 = Desired oil temperature [°C]0 o$ [( d! g/ ^# U
T3 = Ambient temperature [°C]
( e* R) C' T/ U1 }% b- t0 s0 |4 HExample 1:9 \* B5 u8 \# T, n1 c5 k
Measurement of the power loss
* ]9 J* |- Z& F- f; ?2 Pon existing units and machinery." p; _+ t# \6 O$ N0 ^) L* P
For this method the temperature) _' T( `* s; @. \
increase of the oil is measured
; _% o R' O" q2 Y: G! U# P: r- Wover a certain period. The power
: v7 A5 s, n; w$ [: Aloss can be calculated from the1 M8 D; `* v' k6 i
temperature increase. b& a, B/ n8 [" ]5 M P5 J
Parameters:9 |7 K6 F% q) K1 m0 l9 `
The oil temperature increases
& ~9 k0 p7 Y" _3 k7 Lfrom 20 °C to 45 °C over G4 R+ G. w! |3 L% {+ }
15 minutes.
9 @& j. T8 o! D" W {, WThe tank contains 100 l.
; K' F4 c: w& G& C3 \9 O( E4 F' U0 }, GHeat to be dissipated:$ \7 j3 [1 S# D) B
PV =（∆T × C oil × ρoil × V）/t × 60 [kW]
3 S2 d; H( [0 ~9 |7 g9 V4 q! T1 U, k9 i5 P% @
PV =25 × 1.88 × 0.915 × 100 / (15 × 60 )- T" m! @ } F
= 4.78 [kW]2 e$ ~+ J$ e1 V! b
Cooler selection:: v" w9 j, d: s1 t
– Desired oil temperature: 60 °C" }3 P; J5 |( D3 s/ I( E4 p) U6 ~
– Ambient temperature air: 30 °C
" x, u- d S3 ^* f* m6 X2 y. @% `6 M
P01 =PV / (T1 - T3 ) [kW/°C] =0.159
+ I. {7 j; M1 t( o. _$ l; H4 [# ^A 10% safety margin is
& ~! X% i8 k' b3 h" Qrecommended to allow for
, t* e) E- s' b/ L( kelement contamination, and, j. a' e* e7 H# x9 ^
therefore the specific power is:2 V9 Z/ M, j) | t
P01 × 1.1 = 0.175 kW/°C.
; D" Y: X" Z. B$ S. q% sThe power loss 0.175 kW/°C must1 @. b" p6 x$ q6 G8 U
be dissipated by an oil cooler.6 E' H. f- i: z4 f/ i
Suggestion:& T8 F/ x8 M: C0 ~, n* S& U
–Cooler OK-ELH2 - 3000 rpm,
1 b: l4 L6 M6 r oP01 = 0.20 kW/°C at 80 l/min
2 A; d: O! s; h: E4 D2 v, ? e
0 H$ t6 I/ i: U. q- @ {* o9 ~$ O* t* \) G* @0 w' E) M1 L
我发了一段在上面！ |
发表于 2007-11-30 11:19:26
