COOLER
' v0 _" `: N. K& q- b1 n& V, QSELECTION+ I1 H% Y; X: G) F: B- ~' h
Designation:
- c I* E/ r5 o% VPV = Power loss [kW]
, e5 V1 y2 X+ t4 X uP01 = Specific cooling capacity [kW/°C]
4 \9 T: r- B/ g1 a- x5 \V = Tank contents [l]" X4 e% x. Z1 z
ρoil = Density of the oil [kg/l]
- R. E* Y8 Q8 b7 s- X! y1 n. |for mineral oil: 0.915 kg/l: T: X+ P2 E" h% h5 F! z
Coil = Specific heat capacity [kJ/kgK]
4 _1 A5 ?3 u5 J2 |+ W- E" J, Ofor mineral oil 1.88 kJ/kgK1 p/ P3 Q$ t& ~/ u. @
∆T = Temperature increase in the
$ `# b8 [. I7 Hsystem [°C]
, Q( l1 d1 U1 C/ N/ U5 {, S G( u" ^t = Operating time [min]3 G) t" o! t! _9 g) H7 a
T1 = Desired oil temperature [°C]4 k* }! D; z# G3 b! [
T3 = Ambient temperature [°C]& |* a& P( e7 ~" |0 ~$ U5 a
Example 1:
9 \! I# @6 R EMeasurement of the power loss4 z$ F# z' z/ V) M9 q( q; Z: X
on existing units and machinery.6 v2 r, h3 e4 b+ s9 e$ a+ g( [
For this method the temperature3 E- E3 L) N' G, I1 j5 U$ `+ V
increase of the oil is measured
! D* S# q+ Z1 N3 fover a certain period. The power
9 {% c6 y9 c% {# G& hloss can be calculated from the, V3 k% u0 L) y1 w1 c# Z+ C
temperature increase.
( Z2 u: P. f+ ?: [/ e$ D" LParameters:; v$ Z1 x' m+ @% a/ f
The oil temperature increases
; f! I0 l( P Dfrom 20 °C to 45 °C over
! p. z, A8 P3 d$ ^: y* ]/ i15 minutes.
% Q1 N7 x& _& z: f( y# f( EThe tank contains 100 l.
( {9 y+ G( n8 Y& f3 j6 l$ EHeat to be dissipated:
& E+ A' M) Q6 tPV =（∆T × C oil × ρoil × V）/t × 60 [kW]
. Z5 _0 i1 y+ Y+ _7 V2 `$ b( f4 N1 B( R' \
PV =25 × 1.88 × 0.915 × 100 / (15 × 60 )
4 S. x# K/ P; i) q= 4.78 [kW]1 L1 `) l1 U& e$ z+ s# ]" m
Cooler selection:
5 X' L+ [. J/ v6 C– Desired oil temperature: 60 °C( C% A8 E' e: I& j( L
– Ambient temperature air: 30 °C8 p& ^9 s" v; ^( S7 S. K2 Z! V
8 ]# q4 [: c* C/ q* F6 ~$ K
P01 =PV / (T1 - T3 ) [kW/°C] =0.1594 c3 r/ g0 ~2 v( s: {
A 10% safety margin is
5 X2 l6 m) ]* Z' M5 Krecommended to allow for
7 A: i. [! Y9 uelement contamination, and
2 O9 M' `; ~7 t8 N+ x' q% G7 ztherefore the specific power is:5 f% L( N7 p: c& ?
P01 × 1.1 = 0.175 kW/°C.4 \" N$ I6 w6 s ]% d _: ~2 B) C4 Z
The power loss 0.175 kW/°C must
3 F( ~! L( E7 ~1 `: b2 x; V- wbe dissipated by an oil cooler.* L9 w4 P4 T% M$ W. v' ^% c, v1 N
Suggestion:
" w/ g& [% L- A–Cooler OK-ELH2 - 3000 rpm,3 R" p% m: T6 a0 l
P01 = 0.20 kW/°C at 80 l/min
% \2 o6 _/ t8 U" ^" e9 I* c! D) a4 m. p5 r% a
s) D$ q6 D. v% |" }. w我发了一段在上面！ |
发表于 2007-11-30 11:19:26
