COOLER
! F; B. ^5 r1 `& D8 Q, p7 i. ySELECTION4 g' f$ O# ^3 v& D+ M$ a$ C
Designation:
% t1 ?2 y, Y; c8 n8 z tPV = Power loss [kW]$ L0 Z2 w1 [- e1 T4 R! x2 T% O
P01 = Specific cooling capacity [kW/°C]
' b- @9 [& ?; YV = Tank contents [l]
1 y$ r: A! m/ N7 K5 B) rρoil = Density of the oil [kg/l]
! y& i2 ~8 o1 r+ E6 Z. ~9 p, }for mineral oil: 0.915 kg/l9 A7 V* I# _' s2 n+ a8 J7 c! K, u3 d
Coil = Specific heat capacity [kJ/kgK]
. T, n& Z5 l8 c: i. _for mineral oil 1.88 kJ/kgK
7 L. s7 _2 ~7 F' r( f/ { T* h∆T = Temperature increase in the
& {+ o) _2 D, g7 _+ Zsystem [°C]
6 m9 l ?' C! D- b' ct = Operating time [min]
5 i" E; C$ D3 w6 g1 v ]% J' X3 `. uT1 = Desired oil temperature [°C]
, e- I& L c9 OT3 = Ambient temperature [°C]
1 G) P, P/ Q V& N4 l- m% }Example 1:
' @% I) O, h8 Y& j, r; T4 W9 kMeasurement of the power loss
& @" v* A4 |: }$ O& d2 Q( V! P8 Son existing units and machinery.: Y/ D9 |8 g7 t4 G5 x
For this method the temperature& J7 Q6 t1 s) l3 E P
increase of the oil is measured
1 v5 o }" j$ t7 ?! o: {+ rover a certain period. The power$ C/ R" R5 v& l7 ?. k* c
loss can be calculated from the, N3 B3 @: t6 L( t& A
temperature increase.4 O5 E1 G' ~6 }' p
Parameters:
' _$ s: T) C$ N* yThe oil temperature increases
) n( F3 Y& V# [* g+ _: m4 W H1 I) s, Pfrom 20 °C to 45 °C over/ q/ | G; a u, s' _1 z
15 minutes.
* O/ E$ @2 }6 F/ |$ a) `The tank contains 100 l.. C6 B! j, U+ }8 F& _' A) z
Heat to be dissipated:
) ^1 D: @% Z5 u/ Z+ W% P7 CPV =（∆T × C oil × ρoil × V）/t × 60 [kW]' m3 \. `3 Q" k5 ~, B9 V
7 ~ V( h! r6 c: S
PV =25 × 1.88 × 0.915 × 100 / (15 × 60 )
& b0 b D7 f) B5 Q. t- U6 V= 4.78 [kW]! X4 i: v" ~- v# c9 c Q t
Cooler selection:
) a" K5 U f8 M– Desired oil temperature: 60 °C' c" F) T: m, d$ L A. Q% ~
– Ambient temperature air: 30 °C
$ E( s$ B! h! y2 A- W
1 f% s, R( w1 @. Z1 ? bP01 =PV / (T1 - T3 ) [kW/°C] =0.159
# Y7 @7 l6 C8 }7 q2 y9 VA 10% safety margin is& y9 [- P) O1 y3 W- E5 ]7 J- D, X
recommended to allow for: a% I+ F5 V6 \& Z. e+ f
element contamination, and1 B$ D8 X7 d* T8 {8 j1 J( p
therefore the specific power is:- ?9 ~- Q$ Q7 l6 [
P01 × 1.1 = 0.175 kW/°C.) o6 T) Q6 ~% N* d
The power loss 0.175 kW/°C must: f' D% J5 O. T& b" |, }& g
be dissipated by an oil cooler.+ Z( B r M% \
Suggestion:' F ]. x; B6 h) L! _2 n
–Cooler OK-ELH2 - 3000 rpm,- r0 C$ f# t& C
P01 = 0.20 kW/°C at 80 l/min4 w4 N* ~9 _/ A. p, Y" w
) d9 k, {9 I- ]/ f. H# w7 J# y& K' L/ y$ ~) e
我发了一段在上面！ |
发表于 2007-11-30 11:19:26
