COOLER
3 E2 r$ x; D, m# ^% d! wSELECTION, t$ q* g% |8 y9 w
Designation:
! Y5 T0 B l" hPV = Power loss [kW]
( B/ M. ?/ V w @ k* g9 vP01 = Specific cooling capacity [kW/°C]
8 a4 F. E4 D' F( q% iV = Tank contents [l]
- S2 a& D4 I- M, v$ Qρoil = Density of the oil [kg/l]" V2 I! l1 j( C& S% b/ X6 m% l. B
for mineral oil: 0.915 kg/l
! B, L' e' @! d2 E% |+ KCoil = Specific heat capacity [kJ/kgK]
/ J% }4 c- @6 X, G3 I0 S' dfor mineral oil 1.88 kJ/kgK% G J2 {" y1 m7 {1 I* P
∆T = Temperature increase in the# n: I4 N. ^1 i2 f& |' z
system [°C]
) O) R: F* T6 C3 Gt = Operating time [min]) k' X! p6 `5 ~9 c0 A
T1 = Desired oil temperature [°C]* f* E+ z& x$ b6 o- T& T4 p
T3 = Ambient temperature [°C]
3 R! y/ o. w6 f8 h! a& x' |$ aExample 1:4 b1 \6 P& W- g5 J& T
Measurement of the power loss
, E8 q; [0 U( ]on existing units and machinery.* e8 R3 [5 z6 F9 P
For this method the temperature; \+ b% i; m& Z) w
increase of the oil is measured
. o9 f8 N( D' |( X t$ oover a certain period. The power
& Q+ d; [: u9 y" B" Rloss can be calculated from the N' s' H9 P) C$ I" A- R5 \7 @
temperature increase.
2 y& C( y0 e4 N! P cParameters:9 E7 u; X7 s; q1 t9 H, }# S+ ]
The oil temperature increases' ]" P; {' V7 ?; Y- f( ?4 }
from 20 °C to 45 °C over+ c! l3 g7 N5 [3 Z4 }
15 minutes.: P. e1 x! ^6 n& ^' j. |/ Y4 k$ u
The tank contains 100 l.
- K4 v2 x7 F bHeat to be dissipated:
: x) K" Y( m2 i: I0 sPV =（∆T × C oil × ρoil × V）/t × 60 [kW]
8 [0 g0 |1 y/ m! |# t2 |$ H8 Y. n+ _( L
PV =25 × 1.88 × 0.915 × 100 / (15 × 60 )6 ~4 O8 `8 ]/ f7 ]+ J3 `2 z
= 4.78 [kW]
/ @& R$ f+ x5 u+ YCooler selection:
/ K/ V5 M0 ` O, E% i– Desired oil temperature: 60 °C( G" r* S0 D2 M* ?6 @' Z4 ^
– Ambient temperature air: 30 °C
5 b1 \8 J+ w8 |0 b% J7 @( K! q
# J ]" X5 H( c0 o; |: I' |. }P01 =PV / (T1 - T3 ) [kW/°C] =0.159
1 @% G' B) Z O6 \A 10% safety margin is
0 h6 ^' E7 h( ^0 _' D/ ?+ F' y5 {: @recommended to allow for4 f# B- t: K. b7 Y1 ?6 g
element contamination, and9 m. V; S$ |) H; G+ r8 M6 l
therefore the specific power is:9 ?# b# C* A: E# ^9 g3 G" W
P01 × 1.1 = 0.175 kW/°C.$ N' ~: {9 U: z) Z4 N: W$ a, k
The power loss 0.175 kW/°C must
8 z/ E1 ]; \$ K( C e6 Q5 F) abe dissipated by an oil cooler.3 m# E) P2 A$ v; u1 m
Suggestion:
. n+ o# R$ K1 F% M3 u4 J–Cooler OK-ELH2 - 3000 rpm,- `( D3 b9 T; e
P01 = 0.20 kW/°C at 80 l/min7 W8 [# r2 p0 u! V) M
$ ^; _- D* N5 X2 y( \. y7 T
" m5 c% J* m' A1 u我发了一段在上面！ |
发表于 2007-11-30 11:19:26
