COOLER
( `/ U$ M' k4 J w6 L7 k3 jSELECTION* G. [' E* q8 ^' }6 T, e C
Designation:
" v' v5 ^9 z% n3 uPV = Power loss [kW]
- r7 m3 [5 c/ _- i9 kP01 = Specific cooling capacity [kW/°C]
' V8 V! N$ H% X9 SV = Tank contents [l]
% x& k& |9 y9 {) F' M- U1 j9 l; rρoil = Density of the oil [kg/l]4 d! u- o1 \ L% Y
for mineral oil: 0.915 kg/l# C. m; e8 y# ]) t# b
Coil = Specific heat capacity [kJ/kgK]! q8 S5 V" |4 B* [, E# h
for mineral oil 1.88 kJ/kgK. ?/ j5 p3 F( e; S8 x3 A
∆T = Temperature increase in the
1 H2 Y+ s: r' N0 t8 \system [°C]
) ~& R1 D8 r+ B' o+ o b4 d/ L8 ~t = Operating time [min]
}" Y6 Z9 M! W$ S# {T1 = Desired oil temperature [°C]
8 j' P3 A( p+ @7 J. bT3 = Ambient temperature [°C]
+ ~! n: g0 T; c% k0 W% RExample 1:
]# I! D) D1 c" T, Z2 Q4 MMeasurement of the power loss1 G6 g& n4 P7 ]
on existing units and machinery.2 K, K8 M2 y. o: O4 y& T5 g
For this method the temperature* ]0 J% ?4 N5 U1 b
increase of the oil is measured, B* `1 y: x) N) Z% P
over a certain period. The power
/ C/ O9 ~& f k1 J( C% w* Uloss can be calculated from the
0 m' {0 q1 |8 Q2 Atemperature increase.
1 Z. ~4 s& D9 D+ b0 {2 W% Z$ l$ PParameters:
: Z6 E+ Z+ e: k' zThe oil temperature increases- K! V2 i- q5 y, x) g7 x
from 20 °C to 45 °C over5 Y5 y7 y- E( M) t' V0 J4 g
15 minutes.
4 T/ `/ L. k+ B/ \# R& m1 gThe tank contains 100 l.
$ ?' z2 ^, i( k4 j: b- u3 SHeat to be dissipated:
# D" ]( D- E/ o4 T, @9 A) Q4 gPV =（∆T × C oil × ρoil × V）/t × 60 [kW]
9 Z+ j5 m8 J, Q0 i
3 }0 t6 ]6 c4 X. X. O3 HPV =25 × 1.88 × 0.915 × 100 / (15 × 60 )' x1 H' ^5 u L% S% M2 M* ~
= 4.78 [kW]& N! L' D9 }* S# C% [
Cooler selection:
6 b9 }. z5 C) Z" {0 F! S– Desired oil temperature: 60 °C: q) O/ R: @) t* e$ X# F2 n( u
– Ambient temperature air: 30 °C# O X O2 Y/ I$ Q
5 D% o; v. A; Q9 x, ~P01 =PV / (T1 - T3 ) [kW/°C] =0.1593 s# G; A' F7 }/ v9 a# R
A 10% safety margin is
- v- @5 I2 q+ Z/ Y: q8 Drecommended to allow for
5 R8 N+ A5 q- welement contamination, and) q. S- ?2 r! g
therefore the specific power is:; |5 v5 U; H" x" f
P01 × 1.1 = 0.175 kW/°C.
* @# H0 j7 ~6 \4 R1 G& VThe power loss 0.175 kW/°C must
2 c# g: p! L X0 I) a0 z3 w3 w+ z8 wbe dissipated by an oil cooler.
7 @- D$ b- D: g9 USuggestion:2 w; U& U/ n. R; \$ X% k" w* E9 T& Y
–Cooler OK-ELH2 - 3000 rpm,
- a+ V9 o0 a1 t' G& T* f' hP01 = 0.20 kW/°C at 80 l/min
9 o& X/ M4 Q2 q O' U
7 L, j' ~; u; |' P% m' s: \8 k
- h' `3 m! g8 R4 w( {6 N- g( b我发了一段在上面！ |
发表于 2007-11-30 11:19:26
