COOLER8 s: E) \- f U" C
SELECTION
. j; @7 H; \! `. e# n; [/ h1 rDesignation:3 m5 o8 I# Q: ~) ^1 _& ?& \
PV = Power loss [kW]/ W4 ~4 ]0 ~6 n9 X* Y+ l: T
P01 = Specific cooling capacity [kW/°C]9 \% w% E! ~& M+ L ]; x# m
V = Tank contents [l]$ b, O' R0 Z) v* j4 x" ^
ρoil = Density of the oil [kg/l]
9 G4 C% d, T/ M* b3 ?for mineral oil: 0.915 kg/l
! H' c& T3 q, Z$ s% ?5 R5 sCoil = Specific heat capacity [kJ/kgK]; \9 f0 t! `/ @( O% j
for mineral oil 1.88 kJ/kgK
+ S7 D. [- C2 r) N/ M∆T = Temperature increase in the& n r- E* ~- c0 | o3 F/ x
system [°C]
6 X+ e) h; c# F; z& [1 \5 Et = Operating time [min]
, X9 {9 T! i7 @$ P* OT1 = Desired oil temperature [°C]
# h* U# ~& h% H I. H6 Y x9 oT3 = Ambient temperature [°C]( ^" {4 k1 `" a
Example 1:
& G! k8 m) ^% YMeasurement of the power loss
# ^7 V" y2 ^+ {7 A" x+ bon existing units and machinery.
: X) t* B* c" V; cFor this method the temperature1 u( \# Q/ G, o; x0 h
increase of the oil is measured5 U6 B* q8 m8 z. n' F" c# a
over a certain period. The power; a4 Q1 f6 v- g) J
loss can be calculated from the2 z) |8 j i6 U* P- U1 H
temperature increase., H; x! e2 r& U% w
Parameters:
1 p1 x! x& h9 \, G+ N3 W: OThe oil temperature increases
1 N5 B( \& _0 \: x) sfrom 20 °C to 45 °C over
( B j6 ?0 @5 f- ^9 }6 D4 A15 minutes.( F: O0 Z/ U; p- k! b+ v
The tank contains 100 l.
" S# B6 o7 e' v$ ^; @2 } _Heat to be dissipated:
/ d' [3 F0 ^ `PV =(∆T × C oil × ρoil × V)/t × 60 [kW]0 Y; n2 v' [4 q8 E8 ^) Y- a; m
' n: V" _4 E. a( b7 D+ ^: }PV =25 × 1.88 × 0.915 × 100 / (15 × 60 )
K0 } \9 f: j, t= 4.78 [kW]" ~/ f% N' |1 g- G2 ^5 [
Cooler selection:
* l- A# `- L8 S. F! F– Desired oil temperature: 60 °C
' h: K+ W2 n- R– Ambient temperature air: 30 °C3 A3 q! A# Q) v) n5 R t: b: B# R/ T
( B: G, z0 G6 XP01 =PV / (T1 - T3 ) [kW/°C] =0.159. d: |) b. t; s# I. u& q
A 10% safety margin is3 B3 n ~- F2 E. W, a7 X7 P
recommended to allow for
+ d0 G. t( r; Y k0 @element contamination, and
4 C& H2 k2 Z8 ~: o& Mtherefore the specific power is:
/ B7 ?, @# |* h6 RP01 × 1.1 = 0.175 kW/°C.
- ]6 | M3 u* w$ dThe power loss 0.175 kW/°C must
( e6 u: W5 G7 J j" ]be dissipated by an oil cooler.
/ E' L9 I5 T. O- C, L8 Z6 PSuggestion:
( P, F; T4 w1 u–Cooler OK-ELH2 - 3000 rpm,
8 |0 B7 f7 ~2 N. \P01 = 0.20 kW/°C at 80 l/min
1 l" h) R! x! K$ N" m+ k P
f" S+ `7 q+ b, t3 M" E& n, k% I* z8 e: R) r( M
我發了一段在上面! |