COOLER% B2 y' K7 v+ m. t+ ~6 C4 Y# j& u
SELECTION1 K" v; t% C0 v# T2 Q* G s' G7 n
Designation:- @6 v& y# B( I0 r, ^ m/ K
PV = Power loss [kW]- S# \# f5 \& l4 T7 U3 i; v: B5 V
P01 = Specific cooling capacity [kW/°C]. J: u% z0 ~* G u, c
V = Tank contents [l]
. l: h0 S9 e! Z) h! |. uρoil = Density of the oil [kg/l]
( ^# e+ U |2 U* P' Y1 u$ tfor mineral oil: 0.915 kg/l
9 e! ]+ i, a$ M9 B: d0 gCoil = Specific heat capacity [kJ/kgK]
! Y1 i; ~; B3 a! L5 Nfor mineral oil 1.88 kJ/kgK
0 r+ o( c+ _+ M3 w∆T = Temperature increase in the
$ q2 t. T. E, c. m/ `system [°C]
7 W5 C/ G+ s4 B# lt = Operating time [min]0 `( }. |, f8 |7 d8 D9 ~: d/ `: g& I
T1 = Desired oil temperature [°C]7 J4 \+ E$ L; z' b' _3 _
T3 = Ambient temperature [°C]" i \% N1 M) T/ Z* B$ B* e |: Q
Example 1:4 c/ j6 \$ I2 \: q# b( l
Measurement of the power loss3 W% R+ z1 ~: x! o5 g: |& a
on existing units and machinery.7 U( q% \! f2 b" S+ O
For this method the temperature
9 ]6 M8 h6 ^) D2 L) d% j. qincrease of the oil is measured
, `3 [$ f3 W6 c6 b4 U* Sover a certain period. The power0 N4 G3 m: U+ r) N7 i& y
loss can be calculated from the8 a7 F9 T5 R1 @5 t
temperature increase.
- u6 v0 E, q( k f1 yParameters:) L: q \3 ~6 f
The oil temperature increases
+ W. u1 {% W6 y5 mfrom 20 °C to 45 °C over# }2 T6 {2 n8 X8 ~
15 minutes.
% ?4 h4 F0 r- hThe tank contains 100 l.7 ?) ~* t+ Y& ]+ k* T
Heat to be dissipated:; }5 Y( {! W+ o
PV =(∆T × C oil × ρoil × V)/t × 60 [kW]
5 V) _- W0 f7 c0 q! ~3 E/ P& \; T- h; p
" R% y V5 T+ L" cPV =25 × 1.88 × 0.915 × 100 / (15 × 60 )
7 z/ R& I- W4 v; ]9 w= 4.78 [kW]
7 e% d k9 y6 L2 W7 ~$ lCooler selection:
( i2 k7 n _ ?8 K. U7 P– Desired oil temperature: 60 °C0 Y( O! V% u+ @% N2 v5 p
– Ambient temperature air: 30 °C' w7 w" l5 W* w$ F6 I
2 Q6 T. z0 G7 l, a
P01 =PV / (T1 - T3 ) [kW/°C] =0.1591 V) i! _& O7 ]3 k: Y9 M7 n
A 10% safety margin is
. S9 s0 L/ U; C2 w% lrecommended to allow for% x7 z3 v, D- ~5 o# J
element contamination, and1 n; w x1 [9 [' H7 y2 c5 M; D
therefore the specific power is:; ~1 x% h% \. _' L7 R2 v$ C) W
P01 × 1.1 = 0.175 kW/°C.
( Q% y. @3 j; i YThe power loss 0.175 kW/°C must) K( G7 x9 B0 L$ y" X# \. J$ |
be dissipated by an oil cooler.* t/ N# i* T6 T% y J) e" q7 [% w
Suggestion:) M, ~2 Q$ z6 p! g; \4 O
–Cooler OK-ELH2 - 3000 rpm,
7 N2 G( j* {6 [: a$ s2 V9 z: MP01 = 0.20 kW/°C at 80 l/min
6 k- c8 _& U! g7 @
" J$ d- A8 |+ W% y" H1 ]7 F
5 G; T$ ^+ T. `( W3 N0 ~我發了一段在上面! |