COOLER
) r" ~$ P, U" X( ESELECTION
# ~5 Q) ?. h4 I0 {: C& z1 I0 i9 X- X9 ^$ [" QDesignation:4 B5 S# N0 A. E; p
PV = Power loss [kW]# i3 h; D2 B1 I! f: U+ w
P01 = Specific cooling capacity [kW/°C] m/ r# v6 d3 U, }) o h3 M
V = Tank contents [l]
# G( Z+ _. t& W. X3 uρoil = Density of the oil [kg/l]/ G4 @0 e$ g7 [& T0 `
for mineral oil: 0.915 kg/l# j! Z: [# l- k& n
Coil = Specific heat capacity [kJ/kgK]0 ^) f+ i0 W) | g! J4 j
for mineral oil 1.88 kJ/kgK
6 ~: _3 Z3 U/ Z l∆T = Temperature increase in the
: k& @+ X3 l0 Y; w" ?( {! Ksystem [°C]; P" ~7 Y4 R% I7 q
t = Operating time [min]4 V* F8 e0 g* ]7 N% i
T1 = Desired oil temperature [°C]* W/ j/ T7 h y$ _# ~
T3 = Ambient temperature [°C]
6 H' M/ n+ Y- y+ A, J+ }$ |Example 1:' e1 T* [# m" p+ M3 B
Measurement of the power loss+ ^: K! ^3 v0 g5 }5 z$ B
on existing units and machinery.
0 I, m4 A# H+ F$ I: g& ]" S/ R8 f5 ]For this method the temperature8 z, V# o# i8 E- c$ N0 @6 [5 R
increase of the oil is measured
. @9 D( y, Z% J3 f; a- wover a certain period. The power$ Q6 ?8 m2 l1 i: H
loss can be calculated from the
7 ?# I: G, Y; p3 X, Dtemperature increase.
7 T3 I; I; d3 N& A4 ?- ?+ l5 G/ k5 [Parameters:0 c6 E; i6 H% Q0 l
The oil temperature increases
" r D5 T. z4 x8 ]. [. Hfrom 20 °C to 45 °C over
% O2 E( \+ X# Z0 G) k% f15 minutes.: n, h1 g) ?1 _3 l/ w
The tank contains 100 l.* P$ K) h( J# G7 E$ ?' M' t
Heat to be dissipated:
6 m7 p9 g" b- L- h! r: UPV =（∆T × C oil × ρoil × V）/t × 60 [kW]
3 v: I7 `1 w, }0 K; p" b
& Y' g# u/ r: W6 e! L2 @3 nPV =25 × 1.88 × 0.915 × 100 / (15 × 60 )
4 [0 H: v6 g& d6 p* y( L% B= 4.78 [kW]
# z7 l9 b- p$ Y- w/ C) [Cooler selection:
0 O. T' I- m2 X3 Z* p– Desired oil temperature: 60 °C: X4 U/ ~ i3 h8 [ n/ T& K. y
– Ambient temperature air: 30 °C
. j$ D& C! T0 x1 _+ P
0 z% A5 {; {( z% \ G' S6 FP01 =PV / (T1 - T3 ) [kW/°C] =0.159* R5 C0 I% B: n7 p9 `
A 10% safety margin is/ ^; y( K% m" H* [
recommended to allow for
7 n7 F4 j3 E0 M1 F& felement contamination, and, H+ G' }1 y/ }& Q- X
therefore the specific power is:
, _4 b5 [, Y$ T/ p( z& K8 KP01 × 1.1 = 0.175 kW/°C.
8 j! l$ O$ J9 l( P# @The power loss 0.175 kW/°C must; t$ L3 ?, ^0 `0 [5 b) @
be dissipated by an oil cooler./ |# E& c! C( P3 k9 T2 I9 r
Suggestion:- q% u# q5 y7 W+ \
–Cooler OK-ELH2 - 3000 rpm,
' I, M# d( ?) hP01 = 0.20 kW/°C at 80 l/min
/ q* ~8 D; [, U8 l" U3 g) i2 s( |: V. A" n' e2 z7 p
+ R' g/ m9 [$ t$ P5 U3 ]
我发了一段在上面！ |
发表于 2007-11-30 11:19:26
