Transfluid BT 40 Elastic Coupling

SELECTION
8.4 CALCULATION EXAMPLE Pm = 20 kW
PL = 12 kW J = 350 kgm2 T = 25 °C
Assuming:
Tab. C
THERMAL CAPACITY
nm = 1450 giri/min nL = 700 giri/min
size
7 8 9 11 12 13 15 17 19 21 24 27 29 34 D34 46 D46
Trasmission via belts.
From selection graph. on Tab. A, selected size is 12K.
A) Acceleration time From curve Tf 5078-X (supplied on request) slip S = 4%
nu
= 1450 .
Jr
= 350 .
(
(
)
100 - 4 100
= 1392 rpm
) = 88.5 kgm
700 1392
K
CK
CCK
kcal/°C kcal/°C kcal/°C
2
Mm =
9550 . 20 1450
= 131 Nm
ML =
9550 . 12 1392
= 82 Nm
2
1.2 1.5 2.5 3.2 4.2 6 9 12.8 15.4 21.8 29 43 56 92 138 332
3.7 5 6.8 10 14.6 17.3 25.4 32 50 63 99 -
-
10.3 15.8 19.4 27.5 33.8 53.9 66.6 101 175 -
ML = 1,65 . 131- 82 = 134 Nm ta
1392 . 88.5 9.55 . 134
=
Tab. D
FACTOR K
= 96 sec
B) Max allowable temperature .
.
( 88.576.51392
=
C
= 4.2 kcal/°C (Tab.C)
Ta
=
K
= 8.9 (Tab. D)
TL
= 2.4 .
Tf
= 25 + 86 + 13 = 124 °C
104
361 4.2
+
82 . 96 8
)
= 361 kcal
= 86 °C
12 . 4 8.9
FACTOR K
1392
Q
= 13 °C
C) Max working cycles per hour
tL = 103
H
=
361
.
(
86 2
)
= 724 sec
+ 13 . 8.9
3600 96 + 724
Fluid couplings - 1605
= 4 starts per hour
OUTPUT SPEED rpm
12