Partlist Jøtul F 602 USA
Pos | Description | Dim./Spec. |
1 | Hot plate | 240 mm |
2 | Gasket | LD 360 Ø8.7x850 |
3 | Screw, panhead | M6x12 DIN 7985 |
4Air deflector
5 | Nut | M8 |
6 | Gasket | LD |
7 | Gasket | |
8 | Spring for | Ø7.6x13 |
9Door vent baffle
10 | Screw, self thread | M6 x 16 |
11 | Flue outlet |
|
12 | Back plate |
|
13 | Side panel | Ø3x12 |
14 | Rivet | |
15 | Rivet | Ø2,4x4 |
16Label
17Label shield
18Cover for smoke outlet
19 Countersunk screw | M6x16 |
20Baffle plate
21Air manifold
22Air chamber, complete
23Air chamber
24 | Screw, collar head | M6x10 |
25 | Screw, collar head | M6x25 |
26 | Rivet | Ø8x16 mm |
27 | Knob, wood | M6x70 |
28 | Screw, filister head | |
29 | Nut w/ collar | M6 |
30Door catch
31Door handle
32 | Screw, cylinder head | M6x30 DIN 84 S.krom. |
33 | screw, collar head | M 6x35 |
34 | Insulating blanket | 360x185 |
35Burnplate, side, with insulation
36Burnplate, bottom
37 | Insulating blanket | 360x330 |
38 | Screw, hex | M6x25 |
39 | Washer | Ø18x06,4x1.6 |
40Top plate
41Pin, door
42Glass door, complete incl. handle
44 Glass, ceramic | 4x147x154 |
45Air slide vent
46Front plate, complete
47Front plate
48Base plate
49Legs, package of four
50Leg, 1 pcs
51Heatshield - bottom
52 Increaser | Ø 126xØ154 mm |
53Decorative top
54Casing
55Washer
56Washer, black crom.
Appendix A:
Alternate floor protection
All floor protection materials must be
The easiest means of determining if a proposed alternate floor material meets requirements listed in this manual is to follow this procedure.
= thermal resistance | |
= thermal conductivity | |
= thermal conductance |
1.Convert the specification to
a.If
b.If
c.If
2.Determine the
a.Use the formula in step 1 to convert values not expressed as “R”.
b.For multiple layers, add
determine overall
3. If the overall
Example:
The specified floor protector should be 3/4” thick material with a
a
Step A. Use formula above to convert specifications to
Step B. Calculate r of proposed system. 4” brick of
R brick = 1/c = 1/1.25 = 0.80
1/8” mineral board of K = 0.29 therefore R mineral board = 1/.29 x 0.125 = 0.431
Total R = R brick + R mineral board= 0.8 + 0.431=1.231 Step C. Compare proposed system R = 1.231 to specified R of 0.893. Since R is greater than required, the system
is acceptable.
Definitions:
Thermal conductance =
C = btu = W
(Hr)(ft2)(f) (m2)(k)
Thermal conductivity = | = | W | = | (btu) | ||
K = | btu | |||||
| (Hr)(ft2)(f) | (m2)(k) |
|
| (hr)(ft)(f) | |
Thermal resistance = | = | (m2)(k) | = | (btu)(inch) | ||
R = | btu | |||||
| (Hr)(ft2)(f) | W |
| (hr)(ft2)(f) |
For the 3 requires floor protection with a minimum insulating
R-value of 1.1.
Alcove installation require a minimum
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