Jotul f 400 ユーザーズマニュアル

18

1

Ashlip

2

Door

2A

Gasket

2B

Glass clips

3B

Latch

4

Right side panel

5

Front panel

6

Top casting

7

Upper back panel

8

Smoke outlet

9

Back panel

10

Left side panel

11

Leg  (long leg)

12

Upper bottom panel

13

Front door handle

14

Right burn plate

15

Glass panel

16

Airchamber

17

Air wash manifold

18

Top baffle (stainless steel)

19

Baffle cover (cast iron)

20

Air inspection cover

21

Rear burn plate

22

Fire brick (3)

23

Bottom grate

24

Left burn plate

25

Lower bottom panel

26

Ashpan housing

*

Ashpan

27

Air slider/valve

28

Air devider

30

Air control lever (chrome)
Allen head screw

31

Ashpan door

*

Ashpan door pin

32

Ash door handle

33

Baffle plate II

34

Latch compl.

*

Not shown

All floor protection materials must be non-combustible ie. metal,
brick, stone, mineral fiber boards).  Any combustible material may
not be used.

The easiest means of determining if a proposed alternate floor
material meets requirements listed in this manual is to follow this
procedure.

R-value = thermal resistance
k-value = thermal conductivity
C-value = thermal conductance

1.

Convert the specification to R-value;
a.

If R-value is given, no conversion is needed.

If k-value is given with a required thickness (T) in inches:
R=1/k X T.

If C-value is given: R=1/C.

2.

Determine the R-value of the proposed alternate  floor protector.
a.

Use the formula in Step 1 to convert  values not expressed
as “R”.

For multiple layers, add R-values of each layer to determine
overall R-value.

3.

If the overall R-value of the sustem is greater than the R-value of
the specified floor protector, the alternate is acceptable.

The specified floor protector should be 3/4” thick material with a k-
factor of 0.84.  The proposed alternate is 4” brick with a C-factor of
1.25 over 1/8” mineral board with a k-factor of 0.29.

Step A.  Use formula above to convert specifications to R-value.  R=1/
k X T= 1/.84 X .75 = .893

Step B.  Calculate R of proposed system.

•

4” brick of C-1.25,  therefore

•

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)(ft

2

)(F)                  (m

2

)(K)

Thermal conductivity =

k =     Btu        =           W        =     (Btu)

                         (hr)(ft

2

)(F)          (m

2

)(K)       (hr)(ft)(F)

Thermal resistance=

R =      Btu        =       (m

2

)(K)      =     (Btu)(inch)

(hr)(ft

2

)(F)

W

(hr)(ft

2

)(F)