# технология и оборудование лесозаготовок

```ɌȿɏɇɈɅɈȽɂə ɂ ɈȻɈɊɍȾɈȼȺɇɂȿ ɅȿɋɈɁȺȽɈɌɈȼɈɄ
625.711.84+625.31
Ⱥɥɟɤɫɚɧɞɪ ɋɟɪɝɟɟɜɢɱ Ɇɢɥɹɟɜ,
,
-
ɉɊɈɆȿɊɁȺɇɂȿ ȼɈȾɇɕɏ ɉɊȿȽɊȺȾ
ɇȺ ɌɊȺɋɋȺɏ ɁɂɆɇɂɏ ɅȿɋɈȼɈɁɇɕɏ ȾɈɊɈȽ
Ɂɢɦɧɢɟ ɥɟɫɨɜɨɡɧɵɟ ɞɨɪɨɝɢ, ɜɨɞɧɵɟ ɩɪɟɝɪɚɞɵ, ɦɟɬɟɨɪɨɥɨɝɢɱɟɫɤɢɟ ɭɫɥɨɜɢɹ, ɥɟɞɹɧɵɟ ɩɨɤɪɵɬɢɹ, ɪɚɫɱɟɬɵ ɫɤɨɪɨɫɬɢ ɩɪɨɦɟɪɡɚɧɢɹ.
Winter timber roads, water obstacles, meteorological conditions, ice coating,
computation of the freezing process rates.
ȼɜɟɞɟɧɢɟ.
(
,
. .)
,
-
,
.
[1].
,
.
ɉɨɫɬɚɧɨɜɤɚ ɡɚɞɚɱɢ.
-
lx
.
Tair
Qw = 3,35&middot;105
78
.
/ .
Qw
−1
0&deg; .
;
-
lx (
)
.
-
&laquo;
+
&raquo;
.
Ɇɚɬɟɦɚɬɢɱɟɫɤɚɹ ɮɨɪɦɭɥɢɪɨɜɤɚ ɡɚɞɚɱɢ.
,
:
:
Tair = T0 − T1 tb ect + T2 e–at sin(ωt),
(1)
ir
T0 , T1 , T2 , a, b, c, ω –
,
: T0 = –5 &deg;C,
T1 = 7 &middot; 10–5 &deg;C/ , T2 = 5 &deg;C, a = 5,32 &middot; 10–6 &deg;C/ , b = 1, c = –2,31  10–6 &deg;C/ ,
/ ,t−
, .
ω = 7,272 &middot; 10–5
Tair
t (1)
. 1.
4,32  105 ( )
, 
–1,816
1  105
0
2  105
3  105
–16,622
4  105
5  105 t,
. 1.
:
Q = αA(Tair – Tw),
Q−
Tw −
;α−
;A−
(2)
;
.
79
α
−
&laquo;
–
−
&laquo;
–
&laquo;
&raquo; αsnw = 23 v  0,3;
–
&raquo;
[2]:
&laquo;
–
&raquo;
(a)
&raquo; αice = 6 v  0,3,
(b)
v−
, / .
&laquo;
+
&raquo;
-
[3]:
H   T    T    T 
  λx
  λy
  qV ,
   λz
t x  x  y  y  z  z 
H–
,
/ 3; t –
/
3
;T–
x, y, z,
); qV –
/
2
(
(3)
, &deg; ; λx , λy , λz –
λx , λy , λz
,
, .
&laquo;
+
&raquo;
(qV = 0),
Qw
.
,
(3)
:
H   T
  λx
t x  x
x−
   T 
  λy
,
 y  y 
(4)
,y−
,
.
H   ρC (T )dT ,
T
(5)
Tb
Tb, T –
&raquo;; ρ –
,
&laquo;
. 2
; lx −
80
: hice −
3
/ ;C–
).
+
&raquo;
,
&laquo;
/ &deg;C (ρ
, hw −
, lx = 1,51 ; ly −
, ly = 0,01 .
+
. 2.
-
y
hw
hice
ly
x
0
lx
. 2.
:
,
y = ly
x=0
(2);
x = lx
: qy = 0.
x=0
:
T = 2 &deg;C.
x = lx
y=0
T = 2 &deg;C;
T = –1 &deg;C;
ρ
-
:
H  (ρC )k  dT , k  1, 2, ...
(6)
,
T
Tb
(6): Tb , T −
.
– 16…10 &deg;C
.
Ɍɟɩɥɨɮɢɡɢɱɟɫɤɢɟ ɯɚɪɚɤɬɟɪɢɫɬɢɤɢ ɜɨɞɵ ɢ ɥɶɞɚ
ρice ,
/ 3
cice ,
/ 
–16 919
2,4
2080
0
–1
2,2
2260
31,086 &middot; 106
917
ρw,
/ 3
λw,
/ 
λice ,
/ 
T,
&deg;C
cw ,
/ 
H,
/
3
0
1000
0,551
4212
366,086 &middot; 106
10
1000
0,574
4191
407,996 &middot; 106
: T −
; λice , λw −
;H−
; ρice , ρw −
; cice , cw −
.
81
λ,
.
-
,
λ0 ,
0&deg;
λ,
0
.
0
:
λ = λ0f1(T), C =
λ
0 f2(T),
H=
0 f3(T),
(c)
λ
C
H
 f1 (T ), C 
 f 2 (T ), H 
 f 2 (T ).
C0
H0
λ0
(d)
(d)
. 3.
fk(T), k = 1, 2, 3
1
T, C
–15
–1
0
10
. 3.
,
&laquo;
&raquo;(
. 2)
(120
Ɇɟɬɨɞ ɪɟɲɟɧɢɹ ɡɚɞɚɱɢ.
(4) −
,&laquo;
,
&raquo;
&laquo;
&raquo;
.
:
151,
82
+
= 432 000 ).
lx = 1,51 ,
0,01  0,01
− 304.
ly = 0,01 ,
,
-
&laquo;
Δh
λ–
&middot;
λ Δt
1
 ,
2
2
C (Δh)
(7)
,
(7)
.
/ &middot;&deg;C,
ρC  Δh 
λ Δt
2
(7)
(8)
+
&raquo;
T = − 1 &deg;C
,
10
α
,
(b)
αice = 6 5  0,3 = 13,81
&laquo;
-
1
 .
2
&laquo;
5 / .
-
[4]
;C–
,
,
3
/
Δt
&raquo;
(Δtw = 182,25 ).
− 20 ,
:
− 25 ,
&laquo;
[2]:
/(
2
&middot;&deg; ).
(9)
(8),
(Δtice = 21,63 )
&laquo;
− 30 .
Δt
&raquo;
-
&raquo;
&raquo;
:
,
.
,
-
Pentium (R)
2,6
2
&laquo;
12
.
&raquo;
.
&laquo;
&laquo;
&raquo;
+
&raquo;
(432 000 )
,
-
tend = 432 000 .
83
Ɋɟɡɭɥɶɬɚɬɵ ɪɟɲɟɧɢɹ ɡɚɞɚɱɢ.
t = 432 000 .
. 4
&laquo;
+
&raquo;
-
. 1.
T = − 0,378 &deg;C
.4
0,25
,
.
,
0,01 ,
0,24 .
,
-
4,8
–0,4
.
–10 &deg;C.
, 
2
0,808
–0,378
–1,565
–2,752
5
–3,939
–5,126
13,81
/(
2
.
)
–6,313
–7,5
–8,687
–9,874
0
0,12 0,24 0,36
0,48 0,6
0,72 0,84 0,96 1,08
1,2 1,32 1,44 1,51
,
. 4.
.1
ȼɵɜɨɞɵ.
,
(
)
,
84
-
.
,
-
,
.
-
.
,
,
,
.
-
1,51
Tair
(1)
,
−5
10
4,8
3 &deg;C
.
–0,4 &deg;C
-
−16,6 &deg;C
5 /
–10 &deg;C (
).
,
,
.
Ȼɢɛɥɢɨɝɪɚɮɢɱɟɫɤɢɣ ɫɩɢɫɨɤ
я , . .
[
]:
.
2.
2.06.04–82.
(
,
1982. – 37 .
3. И
, . .
. .
. – .:
4.
, .
.
, .
.–
1.
/ . .
)[
.–
., 2006. – 303 .
]/
[
] /
, 1981. – 416 .
.–
. .
,
.:
,
. .
,
[
.:
] /
, 1972. – 418 .
85
(
,
,
,
. .)
-
,
.
.
&laquo;
+
&raquo;
.
,
,
-
.
.
&laquo;
+
&raquo;
c
.
&laquo;
.
+
,
-
&raquo;
−
:
−
;
−
;
-
.
:
–1 &deg;C;
−
+2 &deg;C.
+
&raquo;
&laquo;
:
.
,&laquo;
&laquo;
&raquo;
&raquo;
.
.
0,24 ,
–0,4 &deg;C
− 4,8
−10 &deg;C.
.
,
(
-
)
,
.
.
***
Introduction. The design procedure of the static water freezing of a water obstacles (filled by water open cuts, ponds, lakes, water basins, etc.) during the set duration
of time on lines of the winter timber-carrying roads, constructed on a numerical example is resulted.
86
Statement of a problem. It is considered the freezing of the top layer of standing
fresh water of the set thickness under influence of cold atmospheric air. It is required
to find thickness of ice and distribution of temperature in system &laquo;ice + water&raquo; in view
of allocation of the latent heat of crystallization after several day.
The mathematical formulation of a problem. Is accepted, that change of temperature of air occurs under the law reflecting its daily fluctuations. Transfer of heat from
water to air is carried out by convection on its surface. Distribution of heat to system
&laquo;ice + water&raquo; is defined from the equation of heat conductivity with the factors dependent on temperature. The settlement area &laquo;ice + water&raquo; is chosen as a rectangular.
Boundary conditions: on a surface of a water obstacle − heat exchange convection; on
some depth temperature − a constant; on lateral borders of settlement area − thermal
streams are absent. Entry conditions: on a surface of a water obstacle temperature –1 &deg;C;
on the bottom border of settlement area − temperature +2 &deg;C. Thickness of ice and distribution of temperature in system &laquo;ice + water&raquo; after five day is searched.
Method of the decision of a problem. The numerical method of finite elements,
the &laquo;step-by-step&raquo; decision on time and iterative procedure of approximation each
&laquo;step&raquo; of the decision is used.
Results of the decision of a problem. Values of temperature in units of a certainly –
element grid after five day are received. Depth of the water freezing makes approximately 0,24 m, speed of the water freezing − 4,8 cm in day. The temperature of ice
changes from – 10&deg;C up to −0,4 &deg;C.
The conclusion and conclusions. The design procedure of a freezing process the
water obstacles on lines of the winter timber-carrying roads is submitted, in which are
taken into account actual (not average for the last years) the meteorological data during the current period of time, phase transformations of water into ice and dependence
a thermal properties of water from temperature. This technique can be used for the decision thermal problems at an estimation of bearing ability winter timber-carrying