Тепловые процессы в технологической системе резания. Неумоина Н.Г - 77 стр.

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77
А
р
= 0,67; А
д
= 1; А
к
= 1;
33,1
1066,22
1007,7
2
b
3
3
1
o
=
⋅⋅
⋅
=
⋅
=
−
−
l
η
;
459,28633,12Pe2u
o
=⋅⋅=⋅⋅=
η
.
По рис. 7.2 при
u = 45 находим А
0
= 0,89, А
т
= 2;
.1013,3289,01167,0033,0
q
q1085,7
AAAAAA
q
A
A
6
1
1
5
токдPc
1
M
4
−
−
⋅=⋅⋅⋅⋅⋅
⋅⋅
=
=⋅⋅⋅⋅⋅⋅=
9) Составим код источника q
2Т
и по алгоритму (рис. 7.1), рассчита-
ем коэффициент А
5
.
12
501,22
212
Код +=
:
T2
6
T2
3
1
Т22
М
q1095,2
9,33
q101,0q
A ⋅⋅=
⋅⋅
=
⋅
=
−
−
λ
l
.
Критерий Пекле 4,19
10067,0
101,03,1
a
v
Pe
4
3
1
2
=
⋅
⋅⋅
=
⋅
=
−
−
l
;
128,0
4,1914,3
1
Pe
1
A
C
=
⋅
=
⋅
=
π
;
А
p
= 0,36; А
д
= 1; А
К
= 1;
36,35
101,02
1007,7
2
b
3
3
2
o
=
⋅⋅
⋅
=
⋅
=
−
−
l
η
;
3114,1936,352Pe2u
o
=⋅⋅=⋅=
η
.
По рис. 7.2 при
u = 311 находим А
0
= 0,92, А
т
= 2;
.105,2292,01136,0128,0
q
q1095,2
AAAAAA
q
A
A
7
T2
T2
6
токдpс
T2
M
5
−
−
⋅=⋅⋅⋅⋅⋅
⋅⋅
=
=⋅⋅⋅⋅⋅=
10) Составим код стока q
2
и по алгоритму (рис. 7.1), рассчитаем ко-
эффициент
A
7
. 12
101.22
212
Код −= :
2
6
2
3
1
22
М
q1095,2
9,33
q101,0ql
A ⋅⋅=
⋅⋅
=
⋅
=
−
−
λ
. 
                           Ар = 0,67;      Ад = 1; Ак = 1;
                                 b      7 ,07 ⋅ 10 −3
                          ηo =        =                  = 1,33 ;
                               2 ⋅ l 1 2 ⋅ 2 ,66 ⋅ 10 −3
                      u = 2 ⋅η o ⋅ Pe = 2 ⋅ 1,33 ⋅ 286 ,9 = 45 .
     По рис. 7.2 при u = 45 находим А0 = 0,89, Ат = 2;
                  A
            A 4 = M ⋅ Ac ⋅ AP ⋅ Aд ⋅ Aк ⋅ Aо ⋅ Aт =
                   q1
            7 ,85 ⋅ 10 −5 ⋅ q1
          =                    0 ,033 ⋅ 0 ,67 ⋅ 1 ⋅ 1 ⋅ 0 ,89 ⋅ 2 = 3 ,13 ⋅ 10 −6 .
                    q1
    9) Составим код источника q2Т и по алгоритму (рис. 7.1), рассчита-
                                   212
ем коэффициент А5. Код = +                 12 :
                                 501,22
                         l 2 ⋅ q 2Т       0 ,1 ⋅ 10 −3 ⋅ q 2T
                  AМ =                =                       = 2 ,95 ⋅ 10 −6 ⋅ q 2T .
                            λ1                    33 ,9
                              v ⋅ l 2 1,3 ⋅ 0 ,1 ⋅ 10 −3
     Критерий Пекле Pe =               =                    = 19 ,4 ;
                                a1        0 ,067 ⋅ 10 −4
                                   1                 1
                         AC =              =                  = 0 ,128 ;
                                 π ⋅ Pe         3 ,14 ⋅ 19 ,4
                           Аp = 0,36;          Ад = 1; АК = 1;
                                   b        7 ,07 ⋅ 10 −3
                          ηo =           =                  = 35 ,36 ;
                                2 ⋅ l 2 2 ⋅ 0 ,1 ⋅ 10 −3
                      u = 2η o ⋅ Pe = 2 ⋅ 35 ,36 ⋅ 19 ,4 = 311 .
     По рис. 7.2 при u = 311 находим А0 = 0,92, Ат = 2;
                    A
               A 5 = M Aс ⋅ A p ⋅ Aд ⋅ Aк ⋅ Aо ⋅ Aт =
                    q 2T
               2 ,95 ⋅ 10 −6 ⋅ q2T
             =                     0 ,128 ⋅ 0 ,36 ⋅ 1 ⋅ 1 ⋅ 0 ,92 ⋅ 2 = 2 ,5 ⋅ 10 −7 .
                       q 2T
   10) Составим код стока q2 и по алгоритму (рис. 7.1), рассчитаем ко-
                             212
эффициент A7. Код = −              12 :
                            101.22
                            l2 ⋅ q2     0 ,1 ⋅ 10 −3 ⋅ q 2
                     AМ =             =                    = 2 ,95 ⋅ 10 −6 ⋅ q 2 .
                              λ1               33 ,9

                                               77

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