Q

21575

:=

E

2.1 10

11

⋅

:=

l1

0.147

:=

RE

285 10

6

⋅

:=

lr

2 l1

0.04

+

(

)

⋅

:=

lr

0.374

=

Im

π d1

4

⋅

64

:=

d1

d1

4

64 Q

⋅

6

⋅

lr

2

⋅

π

3

E

( )

⋅

:=

d1

0.021

=

λ

4 lr

⋅

d1

:=

λ

72.835

=

λgr

π

2

E

⋅

285 10

6

⋅

:=

λgr

85.278

=

a

RE

:=

δkr

4 6

⋅

Q

⋅

π d1

2

⋅

:=

δkr

3.907

10

8

×

=

b

a

δkr

−

λ

:=

b

1.451

−

10

6

×

=

D

4

π

2

b

2

⋅

lr

2

⋅

π a

⋅

6

⋅

Q

⋅

+

⋅

:=

D

4.36

10

7

×

=

d3

4π b

⋅

lr

⋅

D

+

2π a

⋅

:=

d3

0.021

=

d

0.028

:=

D1

0.023

:=

Pdop

15 10

6

⋅

:=

A

d

2

D1

2

−

(

)

π

⋅

4

:=

A

2.003

10

4

−

×

=

n

Q

A Pdop

⋅

:=

n

7.182

=

nc

n

1.5

+

:=

nc

8.682

=

P

5

:=

Hn

nc P

⋅

:=

Hn

43.409

=

d3

0.0225

:=

M

1

π d3

2

4

























2.1 10

11

⋅

:=

M

1.198

10

8

−

×

=

Dz

4

π 1.1 10

11

⋅

(

)

⋅

M

⋅

:=

Dz

0.031

=

Given

Pkr

Q 6

⋅

:=

D4

0.0285

:=

A1

Pkr

Pdop

:=

DZ

1

:=

A1

DZ

D4

−

(

)

2

π

⋅

4

=

Find DZ

(

)

0.133

=