Find a critical radius of insulation corresponding to maximum heat loss/length of pipe for a pipe surrounded by asbestos (k = 0,181 W/(moC)) and exposed to air of 10oC with h=3,5 W/(m2oC). Surface temperature of the pipe = 275oC and outer radius of the pipe is 25 mm.

Find heat loss from the pipe with insulation at critical radius

Find heat loss from the pipe without insulation

k = 0.181 W/(m.oC)

h= 3.5 W/(m2.oC)

T air = 10

T pipe = 275

R pipe = 25mm = 2.5 x 102 m

ASBESTOS

HEAT SOURCES

R1

R2

T = 10 C

T = 275

C

h=3.5

W/(m2.oC)

k=0.181

W/(m.oC)

q|r q|r+x

x

+ =

(zero because there is no heat generation in the system) +

= 0

lim0

+

= 0

= 0

=

=1

= 1

1

=

=

= ( )

=

=

=

Qmaxwhen = =

= .

Then, =

=0.181 W/(m.oC)

3.5 W/(m2.oC)=51.7 mm

= 3.5 W/(m2.oC)(. ) = . /

Heat Loss at r=Rcr =

= 446.25

22 +

= 215.056 /

=

= 3.5 W/(m2.oC)(26510)

= 892.5 /2

= .

=892.5

2[ 2 25 103 2]

= 140,19 /

A cylindrical vessel of 0,6 m diameter, which opens at the top, contains a liquid and rotates about its vertical axis. The rotation is steady state.

a. Calculate the height of the vessel so that the liquid reaches the top of the vessel at the height of H and begins to uncover the base at 100 rpm

b. If the speed is now increased to 130 rpm and the height of the liquid is maintained the same as that calculated in a, what area of the base will be uncovered?

= 130 rpm

1 2

vessel

uncovered

base area

the base

of vessel

= 100 rpm

1 2

vessel

Pipe diameter, d = 0,6 m

= 100 rpm when 1 = 0, 2 =

= 130 rpm when 1 = , 2 =

From the appendix, we know that:

the base

of vessel

= 100 rpm

1 2

vessel

Notes:

v has constant value for in and out momentums at r the other velocity components have zero value Gravity doesnt work at r

+

+

2

+

=

+

1

(

+1

222

2

2

+22

+

So,

2

=

(1)

From the appendix, we know that:

the base

of vessel

= 100 rpm

1 2

vessel

Notes:

Pressure doesnt work at There is shear stress at r

+

+

++

= 1

+

1

(

+1

222

+2

2+22

+

So, 0 =

1

()

(2)

From the appendix, we know that:

the base

of vessel

= 100 rpm

1 2

vessel

Notes:

vz=0

p has constant value at z

Gravity affects the system at z

gz = -g

+

+

++

=

+

1

+1

222

+22

+

So, 0 =

(3)

1

() = 0

1

() = 0

1

()

0

1

()

= 1

()

= 1

() = 1

=1

21

2 + 2

=1

21 +

2

r = 0 v = 0

r = R v = r

Boundary Condition 1

=1

21 +

2

0 =2+ 0

2 = 0

So, =1

21 (4)

Boundary Condition 2

=1

21

=1

21

=1

21

So, 1 = 2 (5)

=1

21 +

2

=1

2(2)()

So, = (6)

=

2

=()2

= 2

= 22

=22

2+ 1 (7)

From equation (3), we know that

0 =

=

=

= + 2 (8)

If we make (7) = (8)

=

22

2+ 1 = + 2 (9)

(1)r=0; =0

(2)r=R; z=zR; p=patm

Boundary Condition 1

22

2+ 1 = + 2

0 + 1 = + 21 = + 2 (10)

Boundary Condition 2

=22

2 + 2

2 = 22

2+ (11)

2 = 2

22

2+ =

22

2+

=2

22 2 ( )

=

2

2 2 2 ( )

=2(2 2)

2

=2(22)

2(12)

0 =2 0 2

2

=22

2

=(100)2(0,3)2

2 (9,8)

= 45,92

the base

of vessel

= 100 rpm

1 2

vessel

=2(2 2)

2

45,92 =130 2(0,32 2)

2 (9,8)

0,32 2 = 0,053

= 0,037 = 0,192

= 130 rpm

1 2

vessel

uncovered

base area

TERIMA KASIH