Internal Flow BL

8/19/2019 Internal Flow BL

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Temperature field in duct flow pipe

2

21

x

T k

r

T r

r r k

r

T v

x

T uC p

∂

∂+⎟⎟

⎠

⎞⎜⎜

⎝

⎛ ⎟

⎠

⎞⎜⎝

⎛

∂

∂

∂

∂=⎟

⎠

⎞⎜⎝

⎛

∂

∂+

∂

∂ ρ

Neglect axial conduction

02

2

2

2

→∂

∂>>

∂

∂

x

T

r

T

For fully developed laminar flow v=0

)(r

T r

r r x

T u

r

T r

r r

k

x

T uC p

∂

∂

∂

∂=

∂

∂

⎟

⎠

⎞⎜

⎝

⎛

∂

∂

∂

∂=

∂

∂

α

ρ



Define mixed mean temperature

mw

w

m

r

m

T T T T

rudr A

udA A

u

rdr r T r u AV

T o

−−=

==

=

∫∫∫

θ

π

π

211

2)()(1

0

Thermal layer is fully developed when θ is

independent of x for any set of B.C. θ is f(r)only

Laminar Duct Flow



i.e

oo

mw

w

oomw

o

r r r

k T T

T T

r r r

k

T T

Aq

h

onlyr

r

and x

∂

∂=

−

−

∂

∂−=

−

=

=

=∂

∂

θ

θ θ

θ

)(

)(

0

but θ ≠ θ(x) => h ≠ h(x) and h is defined at thewall r=r o => h =constant ( uniform for fully

developed flow)

Laminar Duct Flow



( )

( )dx

dT

dx

dT

x

T

dx

dT

dx

dT

T T

T T

dx

dT

dx

dT T T dx

d T T

dx

dT

dx

dT

T T

T T

T T

x

x

mw

mw

mw

ww

mw

ww

mw

mw

w

θ θ

θ

+−=∂∂

⇒=⎟ ⎠

⎞⎜⎝

⎛ −⎟⎟

⎠

⎞⎜⎜⎝

⎛

−

−−−

⇒=⎟⎟

⎠

⎞⎜⎜

⎝

⎛

−−+⎟

⎠

⎞⎜

⎝

⎛ −

−

⇒=⎟⎟ ⎠

⎞

⎜⎜⎝

⎛

−

−

∂

∂

⇒=∂∂

1

0

011

0

0

Laminar Duct Flow



Two cases:

A- Constant heat flux (qw/A)=constant

.

.

.

(2)

w

w m

w

w m oo

w m

w m

q A

h const T T

q A k

const r T T r

r

T T const

dT dT or uniform

dx dx

θ

⎛ ⎞⎜ ⎟⎝ ⎠ = =

−

⎛ ⎞⎜ ⎟ − ∂⎝ ⎠ = =∂−

⇒ − =

= =

Laminar Duct Flow



But, from equation (2)

Then usedx

dT

dx

T

x

T mw =∂

=

∂

∂

symmetry

r

T r

xT T r r

r

r

V

u

r

T r

r r x

T u

wo

o

m

00

)(

12

1

2

=

∂

∂=

==

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −=

⎟⎟ ⎠

⎞

⎜⎜⎝

⎛ ⎟ ⎠

⎞⎜⎝

⎛

∂

∂

∂

∂=

∂

∂α

Laminar Duct Flow



Case 2:

B- Constant temperature Tw =const.

⎟ ⎠

⎞⎜⎝

⎛

∂

∂

∂

∂=

∂

∂⎟⎟ ⎠

⎞⎜⎜⎝

⎛

−

−⇒

⎟ ⎠

⎞⎜⎝

⎛

∂

∂

∂

∂=

∂

∂⇒

∂

∂=

∂

∂

=∂

∂

+−=∂∂

r

T r

r r x

r T

T T

T T u

r

T r

r r x

T u

x

T

x

T

x

T

dxdT

dxdT

xT

w

mw

w

m

m

w

mw

α

α θ

θ

θ θ

)(

)1(

0

)1(

L

Laminar Duct Flow

w

w m

T T

T T θ

−=

−



Iterate for solution, use θ(r) of qw= const, integrate

to get new θ(r) and so on.

Nu =3.66

xc

mm

m

m

mw

m

oo

mw

eT T dx

dT c

dx

dT const T T h

dx

dT r cvr

A

q

T T h A

q

10

)()(

2

)(

1

3

−≈⇒=−

=−

=

−=

π ρ π

Laminar Duct Flow



Example

• The sketch shows a cast ironwater pipe running outdoors.The pipe is 50 mm I.D. In a 5

mm wall thickness, and carriesa constant water flow rate of0.65kg/sec. The water enters theoutdoor section of the pipe at

x=0 with a bulk temperatureTv=15oC. The ambient airsurrounding the pipe is at –20oCand the outside heat transfer

coefficient between the pipeand air is 250 W/m2C. Ice startsforming at the pipe wall at adistance of x1.



Example

Assuming steady state, fully developed conditions

and constant water properties:

a) Find the inside heat transfer coefficient betweenwater and pipe inner wall area for x<x1 (no ice).

b) Find the overall heat transfer coefficient U based

on the pipe inner wall area x<x1.c) Sketch the temperature profile across the pipe for

x<x1. Show water, pipe wall, and air temperature.

d) Calculate the distance x1, and corresponding localwater bulk temperature T1 at the point where ice

begins to form.



Example

e) Sketch the ice thickness profile along the pipe wall

for x>x1.

f) At x2>x1, water bulk temperature is 3C. Estimatethe steady state ice thickness.

g) Sketch the temperature profile across the pipe at

point x2. Show all temperatures.h) Sketch the bulk temperature distribution between

x=0 and x2. Clearly show that what happens at x1.



Solution

a)

C mW h

d

Nuk

hk

hd

Nu

Nu

Turbulent AV m

d

m

D

d m

A

d mVd

D

o

ooo

2

31

8.0

2

/6.1041

Pr Re023.0

11823Re

4

4

Re

=

=⇒=

=

=⇐=

====

ρ

µ π µ

π µ µ

ρ

b) Overall heat transfer coefficient

∑== th

i

RUA

1



Solution

K mW U

d h

d

k

r r

d

hU

Ld A Ld A

Ah

A

kL

r r A

h A R

AU

AhkL

r r

Ah R

i

oout

ii

oi

ii

ooii

out out

ii

oi

ii

th

ii

out out

i

o

ii

th

2/02.198

0033.000009.000096.02

ln11

2

ln111

1

2

ln1

=⇒

++=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

++=⇒

==

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡++==

++=

∑

∑

π

π π

π

π



Solution

c)



d) @ x=x1, Ti=0C

Solution

C xT T xT R R

R xT

R xT xT fluxheat

R RT xT

W K m R

W K m R

W K m R x x for

o

iiwo

i

b

i

ib

wo

i

o

w

ii

6.5)())(()(

)()()(

/0033.0

/00009.0

/00096.0,

111

111

2

2

2

=+−+

=⇒

−==+

−

=

=

=<

∞

∞

Now from energy balance,



Solution

[ ]

( )

mU r

C m x

C T x xat

eT T T xT

dxC m

U r T xT T xT d

xT cmdx

d T xT U r x

ii

p

o

o

b

x

C m

U r

ob

p

o

ii

b

b

b p

o

oii

po

ii

45.272015206.5ln

2

6.5

)(

2)(

))((

)()(2

1

1

2

=++−=

⇒=⇒=

−=−⇒

−=−−⇒

∆−=−∆

−

∞∞

∞

∞

∞

π

π

π

π



e)

Solution



Solution

f) Here Tm=0 oC and T b=3 Cq”Ai= heat flux based on inside area of cast iron

wall.

All resistances are also based on inside surface of

pipe wall.



Solution

)4(

1

2

)3()2(

4Pr 023.02

)2(

1

11.

2

)1(12

2ln

/00009.0

/0033.0

8.1

8.1

8.0

31

2

2

L

L

L

K

′+++

−=

′−

=″

⎟ ⎠

⎞⎜⎝

⎛ −

=⎟⎟ ⎠

⎞⎜⎜⎝

⎛

−=

⎟⎟ ⎠ ⎞⎜⎜

⎝ ⎛

−−=′

′⎟⎟

⎠

⎞⎜⎜⎝

⎛ −

=−

=′

−=⇒⎟

⎠ ⎞

⎜⎝ ⎛

−=

=

=

∞

−

iicewo

b

i

mb A

inin

i

i

i

o

i

in

ir

i

ii

ii

r

k R

iice

i

ii

ice

w

o

R R R R

T T

R

T T q

n

hh

d

d

d

m

d

k h

hr

r hd

d R

er K

d d

d

R

W K m R

W K m R

i

i

iceice

δ δ

δ π δ

δ δ

δ δ



Solution

Which gives the necessary equations to solve δ.Unfortunately, it is a transcendental equations. We

can now linearize the equations assuring δ<<di and

then we solve it, or we can solve by iteration. Here

we take second path which is more straight forward.



Solution



Solution

g)



h)

Solution

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