Internal Flow Through Pipes And Ducts.pdf

8/10/2019 Internal Flow Through Pipes And Ducts.pdf

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Module18: Internal Flow

Flow and Heat Transfer through

Pipes and Ducts


2/30

Introduction to Internal Flow

Characteristics primary difference from external flow is the presence of an

opposing, confining surface that constrains the boundary

layer growth entry (entrance) length exists (B.L. is developing)

fully developed region eventually forms

Hydrodynamic Boundary Layer Development

ref. Incropera

& DeWitt,

Chap. 8

follow I&D, Chapter 8

r0 R


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4/30

General Assumptions

We will assume:

steady, incompressible, Newtonian,constant

properties

Kn (= /L)


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Hydrodynamics - Laminar Flow

Entrance Region:

B.C.s: u(x, R) = 0 v(x, R) = 0

u(x = 0, r) = uo(r)

Needs numerical solution

u 1 (rv)0 continuity

x r r

+ =

u u p u

u v r x momentumx r x r r r

v v p vu v r r-momentum

x r r r r r

+ = + + = +

r 0

u

0r =

=Assume axi-

symmetry


6/30

Hydrodynamics - Laminar Flow

Fully Developed Region Poiseuille Flow (parallelflow):

v-momentum equation yields

x-momentum reduces to balance between pressure &shear forces:

B.C.s u(R) = 0

dp d dur

dx r dr dr

=

r 0

u 0r =

=

uv 0 0 u(r)

x

= =

0 ( ) onlyp

p p xr

= =


7/30

Integrate twice and apply B.C.s to get

Mean velocity

Velocity distribution

Velocity profile

d du dpr

r dr dr dx

=

can integrate easily

since dp/dx is

independent of r

2

21 dp r u(r) R 14 dx R

=

R2

0m 2 2

u(2 r)drm R dpu

R R 8 dx

= = =

&

2

m

u(r) r 2 1

u R

=

Notice

dimensionless

velocity

distribution not afunctions of Re

why?


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Fully Developed Laminar Flow

Pressure Drop : Expressed in terms of the Moody (or Darcy)

friction factor

fully developed

laminar2

m D

dpD

64dxf f

u / 2 Re

=

2

mfd

up f L2D

=

D = 2R


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ref. Incropera &

DeWitt, Chap. 8

Note: dP/dx is

constant, but f

is not, due to

funny non-dimensionaliza

tion

For turbulent flow the analysis is not as simple as above, and the pressuredrop is very sensitive to roughness (unlike in laminar flow). For smoothsurfaces,

fturb = 0.316 ReD-0.25 (ReD < 20000)

= 0.184 ReD-0.20 (ReD > 20000)


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Thermal Considerations - Laminar Flow

Characteristics:

ref. Incropera & DeWitt , Chap. 8

r0 R


11/30

Thermal entrance length

(unlike in laminar flow, the entrance length is nearly

independent of Pr in turbulent flow, with Lt / D ~ 10)

The shape of the fully developed profile T(r, x) is different

depending on whether Ts or q is a constant Example:

for engine oil (Pr 6000), say D = 1 cm, um = 1 m/s,

= 550 x 10-6 m2/s:

ReD = (1) (0.01)/(550 x 10-6

) = 18 (laminar)xfd,t = 0.05 Re Pr = 5455 tube diameters!

i.e., a tube length of 54 m!!

(t never reaches the centerline in pipes of reasonable length)

fd,tD

x 0.05 Re PrD

Terms and Notation


12/30

Bulk Mean Temperature

Bulk mean temperature:

Weighted w.r.t. mass flow rate

For a circular cross section, with constant-property flow

rate of

thermalenergytransport

c

t v m v c

AE mC T uC TdA= = &

&

c

v cA

m

v

uC TdA

TmC

=

&

m 2

m

2T T(r)u(r) r dr

u R

=


13/30

Thermally Fully Developed Flow

Define a dimensionless temperature

The relative shape of the temperature profile no longerchanges if field is fully developed

s

s m

T T(x, r)

T T (x)

s

s m

T T0

x T T

=

s r R

s m s mr R

T/ rT Tconstant f (x)

r T T T T

=

=

= =

s s m

r R

Tq h(T T ) k

r =

= =

h

constantk

=

In thermally f.d. flow with const. props.,local h is independent of x !

That is, Nux is independent of x

Can this happen withouthydrodynamically fully-

developed flow?


14/30

in outE E=& & 1v=

Temperature Distribution - Energy BalanceEnergy Balance

specific volumeEnergy Balance:

v mconv v m v m

d(C T pv)dq m(C T pv) m(C T pv) m dx 0

dx

+ + + + + =

& & &

thermal energy flow work


15/30

Energy Balance (contd)

( )

v mconv v m v m

m conv s

v ms

v p p m

ms p

d(C T pv)dq m(C T pv) m(C T pv) m dx 0dx

Perfect gas pv= RT ; also dq q (x)Pdx

d(C R)Tq (x)Pdx m dx 0dx

But C R C and d C T di

For constant properties:

dTq (x)P mC

dxFor a c

+ + + + + = =

+ =

+ = =

=

& & &

&

&

sm

p

ircular pipe P D

Dq (x)dT

dx mC

=

&

=What if its not a perfect gas ?

Neglect pressure work, set Cp =Cv

to get the same result.

Perimeter


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Constant Heat Flux Boundary Conditions

( ) ( )

sm

p

s s m s m

s m

s

m s

m s s

s m

DqdTconstant

dx mC

Furthermore,

q h T T T T constant

dT dTconstant

dx dxT -T(x,r)

Also, recall = =constant with xT T

Since T T =constant, T -T(x,r) is also constant

dT dTTThus, = =

x dx dx

A

= =

= =

= =

&

ll temperatures rise at the same rate axially!

C t t T t B d C diti


17/30

Constant Temperature Boundary Conditions

( )

( ) ( )

( ) ( )

sm

p

s s m

s m

s m

p

s m s m inletp

s

m s

Dq (x)dT not constantdx mC

However, q (x) h T T .Thus :

d T T Dh T T

dx mC

DT T T T exp( hx)

mC

Bulk temperature varies exponentially!

T -T(x,r)

Furthermore,since = =constant with xT T

= =

=

=

=

&

&

&

s m m

s m s s m

s

T T dT dTT 1 T 1constant

x T T dx T T x T T dx

All temperatures tend towards T exponentially with x!

= = =

Careful! h is

constant only in

fully-developed

region!


18/30

Axial Temperature Variation

Is bulk temperature variation linear (or

exponential) through the FD region? What about

the temperature at a point (x,r)?


19/30

Energy Equation:

(f.d. vel. Profile, v=0)

Constant Surface Flux Constant Surface

Temperature

See Sec. 8.4.1 I&D for solution

s s mq constant h(T T ) = = sdT 0

dx =

mT dT

x dx

=

s m

s m

T T T dT

x T T dx

=

T Tu r

x r r r

=

Hydrodynamically and Thermally Fully Developed

Flow Solution


20/30

Solution for Constant Heat Flux BC

T Tu r

x r r r

= LHS known

Integrate twice to

obtain

4 22

m ms

2u R dT 3 1 r 1 r T(r) T

dx 16 16 R 4 R

= +

Is this known?

Since both bc are Neumann-type, temperature can

only be determined up to an additive constant. What

is the physical meaning of this?


21/30

Nusselt Number

Steps:1. Recall

2. Find from temperature solution

3. Find bulk temperature:

4. Hence find h, and thus NuD

( )s mr R

Tk h T T

r =

=

r R

Tr =

m 2

m

2T T(r)u(r) r dr

u R=

Known

D s

hDNu 4.36 for q constant

k= = = Notice Nusselt number

not a function of Re or

Pr!


22/30

Solution for Constant Temperature BC

Solution is a bit more complicated because LHSis not constant.

Solution obtained numerically

Can show that

D sNu 3.66 T constant= =


23/30

For the entire tube (i - inlet, o - outlet), overall energy

balance:

Also

conv p m,o m,iq mC (T T )= &

sq constant =

s m

s m,i p

T T (x) P xexp h

T T mC

=

& s

m m,ip

q P

T (x) T xmC

= + &

s lmq h A T= sq q P L= sA P L=

o ilm

o i

T TTn( T / T ) l

sms m

p p

q PdT P h(T T )dx mC mC= = & &

Other Useful Relationships

Ts=constant


24/30

Developing Flow Terminology

Thermal entry length problem Flow is fully developed, temperature is not

Combined entry length problem

Both flow and temperature are developing

Unheated starting length

There is an insulated length of duct at the entrance sothat the flow has a chance to develop while the

temperature does not

Synonymous with Thermal entry length

T=Ts

T=Ts


25/30

Refer to the course text to find correlations for NuD

for:

entry region (section 8.4.2)

Hansen formula (Eq. 8.56 I&D) - assumes only thermal entry length;

for constant surface temperature

Seider-Tate formula (Eq. 8.57 I&D) - for combined entry length; less

accurate; evaluate properties at mean temperature defined as

average between inlet and outlet

turbulent flow (section 8.5) Colburn relation for friction factor for smooth circular tubes (Eq. 8.58

I&D); Dittus Boelter (Eq. 8.59) and Seider-Tate (Eq. 8.61) correlations

for Nusselt number

non-circular tubes (section 8.6)

(Laminar flow - use Table 8.1)

Turbulent flow, use correlations for circular tubes with hydraulic

diameter: Dh = 4Ac/P, where P is the wetted perimeter

concentric tubes (section 8.7)

Convection Correlations

Entry Length in Circular Pipes


26/30

y g p

NuD at x = 0 = ?

Why the difference between thermal and combined entry lengths?

Notice that curves are independent of Re, Pr if x axis is scaled as

shown Graetz number = Re Pr/(x/D) (some texts use inverse)

Example:


27/30

Example:

Water at 280 K enters a 1-inch diameter tube kept at

a constant surface temperature of 360 K. The tube is2 m long and water velocity = 1 m/s.

Find the heat transfer coefficient & exit temperature.Solution:

First, estimate the exit temperature to evaluate

properties. Try 350 K i om m

m

T T 350 280T 315 K

2 2

+ += = =

6 3

7 2 3631 10 1 106.36 10 m /s 991 kg / m991 v 1.009

= = = = = = Q

Pr 4.16 ; k 0.634 W/mK= =

4

D 7

1* 0.0254Re 3.99 10 turbulent6.36 10

= =

L 278 7 10 f d


28/30

Dittus-Boelter correlation (8.60)

Use Eq. (8.43) to get exit temperature (Eq. 8.41would be for qs = constant)

78.7 10 f .d.D 0.0254

= = >

( )0.40.8

D D s mNu 0.023(Re ) Pr (n 0.4 for heating, T T )= = >Q

2h 4867 W/m K =

s m

s

s m,i p

T T (x) Px hexp for T constantT T mC

= = &

= 195

@ x L T (x) T T= = =


29/30

m m,L m,o@ x L, T (x) T T= = =

2

m

DP D; m u 4

= = &

p m p

PL h 4L h 4LSt where St Stanton #

m C D u C D= = =

&

4

Nu 195St 0.0017

Re Pr 3.99 10 4.16

= = =

p

PL h

m C0.3685=

&

m,o

m,o

360 T0.692 T 304.6 K

360 280

= =

304.6 2802+=

mT

not 350K as assumed!

exercise, notdone here

now recalculate with new

Total heat transfer rate


30/30

Total heat transfer rate,

Alternatively, we can calculate this as: (Eq. 8.44)

giving Same answer!!(with slight calculation difference)

p mQ mC T= &

2

p m m

D

C u T4

=

2

4991 1 (0.0254) 4179 (24.6) 51622 W

= =

Tm

conv s m sq h A T for T constant= =l

i o

m i

o

T T 24.6T

360 280T nn360 304.6T

= =

lll

q 51853 W=

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Internal Flow Through Pipes And Ducts.pdf

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