internal forced convection · 2018. 1. 30. · internal forced convection • hydrodynamic...

INTERNAL FORCED CONVECTION

• Hydrodynamic Considerations

• Thermal Considerations

• Overall Energy Balance

• Convection Correlations: Circular & Non-Circular TubesConcentric Tube Annulus

• Heat Transfer Enhancement

hydrodynamic entrance region

hydrodynamic entrance regionflow acceleration

u

x

r

fully developed region

Hydrodynamic Considerations

inertia force ~ pressure force~ viscous force

ro

hydrodynamically fully developed region

fully developed condition : constant pressure gradient



u

x

rorro

no acceleration (inertia force = 0)pressure force ~ viscous force

0, 0uvx∂

= =∂

0,dudt

→ = ( )u u r=

Characteristic Velocity

rro

x

2 0

2 ( , )or

mo

u u r x rdrr

= ⋅∫ 2

4mDπρ

=

( , )c

cAm u r x dAρ= ∫ 0

( , ) 2or u r x rdrρ π= ⋅∫2

m ou rρ π≡ ⋅ 2

4m c mu A u Dπρ ρ= =

mean velocity over the cross-sectional area

drr

u(r,x)

Reynolds number :

critical Reynolds number :

hydrodynamic entry length:

4Re mD

u D mD

ρμ π μ

= =

,Re 2,300D c ≈

fd,h fd,h

lam turb

0.05Re , 10 60D

x xD D

⎛ ⎞ ⎛ ⎞≈ ≤ ≤⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

Velocity Profile in Fully Developed Regionmomentum eq:

boundary conditions:

0 dp d durdx r dr dr

μ ⎛ ⎞= − + ⎜ ⎟⎝ ⎠

or constantdp d durdx r dr dr

μ ⎛ ⎞= =⎜ ⎟⎝ ⎠

0

( ) 0, 0or

duu rdr =

= =2

11

2du dp rr Cdr dxμ

⎛ ⎞= +⎜ ⎟⎝ ⎠

2

1 21( ) ln

4dp ru r C r Cdxμ

⎛ ⎞= + +⎜ ⎟⎝ ⎠

rro

2

1 210,

4ordpC C

dxμ⎛ ⎞= = − ⎜ ⎟⎝ ⎠

221( ) 14o

o

rdp ru rdx rμ

⎡ ⎤⎛ ⎞⎛ ⎞= − −⎢ ⎥⎜ ⎟⎜ ⎟⎝ ⎠ ⎢ ⎥⎝ ⎠⎣ ⎦

2 0

2 ( , )or

mo

u u r x rdrr

= ⋅∫2

8or dp

dxμ= −

2

2( ) 2 1mo

ru r ur

⎛ ⎞= −⎜ ⎟

⎝ ⎠

(0) 2 ,m cu u u= ≡2

2( ) 1co

ru r ur

⎛ ⎞= −⎜ ⎟

⎝ ⎠

Moody friction factor

friction coefficient

Friction Coefficient

o

sr r

ur

τ μ=

∂= −

∂422 m

mo o

uur r

μμ⎛ ⎞

= − ⋅ − =⎜ ⎟⎝ ⎠

2 / 2s

fm

Cuτ

ρ= 2

4 2m

o m

ur uμ

ρ= ⋅

2 2 2

8( / ) 64 64 4/ 2 / 2 Re

mf

m o m m D

u Ddp dx Df Cu r u D u

μ μρ ρ ρ

−≡ = = = =

16ReD

=

Ti


Thermal entry length : fd,t

lam

0.05Re PrD

xD

⎛ ⎞≈⎜ ⎟

⎝ ⎠

Thermal Considerations



u

x

rorro

rro

thermal entrance regionx

Ts

Mixed Mean Temperature

heat transfer coefficient :

dr

rro

x

u(r,x), T(r,x)

(bulk fluid temperature, mixing-cup temperature)

02or u cT rdrρ π⋅∫ 0

2or

mT u c rdrρ π≡ ⋅∫2

m m ocT u rρ π=

2 0

2 ( , )or

mo

u u r x rdrr

⎛ ⎞= ⋅⎜ ⎟

⎝ ⎠∫2 0

2( ) or

mm o

T x uTrdru r

= ∫

( )o

s s mr r

Tq k h T Tr =

∂′′ = = −∂

sq′′

Thermally Fully Developed Conditionexperimental observation: in the far downstream region, h =const. for a given fluid and tube diameter

= constant (not function of x)

= constant

Suppose alone( ) ( , ) ( )( ) ( )

s

s m

T x T r x rT x T x

φ−=

−

( ) ( , )( ) ( )

o

s

s m r r

T x T r xr T x T x

=

⎡ ⎤⎛ ⎞−∂⎢ ⎥⎜ ⎟∂ −⎝ ⎠⎣ ⎦

( )o

s s mr r

Tq h T T kr =

∂′′ = − =∂

)/

( ) ( )or r

s m

T rhk T x T x

=∂ ∂

→ = −−

)/

( ) ( )or r

s m

T r

T x T x=

∂ ∂= −

−

thermally fully developed condition( ) ( , ) 0( ) ( )

s

s m

T x T r xx T x T x⎡ ⎤−∂

=⎢ ⎥∂ −⎣ ⎦

( ) ( ) 0s s ms m s

dT dT dTT T T T Tdx x dx dx

∂⎛ ⎞ ⎛ ⎞− − − − − =⎜ ⎟ ⎜ ⎟∂⎝ ⎠ ⎝ ⎠

s s s s m

s m s m

dT T T dT T T dTTx dx T T dx T T dx

− −∂= − +

∂ − −

Constant surface heat flux condition

Constant surface temperature condition

s s s s m

s m s m

dT T T dT T T dTTx dx T T dx T T dx

− −∂= − +

∂ − −

const. ( )s s mq h T T′′ = − = const.s mT T→ − =

0s mdT dTdx dx

− = s mdT dTdx dx

→ =

s m

s m

T T dTTx T T dx

−∂=

∂→

−const. sT =

s mdT dTTx dx dx

∂=→ =

∂

Example 8.1

Find: Nusselt number at the prescribed locationAssumptions:Incompressible, constant property flow

Flow

Ts

ro

slug flow: liquid metal, Pr << 1

rro

Velocity profile

1( )u r C=

Temperature profile

rro

( )22( ) 1 /s oT r T C r r⎡ ⎤− = −⎣ ⎦

( )21 2( ) , ( ) 1 /s ou r C T r T C r r⎡ ⎤= − = −⎣ ⎦

, ,NuDs

s m

qh hDTk T

= =−′′

2 0

2 o

m

r

m o

uTrdrTu r

= ∫ ( ){ }21 22 0

2 1 /or

s om o

C T C r r rdru r

⎡ ⎤= + −⎣ ⎦∫( ){ }2

22 0

2 1 /or

s oo

T C r r rdrr

⎡ ⎤= + −⎣ ⎦∫2

2 22 2 22

22 2 4 2o

s o o so

r C C CT r r Tr⎛ ⎞

= + − = +⎜ ⎟⎝ ⎠

or rs

Tkr

q=

′ ∂=

∂′ 2 22

oo r r

rkCr

=

= − 22o

kCr

= −

2

2

2 / 4/ 2

s o

s m o

h q C k r kT T C r

′′ −= = =

− −Thus Nu 8D

hDk

= =

or rs

Tkr

q=

′ ∂=

∂′

x dx

perimeter: P = πD

Overall Energy Balance

Dinq

, mm Toutq

sq′′

p mmc T ( )p m p mdmc T mc T dxdx

= +

( ) convs p m

dqdq P mc Tdx dx

′′ = ≡

sq Pdx′′+

( )m ss m

p p

dT q P P h T Tdx mc mc

′′= = −

( )conv

o o

p mi idq d mc T=∫ ∫

or ( )conv , ,p m o m iq mc T T= −

x dx

Dinq

, mm Toutq

sq′′


= constant in fully developed region

( ) ,m ss m

p p

dT q P P h T Tdx mc mc

′′= = − const.sq′′ =

sm

p

q PdT dxmc′′

=

0 0

x xs

mp

q PdT dxmc′′

=∫ ∫,( ) s

m m ip

q PT x T xmc′′

= +

( ) ( ) ss m

qT x T xh′′

= +

( ) ( ) ss m

qT x T xh′′

− =


( )ms m

p

dT P h T Tdx mc

= − or m

m s p

dT P hdxT T mc

= −−

0 0

x xm

m s p

dT P hdxT T mc

= −−∫ ∫

( )0 0

lnxx

m sp

PT T hdxmc

⎡ ⎤− = −⎣ ⎦ ∫

0

1 x

xp p

Px Pxhdx hmc x mc

= − = −∫

,

( ) exps mx

s m i i p

T T x T Px hT T T mc

⎛ ⎞− Δ= = −⎜ ⎟⎜ ⎟− Δ ⎝ ⎠

For a tube of length L

,

,

s m o o

s m i i

T T TT T T− Δ

=− Δ

exp Lp

PL hmc

⎛ ⎞= −⎜ ⎟⎜ ⎟

⎝ ⎠

Log Mean Temperature Difference (LMTD)

Let , ,, ,i s m i o s m oT T T T T TΔ = − Δ = −

then ( )conv p i oq mc T T= Δ − Δ

,

,

exps m o oL

s m i i p

T T T PL hT T T mc

⎛ ⎞− Δ= = −⎜ ⎟⎜ ⎟− Δ ⎝ ⎠

lnln( / )

o LL p

p i i o

TPL PLhh mcmc T T T

Δ− = → =

Δ Δ Δ

( )conv , ,p m o m iq mc T T= −

( ) ( ), ,p s m i s m omc T T T T⎡ ⎤= − − −⎣ ⎦

( )conv p i oq mc T T= Δ − Δ

ln( / )L

pi o

PLhmcT T

=Δ Δ

( ) ( )conv ln /L

i oi o

PLhq T TT T

= Δ − ΔΔ Δ

( ) lmln /o i

L L so i

T TPLh h A TT T

Δ − Δ= ≡ Δ

Δ Δ

lm ln( / )o i

o i

T TTT T

Δ − ΔΔ =

Δ Δ

WaterD = 50 mm

L = 6 m

Example 8.3

Find: Average convection heat transfer coefficient

0.25 kg/sm =

, 15 Cm iT = °100 CsT = ° , 57 Cm oT = °

WaterD = 50 mm

L = 6 m

0.25 kg/sm =

, 15 Cm iT = °100 CsT = ° , 57 Cm oT = °

( )conv , ,p m o m iq mc T T= − lmshA T= Δ( ), ,

lm

p m o m i

s Tmc T

hT

A−

Δ→ =

( ) ( )( ) ( )

, ,

, ,lm 61.6 C

ln /s m o s m i

s m o s m i

T T T T

T T T TT

− − −= = °

⎡ ⎤− −⎣ ⎦Δ

( ) 20.25 4178 57 15756 W/m K

0.05 6 61.6h

π× × −

= =× × ×

in the cylindrical coordinate system (r,x)

for fully developed flow

Laminar Flow in Fully Developed RegionConvection Correlations

2

2

1T T T Tu v rx r r r r x

α⎡ ⎤∂ ∂ ∂ ∂ ∂⎛ ⎞+ = +⎜ ⎟⎢ ⎥∂ ∂ ∂ ∂ ∂⎝ ⎠⎣ ⎦

2

2

10 T T Tv u rx r r r x

α⎡ ⎤∂ ∂ ∂ ∂⎛ ⎞= → = +⎜ ⎟⎢ ⎥∂ ∂ ∂ ∂⎝ ⎠⎣ ⎦2

2( ) 2 1mo

ru r ur

⎛ ⎞= −⎜ ⎟

⎝ ⎠

Scaling , , ,u T r x2

2~ , ~ , ~ m om

o

u rT T T Tu r u xx r r r x r

α αα

∂ ∂ ∂ Δ Δ⎛ ⎞⎜ ⎟∂ ∂ ∂⎝ ⎠

2

/ ,Re Pr

o

o

m o o m o r

x rx x vxu r r v u rα α∗ = = =

,

, ,s

s m i m o

T T u ru rT T u r

θ ∗ ∗−= = =

−

Then,2 2

2 2 2 2

1 12 Re Pr

or

ux x r r rθ θ θ θ∗

∗ ∗ ∗ ∗ ∗

∂ ∂ ∂ ∂= + +

∂ ∂ ∂ ∂

When Re Pr Pe 100,or

= >T Tu rx r r r

α∂ ∂ ∂⎛ ⎞= ⎜ ⎟∂ ∂ ∂⎝ ⎠


boundary conditions

s mdT dTTx dx dx

∂= =

∂

Thus, mdT Tu rdx r r r

α ∂ ∂⎛ ⎞= ⎜ ⎟∂ ∂⎝ ⎠

or 2

22 1 mm

o

dTr Tu rr dx r r r

α⎛ ⎞ ∂ ∂⎛ ⎞− =⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠⎝ ⎠

( , ) ( ),o sT r x T x=0

0r

Tr =

∂=

∂o

sr r

Tq kr =

⎛ ⎞∂′′=⎜ ⎟⎜ ⎟∂⎝ ⎠

,T Tu rx r r r

α∂ ∂ ∂⎛ ⎞= ⎜ ⎟∂ ∂ ∂⎝ ⎠

from overall energy balance :

Integrating energy equation with boundary conditions

4m s s

p m p

dT q P qdx mc u Dcρ

′′ ′′= = = const.

4 22

2

2 3 1( , ) ( )16 16 4

m ms s

s

u dT r rT r x T x rdx rα

⎛ ⎞= − + −⎜ ⎟

⎝ ⎠

22 0

22 11( ) ( )96

or m mm s o

o m

u dTT x uTrdr T x rr u dxα

= = −∫11( )24

s os

q rT xk′′

= −

( ) 1124

s os s m

q rq h T T hk′′

′′ = − = ⋅

24 4811 11o

k khr D

= =

4811

Nu 4.364DhDk

= = =


,s m

s m

T T dTTx T T dx

−∂=

∂ −s m

s m

T T dT Tu rT T dx r r r

α− ∂ ∂⎛ ⎞= ⎜ ⎟− ∂ ∂⎝ ⎠2

22 1 s mm

o s m

T T dTr Tu rr T T dx r r r

α⎛ ⎞ − ∂ ∂⎛ ⎞− =⎜ ⎟ ⎜ ⎟− ∂ ∂⎝ ⎠⎝ ⎠

Assume2

20

n

sn

ns m o

T T rCT T r

∞

=

⎛ ⎞−= ⎜ ⎟− ⎝ ⎠∑

( )2 20 0

0 2 2 2 4 2 221, 1.828397,4 (2 )n n nC C C C C

nλ λ

− −= = − = − = −

0 2.704364λ =

20Nu

23.657D

hDk

λ= = =

Example 8.4

absorber tube

concentrator

Conditions:

Find: 1) Length of tube L to achieve required heating2) Surface temperature Ts,o at the outlet section, x = L

Assumptions:1) Incompressible flow with constant properties2) Fully developed conditions at tube outlet

water22000 W/msq′′ =

0.01 kg/sm =

60 mmD =

, 20 Cm iT = °

, 80 Cm oT = °

( )s Lq Dπ′′=

1) From energy balance

2) surface temperature at the outlet

→ laminar flow

Nu 4.36DDk

h= =fully developed condition

24.36 48.7 W/m KDk

h = =

Conditions:22000 W/msq′′ =

0.01 kg/sm =60 mmD =

, 20 Cm iT = °

, 80 Cm oT = °

( )conv , ,p m o m iq mc T T= −

( ), , 6.65 mp m o m i

s

mc T TDq

Lπ

−= =

′′

( ), ,s os m oTq Th′′ = − ,,s

os o mqT Th′′

→ = +

6

4 4 0.01Re 6030.06 352 10D

mDπ μ π −

×= = =

× × ×

, ,2000 80 121 C48.7s o

sm oT

hT q′′

= + = + = °

Laminar Flow in the Entry RegionThermal entry length

Ts = constant,

properties at

( )( ) 2 / 3

0.0668 / Re PrNu 3.66

1 0.04 / Re PrD

D

D

D L

D L= +

⎡ ⎤+ ⎣ ⎦

lmNu ,D shD q hA Tk

= = Δ

( ), , / 2m m i m oT T T= +

Combined entry length(hydrodynamic + thermal)

properties at

Ts = constant

0.141/ 3Re PrNu 1.86/D

DsL Dμμ

⎛ ⎞⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠

0.48 Pr 16,700

0.0044 9.75s

μμ

< <⎡ ⎤⎢ ⎥⎛ ⎞⎢ ⎥< <⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

( ), , / 2m m i m oT T T= +

Gr: Gratez number

Turbulent flow in circular tubesFully developed region

• Dittus-Boelter equation (smooth wall)

moderate temperature differenceproperties at Tm

4 / 5Nu 0.023Re PrnD D=

( ) ( )0.4 , 0.3 s m s mn T T n T T= > = <

0.7 Pr 160, Re 10,000, 10DLD

≤ ≤ ≥ ≥

• Sieder & Tate (1936)

properties at Tm

large temperature difference0.14

4 / 5 1/ 3Nu 0.027Re PrD Ds

μμ

⎛ ⎞= ⎜ ⎟

⎝ ⎠

0.7 Pr 16,700, Re 10,000, 10DLD

≤ ≤ ≥ ≥

• Petukhov (1970)

f from Moody diagram

properties at Tm

( )( ) ( )1/ 2 2 / 3

/ 8 Re PrNu

1.07 12.7 / 8 Pr 1D

D

f

f=

+ −

4 60.5 Pr 2,000, 10 Re 5 10D< ≤ < < ×

• Gnielinski (1976)

properties at Tm

( )( )( ) ( )1/ 2 2 / 3

/ 8 Re 1000 PrNu

1 12.7 / 8 Pr 1D

D

f

f

−=

+ −60.5 Pr 2,000, 3,000 Re 5 10D< ≤ < < ×

Entry regionsince

For short tubes :

,fdNu NuD D= ( )fd10 / 60x D≤ ≤

,fd

Nu 1Nu ( / )

Dm

D

Cx D

= +

Liquid metals

constant

Ts = constant

fully developed turbulent flow0.827Nu 4.82 0.0185Pe ,D D= + sq′′ =

3 5

2 4

3.6 10 Re 9.05 1010 Pe 10

D

D

⎡ ⎤× < < ×⎢ ⎥< <⎣ ⎦

0.8Nu 5.0 0.025Pe ,D D= +

Hot air D = 0.15 m

L = 5 m

Cold ambient airx

Assumptions:1) Steady-state, constant properties, ideal gas behavior2) Uniform convection coefficient at outer surface of duct

Example 8.6

0.05 kg/sm =

,0 103 CmT = °

20 C, 6 W/m KoT h∞ = ° =

, 77 Cm LT = °

Find: 1) Heat loss from the duct over the length L, q[W]2) Heat flux and surface temperature Ts,L at x = L( )sq L′′

1) Energy balance for the entire duct :

2) Heat flux and surface temperature at x = L

Hot air D = 0.15 m

L = 5 m

Cold ambient airx

0.05 kg/sm =

,0 103 CmT = °

20 C, 6 W/m KoT h∞ = ° =

, 77 Cm LT = °

( ), ,0p m L mTq mc T= − ( )0.05 1010 77 103 1313 W= × × − = −

( )sq L′′,m LT ,s LT T∞

1/ x Lh = 1/ oh,

1 / /(

1)

x L

m L

os

T TL

hhq

=

∞−=

+′′

20 C, 6 W/m KoT h∞ = ° =

, 77 C, m L x LT h == °,s LT

( ) ( ), , ,( ) m L s Lx sL os Lq TL T Th h T= ∞′ =′ − = −or

turbulent flow


7

4 4 0.05Re 20,4040.15 208 10D

mDπ μ π −

×= = =

× × ×

/ 5 / 0.15 33.3 10L D = = >

4/ 5 0.3Nu 0.023Re Prx LD D

h Dk== = 57.9=

20.03Nu 57.9 11.6 W/m K0.15x L D D

h k= = = × =

Hence, , 277 0 304.5 W/m1/ 1/ 1 / 11.6 1/ 6.0

( ) m L

x L os

T Th

qh

L ∞

=

− −=

+′ = =

+′

( ) ( ), , ,( )s x L m L s L s LoT Tq L h T h T= ∞′′ = − = −

,,( ) 304.577 50.7 C

11.6s

m Lx L

s L TT q Lh =

′′= − = − = °

Convection Correlations: Noncircular Tube

hydraulic diameter :

Ac : flow cross-sectional area

P : wetted perimeter

4 ch

ADP

≡

Nusselt numbers and friction factors for fully developed laminar flow in tubes of different cross section

Concentric Tube Annulusinner wall :

outer wall :

Nusselt numbers

( ),i i s i mq h T T′′ = −

( ),o o s o mq h T T′′ = −

Nu , Nui h o hi o

h D h Dk k

≡ ≡

( )( )2 24 / 44 o ich o i

o i

D DAD D DP D D

π

π π

−= = = −

+

( ) ( )Nu NuNu , Nu

1 / 1 /ii oo

i oo i i i o oq q q qθ θ∗ ∗= =′′ ′′ ′′ ′′− −

For fully developed laminar flow1) One surface insulated and the

other at constant temperature

2) Uniform heat flux maintained at both surfaces

( ) ( )Nu NuNu , Nu

1 / 1 /ii oo

i oo i i i o oq q q qθ θ∗ ∗= =′′ ′′ ′′ ′′− −

Heat Transfer EnhancementInternal flow heat transfer enhancement schemes

Coil-spring wire insert Twisted tape insert

Longitudinal fins Helical ribs

Helically coiled tube and secondary flow

internal forced convection · 2018. 1. 30. · internal forced convection • hydrodynamic...

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