Transport processes

7. Semester

Chemical Engineering

Civil Engineering

Course plan

1. Elementary Fluid Dynamics2. Fluid Kinematics3. Finite Control Volume Analysis4. Differential Analysis of Fluid Flow5. Viscous Flow and Turbulence6. Turbulent Boundary Layer Flow7. Principles of Heat Transfer8. Internal Forces Convection9. Unsteady Heat Transfer10. Boiling and Condensation11. Mass Transfer12. Porous Media Flow13. Non-Newtonian Flow

Today's lecture

• Internal Forced Convection– The thermal boundary layer

– Thermal entrance length

– Forced convection in pipes

Newtons law of cooling

• Calculation of convective heat transfer

– h depends on:• Geometry

• Fluid properties

• Flow properties

( ) [ ]conv s sQ hA T T W∞= −

Thermal boundary layer

• Thermal boundary thickness– 99% of the free stream temperature, that is when:

( )0.99s sT T T T∞− = −

Thermal boundary layer

• Turbulence enhances momentum and heat transfer

• Similar relations apply

Thermal boundary layer

• The thermal boundary layer does not necessarily develop at the same rate as the momentum b. layer

Molecular diffusivity of momentumPrMolecular diffusivity of heat

pckµν

α= = =

Solution to the energy equation for boundary layer flow

• We can get the similar solution as Blasius for the thermal b. layer

• For Pr=1, exactly the same

3 20.332w Uxµρτ = 1 30.332Prx

uhx

ρµ

∞=

Momentum transfer Heat transfer

Thermal boundary layer

• The heat transfer rate is proportional to the temperature gradient

0conv cond

y

Tq q ky =

∂= = −

∂

Nusselt number

• Heat transfer is by conduction when the fluid is motionless and by convection when the fluid moves

• Taking there ratios:

• For Nu=1 we have pure conduction

convq h T= ∆ condTq kL∆

=

Nu/

conv

cond

q h T hLq k T L k

∆= = =

∆

Entrance length

Entrance region

• Larger pressure drop

• Larger heat flux

• Entrance lengths– Laminar flow

– Turbulent flow

0.05RevelocityL D=

0.05Re Pr Prthermal velocityL D L= =

10velocity thermalL L D≈ ≈

Intermezzo

• Use the next 2 minutes to discus with the person next to you what kind of heat transfer enhancement mechanisms are employed in this plate heat exchanger

Boundary conditions

• For empirical relations we have two options:– Constant heat flux

– Constant surface temperature

General thermal analysis

• From energy-balance:

• From surface heat flux

– If ts is constant, qs must change

– if qs is constant, ts must change

( ) (W)p e iQ mC T T= −

( ) 2(W/m )x s mq h T T= −

Bulk mean temperature, Tm

• Note: the mean temperature of the fluid must change during heating or cooling

• The energy transported by the fluid through a cross section in actual flow must be equal to the energy that would be transported through the same cross section if the fluid were at a constant temperature Tm

– Use either inlet temperature or arithmetic average, (Te+Ti)/2, to determine fluid properties

– Use logarithmic mean temperature difference if constant surface temperature (will be show next)

Constant surface flux

• If qs= constant then:

thus

( ) (W)s s p e iQ q A mC T T= = −

s se i

p

q AT TmC

= +

Constant heat flux = Constant temperature gradient

perimeter constants

p

qdTdx mC

×= =

Constant temperature

• If Ts=constant then:

• The logarithmic mean temperature:

• The mean fluid temperature at exit:

( ) ln (W)s s m saveQ hA T T hA T= − = ∆

( ) ( )ln lni e

s e s i

T TTT T T T

−∆ =

− −

( ) ( )expe s s i s pT T T T hA mC= − − −

Laminar flow in pipes

• Fully developed flow, circular pipe, constant surface flux:

• Fully developed flow, circular pipe, constant surface flux:

4.36hDNuk

= =

3.66hDNuk

= =

Laminar flow in pipes – empirical correlations

• Its basically a matter of finding an appropriate formulae in a book– Constant heat flux / constant surface temperature

– Non-circular ducts/ circular pipes / flat plate

– Entrance region / fully developed

• It could look something like: (Sieder and Tate)

( )0.141

3av b

Nu re Pr Reavw

= =1.86 <2100h D DN N N Nk L

µµ

Turbulent flow in pipes – empirical correlations

• Again, mostly it is a matter of finding a usable expression..

– The most simple correlation is the Colburn equation:

– This can be improved by the Dittus-Boelter equation:

– This is only around ±25% accurate and more precise but increasingly more complex equations exists.

0.8 1 30.023Re PrNu =

0.80.023Re PrnNu = n = 0.3 for coolingn = 0.4 for heating

Example

• Calculate the heat loss from a oil pipe running through a icy lake.

1. Fluid properties evaluated at 20 °C initially2. Calculate the Reynolds number:

3. Calculate the entrance length:

4. Find suitable Nu correlation, calculate the Nusselt number:laminar flow, thermally developing flow, small temperature difference - use correlation by Edwards et al 1979:

Note, this is the average Nu for the pipe flow

3

Oil @ 20 C: 888 /0.145 /0.8 /1880 /

Pr 10400p

kg mk W m K

kg m sC J kg K

ρ

µ

== ⋅= ⋅= ⋅

=

888 2 0.3Re 666 Laminar0.8

UDρµ

⋅ ⋅= = = →

0.05Re Pr 0.05 666 10400 0.3 104thermalL D km= = ⋅ ⋅ ⋅ =

( )( ) 2 3

0.065 Re Pr3.66 37.3

1 0.04 Re Pr

D LhDNuk D L

= = + =+

Example cont.

5. Calculate the convective heat transfer coefficient:

6. Calculate the exit temperature

Note, this makes the bulk mean temperature Tm=(20+19.71)/2=19.85°C. This low temperature difference makes it acceptable to evaluate fluid properties at 20°C

7. Calculate the logarithmic temperature difference:

3

Oil @ 20 C: 888 /0.145 /0.8 /1880 /

Pr 10400p

kg mk W m K

kg m sC J kg K

ρ

µ

== ⋅= ⋅= ⋅

=

20.145Nu 37.3 18.0 /0.3

kh W m KD

= = = ⋅

( ) ( )exp 19.71e s s i s pT T T T hA mC C= − − − =

2188.5sA DL mπ= =2

4888 0.3 2 125.5 /c meanm A U kg sπρ= = ⋅ ⋅ =

( ) ( ) ( ) ( )ln20 19.71 19.86

ln ln 0 19.71 0 20i e

s e s i

T TT CT T T T

− −∆ = = = −

− − − −

Example cont.

8. Calculate the heat loss:

a) Calculate the pressure loss:

b) Calculate the required pump work:3

Oil @ 20 C: 888 /0.145 /0.8 /1880 /

Pr 10400p

kg mk W m K

kg m sC J kg K

ρ

µ

== ⋅= ⋅= ⋅

=

( )ln 18.0 188.5 19.85 67.4 WsQ hA T k= ∆ = ⋅ ⋅ − = −

2 2564 200 888 2 1.14 10 1.14

2 Re 0.3 2L UP f Pa barDρ ⋅

∆ = = = ⋅ =

16.1pumpm PW kWρ∆

= =

Excercises

• Time to wake up!