Supercritical Fluid Process Lab

21CONVECTIVE

HEAT-TRANSFER COEFFICIENT

Introduction

In most transport processes heat transfer in fluids is accompanied by some form of fluid motionso that the heat transfer does not occur by conduction alone.

Forced convection: fluid motion arises principally froma pressure gradient caused by a pump or blower

Natural convection: fluid motion arises only fromdensity differences associated with the temperature field

Supercritical Fluid Process Lab

Introduction

Whether the heat-transfer mechanism is natural or forced convection, the fluid motion can be described by the equations of fluid mechanics.

@ low velocities → the flow is laminar throughout the system@ high velocities → laminar near the heating surface

+ turbulent some distance away

Although fluid velocities in natural convection are usually lower than in forced convection, it is incorrect to think of natural convection as causing only laminar flow; turbulence dose occur when critical Reynolds number for the system is exceeded.


Heat transfer between a fluid and a flat platewhen the fluid is flowing parallel to the plate.

(Fig. 21-1)

All problems of convective heat transfer can be expressed in terms of the differential mass, energy, and momentum balances. However, the mathematical difficulties connected with integration of these simultaneous nonlinear partial differential equations are such that analytical solutions exist only for simplified cases.

One of the problems simplest to deal with analytically is that of


Analytical Solution

Heat transfer between a fluid and a flat platewhen the fluid is flowing parallel to the plate.

(Fig. 21-1)


x0 x1 x2 x3tS tS tS tS

t0

t0

t0t0

x

y

flat plate

Edge of thermalboundary layer

Fig. 21-1Development of a thermal boundary layer for flow over a flat plate

Direction of flow

t0

t0 t0>ts

Leading edge of the plate


Temperature profiles in a developing thermal boundary layer on a flat plate

y, distance normal to flat plate

t, flu

id te

mpe

ratu

re

t0

ts

x1 x2 x3

x0

x0 x1 x2 x3tS tS tS tS

t0

t0

t0t0

x

y

flat plate


Supercritical Fluid Process LabSupercritical Fluid Process Lab

Heat transfer systemHeat is transferred between a fluid and the wall of a pipe


Thermal-entrance length

Direction of flow


t0

t0t0

t0

t0

t0

tS tS tS tS tS tS tS

tS tS tS tS tS tS tS

Temperature profile become flatter.


Temperature profiles near the entrance of a pipe

ri rir=0

ts

t0 x0x1x2x3

Figure 21-4


Temperature profiles vs Velocity profiles

The temperature profile at a point in a flow system is influenced by the velocity profile.

The velocity profile is influenced by the temperature profile. (The velocity profile of an isothermal system may differ substantially from the velocity of a system in which heat is being transferred.)


(8-11)

Differential energy balance

Navier-Stokes equation


Simplification

The non-isothermal velocity profile can be used in solving the differential momentum and energy balances for a non-isothermal system.

This simplification may introduce a serious error when the viscosity of the fluid is strongly dependent on temperature.

Frequently the temperature gradient is greatest near the wall, and it is in this region that the velocity gradient is also greatest. The effect of temperature on the viscosity of the fluid at the wall may therefore have a pronounced effect on both the velocity and temperature profiles of the system.

ts

t0

t0


m


Individual heat transfer coefficient, h

h is a function of

the properties of the fluid,the geometry and roughness of the surface, andthe flow pattern of the fluid.


Evaluation of individual heat transfer coefficient, h

Several methods are available for evaluating hLaminar flow systems→ Analytical methods

Turbulent systems → Integral methods→ Mixing-length theory→ Dimensional analysis


Evaluation of individual heat transfer coefficient, h

dAtthdydtkdAdq s

y

)( 00

−=⎟⎟⎠

⎞⎜⎜⎝

⎛=

=

[ ]0

0

00

)/()(

== ⎭⎬⎫

⎩⎨⎧ −−

=⎟⎟⎠

⎞⎜⎜⎝

⎛−

=y

ss

ys dyttttdk

dydt

ttkh

Velocity of flow past the heated surface ↑Temperature gradient at the wall ↑Heat transfer coefficient, h ↑


Individual film coefficient, h

exkhΔ

=

Fully developedTurbulent flow

Bufferzone

Laminarsublayer

t

r

If the resistance to heat flow is through of as existing only in a laminar film, h is k/Δxe where Δxe is the equivalent thickness of a stationary film just thick enough to offer the resistance corresponding to observed value of h. Since there is often an appreciable resistance in the turbulent core and since the stationary film has not even an approximate physical counterpart in laminar flow, boiling and radiation, we shall refer h as an individual heat transfer coefficient.

exΔFictitious film

∫=A

xav

b tdAuAu

t 1


Definition of h


Example 21-1

The analysis of unsteady-state heating or cooling of solid objects can

be often be simplified by assuming that the resistance to heat

conduction inside the solid is negligible compared to the convective

heat-transfer resistance in the surrounding fluid.

The circumstances necessary to justify this assumption are a high

thermal conductivity for the solid and a low convective heat transfer

coefficient in the adjacent fluid. A metallic object being heated or

cooled in air often constitute such a system.


Example 21-1

To illustrate, we shall calculate the time required for a mercury

thermometer initially at 70oF, placed in an oven at 400oF, to reach

279oF. The thermometer bulb has a diameter of 0.24 in and will be

assumed to be adequately represented as an infinitely long cylinder

with negligible resistance to heat transfer either in the glass which is

very thin or in the mercury, which has an adequately high thermal

conductivity [k=5.6 Btu/(h)(ft)(oF) at 140oF]. The mean convective

heat-transfer coefficient between the outside of the thermometer and

the air in the oven will be taken as h=2 Btu/(h)(ft2)(oF).

Supercritical Fluid Process LabFor an infinite cylinder, A/V=(πDL)/(πD2/4)= 4/D

= 4.24 min


For an infinite cylinder, L=V/A=D/4

FoBieY ⋅−=

τθ

−= eY

min24.4=τ


θ


τ

θ

Temperature response, t =time constant

50 20 10 5 1 0.5


50=τ20

10510.5

min,θ

t

Temperature response, t, with time


BiotBiot Number (Bi)Number (Bi)

skhLBi =

resistance mallayer therboundary the solid a of resistance thermalinternal the

=Bi


Fourier Number (Fourier Number (FoFo))

2LFo αθ

=

solid ain storageenergy thermalrate theconductionheat of rate the

=OF


Reynolds Number (Re)Reynolds Number (Re)

ν= ∞LuRe

forces viscoustheforces inertia theRe =


Prandtl Number (Pr)Prandtl Number (Pr)

αν

=μ

=k

C pPr

ydiffusivit thermalthemomentummolecular thePr =


overalltUAq Δ=

individualthAq Δ=

Individual heat transfer coefficient, hOverall heat transfer coefficient, U


hi

h0


0054

,43

,32

21

1

1

Ahqtt

Akrqtt

Akrqtt

Ahqtt

lmcc

c

lmbb

b

ii

=−

Δ=−

Δ=−

=−

Individual temperature drops


Overall temperature drops

⎟⎟⎠

⎞⎜⎜⎝

⎛+

Δ+

Δ+=

+Δ

+Δ

+=−

−=Δ

00,,

00,,51

51

11

11

AhAkr

Akr

Ahq

Ahq

Akrq

Akrq

Ahqtt

ttt

lmcc

c

lmbb

b

ii

lmcc

c

lmbb

b

ii

overall

⎟⎟⎠

⎞⎜⎜⎝

⎛+

Δ+

Δ+

−=

00,,

51

11AhAk

rAkr

Ah

ttq

lmcc

c

lmbb

b

ii

Rtq overall

ΣΔ

=


⎟⎟⎠

⎞⎜⎜⎝

⎛+

Δ+

Δ+

Δ=

0,

0

,

00

0

1hAk

rAAk

rAAh

AtAq

lmcc

c

lmbb

b

ii

overall

Overall coefficients of heat transferbased on the outside area U0

0,

0

,

00

0

11hAk

rAAk

rAAh

AU lmcc

c

lmbb

b

ii

+Δ

+Δ

+=

overalltAUq Δ= 00


⎟⎟⎠

⎞⎜⎜⎝

⎛+

Δ+

Δ+

Δ=

00,,

1Ah

AAk

rAAk

rAh

tAqi

lmcc

ci

lmbb

bi

i

overalli

Overall coefficients of heat transferbased on the inside area U0

i

i

lmcc

ci

lmbb

bi

ii AhA

AkrA

AkrA

hU 0,,

11+

Δ+

Δ+=

overallii tAUq Δ=


Overall coefficients of heat transfer

overallii

overall

tAUqtAUq

Δ=Δ= 00


Rtq overall

ΣΔ

=


ii AUAUR 11

00

==Σ

ii AUAU =00


Approximation of overall coefficients of heat transferbased on the inside area Ui

i

i

lmcc

ci

lmbb

bi

ii AhA

AkrA

AkrA

hU 0,,

11+

Δ+

Δ+=

0

111hk

rkr

hU c

c

b

b

ii

+Δ

+Δ

+=

Approximation0

1,,1hk

rkr

h c

c

b

b

i

ΔΔ>>


Overall coefficients of heat transferof the flat parallel walls

00

1111hk

xkx

hUU c

c

b

b

ii

+Δ

+Δ

+==



U, Btu/(h)(ft2)(oF)

Stabilizer reflux condenser 94Oil pre-heater 108Reboiler (condensing steam to re-boiling water) 300-800Air heater (molten salt to air) 6Steam-jacketed vessel evaporating milk 500

Table 21-2

Fouling in heat exchanger



Fouling Coefficients

Fouling: deposits on heat transfer surface

- porous deposits (mud, soot, vegetable)

- hard scale (boiler or evaporator)

- coke (oil heater in a refinery)

Heat thermal conductivityTransfer =Coefficient thickness of scale

Sandblasting, pneumatic cleaningtools, chemical cleaning

Thermal conductivity itself may be highBut effective thermal conductivity may bealmost as low as that of the fluid.

Steam (air, hot water) blowing



olmc

o

c

c

lmb

o

b

b

idi

o

ii

oo

scaled

hAA

kr

AA

kr

AhA

AhAU

tAhq

11

,,

+Δ

+Δ

++=

Δ=

Outsidefluid Outside

fluid

Insidefluid

Outsidefluid

kc kb

Insidefluid

fouling

ho

Outsidefluid

hi

hdi



odolmc

o

c

c

lmb

o

b

b

idi

o

ii

oo

scaled

hhAA

kr

AA

kr

AhA

AhAU

tAhq

111

,,

++Δ

+Δ

++=

Δ=


fluid

Insidefluid

Outsidefluid

kc kb

Insidefluid

Insidefouling

ho

Outsidefluid

hi

hdi

outsidefouling

hdo


Fouling coefficients

hd, Btu/(h)(ft2)(oF)

Overhead vapors from crude-oil distillation 1000Dry crude oil (300-100oF):

Velocity under 2 ft/sec 250Velocity 2-4 ft/sec 330Velocity over 4 ft/sec 500

Air 500Steam (non-oil-bearing) 200Water, Great Lakes, over 125oF 500

Table 21-3

Tubular Exchanger Manufacturers Association, New York, 1949


Example 21-2

A reflux condenser contains ¾ in. 16-gauge copper tubes in which cooling water circulates. Hydrocarbon vapors condense on the exterior surfaces of the tubes. Find the overall heat-transfer coefficient Uo. The inside convective coefficient can be taken as 4500 W/m2·K, and the outside coefficient as 1500 W/m2·K.

Approximately fouling coefficients from Table 21-3 are

hd0 = 5700 W/m2·Khdi = 2840 W/m2·K

Overhead vapors from crude-oil distillation 1000

Water, Great Lakes, over 125oF 500

Btu/(h)(ft2)(oF)



fluid

Coolingwater

Outsidefluid

k

Insidefluid

insidefouling

ho=1500

Condensingvapor

hi =4500

hdi =2840outsidefouling

hdo =5700

Example 21-2 Find Uo

¾” 16-gauge copper tuberi=0.0157ro=0.0191

unit: W/m2·K



odolmc

o

c

c

lmb

o

b

b

idi

o

ii

oo

hhAA

kr

AA

kr

AhA

AhAU 11

1

,,

++Δ

+Δ

++=


fluid

Insidefluid

Outsidefluid

kc kb

Insidefluid

Insidefouling

ho

Outsidefluid

hi

hdi

outsidefouling

hdo


Example 21-2

00067.01500

11,00018.05700

11

0000047.0)0175.0)(380(

)0191.0)(00165.0(

00043.0)0157.0)(2840(

0191.0

00027.0)0157.0)(4500(

0191.0

0175.0

0157.00191.0ln

0157.00191.0

ln

0191.0,0157.0,/380

/2840678.5500

/5700678.51000

/1500,/4500

0

0

0

0

0

0

0

2

2

20

2

0

0

====

==⋅Δ

==

==

=−

=−

=

==⋅=

⋅=×=

⋅=×=

⋅=⋅=

hh

rkrr

rhrrh

r

m

rrrrr

mrmrKmWk

KmWh

KmWh

KmWhKmWh

d

lmb

idi

ii

i

il

i

d

d

i

m

i

0

0000 11

1

0hhrk

rrrh

rrh

rU

dlmbidii i

++⋅Δ

++=

hi

hdihdo

k

ho


Example 21-2

KmWU

AUAU

KmW

hhrkrr

rhr

rhrU

i

ooii

dlmbidii i

⋅==

=

⋅==

++++=

++⋅Δ

++=

2

2

0

0000

/785)0157.0/(()0191.0)(645(

/64500155.0

1

00067.000018.00000047.000043.000027.01

111

0

hi

hdihdo

k

ho

Resistance of metal wall is negligible

Fouling Resistance is significant~39%


Heat Exchangers


Shell–and–Tube Heat Exchangers


Heat Exchangers


Crossflow heat exchangers

(a) Finned with both fluid unmixed(b) Unfinned with one fluid mixed and the other unmixed

(a) (b)


Compact heat exchangers


Double-pipe heat exchanger



HotFluid

InWh lb/hr

HotFluidOut

Wh lb/h

Cold Fluid inWc lb/hr

Cold Fluid outWc lb/hr

Differential section

“1” “2”

tc

tc th

The choice of path for each fluid depends on considerations such as

corrosion, plugging, fluid pressure, and permissible

pressure drop.



t

Δt1

t Δt

Δt2

Δt= Δt1= Δt2= constant



t t



HotFluid InWh lb/hr

HotFluid OutWh lb/h




“1” “2”

tc

tc th

∫ ∫=Δ

q A

overalldA

tUdq

0 00

0

0

(21-23)

The determination of the required heat transfer area is one of the principal objective in the design of heat exchangers. For a differential segment of the exchanger for which the outside tube area is dA0,

ii

overall

overall

tdAUtUdAtUdq

GenerallytAUq

Δ=Δ←Δ=

Δ=

varingare&,

000

00



∫ ∫=Δ

q A

overalldA

tUdq

0 00

0

0

The overall coefficient of heat transfer U0

Outside heat transfer area, A0

The difference between the mixing-cup temperatures of the hot and cold fluids, Δt

Under steady conditions, the mixing-cup temperature of the hot and cold fluids in a heat exchanger are assumed to be fixed at any cross section normal to the flow. The overall temperature difference is Δt = th - tc.


Countercurrent vs. Concurrent

HotFluid InWh lb/hr

HotFluid OutWh lb/h



“1” “2”

HotFluid InWh lb/hr

HotFluid OutWh lb/h



“1” “2”

Countercurrent

Concurrent(Parallel-flow)


Mix

ing-

cup

tem

pera

ture

Hot fluid, th

Cold fluid, tc

1tΔ

2t

“1” “2”Length

Δ

Mix

ing-

cup

tem

pera

ture

ch ttt −=Δ

Hot fluid, th

Cold fluid, tc

1t

“1” “2”q

Δ2tΔ

Double-pipe heat exchangerCountercurrent

ch ttt −=Δ

dqtd )(Δ

If Cp=constant


Heat balance for the differential section

HotFluid

InWh lb/hr

HotFluidOut

Wh lb/h




“1” “2”

tc

tc th

00 )( dAttUdtCwdtCwdq chhphhcpcc −===

dA0

dq


Double-pipe heat exchangerM

ixin

g-cu

p te

mpe

ratu

re

ch ttt −=Δ

Hot fluid, th

Cold fluid, tc

“1” “2”q

1tΔ2tΔ

dqtd )(Δ

If Cp=constant

qtt

dqtd 12)( Δ−Δ=

Δ

The difference between the mixing cup temperatures, Δt, is also linear w.r.t. q.


Heat Balance in the differential section of the exchanger

qtt

dqtd

tdAUdAttUdtCwdtCwdq chhphhcpcc

12

0000

)(

)(

Δ−Δ=

Δ

Δ=−=== (21-25)

(21-26)


Heat Balance in the differential section of the exchanger

( )⎥⎦⎤

⎢⎣

⎡ΔΔΔ−Δ

=

Δ−Δ=

ΔΔ

Δ−Δ=

ΔΔ

=Δ

∫∫Δ

Δ

12

12

012

12

/ln

)(

)()(

2

1

ttttAUq

dAq

tttUtd

qtt

tdAUtd

dqtd

oo

A

o

t

to

oo

o

If Uo=constant

(21-27)

(21-28)

(21-29)LMTDLogarithmicMeanTemperatureDifference

00 tdAUdq Δ=


Heat Exchange in double pipe heat exchanger

( ) hphhcpccoo tCwtCwtt

ttAUq Δ=Δ=⎥⎦

⎤⎢⎣

⎡ΔΔΔ−Δ

=12

12

/ln


Example 21-3

Crude oilWc=2000 lb/hr90oF

Kerosene450oF

If the temperature of approach (minimum temperature difference between fluids) is 20oF, determine the Ao, wh for (a) concurrent flow and (b) countercurrent flow.

wh

countercurrent flow

“1” “2”

Overall coefficient Uo=80 Btu/(h)(ft2)(oF)Specific heat of crude oil = 0.56 Btu/(ib)(oF)Specific heat of kerosene = 0.60 Btu/(ib)(oF)

200oF


Example 21-3 (a) Concurrent flow

The temperature distribution

450

90

200220

The approach=20oF

2

12

12

1.13

)20/360ln(2036080

000,123)/ln(

/891)220450)(60.0(

000,123

/000,123)90200)(56.0)(2000(

ft

ttttU

qA

hlbtC

qw

hBtutCwq

o

o

hphh

cpcc

=

⎥⎦

⎤⎢⎣

⎡ −=

⎥⎦

⎤⎢⎣

⎡ΔΔΔ−Δ

=

=−

=Δ

=

=−=Δ=Total heat load:

3601 =Δt

202 =Δt

Crude oil

kerosene


Example 21-3 (b) Countercurrent flow

2

12

12

9.16

)20/250ln(2025080

000,123)/ln(

/603)110450)(60.0(

000,123

/000,123)90200)(56.0)(2000(

ft

ttttU

qA

hlbtC

qw

hBtutCwq

o

o

hphh

cpcc

=

⎥⎦

⎤⎢⎣

⎡ −=

⎥⎦

⎤⎢⎣

⎡ΔΔΔ−Δ

=

=−

=Δ

=

=−=Δ=

The temperature distribution

Total heat load:

11090

200

450

The approach=20oF

201 =Δt 2502 =Δtkerosene

Crude oil


If Uo is a function of temperaturetbaU o Δ+=

12

21

1

1

2

2

120

12

ln1

ln1

ln1

)ln(1ln1

)(//1)()(

)()()(

2

1

2

1

2

1

2

1

2

1

2

1

tUtU

a

ttba

tbat

a

tbat

a

tbaa

ta

tdtba

abta

ttbatd

Aq

ttttba

tddAq

tttUtd

o

o

t

t

t

t

t

t

t

t

o

t

t

A

o

t

to

o

ΔΔ

=

⎟⎟⎠

⎞⎜⎜⎝

⎛ΔΔ+

⎟⎟⎠

⎞⎜⎜⎝

⎛Δ+

Δ=

⎥⎦⎤

⎢⎣⎡

Δ+Δ

=

⎥⎦⎤

⎢⎣⎡ Δ+−Δ=

Δ⎟⎠⎞

⎜⎝⎛

Δ+−

+Δ

=ΔΔ+

Δ

Δ−Δ=

ΔΔ+Δ

→Δ−Δ

=ΔΔ

Δ

Δ

Δ

Δ

Δ

Δ

Δ

Δ

Δ

Δ

Δ

Δ

∫∫

∫∫∫(21-30)

(21-31)


12

1221

12

12

1212

11

22

)(

tttUtUa

ttUUb

ttbUU

tbaUtbaU

oo

oo

oo

o

o

Δ−ΔΔ−Δ

=

Δ−Δ−

=

Δ−Δ=−

Δ+=Δ+=

If Uo is a function of temperaturetbaU o Δ+=

(21-32)


If Uo is a function of temperaturetbaUo Δ+=

12

2112 ln1tUtU

aA

qtt

o

oo Δ

Δ=

Δ−Δ

12

1221

tttUtUa oo

Δ−ΔΔ−Δ

=

o

o

o

oo A

tUtU

tUtUq

12

21

1221

lnΔΔ

Δ−Δ=

(21-34)


o

o

o

oo A

tUtU

tUtUq

12

21

1221

lnΔΔ

Δ−Δ=

(21-34)

⎟⎟⎠

⎞⎜⎜⎝

⎛ΔΔΔ−Δ

=

1

2

12

lntt

ttAUq oo(21-29)

Heat transfer rate in double pipe heat exchanger

tbaU o Δ+=

constU o =


Homework

PROBLEMS21-121-621-1021-1521-17

Due on November 2

Supercritical Fluid Process Lab

Documents

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