高等輸送二 — 熱傳

Lecture 10Fundamentals of Heat Transfer

郭修伯 助理教授

Heat transfer

• Three modes of energy transfer– Conduction

• Fourier’s law for conduction

• Energy transport through a thin film

• Energy transport in a semiinfinite slab

– Convection– Radiation

Conduction

• Two ways– Molecular interaction:

• greater motion of a molecule at a higher energy level (temperature) imparts energy to adjacent molecules at lower energy lever.

– By “free” electrons:• The ability of solids to conduct heat varies directly

with the concentration of free electrons.

Conduction - Molecular phenomenon

• Fourier (1807):

– Where qx is the heat-transfer rate in the x direction (Watts or Btu/hr); A is the area normal to the direction of heat flow (m2 or ft2), dT/dx is the temperature gradient in the x direction (K/m or °F/ft) and k is the thermal conductivity (W/mK or Btu/hr ft °F)

– More general form:

dx

dTk

A

qx

Tkq Fourier’s first law of heat conduction

Primarily a function of temperature, vary significantly with pressure only in the case of gases subjected to high pressures

Steady heat conduction across a thin film

• On each side of the film is a well-mixed solution of one solute, T0 > Tl

T0

Tlz

l

z

Energy balance in the layer z

Energy conducted out of the layer at z + z

Energy conducted into the layer at z

Energy accumulation =

s.s.

zzz qqA 0

zzz qqA 0

Dividing A z

zzz

qq zzz

)(0

z 0

dz

dq0

z

Tkq

2

2

0dz

Tdk

2

2

0dz

Tdk

B.C.z = 0, T = T0

z = l, T = Tl

l

zTTTT l )( 00

z

Tkq

lTTl

kq 0

T0

Tlz

l

z

linear concentration profile

Since the system is in s.s., the flux is a constant.

Questions

• How are the results changed if the fluid at z = 0 and T0 is replaced by a different liquid that is at the same temperature?– There is no change as long as the interfacial temperature

is constant.

• What will the temperature profile look like across two thin slabs of different materials that are clamped together?– In steady state, the heat flux is constant. Thus the

temperature drop across the poorly conducting slab will be larger than that across the better conductor.

An energy balance on the fluid at z = l:

)(ˆ0 llzlv TT

l

kAAqTCV

dt

d

Mass of fluid located at z = l

Specific heat capacity of the liquid

B.C.t = 0, Tl = Tl0

t

CVlkATTTT

vlll )ˆexp(1)( 000

The temperature rises to a limit of T0.

Imagine that for the system the fluid at z = l has a small volume, V, but the fluid at z = 0 has a very large volume. How will Tl change with time?

Unsteady heat conduction into a thick slab

• Any heat conduction problem will behave as if the slab is infinitely thick at short enough times.

T0

Too

position z

time

At time zero, the temperature at z = 0 suddenly increases to T0

Energy conducted out of the layer at z + z

Energy conducted into the layer at z

Energy accumulation =

zzzv qqATCzAt ˆ

Energy balance on the thin layer Az

Dividing

zzz

qq

Ct

T zzz

v)(ˆ

1

z 0

z

Tkq

2

2

z

T

t

T

The heat conduction equation

ztatTT

ztatTT

zallfortatTT

,0

0,0

,0

0

Boundary conditions

erf0

0

TT

TT

0

22erf dse s

z

Tkq

TTetz

Tkq t

z

04

2

TTt

q z 00

t

z

4

zzzv qqATCzAt ˆ

vCzA ˆ

z

q

Ct

T

v

ˆ1

pC

kˆ

vp CC ˆˆ

2

2

z

T

t

T

Thermal diffusivity

Questions

• To what depth does the temperature change penetrate in a steel slab?– equals unity. For steel, α ~ 0.15 cm2/sec; t = 10

min, z = 15 cm

• How does the flux vary with physical properties for the thick slab as compared with the thin film?– Doubling the temperature difference doubles the heat flux

in both cases. Doubling the thermal conductivity increases the flux by 2^0.5 for the thick slab and by 2 for the thin film. Doubling the heat capacity decreases the flux by 2^0.5 for the thick slab, but has no effect for the thin film.

t

z

4

2

Imagine a well-insulated pipe used to transport saturated steam. How much will the heat loss through the pipe’s walls be reduced if the insulation thickness is doubled? Assume that the thermal conductivity of the pipe’s walls is much higher than that of the insulation.

Steady-state energy balance on cylindrical shell of insulation of volume 2πrΔrL:

Energy conducted out of the layer at r + r

Energy conducted into the layer at r

Energy accumulation =

rrr rLqrLq 220

rqdr

d0

integration

00qRrq

r

Tkq

00qR

dr

dTrk

)/ln(

)(

00

0

00

0

0

i

i

R

R

T

T

RRR

TTk

rdr

dT

R

kq

i

i

)21ln()21(

)1ln()1(

insulationesinglwithflux

insulationdoublewithflux

i

i

R

RR 0

General energy balances

• Energy balance can be difficult because energy and work can take so many different forms.– Internal, kinetic, potential, chemical and surface

energies are all important.– Work can involve forces of pressure, gravity,

and electrical potential.

• As a result, a truly general balance is extraordinary complicated (Slattery, 1978).

Energy balances for a single pure component that has internal and kinetic energy:

][2

1ˆ2

1ˆ 22 vτvgvqv

p

vUvUt

Energy accumulation

Energy convection in minus that out

conduction=Work by gravity

Work by pressure forces

Work by viscous forces

vτvqv

:ˆˆ pUUt

Energy accumulation

Energy convection in minus that out

conduction=Reversible work

Irreversible work

Subtracting the mechanical energy balance

vτvqv

:ˆˆ

pHt

U

p

UH ˆ

vτvqv

:ˆˆ pUUt

Batch system and restricted to changes in internal energy only

vτvq

:ˆ

pt

U WQU

vτvqv

:ˆˆ

pHt

U

Steady-state open system of fixed volume

vτqv :ˆ0 H sWQH

Conduction in a thin film at steady state:

vτvqv

:ˆˆ pUUt

Steady state No energy convection No flow work

0 q

One dimensional

0 qdz

d

z

Tkq

02

2

Tdz

dk

Conduction in a thick film at unsteady state:

vτvqv

:ˆˆ pUUt

No energy convection No flow workOne dimensional

z

Tkq

q

t

U

)(ˆˆ etemperaturreferenceTCU v

vp CC ˆˆ

qzt

U

ˆ

2

2ˆ

z

Tk

t

TC p

Heating a flowing solutionA viscous solution is flowing laminarly through a narrow pipe. At a known distance along the pipe, the pipe’s wall is heated with condensing steam. Find a differential equation from which the temperature distribution in the pipe can be calculated.

vτvqv

:ˆˆ

pHt

U

Steady state heating due to viscous dissipation is smallno reversible work

qv H0

Energy transfer along the pipe axis is largely by convectionEnergy transfer in the radial direction is largely by conduction

rqrr

Hvz

1ˆ0

)(ˆˆ etemperaturreferenceTCH p

r

Tr

rrvC

kT

zp

1

ˆ

r

Tkq

Heat transfer coefficient

• Fourier’s law of heat conduction– useful for heat conduction in solids

– difficult to use in fluid systems, especially when heat is transferred across phase boundaries.

• Heat transfer across interfaces:– isothermal in the separated phases.

– the temperature gradients are close to the interface

– heat flux: TUq

overall heat transfer coefficient

TUq

Common choice of temperature difference at some position z: )()( zTUzq

Another choice(for full size industrial equipment):2

)()( outletTinletTUq

Solid wall T1

T1i T3i

T3

Hot fluid Cold fluid

The heat flux:

)()()( 333312111 TThTThTThq iiii ),,( 321 hhhfU ???

lTTl

kq 0

2

22 l

kh

)()()( 333312111 TThTThTThq iiii

2

22 l

kh

32

2

1

31

11hk

lh

TTq

TUq

32

2

1

11

1

hkl

h

U

Harmonic average

• The overall heat transfer coefficient, U, Vs. the overall mass transfer coefficient, k :– U is simpler than k

• the hot face at the wall = the temperature of the solid wall in contact with the hot fluid (c.f. mass transfer)

• U: a sum of resistance

• k: involve weighting factors

Finding the overall heat transfer coefficientA total of 1.8 x 104 liters/hr crude oil flows in a heat exchanger with forty tubes 5 cm in diameter and 2.8 m long. The oil, which has a heat capacity of 0.43 cal/g-°C and a specific gravity of 0.9 g/cm3, is heated with 240°C steam from 20°C to 140°C. The steam is condensed at 240 °C but is not cooled much below this temperature. What is the overall heat transfer coefficient based on the local temperature difference? What is it when based on the average temperature difference?

Fig 19.3.2

Energy balance: Energy conducted through walls

Energy in minus energy out by conduction

Energy accumulation =

TUzdTCvTCvd zzpzp

240)(ˆˆ

40 2

TUzdTCvTCvd zzpzp

240)(ˆˆ

40 2

Dividing by zd 2

4

CTz 20,0

TdCv

U

dz

dT

p

240

ˆ4

B.C.

0z

dCv

U

Tp

ˆ4

240

20240ln

scmv 4.6

)5)(4/(40

3600/108.12

7

C140T

KscmcalU

2

3107.8

Based on the local temperature difference

Based on the average temp.

2

)()( outletTinletTUq

C

outletTinletTT

1602

)100220(2

)()(

scmcal

rl

dCQ

rl

TdCQq

p

p

232.1

)20140(240

ˆ240

ˆ

KscmcalU

2

3102.8

z

Tkq

The time for tank coolingA 100-gallon tank filled with water initially at 80°F sits outside in air at 10°F. The overall heat transfer coefficient for heat lost from the water-containing tank is 3.6 Btu/hr-ft2-°F, and the tank’s area is 27 ft2. How long can we wait before the water in the tank starts to freeze?

Energy balance on the tank:

Heat lost from tank

Energy accumulation =

FTUATCVdt

dv

10ˆ

i.c.

FT 32

vCVUAt

FF

FTˆexp

1080

10

FTt 80,0

hrt 10

The effect of insulationInsulation advertisements claim that we can save 40% on our heating bills by installing 10 inches of glass wool as insulation. The glass wool has a thermal conductivity of about 0.03 Btu/hr-ft2-°F; the average winter temperature is 15°F and the house temperature is 68°F. If the advertisements are true, and if heat loss from doors and windows is minor, how much can we save with 2 ft of insulation?

Heat loss in our current home:

FFT 1568

Thq

Adding 10 inches of glass wool: T

h

q

3.01

12101

16.0

FhrftBtuh

2024.0

TTq

0092.0

3.012024.0

11

q

q

?

%40%36?

Heat loss from a barIn 1804, Biot carried out an experimental investigation of the conductivity of metal bars by maintaining one end at a high known temperature and taking readings of thermometers places in holes along the bar. He found that the steady state temperature decreased exponentially along the bar. Why?

T0

Air at Too

T(z)Energy in minus energy out by conduction

Energy accumulation =

Energy lost to the surroundings

)()(20 TTzhlWWlqWlq zzz

Dividing by zWl0zz

Tkq

)(

)(2

02

2

TT

lWWl

h

dz

Tdk

)(2 lWWlL

kLz

hbkLzhaTT expexp

B.C.

TTz

TTz

,

,0 0

kLz

hTT

TTexp

0

Thermal conductivity, thermal diffusivity, and heat transfer coefficient

• The thermal conductivity k, the thermal diffusivity α, and the heat transfer coefficient h:– Values for gases can be predicted from kinetic theory– Thermal conductivities of gases:

– Values for liquids and solids are found by experiment

k

MTk

2

4 ~1099.1

(Hirschfelder, Curtiss, and Bird, 1954)

Collision diameter A

Molecular weight

Of order 1, a weak function of TkB

Table 5.1-2 and Table 19.4-2

Values of k, α and h

• Typical values of k and α in gases, liquid and solids are in Table 19.4-2.– Thermal conductivities of metals are much higher than

those of liquids or gases.

– Thermal conductivities of nonmetallic solids and liquids are comparable.

– The effective thermal conductivity of composite materials tends to be dominated by the continuous phase.

• Common correlations of h are given in Table 19.4-3.– All refer to heat transfer across a solid-fluid interface.

Dimensionless groups in heat transfer

• Nusselt number– c.f. Sherwood number for mass transfer

–

• Prandtl number– c.f. Schmidt number for mass transfer

– viscosity - thermal conductivity

–

k

hlNu

k

C pˆ

Pr

The overall heat transfer coefficient of a heat exchangerAs part of a chemical process, we plan to use a shell-tube heat exchanger of 20 banks of 5 cm outside-diameter steel tubes with 0.3 cm walls. Outside the tubes, we plan to use 400°C flue gas; inside, we expect to be heating aromatics like benzene and toluene at around 30°C. The gas flow will be 17 m/s, and the liquid flow will be 2.7 m/s. Determine the overall heat transfer coefficient.

32

2

1

11

1

hkl

h

U

hot flue gas

steel wallliquid

Cscmcal

A

K

MTk

k

42

4

2

4

1077.0)87.0()8.3(

286731099.1

~1099.1

Table 19.4-3, flow over tube banks:

Cscmcal

k

Cdv

d

kh p

23

3.06.0

1 1056.1ˆ

33.0

Cscmcal

lkh

22

22 33.0

3.0

1.0

Cscmcal

k

Cdv

d

kh p

2

33.08.0

1 068.0ˆ

027.0

Table 19.4-3, liquid inside the tube:Cscmcal

hkl

h

U

3

32

2

1

1052.1

11

1