Chapter 2: Overall Heat Transfer Coefficient

Heat transfer by conduction:Fourier’s law: “The heat flux is directly proportional to the temperature gradient”.

k = thermal conductivity of material [W/m.K]

dxdTq

dxdTkq

AQq

.

ordinates)-co (cartesian .

dxdTkAQ

ordinates)-co al(cylindric .

drdTkAQ

General heat conduction equation: za) In Cartesian co-ordinates (x,y,z): x yb) In cylindrical polar co-ordinates (r,,z):

T

kq

zT

yT

xT v 1

.

2

2

2

2

2

2

T

kq

zTT

rrT

rrT v 111

.

2

2

2

2

22

2

][W/m volumeunit per generationheat 3.

vq

]/[m 2 sydiffusivitthermaloftcoefficien

[s] time

Plane wall: For one dimensional steady-state heat transfer by conduction in a plane wall without heat generation, the general heat conduction is reduced to

By integration

By another integration

The boundary conditions are the known temperatures. That isand,

02

2

dx

Td

1CdxdT

21 CxCT

.Q

0xat 1 TTLxat 2 TT

When these boundary conditions are applied to the equation for temperature distribution, we obtain

Accordingly, the expression for the temperature distribution (temperature profile) becomes:

The temperature distribution is thus linear (straight line) across the plane wall.

12 TC

xCTT 11

LCTT 112

LTTC 21

1

xL

TTTT

211

Rate of heat Transfer:Fourier’s law

Evidently for heat conduction in a plane wall, the thermal resistance takes the form

dxdTkAQ

.

LTTC

dxdT 21

1

kAL

TTL

TTkAQ 2121.

kALR wallplane

.Q 1T 2T

kALR wallplane

Cylindrical pipes: For one dimensional steady-state heat transfer by conduction in a cylindrical pipe without heat generation, the general heat conduction is reduced to

By integration

By another integration

012

2

drdT

rdrTd

02

2

drdT

drTdr

0

drdTr

drd

rC

drdTC

drdTr 1

1 or

Incorporating the relevant boundary conditions that1) 2) The constants are determined as follows

By subtraction

11 at rrTT

22 at rrTT

21 ln CrCT

2111 ln CrCT

2212 ln CrCT

2

1121 ln

rrCTT

1

2

21

2

1

211

lnlnrrTT

rrTTC

The substitution in the equation of gives

When the values of the constants are substituted into the equation of the temperature distribution, one obtains the following expression for temperature distribution in the pipe wall

1

1

2

2112 ln

lnr

rrTTTC

2C

1

1

2

211

1

2

21 ln ln

lnln

r

rrTTTr

rrTTT

1

1

2

211 ln

lnrr

rrTTTT

Rate of heat flow:Fourier’s law

evaluated at r1or r2

Then, the thermal resistance for heat conduction in a cylindrical pipe takes the form

21

.

rrrr drdTAk

drdTAkQ

rrrTT

rC

drdT 1

ln1

2

211

kLrrTT

rrTTkl

rrrTTlrkQ

2lnln21

ln2

12

21

12

21

112

211

.

kl

rrRpipe 2ln 12

Equivalent thermal circuit for heat flow by conduction through the walls of a cylindrical pipe is shown in the following figure:  

Heat transfer by convection:Newton-Rikhman’s law of convection: “The heat flux is directly proportional to the temperature difference between the wall and the fluid”.

.Q 1T 2T

kl

rrRpipe 2ln 12

Tq

TTT w

In case of heat convection from/to a cylindrical pipe, the thermal resistance takes the form

The equivalent thermal resistance circuit for heat transfer by convection is shown in the following figure:

TThq w

hA

TTTThAQ ww 1

.

.Q

1T2T

hARconvection 1

hARconvection

1

Overall heat transfer coefficient through a plane wall (U):Consider the plane wall, shown in the figure, exposed to a hot fluid A on one side and a cooler fluid B on the other side. The rate of heat transfer is expressed by these three expressions 1) Heat transfer by convection from fluid A to wall surface (1):

2) Heat transfer by conduction the plane wall:

11

.TTAhQ A

kALTTQ 21

.

3) Heat transfer by convection from wall surface (2) to fluid B:

The three previous equations can be rewritten as follows:

By addition

BTTAhQ 22

.

AhQTTA

1

.

11

kALQTT

.

21

AhQTT B

2

.

21

AhkAL

AhQTT BA

21

. 11

The equivalent thermal resistance circuit for heat transfer through the plane wall with convective boundaries is shown in the following figure:Let

U = overall heat transfer coefficient [ W/m²K] Comparing these two equations, one obtains

AhkAL

Ah

TTQ BA

21

.

11

.Q

Ah1

1kAL

Ah2

1 BA TTUAQ .

h1 = heat transfer coefficient of surface (1) [W/m2.K] = heat transfer film coefficient of wall surface (1) = individual heat transfer coefficient of wall surface (1)

h2 = heat transfer coefficient of surface (2) [W/m2.K] = heat transfer film coefficient of wall surface (2) = individual heat transfer coefficient of wall surface (2)

k = thermal conductivity of the wall material [W/m.K]

L = wall thickness [m]

21

111

hkL

h

U

The wall conduction term may often can be neglected, since a thin wall of large thermal conductivity is generally used in heat exchangers. Also, one of the convection coefficients is often much smaller than the other and hence dominates determination of the overall heat transfer coefficient. For example, if one of the fluids is a gas and the other is a liquid or a liquid-vapor mixture such as boiling or condensation, the gas-side convection coefficient is much smaller.

h is small, in case of gases (low viscosity and low specific heat) and in case of laminar flow (low velocity).h is big, in case of liquids (high viscosity and high specific heat) and in case of turbulent flow (high velocity).

kL

small

bigsmall

h

hkL

h

U

111

Example 1:

An iron plate of thickness L with thermal conductivity k is subjected to a constant, uniform heat flux qo (W/m²) at the boundary surface at x = 0. From the other boundary surface at x = L, heat is dissipated by convection into a fluid at temperature T∞ with a heat transfer coefficient h. The figure shows the geometry and the nomenclature.Develop an expression for the determination of the surface temperatures T1 and T2 at the surfaces x = 0 and x = L. Also, develop an expression for the overall heat transfer coefficient U.Calculate the surface temperatures T1 and T2 and the overall heat transfer coefficient U for L = 2 cm, k = 20 W/m.K, qo = 105 W/m2 , T∞ = 50oC, and h = 500 W/m2.K.

Data: L = 2 cm, k = 20 W/m2 °C, qo = 105 W/m2 , T∞ = 50oC, and h = 500 W/m2 °C.Find: T1, T2, U Solution: Applying the thermal resistance concept:

.

Q

kAL

hA1

hAkAL

TThA

TT

kAL

TTQ

1

1

1

221.

111221

hkL

TT

h

TT

kL

TTqo

By equating the first and the last expression, T1 is found

and by equating the first and the third expressions, T2 is found:

and since there is no convective heat transfer on surface (1),

ThqT o

2

21

111

hkL

h

U

h 012

1

handh

hkL

U1

1

T

UqTq

hkLT o

o 11

Introducing The numerical values of various quantities in the above results, we obtains

Note if the wall thickness is 2 mm, then T1 = 260 °C, T2 = 250°C and U = 476.2 W/m2 °C.

CTqhk

LT oo 3505010

5001

2002.0 1 5

1

CThqT oo 25050

500105

2

CmW

hkL

U o./ 33.333

5001

2002.0

11

1 2

Overall heat transfer coefficient in pipes: Consider a pipe exposed to a hot fluid on the inner side and a cooler fluid on the outer side, as shown in the figure. The area of convection is not the same for both fluids in this case, these areas depend on the inside pipe diameter and wall thickness.The heat transfer is expressed by the following relations:1) Heat transfer by convection from the hot fluid on the inner side to the inner wall surface of the pipe:

ii

i

Ah

TTQ 11

.

2) Heat transfer by conduction through the pipe wall itself:

3) Heat transfer by convection from the outer wall surface of the pipe to the cold fluid on the outer side:

The three previous equations can be rewritten as follows:

kL

rrTTQ

io

2ln

21.

oo

o

Ah

TTQ 12

.

iii Ah

QTT 1.

1

kL

rrQTT io

2ln.

21

By addition

The equivalent thermal resistance circuit for heat transfer through the pipe wall with convective boundaries is shown in the following figure:

ooo Ah

QTT 1.

2

oo

io

iioi AhkL

rrAh

QTT 12

ln1.

oo

io

ii

oi

AhkLrr

Ah

TTQ1

2ln1

.

.Q ii Ah1 kLrr io 2ln oo Ah1

Let

Uo = overall heat transfer coefficient based on the outer area of pipe.Ui = overall heat transfer coefficient based on the inner area of pipe.Ao = 2ro L, is the outer surface area of the pipe.Ai = 2ri L, is the inner surface area of the pipe. Upon comparing the two equations of , one obtains

oiiioioo TTAUTTAUQ .

.Q

oo

io

ii

oo

AhkLrr

Ah

AU 12

ln11

iioo AUAU

and

o

ioo

ii

oo

hkrrr

rhrU 1ln

1

oo

io

ii

ii

AhkLrr

Ah

AU 12

ln11

oo

iioi

i

i

rhr

krrr

h

U

ln11

Example 2:Steam at 120oC flows in an insulated pipe. The pipe is made of mild steel (kp =45 W/m.K) and has an inside radius of 5 cm and an outside radius of 5.6 cm. The pipe is covered with 2.5 cm layer of magnesia insulation (kin = 0.071 W/m.K). The inside heat transfer coefficient is 85 W/m2.K and the outside heat transfer coefficient is 12.5 W/m2.K. Determine the overall heat transfer coefficients Uo and Ui and the heat transfer rate from the steam per meter of pipe length, if the surrounding air temperature is 20oC. Data: Ti = 120oC, kp = 45 W/m.K, r1 = 0.05 m, r2 = 0.056 m,

r3 = 0.081 m, kin = 0.071 W/m.K, hi = 85 W/m2.K, ho = 12.5 W/m2.K, To = 20oC.Find: Uo , Ui ,

LQ.

Solution:

oin

233

p

123

1i

3o

h1

krrlnr

krrlnr

rhr

1U

5.12

1071.0

056.0081.0ln081.045

05.0056.0ln081.005.085

081.01

oU

08.04211.000020.001906.01

oU

KmWUo ./ 944.15144.01 2

iioo AUAU

KmWUrrU

AAU oo

i

oi ./ 149.3944.1

05.0081.0 2

1

3

oio TTrULQ

3

.

2

mWLQ / 94.9820120081.02944.1

.

oioo TTAUQ .

oio TTLrUQ 3

.2

Values of the overall heat transfer coefficient

U (W/m².K) Fluid combination

850 – 1700 Water to water heat exchanger

110 – 350 Water to oil heat exchanger

1000 – 6000 Steam condensers (water in tubes)

800 – 1400 Ammonia condenser (water in tubes)

250 – 700 Alcohol condenser (water in tubes)

25 - 50 Finned-tube heat exchanger (water in tubes, air in cross flow)

10 – 40 Gas to gas heat exchanger