BKF2422 HEAT TRANSFERBKF2422 HEAT TRANSFERCHAPTER 3CHAPTER 3Part 1 Principles of steady-state heat Principles of steady-state heat transfer in convectiontransfer in convection

FORCED CONVECTIONFORCED CONVECTION

TOPIC OUTCOMESTOPIC OUTCOMES

Define and differentiate between forced convection and natural convection

Solve problems involve with forces convection inside pipe Determine the heat transfer coefficient and solve problems

for heat transfer in system with fluid flow across plate, tube cylinder, sphere and bank of tubes.

CONTENTCONTENT Forced Convection Heat Transfer Inside Pipe

For laminar flow inside pipeFor turbulent flow inside pipeFor transition flow inside pipeEntrance-region effect on heat transfer coefficientLiquid-metal heat transfer coefficient

Log mean Temperature Difference Heat Transfer Outside Various Geometries In Forced Convection

Flow Parallel to Flat PlateFlow Past Cylinder With Axis PerpendicularFlow Past Single SphereFlow Past Banks of Tubes or CylindersFlow Past in Packed Beds

Natural convectionFrom planes and cylinders In enclosed spaces

BoilingNucleate boilingFilm boiling

CondensationFilm-condensation coefficient for vertical surfacesFilm-condensation coefficient for horizontal cylinders

CONTENTCONTENT

Convection: Heat transfer using movement of fluids. Heat transfer is considered as convection with the presence of

bulk fluid motion. Fluid motion enhances heat transfer where the higher the fluid velocity, the higher the rate of heat transfer.

2 main classification of convective heat transfer;

1. Forced Convection : fluid flow by pressure differences, a pump, a fan and so on

2. Natural Convection: motion of fluid results from the density changes in heat transfer

CONVECTION HEAT TRANSFERCONVECTION HEAT TRANSFER

The rate of heat transfer :

Tw = 80 oC

To = 30 oCq

q Ah(Tw To)The convection coefficient is a measure of how effective a fluid is at carrying heat to and away from the surface.h = heat transfer coefficient

(W/m2.K)A= surface area (m2)

Fluid flow

CONVECTION HEAT TRANSFER

Metal wall

Warm fluid A

Cold fluid B

q

Turbulence absent

T2

T3 Turbulence region

T1

q = hA (T-Tw)

FKKSA

FORCED CONVECTION INSIDE PIPESFORCED CONVECTION INSIDE PIPES

Forced convection – fluid forced to flow by pressure differences

Types of fluid, laminar or turbulent

– great effect on heat-transfer coefficient

More turbulent– greater heat-transfer coefficient

Reynolds number, NRe

NRe

D

wherev = velocity of fluid (m/s) = viscosity of fluid (Pa.s)= density of fluid (kg/m3)

D = diameter of pipe (m)

FKKSA

FORCED FORCED CONVECTIONCONVECTION

where

= viscosity of fluid (Pa.s)= density of fluid (kg/m3)k = thermal conductivity of fluid (W/m.K)cP = heat capacity of fluid (J/kg.K)h = heat transfer coefficient (W/m2.K)

D = diameter of pipe (m)

Prandtl number, NPr

Dimensionless numbers:

Nusselt number, NNu

kμc

ρckρ

μN P

P

Pr

khDN

Nu

FKKSA

LAMINAR FLOW INSIDE HORIZONTAL PIPELAMINAR FLOW INSIDE HORIZONTAL PIPE

whereD = inside diameter of pipe (m)

L = length of pipe (m)

b = viscosity of fluid at bulk temperature (Pa.s)

w = viscosity of fluid at wall temperature (Pa.s)ha = average heat transfer coefficient (W/m2.K)

NNu

ah

aD

k1.86 N

ReN

PrDL

13

b

w

0.14

NRe 2100 & NReNP r 100 :L

D

All physical properties at except w 2bi

Tbo

T

mean bT

q = haA∆Ta where 2

boT

wT

biT

wT

aΔT

100

2100

PrRe

Re

L

DNN

N

Limitations

FKKSA

TURBULENT FLOW INSIDE HORIZONTAL PIPE

where

NNu

hLD

k0.027 N

Re0.8N

Pr

13

b

w

0.14

NRe 6000 , 0.7 ≤ NP r ≤ 16000 & 60:DL

Rate of heat transfer is greater

cP = heat capacity of fluid (J/kg.K)

D = inside diameter of pipe (m)

k = thermal conductivity of fluid (W/m.K)

b = viscosity of fluid at bulk average temperature (Pa.s)

hL = heat transfer coefficient based on the log mean driving force ∆Tlm (W/m2.K)

w = viscosity of fluid at wall temperature (Pa.s)

Many industrial heat transfer processes in the turbulent region

60

160007.0

6000

Pr

Re

D

LL

DN

NLimitations

FKKSA

TURBULENT FLOW INSIDE HORIZONTAL PIPE

where

D = inside diameter of pipe (m)

= velocity of fluid (m/s)

hL3.52

0.8

D0.2

hL1429 10.0146 ToC

0.8

D0.2

hL423

0.8

D0.2

Air at 1 atm total pressure (NRe 2100) :

Water at T = 4 to 105oC :

Organic liquids :

Flow inside helical coils :

Hcoil = hstraight pipes + (1 + 3.5D/Dcoil)

2100Re N

Limitations

CT o105 to4

Limitations

4Re 10N

Limitations

EXAMPLE 4.5-1EXAMPLE 4.5-1 Page 262: Heating of Air in Page 262: Heating of Air in Turbulent FlowTurbulent Flow

Air at 206.8 kPa and an average of 477.6 K is being heated as it flows through a tube of 25.4mm inside diameter at velocity of 7.62 m/s. The heating medium of 488.7 K steam condensing on the outside of the tube. Since the heat-transfer coefficient of condensing steam is several thousand W/m2.K and the resistance of the metal wall is very small, it will be assumed that the surface wall temperature of the metal in contact with the air is 488.7 K. Calculate the heat-transfer coefficient for an L/D > 60 and also the heat-transfer flux q/A.

boT

K 7.488 Steam, wT

steamo hh

L

mm 5.42air

biTkPa 8.206

m/s 62.7

K 6.477

P

v

Tave

3

Pr

5

kg/m 74.0

W/m03894.0

686.0

Pa.s 106.2

K 477.6 kPa, 101.32at A.3,Appendix From

k

N

TTP

b

bmave

Pa.s 1064.2

K 7.884at A.3,Appendix From5

w

wT

3kPa 8.206

21

1

2

1

212

kg/m 509.135.101

8.20674.0

on depend is

,For

TT

T

T

P

P

P & TRT

PM

RTPMRTV

mPM

RTM

mPVnRTPV

)6000(10122.1

106.2

)509.1)(62.7(104.25

4

5

3

Re

D

N

.K W/m2.63

0264.0

0260.0686.010122.1027.0

03894.0

)104.25(

027.0

2

14.0

3

18.043

14.0

3

1

Pr8.0

Re

L

L

w

bLNu

h

h

NNk

DhN

2 W/m1.701

6.4777.4882.63

bmwL TThA

q

FKKSA

TRANSITION FLOW INSIDE A PIPETRANSITION FLOW INSIDE A PIPE

2100 NRe 6000 :

whereG = mass velocity of fluid (kg/s.m2) =

EXAMPLE 4.5-2 Page 264: Water Heated by Steam and Trial and error Solution

Water is flowing in a horizontal 1-in schedule 40 steel pipe at an average temperature of 65.6oC and a velocity of 2.44 m/s. it is being heated by condensing steam at 107.8oC on the outside of pipe wall. The steam side coefficient has been estimated as ho = 10500 W/m2.K.

a) Calculate the convective coefficient hi for water inside the pipe

b) Calculate the overall coefficient Ui based on the inside surface area

c) Calculate the heat-transfer rate q for 0.305m of pipe with the water at an average temperature of 65.6oC

boT

C 8.107 Steam, owT

.K W/m10500 2oh

m 305.0

mm 6.642water

biT m/s 44.2

C 6.65 o

v

Tbm

mm 3.43

3

Pr

3

o

kg/m 9.981

W/m6629.0

72.2

Pa.s 10432.0

C, 5.56at A.3,Appendix From

k

N

T

b

bm

Pa.s 1056.3C 80at

C 80 l,first triafor C80or 107.8 and 65.5between way the

thirdoneabout as assumed be willand needed is metal inside of re temperatuThe

4o

oo

ww

w

T

T

)6000(10473.1

1032.4

)980)(44.2(0266.0

5

4

Re

D

N

.K W/m13324

1056.3

1032.472.210473.1027.0

663.0

)0266.0(

027.0

2

14.0

4

4

3

18.05

14.0

3

1

Pr8.0

Re

L

L

w

bLNu

h

h

NNk

DhN

again. re temperatufor the make weassumption check the tohave We

2m 0255.0

)305.0)(0266.0(

LDA ii

2m 032.0

)305.0)(0334.0(

LDA oo

:follow as are areas variousThe

002633.0)305.0()45( 2

)0266.0/0334.0ln(

2

)ln( 1/2 kL

rrRpipe

002976.0)032.0(10500

11

ooo Ah

R

002943.0)0255.0(13324

11

iii Ah

R

008552.0 R

necessary.not is trialSecond

C. 80 of estimate original the toclose quite is C.This 1.805.145.65T Hence,

C 14.5K 5.14)2.42(008852.0

002943.0)2.42( drop eTemperatur

is film water theacross drop re temperatuThe

C 42.265.6) (107.8 is difference re temperatuoverall The

oow

o

o

R

Ri

(b)

.K W/m4586

)008552.0(0255.0

1

1

2

RAU

TTAUR

TTq

ii

ioiiio

(c)

W4935

)2.42)(0255.0(4586

C 2.426.658.107

C, 65.6 of re temperatuaveragean at with watero

o

ioii

io

TTAUq

TT

FKKSA

ENTRANCE-REGION EFFECT ON HEAT-ENTRANCE-REGION EFFECT ON HEAT-TRANSFER COEFFICIENTTRANSFER COEFFICIENT

At entrance, h = ∞

0.7

LLD1

hh

LD61

hh

L

2 20DL

20 60DL

h = average heat transfer coefficient for a tube of finite length L

hL= heat transfer coefficient for a very long tube

where

At L/D 60, h = hL

202 D

L

Limitations

6020 D

LLimitations

FKKSA

LIQUID METAL HEAT-TRANSFER COEFFICIENT

NPe = Peclet number = NReNPr

where

Turbulent flow, NRe 6000 :

0.4Pe

N 6250k

DL

h

NuN .

0.8Pe

N 02505.0k

DL

h

NuN .

Constant wall temperature ( 60, NPe 100 ) :DL

DL

Uniform heat flux, 100 NPe 104 & 60 :

4

Re

10100

60

2100

PeN

D

L

N

Limitations

100

60

PeND

L

Limitations

EXAMPLE 4.5-3EXAMPLE 4.5-3 Page 266: Liquid-metal Heat Page 266: Liquid-metal Heat Transfer Inside a TubeTransfer Inside a Tube

A liquid metal flows at a rate of 4.00kg/s through a tube having an inside diameter of 0.05m. The liquid enters at 500K and is heated to 505K in the tube. The tube wall is maintained at a temperature of 30K above the fluid bulk temperature and constant heat flux is maintained. Calculate the required tube length. The average physical properties are as follows: μ = 7.1 x 10-4 Pa.s, ρ = 7400 kg/m3, cp = 120 J/kg.K, k = 13 w/m.K

kg/s 4m

K 30 bmw TT

metal liquidK 500biTK 505boT

?L

m 05.0iD

3

4

kg/m 7400

W/m.K13

kJ/kg.K 120

Pa.s 101.7

K, 5.502at A.3,Appendix From

k

c

T

p

b

bm

K 5.5022

505500

2

bobi

bmean

TTT

wT

bmT

q

corhbibopbmeanw TTcmTThAq

LDA ii

K 30 bmw TT

)2000(10435.1

05.04

101.7

)4(05.0

5

24

Re

xA

mDN

bmT

wT

4

2i

x

DA

)10100(940

)1055.6(10435.1 4

35

PrRe

Pe

Pe

N

NNN

3

4

Pr

1055.6

13

)101.7(120

k

cN p

.K W/m2512

)940(625.013

)05.0(

625.0

2

4.0

4.0

i

i

Peii

Nu

h

h

Nk

DhN

W2400

)500505)(120(4

bibop TTcmq

2m10185.3

2400)30(2512

2400

2

i

ibmeanwi

i

A

A

ATTh

A

q

m 203.0

)05.0(10185.32

L

L

LDA ii

bmoobmii TAUTAUq

cicocphihohp TTcmTTcmq

1

2

12

2

1

21

lnlnTT

TT

TT

TTTlm

flowrent countercur

flow parallel

hoThiT

hiT hoT

coT

coTciT

ciT

1T

1T

2T

2T

hiT

hiT

hoT

hoT

ciT

ciT

coT

coT

2

bobiwiaiaa

TTTAhTAhq

2

1

21

lnTT

TTAhTAhq iailma

EXAMPLE 4.5-4 Page 268: Heat Transfer Area and EXAMPLE 4.5-4 Page 268: Heat Transfer Area and Log Mean Temperature DifferenceLog Mean Temperature Difference

A heavy hydrocarbon oil which has a cpm = 2.30kJ/kg is being cooled in a heat exchanger from 371.9 K to 349.7 K and flows inside the tube at a rate of 3630 kg/h. A flow of 1450kg water/h enters at 288.6K for cooling and lows outside the tube.

a) Calculate the water outlet temperature and heat-transfer area if the overall Ui = 340 W/m2.K and the streams are countercurrent

b) Repeat for parallel flow

flowrent countercur (a)

1T

2T

hiTK 1.397

K7.349hoT

K 6.288ciT

coT

kg/h 3630 oil, m

kg/h 1450 water, m

W51490

3600/)7.3499.371)(3.2(3630

hihohp TTcmq

?

.K W/m340

kJ/kg.K 187.4

kJ/kg.K 3.2

2

i

i

waterp

oilp

A

U

c

c

K 1.319

)6.288)(187.4(145051490

co

co

cicocp

T

T

TTcmq

K 9.56

8.521.61

ln

8.521.61

ln2

1

21

TT

TTTlm

2m 66.2

)9.56(34051490

i

i

lmii

A

A

TAUq

K 1.616.2887.349

K 8.521.3199.371

2

1

T

T

flow parallel (b)

K 3.836.2889.371

K 6.301.3197.349

2

1

T

T

K 7.52

6.303.83

ln

6.303.83

ln2

1

21

TT

TTTlm

2m 87.2

)7.52(34051490

i

i

lmii

A

A

TAUq

forces. driving peraturelarger tem

gives wscounterflo because occurs This w.counterflofor than arealarger a is This

EXAMPLE 4.5-5EXAMPLE 4.5-5 Page 269 : Laminar Heat Page 269 : Laminar Heat Transfer and Trial and ErrorTransfer and Trial and Error

A hydrocarbon oil at 150oF enters inside a pipe with an inside diameter of 0.0303 ft and a length of 15 ft with a flow rate of 80 Ibm/h. The inside pipe surface is assumed constant at 350oF since steam is condensing outside the pipe wall and has a very large heat-transfer coefficient. The properties of the oil are cpm= 0.5 btu/Ibm.oF and km = 0.083 btu/h.ft.oF. The viscosity of the oil varies with temperature as follows: 150oF, 6.50 cp; 200oF, 5.05 cp; 250oF, 3.80 cp; 300oF, 2.82 cp; 350oF, 1.95cp. Predict the heat-transfer coefficient and the oil outlet temperature, Tbo

Ibm/h 80m

F 350 owT

oil

F 150 obiT?boT

ft 15

ft 0303.0

4

2DAx

Fbtu/h.ft. 083.0

Fbtu/Ibm. 5.0o

o

m

oilp

k

c

Ib/ft.h 23.12cp1

Ib/ft.h 2.41915.05cp

F 2002

150250

F 250 if valueAssume

o

o

bmean

bo

T

T

)2100(5.275

0303.04

23.12

)80(0303.0

2

Re

xA

mDN

7.73 083.0

)23.12(5.0

Pr

k

cN p

4115

0303.0)7.73(5.275PrRe

L

DNN

F.btu/h.ft 1.20

4191.295.1

23.124186.1

083.0

)0303.0(

86.1

o2

14.0

3

1

14.03

1

PrRe

h

h

L

DNN

k

hDN

w

bNu

again. re temperatufor the make weassumption check the tohave We

150)5.0(0.80 bobibopm TTTcmq

aa TAhq

bobobobi

wa TTTT

TT 5.02752

150350

2

F 255

5.0275150)5.0(0.80o

bo

bobo

aabibopm

T

TT

TAhTTcmq

correct. is F 255 of peratureoutlet tem theHence, l.first tria

for 5.05 with compared cp 5.0 is viscositynew The F. 202.5or 255)/2(150 be wouldboil theof

raturebulk tempemean the trial,second For the F. 250 of valueassumed n thehigher tha is This

o1

o

o

T

)2100(5.275

0303.04

23.12

)80(0303.0

2

Re

xA

mDN

7.73 083.0

)23.12(5.0

Pr

k

cN p

4115

0303.0)7.73(5.275PrRe

L

DNN

F.btu/h.ft 1.20

4191.295.1

23.124186.1

083.0

)0303.0(

86.1

o2

14.0

3

1

14.03

1

PrRe

h

h

L

DNN

k

hDN

w

bNu

again. re temperatufor the make weassumption check the tohave We

150)5.0(0.80 bobibopm TTTcmq

aa TAhq

bobobobi

wa TTTT

TT 5.02752

150350

2

F 255

5.0275150)5.0(0.80o

bo

bobo

aabibopm

T

TT

TAhTTcmq

correct. is F 255 of peratureoutlet tem theHence, l.first tria

for 5.05 with compared cp 5.0 is viscositynew The F. 202.5or 255)/2(150 be wouldboil theof

raturebulk tempemean the trial,second For the F. 250 of valueassumed n thehigher tha is This

o1

o

o

T

3

1

PrRe NCNN mNu 1-4.6 tablefrom m &C

re temperatufluidbulk average the and re temperatuor wall surface theis where

,2/ re, temperatufilm at the evaluated are properties fluid The

bw

bwf

TT

TTT

3

1

Pr5.0

Re664.0 NNN Nu 7.0

103

Pr

5Re

N

N

3

1

Pr8.0

Re0366.0 NNN Nu 7.0

103

Pr

5Re

N

N

Limitations

Limitations

EXAMPLE 4.6-1 Page: 272 Cooling a Copper Fin

A smooth, flat, thin fin of copper extending out from a tube is 50 mm by 51 mm square. Its temperature is approximately uniform at 82.2oC. Cooling air at 15.6oC and 1 atm abs flows parallel to the fin at a velocity of 12.2 m/s.

a) For laminar flow, calculate the heat-transfer coefficient, h

b) If the leading edge of the fin is rough so that all of the boundary layer o film next to the fin is rough so that alls o the boundary layer or film next to the fin is completely turbulent, calculate h

mm 15

mm 15

q C 2.82 owT

C 6.15 obiTm/s 2.12 air, v

atm 1P

C 49C 9.482

6.152.82

2oo

bw

f

TTT

3

Pr

5

o

kg/m 097.1

W/m.K028.0

704.0

Pa.s 1095.1

C, 49at A.3,Appendix From

k

N

T

b

f )103(1049.3

1095.1

)097.1)(2.12(1051

54

5

3

Re

L

N

.K W/m7.60

)704.0()1049.3(664.0028.0

)1051(

664.0

2

3

15.04

3

3

1

Pr5.0

Re

h

h

NNk

hLNNu

W51.10

6.152.82)1051(7.60 6

bmw TThAq

3

1

Pr5.0

Re60.02 NNN Nu 400 to6.0

00007 to1

Pr

Re

N

N

Limitations

EXAMPLE 4.6-2 Cooling of a Sphere

Using same condition as Example 4.6-1, where air at 1 atm abs pressure and 15.6oC is flowing at velocity of 12.2 m/s, predict the average heat-transfer coefficient for air flowing by a sphere having a diameter of 51mm and an average surface temperature of 82.2oC. Compare this with the value of h = 77.2 W/m2.K for the flat plate in turbulent flow.

mm 15

C 2.82 owT

3

Pr

5

o

kg/m 097.1

W/m.K028.0

704.0

Pa.s 1095.1

C, 49at A.3,Appendix From

k

N

T

b

f

C 49C 9.482

6.152.82

2oo

bw

f

TTT

.K W/m56.1

)704.0()1049.3(60.02028.0

)1051(

60.02

2

3

15.04

3

3

1

Pr5.0

Re

h

h

NNk

hDN Nu

)103(1049.3

1095.1

)097.1)(2.12(1051

54

5

3

Re

L

N

EXAMPLE 4.6-3 Page 273: Heating Air by a Bank of Tubes

Air at 15.6oC and 1 atm abs flows across a bank of tubes containing four transverse rows in the direction of flow and 10 rows normal to the flow at a velocity of 7.62 m/s as the air approaches the bank of tubes. The tube surfaces are maintained at 57.2oC. The outside diameter of the tubes is 25.4mm and the tubes are in-line to the flow. The spacing Sn of the tubes normal to the flow is 38.1mm and also Sp is 38.1mm parallel to the flow. For a 0.305m length of the tube bank, calculate the heat-transfer rate

mm 1.38nS

mm 1.38pS

mm 4.25D

C 2.57 owT

1

2

3

9

10

C 6.15 obiTm/s 6.7 air, v

atm 1P

C 7.372

3.182.57

2o

bw

f

TTT 3

Pr

5

o

kg/m 137.1

W/m.K027.0

705.0

Pa.s 10904.1

C, 7.37at A.3,Appendix From

k

N

T

b

f

2 3 4

)103(1047.3

1090.1

)137.1)(86.22(02547.0

54

5

maxRe

L

N

m/s 86.2247.251.38

1.386.7max

DS

vSv

n

n

620.0 ,278.05.1 line,In

251) (pg 2-4.6 Table From

mcD

S

D

S pn

.K W/m8.171

)705.0()1047.3(278.0027.0

)0254.0(

2

3

1620.04

3

1

PrRe

h

h

NcNk

hDN m

Nu

3-4.6 Tablein given as

0.9,by multiplied bemust thedirection, e transversin the rows 4only For rows. 10for is This hh

.K W/m62.154)8.171(9.0 2h

2m 973.0)305.0)(0254.0(4040 DLA

kg/s 084.1)1162.0)(224.1(6.7)3600( tAvm

2m 1162.0)305.0)(0381.0(1010 LSA nt

C 20

)6.15)(100048.1(084.12

6.152.57)973.0(62.154

2

o

3

bo

bobo

bibopbibo

w

T

TT

TTcmTT

ThAq

C 17.8or 20)/2(15.6 be woulduse be to average new themade, be to ware trialsecond If obT

35.0ReRe

3

2

3023.0876.2

'

NNk

c

c

h

f

p

p

00001 to10

Gases

Re N

Limitations

Note : All properties are at Tbmean, except for those with subscript f