Heat and Mass Transfer: Free Convection : Formulas and solved examples... Use of Heat and Mass transfer data book is necessary in order to obtain certain values.

1. Chapter 3
Free convection
(Formulae & Problems)

2. Forced Convection – Internalflow – Home work
Air at2 bar pressure and 60oC isheated asitflowsthrough a tube ofdiameter 25 m m ata
velocityof15 m/s.Ifthewalltemperature ismaintained at100oC, findtheheattransferper
unitlength ofthe tube.How much would be the bulk temperature increase over one meter
lengthofthetube..
Given:
Pressure,p = 2x 105N/m2
Velocity,U = 15m/s
Diameter,D = 0.025m
Inletairtemperature,Tmi = 60oC
Length,L = 1 m
Walltemp = 100oC
To find: i)Heat transferrate(Q ) ii)Riseinbulktemperature (ΔT)
Solution:
Given pressureisabove patm. So ρ and γvarieswithpressure.
(ValueofPr,k,Cp willremain same)
γ= γatm x(patm/pgiven)
γ= 9.485x10-6m2/s
ρ = (p /RT)
ρ = 2.092kg/m3
Propertiesofairat60oC
ρ = 1.060kg/m3
γ= 18.97x10-6m2/s
Pr= 0.696
k = 0.02896W/mK
Re = (UD/γ)= 3.95x 104
SinceRe > 2300,flowisturbulent.
Re > 10000and Pr valueisinbetween 0.6and 160.
So, Nu =0.023(Re)0.8(Pr)n
The processinvolvedisheating,hence n = 0.4
Nu = 94.7
Heat transfercoefficient,h = Nu k /D
h = 109.7W/m2K

3. Forced Convection – Internalflow – Home work
Mass flowrate,m = ρAU
m = 0.015kg /s
Heat transfer,Q = m Cp ∆T
Q = m Cp (Tmo -Tmi) = 0.015x1005x (Tmo –60) (Cp forairis1005J/kgK)
Heat transfer,Q = hA ∆T = h πDL (Tw-Tm)
Q = 109.7xπ x0.025x1 x(100–Tm)
Q = 109.7xπ x0.025x1 x(100–[Tmi + Tmo) /2])
Q = 109.7xπ x0.025x1 x(100–[60+ Tmo) /2])
EquatingtheequationsforQ and solvingforTmo, we getTmo = 77.78oC
Riseinbulktemperature (ΔT) = (Tmo –Tmi)
ΔT = 17.78oC
Heat transfer,Q = m Cp ∆T = m Cp (Tmo -Tmi)
Q = 268.03W

4. Free convection
Free convection – If the fluid motion is produced due to change in density
resulting from temperature gradients, the mode of heat transfer issaid to be
freeornaturalconvection.
Exam ples:
Heatingofrooms by useofradiators
Cooling of transmission lines,electric transformers and rectifiers
Wallofairconditioninghouse,Condenser ofsome refrigerationunits.
The rateofheattransferiscalculatedusingthegeneralconvectionlaw
Q = hA (Tw–T∞)
Where Q = heattransferrateinWatts
A =Areainm2
Tw = SurfacetemperatureinoC
T∞ = FluidtemperatureinoC

5. 1. Film temperature,Tf= (Tw+ T∞) /2,
whereTw– Surfacetemperature inoC and T∞– Fluidtemperature inoC
2. Coefficientofthermal expansion
β = 1/Tf (TfinK)
3. NusseltNumber, Nu = hL/k,
where h – Heat transfercoefficient,W/m2K
L –Length,m,
k –Thermal conductivity,W/mK
4. Grashof Number forverticalplate,Gr = (gxβ x L3 x ∆T) /γ2
(HMT Data book,Pg:134)
where L – Lengthoftheplate,
∆T –(Tw–T∞)
γ– Kinematic viscosity,m2/s,
β –Coefficientofthermal expansion
5. IfGrPr valueislessthan109,flowislaminar.IfGrPr valueisgreaterthan109,flowis
turbulent.
i.e.,Gr Pr < 109,->Laminar flow
Gr Pr > 109,->Turbulentflow
Formula used forFree Convection

6. 6. Forlaminarflow(Verticalplate):
NusseltNumber, Nu = 0.59(GrPr)0.25
Thisexpressionisvalidfor,104< Gr Pr < 109 (HMT Data book,Pg:135)
7. Forturbulentflow(Verticalplate):
NusseltNumber, Nu = 0.10[GrPr]0.333 (HMT Databook,Pg:135)
8. Heat transfer(Verticalplate):Q = hA (Tw–T∞)
9. Grashof Number forHorizontalplate:Gr = (gx β x Lc
3 x
∆T) /γ2 where,Lc –Characteristiclength= W /2,
W –Width oftheplate (HMT Data book,Pg:135)
10. Forhorizontalplate,upper surfaceheated,
NusseltNumber, Nu = 0.54[Gr Pr]0.25
Thisexpressionisvalidfor,2x104< Gr Pr < 8x106 (HMT Databook,Pg:135)
NusseltNumber, Nu = 0.15[GrPr]0.333
Thisexpressionisvalidfor,8x106< Gr Pr < 1011 (HMT Data book,Pg:135)
11. Forhorizontalplate,lowersurfaceheated,
NusseltNumber, Nu = 0.27[GrPr]0.25
Thisexpressionisvalidfor,105< Gr Pr < 1011 (HMT Data book,Pg:136)
Formula used forFree Convection

7. Formula used forFree Convection
12. Heat transfer(Horizontalplate),Q = (hu + hl)xA x (Tw–T∞)
where hu –Upper surfaceheated,heattransfercoefficientW/m2K,
hl–Lower surfaceheated,heattransfercoefficientW/m2K
13. Forhorizontalcylinder,Nu = C [Gr Pr ]m (HMT Data book,Pg:137)
14. Forhorizontalcylinder,
Heat transfer,Q = hA (Tw–T∞)
whereA = π D L
• (HMT Data book,Pg:137)For sphere,NusseltNumber, Nu = 2+ 0.43[Gr Pr]0.25
Heat transfer,Q = h xA x(Tw–T∞)
whereA = 4π r2
GrDPr C m
10-10 to 10-2 0.675 0.058
10-2 to 102 1.02 0.148
102 to 104 0.85 0.188
104 to 107 0.48 0.25
107 to 1012 0.125 0.333

8. Formula used forFree Convection
16. Boundary layerthickness
δx= [3.93x(Pr)-0.5(0.952+ Pr)0.25x(Gr)-0.25] (HMT Data book,Pg:134)
17. Maximum velocity,
umax = 0.766x γx (0.952+ Pr)-1/2x[{gβ (Tw–T∞)} /γ2]1/2 x χ1/2

9. Free Convection
1)A verticalplateof0.75m heightisat170oC and isexposedtoairatatemperature of
105oC and one atmosphere. Calculate:
(i)Mean heattransfercoefficient,
(ii)Rate ofheattransferperunitwidthoftheplate
Given:
Length,L = 0.75m Wall temperature,Tw = 170oC Fluidtemperature,T∞ = 105oC
To find:1.Heat transfercoefficient,(h )2.Heat transfer(Q )perunitwidth
Solution:
Velocity(U )isnotgiven.So
this isnatural convection
typeproblem.
FilmTemp, Tf = (Tw+ T∞) /2
Tf= 137.5oC
Propertiesofairat
137.5oC ~ 140oC
(HMT Data book,Pg:33)
ρ = 0.854kg/m3
γ= 27.80x10-6m2/s
Pr= 0.684
k = 0.03489W/mK
Coefficientofthermal expansion,
β = 1/(TfinK) = 1 /(137.5+ 273)= 2.4x 10-3K-1
GrashofNumber,
Gr = (gxβ x L3 x∆T) /γ2(HMT Data book,Pg:134)
8Gr = 8.35x10
Gr Pr = 5.71x108
SinceGr Pr < 109,flowislaminar.Gr Pr valueisinbetween
104and 109i.e.,104< Gr Pr < 109
(HMT Data book,Pg:135)
Q = hA (Tw–T∞ )= h xL W x(Tw–T∞ ) (Given,W = 1m)
Heat transfer,Q = 206.8W
So Nu = 0.59(GrPr)0.25
Nu = 91.21
Heat transfercoefficient,h = Nu k /L
h = 4.24W/m2K

10. Free Convection
2)A verticalplateof0.7m wideand 1.2m heightmaintained atatemperature of90oC ina
room at30oC.Calculatetheconvectiveheatloss.
Given:
Walltemperature,Tw = 90oC RoomWidth,W = 0.7m Height(or)Length,L = 1.2m
temperature,T∞ = 30oC
To find:Convectiveheatloss(Q )
Solution:
Velocity(U )isnotgiven.So
this isnatural convection
typeproblem.
FilmTemp, Tf = (Tw+ Tά) /2
Tf= 60oC
Properties of air at 60oC
(HMT Data book,Pg:33)
ρ = 1.060kg/m3
γ= 18.97x10-6m2/s
Pr= 0.696
k = 0.02896W/mK
Convectiveheattransfercoefficient,h = Nu k /L
h = 4.32W/m2K
Q = hA (Tw–T∞ )= h xL W x(Tw–T∞ )= 218.16
Convectiveheatloss,Q = 218.16W
Coefficientofthermal expansion,
β = 1/(TfinK) = 1 /(60+ 273)= 3 x 10-3K-1
Grashof Number, Gr = (gxβ x L3 x∆T) /γ2
Gr = 8.4x109
Gr Pr = 5.9x 109 SinceGr Pr > 109,flowisturbulent.
So Nu = 0.10(GrPr)0.333 (HMT Data book,Pg:135)
Nu = 179.3

11. Free Convection
3)A horizontalplateof800m m long,70m m wideismaintained ata temperature 140oC ina
largetankoffullofwaterat60oC.Determine thetotalheatlossfrom theplate.
Given:
Platetemperature,Tw = 140oCHorizontal platelength,L = 0.8m Wide,W = 0.070m
Fluidtemperature,T∞ = 60oC
To find:Totalheatlossfrom theplate.
Solution:
FilmTemp, Tf = (Tw+ T∞ )/2
Tf= 100oC
Propertiesofwaterat100oC
(HMT Data book,Pg:21)
ρ = 961kg/m3
γ= 0.293x10-6m2/s
Pr= 1.740
k = 0.6804W/mK
For horizontalplate,Lc –Characteristiclength= W /2 = 0.035m
(Pg:135)
βwater = 0.76 x 10-3 K-1
(HMT Data book,Pg:29)
cGrashof Number, Gr = (gx β xL 3 x∆T) /γ2 (Pg:134)
Gr = 0.297x109
Gr Pr = 0.518x109
Gr Pr valueisinbetween8x106and 1011
i.e.,8x106< Gr Pr < 1011
So,forhorizontalplate,uppersurfaceheated,
Nu = 0.15(GrPr)0.333
Nu = 119.66 Heat transfercoefficientforuppersurfaceheated,
hu = Nu k /Lc = 2326.19W/m2K

12. Free Convection
For horizontalplate,Lower surfaceheated,
Nu = 0.27[GrPr]0.25 (HMT Data book,Pg:136)
Nu = 40.73
Heat transfercoefficientforlowersurfaceheated,hl= Nu k /Lc
hl= 791.79W/m2K
Totalheattransfer,Q = (hu + hl)A ∆T
= (hu + hl)xL W x(Tw–T∞ )
Q = 13968.55W
Totalhealloss,Q = 13,968.55W

13. Free Convection
4)Airflowsthrougha longrectangular300 m m heightx 800 m m widthair-conditioningduct,
theouterduct surfacetemperature isat20oC. Iftheduct isuninsulatedand exposed toairat
40oC. Calculatetheheatgainedby theduct,assuming ducttobehorizontal.
Given:
Width,W = 0.8m Surfacetemperature,Tw = 20oCLength (or)Height,L = 0.3m
Fluidtemperature,T∞ = 40oC
To find:Rateofheatgained(Q)
Solution:
FilmTemp, Tf = (Tw+ T∞ )/2
Tf= 30oC
Properties of air at 30oC
(HMT Data book,Pg:33)
ρ = 1.165kg/m3
γ= 16x10-6m2/s
Pr= 0.701
k = 0.02675W/mK
β = 1/(TfinK)
β = 3.3x10-3K-1
Sincetheductislaidhorizontally,theheatgainisby free
convection from theverticaland thehorizontaltopand
bottomsides.
Free convection from the vertical sides:
Grashof Number, Gr = (gxβ x L3 x∆T) /γ2
Gr = 6.8x107
Gr Pr = 4.7x107
SinceGr Pr < 109,flowislaminar.
Gr Pr valueisinbetween104and 109i.e.,104< Gr Pr < 109
So,Nu = 0.59(Gr Pr)0.25 (HMT Data book,Pg:135)
Nu = 48.85

14. Free Convection
Average heattransfercoefficient,h = Nu k /L
hv= 4.35W/m2K
Heat transferfrom verticalside,Qv = hvA (T∞–Tw) = h xL W x (T∞–Tw) = 20.88W
Heat transferfrom bothsideofverticalsides,Qv = 2xQv = 41.76W
Heat transferfrom horizontalsides:
For horizontalplate,Characteristiclength,Lc = W /2 = 0.4
Grashof Number, Gr = (gxβ xLc
3 x∆T) /γ2
Gr = 1.6x108
Gr Pr = 1.13x108
For horizontalplate,uppersurfaceheated, 8 x 106< Gr Pr < 1011.
Nu = 0.15[GrPr]0.333= 72.17
Heat transfercoefficientforuppersurfaceheated, hhu = Nu k /Lc = 4.82W/m2K
For horizontalplate,lowersurfaceheated, 105< Gr Pr < 1011
Nu = 0.27[GrPr]0.25 = 27.8
Heat transfercoefficientforlowersurfaceheated,hhl= Nu k /Lc = 1.85W/m2K
Heat transferfrom horizontalplate,Q H = (hu + hl)A ∆T = (hu + hl)x L W x(T∞–Tw) = 32.05W
Totalheattransfer,Q = {Heat transferfrom verticalsides}+ {Heat transferfrom horizontalsides}
Q = 73.8W

15. 5)A horizontalwireof3m m diameter ismaintained at100oC and isexposedtoairat20o
C. Calculatethefollowing:
1.Heat transfercoefficient,
2.Maximum current.Take resistanceofwireis7 ohm/m.
Free Convection
G iven:
Horizontalwirediameter,D = 3x10-3m
Airtemperature,T∞ = 20oC
Surface temperature, Tw = 100o C
Resistanceofthewire,R = 7ohm/m
To find:1.Heat transfercoefficient(h),2.Maximum current(I).
Solution:
FilmTemp, Tf = (Tw+ T∞) /2
Tf= 60oC
Propertiesofairat60oC
ρ = 1.060kg/m3
γ= 18.97x10-6m2/s
Pr= 0.696
k = 0.02896W/mK
(Pg:137)
(Pg:134)
β = 1/(TfinK)
β = 3 x10-3K-1
Grashoff Number, Gr = (gxβ xD3 x∆T) /γ2
Gr = 176.64
For horizontalcylinder,Nu = C [Gr Pr]m
Gr Pr = 122.9, CorrespondingC =0.85and m = 0.188
Nu = 2.1
Heat transfercoefficient,h = Nu k /D
h = 20.27W/m2K
Heat transfer,Q = h A (Tw–Tα) =h x π DL x(Tw–T∞ )= 15.2W/m
Heat transfer,Q = I2R
Maximum current,I= (Q/R)1/2= 1.47Amps /m

16. Free Convection
6)A Sphere ofdiameter 20 m m isat300oC isimmersed inairat25oC.Calculatetheconvective
heatloss.
Given:
Diameter ofsphere,D = 0.020m Surfacetemperature,Tw = 300oC Fluidtemp,T∞ = 25oC
To find:Convectiveheatloss,(Q )
Solution:
FilmTemp, Tf = (Tw+ T∞) /2
Tf= 162.5oC
Propertiesofairat
162.5oC ~ 160oC
ρ = 0.815kg/m3
γ= 30.09x10-6m2/s
Pr= 0.682
k = 0.03640W/mK
(Pg:137)
Heat transfercoefficient,h = Nu k /D
h = 14.51W/m2K
(Pg:134)
Heat transfer,Q =h A (Tw–T∞) =h x 4π r2x(Tw–T∞)
Convectiveheatloss,Q = 5.01W
β = 1/(TfinK)
β = 2.29x10-3K-1
Grashof Number, Gr = (gxβ xD3 x∆T) /γ2
Gr = 54,734.2
Gr Pr= 37,328.7
For sphere,[1< Gr Pr < 105]
Nu = 2+0.43[Gr Pr]0.25
Nu = 7.97

17. A vertical plate of 40 cm long ismaintained at 80 oC and isexposed to air at 22 oC.
Calculatethefollowing
i)Boundary layerthicknessatthetailingedgeoftheplate.
ii)The same plateisplacedina wind tunneland isblown overitata velocityof5
m/s. Calculatetheboundary layerthickness.
iii)Average heattransfercoefficientfornaturaland forcedconvectionfor
above mentioneddata.
A horizontal pipe of 15 cm diameter ismaintained at wall temperature of 200 oC and is
exposed toairat37 oC. Calculate the heat loss(including radiativeloss)per meter length
ifemissivityofpipeis0.92.
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