chapter 2 convection dr. r. velraj, professor anna university chennai heat transfer
TRANSCRIPT
CHAPTER 2
CONVECTION
Dr. R. VELRAJ, PROFESSORANNA UNIVERSITY CHENNAI
HEAT TRANSFER
1. Introduction to Convection
2. Boundary Layer Concepts
CHAPTER 2 (CONVECTION) – SESSION 1
IN THIS SESSION
Newton’s Law of Cooling
Q = h A (Tw – T∞)
Convection 1
GOVERNING LAW
h – convective heat transfer coefficient
A – surface area over which convection occurs
(Tw – T∞) – temperature potential difference
Flow Regimes on a flat plate
FLAT PLATE
LAMINARREGION
TRANSITION TURBULENTREGION
uu∞
u∞
u
x
y
Convection 2
CONCEPT OF BOUNDARY LAYER
u = 0 at y = 0 u = u∞ at y = δ
Laminar Region (Re < 5 x 105)FLOW REGIMES ON A FLAT PLATE
FLAT PLATE
x
y
u∞
u
LAMINAR BOUNDARY LAYER
dydu
τ - Shear stressµ - Dynamic viscosity (proportionality constant)
Reynolds’ no.
xuRe
Convection 3
Laminar Region REYNOLDS’ NUMBER
ρ Density, kg / m3
u∞ Free Stream Velocity, m / sx Distance from leading edge, mµ Dynamic viscosity, kg / m-s
Re < 5 x 105 FLOW OVER FLAT PLATERe < 2300 FLOW THROUGH PIPE
xuRe
Convection 4
Transition RegionFLOW REGIMES ON A FLAT PLATE
5 x 105 < Re < 106 FLOW OVER FLAT PLATE2000 < Re < 4000 FLOW THROUGH PIPE
FLAT PLATE
TRANSITION
Convection 5
Turbulent RegionFLOW REGIMES ON A FLAT PLATE
FLAT PLATELAMINAR SUB LAYER
BUFFER ZONE
TURBULENT CORETURBULENT BOUNDARY LAYER
x
y
Convection 6
u∞
u
Flow DevelopmentFLOW THROUGH TUBE
x
y
Convection 7
UNIFORM INLET FLOW
BOUNDARY LAYER
FULLY DEVELOPEDFLOW
STARTING LENGTH
x
y
Convection 8
THERMAL BOUNDARY LAYER
FLAT PLATE
T∞
δtTW
TEMPERATURE PROFILE
Convection 9
Dimensional Analysis
• Reduces the number of independent variables in a problem.
• Experimental data can be conveniently presented in terms of dimensionless numbers.
• Buckingham’s Pi theorem is used a rule of thumb for determining the dimensionless groups that can be obtained from a set of variables.
Number of independent dimensionless groups that can be formed from a set of ‘m’ variables
having ‘n’ basic dimensions is (m – n)
Convection 10
Buckingham’s Pi theorem
End of Session
QUESTIONS FOR THIS SESSION
1. What is Newton’s Law of Cooling ?2. Draw the boundary layer for a flow over a flat
plate indicating the velocity distribution in the laminar and turbulent flow region.
3. Draw the boundary layer for flow over through tube.
4. Define Buckingham’s π theorem
Convection 9
Dimensional Analysis
• Reduces the number of independent variables in a problem.
• Experimental data can be conveniently presented in terms of dimensionless numbers.
• Buckingham’s Pi theorem is used a rule of thumb for determining the dimensionless groups that can be obtained from a set of variables.
Number of independent dimensionless groups that can be formed from a set of ‘m’ variables
having ‘n’ basic dimensions is (m – n)
Convection 10
Buckingham’s Pi theorem
Convection 11
Dimensional Analysis for Forced ConvectionConsider a case of fluid flowing across a heated tubeS No. Variable Symbol Dimension
1 Tube Diameter D L2 Fluid Density ρ M L-3
3 Fluid Velocity U L t-1
4 Fluid Viscosity µ M L-1 t-1
5 Specific Heat Cp L2 t-2 T-1
6 Thermal Conductivity k M L t-3 T-1
7 Heat Transfer Coefficient h M t-3 T-1
Convection 12
Dimensional Analysis for Forced Convection• There are 7 (m) variables and 4 (n) basic
dimensions.• 3 (m-n) dimensionless parameters symbolized
as π1 ,π2, π3 can be formed.• Each dimensionless parameter will be formed
by combining a core group of ‘n’ variables with one of the remaining variables not in the core.• The core group will include variables with all
of the basic dimensions
Convection 13
Dimensional Analysis for Forced Convection
Choosing D, ρ, µ and k as the core (arbitrarily), the groups formed is represented as:
π1 = Da ρb µc kd Uπ2 = De ρf µg kh Cp
π3 = Dj ρl µm kn h
Since these groups are to be dimensionless, the variables are raised to certain exponents (a, b, c,….)
Convection 14
Dimensional Analysis for Forced Convection
Starting with π1
Equating the sum of exponents of each basic dimension to 0, we get equations for:M 0 = b + c + dL 0 = a – 3b + d + 1 + eT 0 = -dt 0 = -c -3d -1
tL
TtML
LtM
LM
LtTLMdcb
a33
0000 )(1
Convection 15
Dimensional Analysis for Forced Convection
Solving these equations, we get: d = 0, c = -1, b = 1, a = 1 giving
Similarly for π2
number) (Reynolds Re
UD1
TtL
TtML
LtM
LM
LtTLMigf
e3
2
330000 )(1
Convection 16
Dimensional Analysis for Forced Convection
Equating the sum of exponentsM 0 = f + g + IL 0 = e – 3f – g + i + 2T 0 = -i – 1t 0 = -g – 3i -2Solving, we get e = 0, f = 0, g = 1, i = 1 giving
number) (Prandtl PrkCp
2
Convection 17
Dimensional Analysis for Forced Convection
By following a similar procedure, we can obtain
The relationship between dimensionless groups can be expressed as F(π1, π2, π3) = 0. Thus,
number) (Nusselt Nuk
hD3
Pr)(Re, Nu
Convection 18
Dimensional Analysis for Forced Convection
Influence of selecting the core variables•Choosing different core variables leads to
different dimensionless parameters.• If D, ρ, µ, Cp were chosen, then the π groups
obtained would be Re, Pr and St.• St is Stanton number, a non dimensional form
of heat transfer coefficient.
p.U.Ch
Re.PrNu
St
Convection 19
Dimensional Analysis for Free Convection
TS (SURFACE)
T∞ (FLUID)g
LFLUID PROPERTIES
ρ,µ, CP, k, βg
Free Convection on a Vertical Plate
Convection 20
Dimensional Analysis for Free ConvectionFree Convection on a Vertical Plate
In free convection, the variable U is replaced by the variables ΔT, β and g.
Pertinent Variables in Free ConvectionS.No. Variable Symbol Dimension
1 Fluid Density ρ M L-3
2 Fluid Viscosity µ M L-1 t-1
3 Fluid Heat Capacity Cp L2 t-2 T-1
4 Fluid Thermal Conductivity k M L t-2 T-1
Convection 21
Dimensional Analysis for Free Convection
Pertinent Variables in Free Convection (contd.)S.No. Variable Symbol Dimension
5 Fluid Coefficient of Thermal Expansion β T-1
6 Gravitational acceleration g L t-2
7 Temperature difference ΔT T8 Significant length L L9 Heat Transfer Coefficient h M t-2 T-1
Convection 22
Dimensional Analysis for Free Convection
Choosing L, ρ, µ and k as the core (arbitrarily), the groups formed is represented as:
π1 = La ρb µc kd ΔTπ2 = Le ρf µi kj βgπ3 = Ll ρm µn ko Cp
π4 = Lp ρq µr ks h
Convection 23
Dimensional Analysis for Free Convection
Following the procedure outlined in last section,we get:
π1 = (L2 ρ2 k ΔT) / µ2
π2 = (Lµβg) / kπ3 = (µCp) / k = Pr (Prandtl number)π4 = (hL) / k = Nu (Nusselt number)
Grashof Number2
32
21TgL
.
Gr
FORCED CONVECTION
FREE CONVECTION
Convection 24
Dimensional Analysis
(Gr.Pr) Nu F
Pr)(Re, Nu
PRANDTL NUMBER
Convection 25
kpc
Pr
Multiplying with ρ in the numerator and denominator,
ydiffusivit Thermalydiffusivit Molecular
kpc
Pr
Prair = 0.7 Prwater = 4.5 Prliquid Na = 0.011
PRANDTL NUMBER
Convection 26
Pr = 1
δt
δh
Pr >> 1Pr << 1
δt = δh
δt
δh
δh = Hydrodynamic thicknessδt = Thermal Boundary layer thickness
End of Session
QUESTIONS FOR THIS SESSION
1. What are the dimensionless numbers involved in forced convection and free convection ?
2. Define Prandtl number.3. List the advantages of using liquid metal as
heat transfer fluid.4. Draw the hydrodynamic and thermal
boundary layer (in the same plane) for Pr << 1, Pr >> 1 & Pr = 1.
1. Continuity Equation
2. Momentum Equation
3. Energy Equation
Convection 27
What is …
Convection 28
Laminar – Momentum Equation –Flat Plate
FLAT PLATE
dydx
u∞ x
y
Momentum EquationAssumptions1. Fluid is incompressible2. Flow is steady3. No pressure variations in the direction
perpendicular to the plate4. Viscosity is constant5. Viscous-shear forces in ‘y’ direction are
negligible.
Convection 29
Laminar Boundary Layer on a Flat Plate
Convection 30
Continuity Equation
x
y
u - Velocity in x direction v - Velocity in y direction
Velocitydy
yv
v
dxxu
u
v
u
Convection 31
Continuity Equation – Laminar – Flat Plate
Mass flowdxdy
xv
v
vdx
udy dydx
xu
u
Convection 32
Continuity Equation
Mass balanceMass balance on the element yields:
OrMass Continuity Equation
0yv
xu
dxdyyv
vdydxxu
uvdxudy
Convection 33
Momentum Equation – Laminar – Flat Plate
x
y
Pressure Forces
p - Pressure
dydxxp
p
dy.p
Convection 34
Momentum Equation – Laminar – Flat Plate
x
y
dy
yu
yyu
dx.Shear
Stresses
yu
.dx.
µ - Dynamic viscosityu - Velocity in x direction v - Velocity in y direction
Convection 35
Momentum Equation – Laminar – Flat Plate
Newton’s 2nd Law
dmVd
F xx
Momentum flux in x direction is the product of mass flow through a particular side of control
volume and x component of velocity at that point
Convection 36
Momentum Equation – Laminar – Flat Plate
Momentum flux
dydxxu
u2
. uudy
dxdyyu
udyyv
v
. uvdx
Convection 37
Momentum Equation – Laminar – Flat Plate
Momentum and Force AnalysisNet pressure force
Net Viscous-Shear force
dxdyxp
dxdyyu2
2
Convection 38
Momentum Equation – Laminar – Flat Plate
Equating the sum of viscous-shear and pressure forces to the net momentum transfer in x direction, making use of continuity relation and neglecting second order differentials:
xp
yu
yv
vxu
u 2
2
Convection 39
Energy Equation – Assumptions1. Incompressible steady flow2. Constant viscosity, thermal conductivity and
specific heat.3. Negligible heat conduction in the direction of flow
(x direction).
FLAT PLATE
dydx
u∞
x
y
Convection 40
Energy Equation – Laminar – Flat Plate
x
y
dy
dx
Energy convected in (left face + bottom face) + heat conducted in bottom face + net viscous work done on element
Energy convected out in (right face + top face)
+ heat conducted out from top face
Convection 41
Energy Equation – Laminar – Flat Plate
x
y
u - Velocity in x direction v - Velocity in y direction
Energy Convected
dydxxT
Tdxxu
ucp
uTdycp
vTdxcp
dxdyyT
Tdyyv
vcp
Convection 42
Energy Equation – Laminar – Flat Plate
x
y
u - Velocity in x direction v - Velocity in y direction
Heat Conducted
yT
kdx
dy
yT
yyT
kdx
dy
yu
.dxyu
Net Viscous Work
Convection 43
Energy Equation – Laminar – Flat Plate
u - Velocity in x direction v - Velocity in y direction
Net Viscous Work
dy
yu
.dxyu
dxdyyu 2
Convection 44
Energy Equation – Laminar – Flat Plate
Writing energy balance corresponding to the quantities shown in figure, assuming unit depth in the z direction, and neglecting second-order differentials:
dxdyyv
xu
TyT
vxT
uc p
dxdyyu
dxdyy
Tk
2
2
2
Convection 45
Energy Equation – Laminar – Flat Plate
Using the continuity relation and dividing the whole equation by ρcp
for Low Velocity incompressible flow
2
p2
2
yu
cyT
yT
vxT
u
2
2
yT
yT
vxT
u
Convection 46
Energy Equation & Momentum Equation
2
2
yT
yT
vxT
u
2
2
yu
yu
vxu
u
Energy Equation
Momentum Equation(constant pressure)
The solution to the two equations will have exactly the same form when α = ν
End of Session
QUESTIONS FOR THIS SESSION
1. What is the momentum equation for the laminar boundary layer on a flat plate?
2. What are the assumptions involved in derivation of momentum equation?
3. Write the energy equation for laminar boundary layer on a flat plate
4. Explain the analogy between momentum and energy equation.
Convection 47
Integral form of Momentum Equation
Integral form of Momentum equation can be obtained using Von Kármán method:
(for constant pressure condition)
0yw0 y
uudy)uu(
dxd
Convection 48
Integral form of Momentum Equation
Polynomial equation for velocity
Boundary Conditions
34
2321 yCyCyCCu
0 yat 0u yat uu
yat 0yu
00yu2
2
yat
Convection 49
Integral form of Momentum Equation
Applying the boundary conditions, we get
Substituting,
Velocity Equation3y
21y
23
uu
u
23
C20C1 0C3 34u
21
C
Convection 50
Integral form of Momentum EquationUsing expression for velocity in integral equation,
Carrying out integration leads to
0y0
332
yu
dyy
21y
23
1y
21y
23
udxd
u
23
u28039
dxd 2
Convection 51
Integral form of Momentum EquationSince ρ and u∞ are constants, the variables may be separated to give
dxu13
140dx
u13140
d
constu
x13
1402
2
Convection 52
Integral form of Momentum EquationAt x=0, δ=0; so
Writing in terms of Reynolds number
BL thickness in terms of Reynolds number
Exact solution of BL equation
2/1xRe64.4
x
u
x64.4
x
2/1xRe0.5
x
Convection 53
Integral form of Energy Equation
FLAT PLATE
T∞
δtTW
TEMPERATURE PROFILE
wallyT
kAq
Convection 54
Integral form of Energy Equation
Polynomial equation for temperature
Boundary Conditions
34
2321 yCyCyCC
0 yat wTT tTT yat
t0yT
yat 00y
T2
2
yat
Convection 55
Integral form of Energy Equation
3
tt
y21y
23
Applying boundary conditions
Integral form of Energy Equation is given by:
0y
H0 y
T udy)TT(
dxd
Convection 56
Integral form of Energy EquationIntegral form of Energy Equation is given by:
Writing in terms of θ,
0y
H0 y
T udy)TT(
dxd
0y
H0 y
udy)(dxd
Convection 57
Integral form of Energy EquationWhere,
Using temperature & velocity profile equation in LHS
dyy
21y
23y
21y
23
1(dxd
u 3
H0
3
tt
t0y 23
yT
Convection 58
Integral form of Energy EquationPerforming algebraic manipulation and making the substitution ζ (zeta) = δt / δ
23
2803
203
dxd
u 42
3/1t Pr026.11
Convection 59
Heat Transfer Coefficient
wall
"
yT
kqAq
)TT(hq w"
Combining these equations,
TT
)y/T(kh
w
wall
Convection 60
Making an energy balance at the surface,
solving,
Local Nusselt Number
k23k
23
TT)y/T(k
htw
w
2/1x
3/1x RePr332.0Nu
End of Session
QUESTIONS FOR THIS SESSION
1. What is the assumption made by Von Karmen to solve the integral momentum equation ?
2. Write the velocity profile and the temperature profile equation used by Von Karmen in solving the momentum and energy equation
3. Write the equation to determine hydrodynamic & thermal boundary layer thicknesses
Part 14
CONVECTION
FORCED CONVECTION(FLOW OVER A FLAT PLATE)
CORRELATIONS
Convection 61
Heat Transfer Coefficient
wall
"
yT
kqAq
)TT(hq w"
Combining these equations,
TT
)y/T(kh
w
wall
Convection 62
Nusselt NumberMaking an energy balance at the surface,
Using expression for δT
Introducing Nusselt no. Local Nusselt Number
tw
w k23
TT)y/T(k
h
2/1x
3/1x RePr332.0Nu
2/13/1
x xu
Prk332.0h
kxh
Nu xx
Convection 63
Nusselt Number
Ratio of temperature gradients by conduction and convection at the surface
khL Nu Number, Nusselt
Nusselt Number is an indicative of temperature gradient at the wall in the normal direction
Convection 64
Nusselt NumberAverage Nusselt number is obtained from
LxL0
L0 x h2
dx
dxhh
Lxx Nu2kLh
Nu
Average Nusselt Number
2/1L
3/1x RePr664.0
kLh
Nu
Convection 65
Use of Correlations1. External Flow
Flow over a Flat PlateFlow across cylinderFlow across sphereFlow across bank of tubes
2. Internal FlowFlow through tubes & ducts
Convection 66
Use of Correlations
Separate correlations are available for • Laminar• Constant temperature surfaces• Constant heat flux boundary condition
• Turbulent• Constant temperature surfaces• Constant heat flux boundary condition
• Combined laminar & turbulent conditions
Special correlations are available for liquid metals
Convection 67
Fluid Friction and Heat Transfer
Shear stress at the wall may be expressed in terms of friction coefficient Cf :
Also,
Using velocity distribution equation,
2u
C2
fw
ww y
u
u
23
w
Convection 68
Fluid Friction and Heat TransferMaking use of relation for boundary layer thickness:
Combining equations,
2/1
w xu
64.4u
23
2/1x2
2/1fx Re332.0
u1
xu
64.4u
23
2C
Convection 69
Fluid Friction and Heat Transfer
The equation may be rewritten as:
Where,
2/1x
3/2
p
x
x
x RePr332.0uc
hPrRe
Nu
2/1x
3/2x Re332.0PrSt
uch
Stp
xx
2/1x
3/1x RePr332.0Nu
Convection 70
Fluid Friction and Heat Transfer
Reynolds-Colburn Analogy2
CPrSt fx3/2
x
2/1x
fx Re332.02
C
End of Session
QUESTIONS FOR THIS SESSION
1. What is the significance of Nusselt Number 2. What is the relationship between local and average
Nusselt number for a flow over a flat plate in the laminar region ?
3. What is drag coefficient ?4. Why separate correlations are available for liquid
metals ?5. What is Reynolds-Colburn analogy ?
Part 15
CONVECTION
FORCED CONVECTION(FLOW OVER A FLAT PLATE)
PROBLEMS
Convection 71
Example – Mass flow and BL thickness
CalculateBoundary Layer Thickness at x = 20 cm & 40 cmMass flow which enters the boundary layer between x=20 cm and x = 40 cm.Assume unit depth in z direction.
FLAT PLATE
AIR2 m/s, 27 °C, 1 atm
1.85x10-5 kg/m.sx
y
Convection 72
Example – Mass flow and BL thickness
Density of Air
Reynolds number
When x = 20 cm, Re = 27,580When x = 40 cm, Re = 55,160
3m/kg77.1RTp
p = 1.0132 x 105 R = 287 T = 300 Kρ = 1.177 kg/m3 u = 2 ms-1 µ = 1.85x10-5
ux
Re
Convection 73
Example – Mass flow and BL thicknessBoundary Layer Thickness
When x = 20 cm, δ = 0.00559 mWhen x = 40 cm, δ = 0.0079 m
Re = 27,580 when x = 20 cm (calculated)Re = 55,160 when x = 40 cm (calculated)
2/1xRe64.4
x
Convection 74
Example – Mass flow and BL thicknessMass flow entering the Boundary Layer
Velocity, u is given by
Evaluating the integral with this velocity distribution,
0
dy.u.
3y21y
23
uu
u85
dyy
21y
23
u0
3
Convection 75
Example – Mass flow and BL thicknessMass flow entering the Boundary Layer
2040u85
m
ρ = 1.177 kg/m3 u∞=2 m/sδ40 = 0.0079 m δ20 = 0.00559 m
kg/s 310x399.3m
Convection 76
Example – Isothermal flat plate (heated)
Plate is heated over its entire length to 60 °CCalculateHeat Transferred (a) at the first 20 cm of the plate(b) at the first 40 cm of the plate
Flat Plate, T = 60 °C
AIR2 m/s, 27 °C, 1 atm
µ = 1.85x10-5 kg/m.sx
y
Convection 77
Example – Isothermal flat plate (heated)Formulae Used
Heat Flow
Nusselt No.
Reynolds No.
All properties are evaluated at film temperature
2/1x
3/1xx RePr332.0
kxh
Nu
xu
Re
)TT.(A.hq w
Convection 78
Example – Isothermal flat plate (heated)Film Temperature
ν=17.36x10-6 m2/s Pr = 0.7k=0.02749 W/m°C cp=1.006 kJ/kg K
K5.316C5.432
6027Tf
Properties of air at Film Temperature:
Convection 79
Example – Isothermal flat plate (heated)At x = 20 cmReynolds No.
Nusselt No.
Heat Transfer Coefficient
u∞= 2 m/s Tf = 316.5 Kν = 17.36x10-6 m2/s Pr = 0.7k = 0.02749 W/m°C cp=1.006 kJ/kg K
Substituted Values
041,23xu
Re
74.44RePr332.0Nu 2/1x
3/1x
Cm/W15.6xk
Nuh 2xx
Convection 80
Example – Isothermal flat plate (heated)At x = 20 cm
Heat Flow
h = 6.15 W/m2 K Tw = 60 °CA = 0.2 m2 T∞ = 27 °C
Substituted Values
)TT.(A.hq w
W 81.18q
K.m/W3.12h.2h 2L
Convection 81
Example – Isothermal flat plate (heated)At x = 40 cmReynolds No.
Nusselt No.
Heat Transfer Coefficient
u∞= 2 m/s Tf = 316.5 Kν = 17.36x10-6 m2/s Pr = 0.7k = 0.02749 W/m°C cp=1.006 kJ/kg K
Substituted Values
082,46xu
Re
28.63RePr332.0Nu 2/1x
3/1x
Cm/W349.4xk
Nuh 2xx
Convection 82
Example – Isothermal flat plate (heated)At x = 40 cm
Heat Flow
h = 4.349 W/m2 K Tw = 60 °CA = 0.4 m2 T∞ = 27 °C
Substituted Values
)TT.(A.hq w
W 114.8q
K.m/W698.8h.2h 2L
End of Session
QUESTIONS FOR THIS SESSION
Calculate:Boundary Layer Thickness & Drag Coefficient at a distance of 0.61 m from leading edge of plate
AIRT = 37.8 °Cu = 0.915 m/sρ = 1.126 kg/m3
ν = 0.167x10-4 m2/sFlat Plate
X = 0.61 m
yLeading
Edge
1
End of Session
QUESTIONS FOR THIS SESSION
Calculate:Local heat transfer coefficient and the heat transfer for 0.61 m length taking width of plate as 1 m
2AIR
T = 65.6 °Cu = 0.915 m/sν = 0.223x10-4 m2/sk = 0.0313 W/mK
Flat Plate at 121.1 °C
y
X = 0.61 m
Part 21
CONVECTION
FORCED CONVECTION(EXTERNAL FLOW)
CORRELATIONS & PROBLEMS
Convection 83
CORRELATIONS – EXTERNAL FLOW
FLAT PLATE
Laminar Flow
333.00
333.05.0xx ]75.0)x/x(1.[PrRe332.0Nu
xL Nu2Nu
Flat PlateLeading
Edge
X0
δt
δh
FLAT PLATETurbulent Flow (Fully turbulent from leading edge)
Combined Laminar and Turbulent Flow
Convection 84
CORRELATIONS – EXTERNAL FLOW
33.08.0xx PrRe0296.0Nu
33.08.0L PrRe037.0Nu
333.08.0LL PrARe037.0Nu
5.0cr
8.0cr Re664.0Re037.0A
CYLINDERGeneralised EquationNuD – Nusselt number based on diameterAll properties to be taken at film temperature
Convection 85
CORRELATIONS – EXTERNAL FLOW
333.0mDD PrReCNu
Re D C m0.4 – 4 0.989 0.3304.0 – 40 0.911 0.385
40 – 4000 0.683 0.4664000 – 40000 0.193 0.618
TUBE BANKS
Convection 86
CORRELATIONS – EXTERNAL FLOW
INLINE
St
STAGGERED
SL
St
D
SL
D
SL
TUBE BANKS
Convection 87
CORRELATIONS – EXTERNAL FLOW
333.0n PrRec13.1Nu For N ≥ 10
10N1NucNu 1 ≤ N ≤ 10Re to be calculated based on max. fluid velocity Vmax
INLINE
STAGGERED
where
u.)DS/(SV TTmax
u)DS(2/SV DLmax
5.02T
2LD )2/S(SS
TUBE BANKS (INLINE)
Convection 88
CORRELATIONS – EXTERNAL FLOW
For 10 ROWS or MORE
ST / D
SL / D
1.25 1.5 2.0 3.0
C n C n C n C n
1.25 0.35 0.59 0.28 0.608 0.1 0.704 0.063 0.751.5 0.37 0.586 0.25 0.62 0.1 0.702 0.068 0.742 0.42 0.57 0.29 0.60 0.23 0.632 0.198 0.653 0.29 0.60 0.357 0.584 0.37 0.581 0.286 0.61
TUBE BANKS (STAGGERED)
Convection 89
CORRELATIONS – EXTERNAL FLOW
For 10 ROWS or MORE
ST / D
SL / D
1.25 1.5 2.0 3.0
C n C n C n C n
0.6 - - - - - - .213 .6361 - - .497 .558 - - - -
1.5 .451 .568 .46 .562 .452 .568 .488 .5683 .31 .592 .356 .58 .44 .562 .421 .574
TUBE BANKS ( C1 values )
Convection 90
CORRELATIONS – EXTERNAL FLOW
For LESS than 10 ROWS
ST – STAGGEREDIN – INLINE
N 1 2 3 4 5 6 7 8 9 10ST .68 .75 .83 .89 .92 .95 .97 .98 .99 1IN .64 .8 .87 .9 .92 .94 .96 .98 .99 1
10N1NucNu
CALCULATEHeat Transfer Coefficient for full length of plateRate of Energy Dissipation from the plateConvection 91
Example – Heated Flat Plate
Flat Plate at 90 °C
y
XAIR
T = 0 °Cu = 75 m/s
45 cm LONG, 60 cm WIDEAssume transition takes place at Re X, C = 5 x 105
Properties of air at Film Temperature
Critical Length (distance at which transition takes place)
Convection 92
Film Temperature
u∞ = 75 m/s ν=17.45x10-6 m2/sk=2.8 x 10-2 W/m°C Pr = 0.698
K318C452
090Tf
Example – Heated Flat Plate
5cc 105
xuRe
cm 11.6xc
Convection 93
Heat Transfer Coefficient
u∞= 75 m/s L = 0.45 mν = 17.45x10-6 m2/s Pr = 0.698k = 2.8 x 10-2 W/m°C
Substituted Values
6L 1093.1
LuRe
2732Pr870Re037.0Nu 3/15/4LL
Cm/W170Lk
Nuh 2LL
Example – Heated Flat Plate
Convection 94
RATE OF ENERGY DISSIPATION FROM THE PLATE
hL = 170 W/m2 K A = 0.45 x 0.6 m2
TS = 90 °C T∞ = 0 °CSubstituted
Values
)TT(Ah2Q SL
kW 8.262Q
Example – Heated Flat Plate
End of Session
QUESTIONS FOR THIS SESSION1. Air at 1 atm and 350C flows across 5.9 cm diameter
cylinder at a velocity of 50m/s. The cylinder surface is maintained at a temperature of 1500C. Calculate the heat loss per unit length of the cylinder.
2. A fine wire having a diameter of 3.94 X 10-5 m is placed in a 1 atm airstream at 250C having a flow velocity of 50 m/s perpendicular to the wire. An electric current is passed through the wire, raising its surface temperature to 500C. Calculate the heat loss per unit length.
Part 22
CONVECTION
FORCED CONVECTION
CORRELATIONS & PROBLEMS
CALCULATEHeat lost while standing in the wind
Convection 95
Example – Flow over Cylinder
D = 30 cmAIRT = 10 °Cu = 36 km/h
H = 1.7 mTS = 30 °C
Assume a man (represented as a cylinder) standing in the direction of wind
Properties of air at Film Temperature
Reynolds Number
Convection 96
Film Temperature
ν = 15x10-6 m2/s k = 2.59 x 10-2 W/m°CPr = 0.707
K393C202
1030Tf
5D 102
DuRe
u∞ = 10
m/sD = 0.3 m
Example – Flow over Cylinder
Convection 97
Rate of Heat Lost
ReD= 2 x 105 Pr = 0.707k = 2.59 x 10-2 W/m°CTS = 30 °C T∞ = 10 °C
Substituted Values
7.444PrRe027.0Nu 333.0805.0DD
Cm/W39.38Dk
Nuh 2DD
kW2.1)TT(AhQ SD
Example – Flow over Cylinder
CALCULATETotal heat transfer per unit length for tube bank and the exit air temperatureConvection 98
Example – Flow through Tube Banks
AIR1 atm, 10 °C
u = 7 m/s
15 ROWS HIGH5 ROWS DEEPSL = ST = 3.81 cm
Heating of air with in-line tube bank
SL Tsurface = 65°C
2.54 cm
ST
Properties of air at Film Temperature
Constants for use ( C & n ) from table
Convection 99
Film Temperature
µ = 1.894 x 10-5 kg/ms ρ = 1.137 kg/m3
k = 0.027 W/m°C Pr = 0.706
K5.310C5.372
1065Tf
5.154.281.3
DSL C =
0.25n = 0.62
5.154.281.3
DST
Example – Flow through Tube Banks
Convection 100
Maximum Velocity
D = 0.0254 m ST = 3.81 u∞ = 7 m/s n = 0.62µ = 1.894 x 10-5 kg/ms ρ = 1.137 kg/m3
c = 0.25 k = 0.027 W/m°C
s/m21u.)DS/(SV TTmax
020,32DV
Re max
35.155RecNu n
Cm/W14.165D/)k.Nu(h 2D
Example – Flow through Tube Banks
Convection 101
Heat Transferred
m/m985.5DLNA 2
)TT(hAQ w
Correction Factor ( C1 ) = 0.92 (from table) Total heat transfer surface area (assuming unit length)
N = 15 D = 0.0254 m L = 1m
Example – Flow through Tube Banks
Convection 102
)TT(mc2
TTThAQ 1,2,p
2,1,w
Subscripts 1 & 2 denote entrance & exit temperatures
Substituting
s/kg99.4S)15(um L
3m/kg246.1)RT/(p
Example – Flow through Tube Banks
C08.19T 2,
Convection 103
Heat Transferred
m/W6.45)TT(mcQ 1,2,p
Example – Flow through Tube Banks
Convection 104
Mixing Cup Temperature / Bulk Mean Temperatureis the temperature, the fluid would assume if placed in a
mixing chamber and allowed to come to equilibrium.
INTERNAL FLOWUNIFORM INLET FLOW
BOUNDARY LAYER
FULLY DEVELOPED FLOW
STARTING LENGTH
A
pzmpm TdAcvTc)Av(
INTERNAL FLOW
A
zm dAvA1
v
Az
Az
m dAv
TdAvT
0
0
r0 z
r0 z
mdr.r2.v
dr.r2.TvT
Where, A
zm
m TdAvAv
1T
A
pzmpm TdAcvTc)Av(
forCIRCULAR
DUCT
R0 z
m2m dr.r.T.vuR2
T
Convection 105
Convection 106
INTERNAL FLOW
2
2
rru
ru
vxu
u
Momentum Equation
(constant pressure)
Energy Equation
rT
r1
rT
ck
xT
u 2
2
p
For Slug flow…
rT
r1
rT
xT
u 2
2
CALCULATEAir side heat transfer coefficient across the tube bundleConvection
Example – Flow through Tube Banks
AIRT∞ = 15 °C u∞ = 6 m/s
7 ROWS in direction of flow
SL = ST = 20.5 mm
Water passing through Staggered tube bankTsurface = 70°C
1.64 cm
SL
1
End of Session
QUESTIONS FOR THIS SESSION
2. What is ‘bulk mean temperature or mixing cup temperature’ ?
3. What is slug flow ?4. Write the momentum and energy
equation for the flow through a tube.
Part 23
CONVECTION
FORCED CONVECTION(INTERNAL FLOW)
CORRELATIONS & PROBLEMS
Convection 107
CORRELATIONS – INTERNAL FLOWProperties to be evaluated at Bulk Mean Temperature
Tm = (Tmi + Tmo) / 2Tmi – Mean Temperature at inletTmo – Mean Temperature at outlet
LAMINAR FLOWFully developed Thermal Layer
Constant Wall Temperature
Constant Heat Flux
D L ,66.3Nu0.6Pr ,36.4Nu
Convection 108
CORRELATIONS – INTERNAL FLOWLAMINAR FLOW (contd.)Entry region (Hydrodynamic layer fully developed, thermal layer developing)
Simultaneous development of hydrodynamic & thermal layers
67.0D
D
Pr]Re)L/D[(04.01PrRe)L/D(0668.0
66.3Nu
8.0D
D
]x/D.Pr[Re16.01)x/D.Pr(Re104.0
66.3Nu
0.7Pr
0.6Pr
Convection 109
CORRELATIONS – INTERNAL FLOWTURBULENT FLOWFully Developed flow (Dittus-Boelter equation)
n = 0.4 for heating of fluids / n = 0.3 for cooling of fluids0.6 < Pr < 100, 2500 < Re < 1.25 x 106 ; L/D > 60
Fully Developed flow (Sieder-Tate equation)
0.7 < Pr < 16,700 ; ReD ≥ 10,000 ; L / D ≥ 60
n8.0D PrRe023.0Nu
14.0wm
n8.0D /PrRe027.0Nu
CALCULATE1. Reynolds number2. Heat Transfer Coefficient3. Difference between wall temperature and bulk (mean)
temperature.
Convection 110
Example 1
Water flowing through pipe with constant wall heat flux
Douter = 2 cm
WATERT = 25 °C
m = 0.01 kg/s
Constant Wall Heat Fluxqs = 1 kW/m2
Properties of water at 25 °C
< 2300. Flow is LAMINAR
For Constant Heat Flux
Convection 111
µ = 8.96 x 10-4 kg/ms k = 0.6109 W/m°C
364.4NuD
Example 1
709uD
Re
skg/m 28.31)01.0(
01.0Am
u 2
Cm/W3.133D/)k.Nu(h 2DD
D = 0.02 m
Difference between Wall Temperature and Bulk (mean) Temperature
Convection 112
Example 1
)TT(hq ms''s
C5.73.133
1000hq
TT''s
ms
CALCULATEAverage heat transfer coefficient by using Sieder-Tate equationConvection 113
Example 2
Water flowing through Copper Tube with constant wall temperature
Douter = 2.2 cm
WATERTinitial = 15 °CTfinal = 60 °Cu = 2 m/s
Constant Wall TemperatureTs = 95 °C
Properties of water
Convection 114
Bulk (mean) Temperature
µ = 0.69 x 10-3 N.s/m2 ρ = 990 kg/m3
k = 0.63 W/m°C cp = 4160 J/kg.KD = 0.022 m
K5.310C5.372
6015Tf
Example 2
63213uD
Re
56.4kc
Pr p
Convection 115
µ = 0.69 x 10-3 N.s/m2 ρ = 990 kg/m3
k = 0.63 W/m°C cp = 4160 J/kg.Kµs = 0.3 x 10-3 N.s/m2 D = 0.022 m
Example 2
84.346(Pr))(Re027.0Nu14.0
s
3/15/4DD
Cm/W9932D/)k.Nu(h 2DD
Estimate the heat leakage per meter length per unit temperature difference.
Convection 116
Example 3
Heat Leakage from an air conditioning duct
400 X 800 mm
AIRT = 20 °Cu = 7 m/s
Properties of air
Equivalent or Hydraulic Diameter
Assuming pipe wall temperature to be higher than air temperature, then Nusselt number is given by:
Convection 117
ν = 15.06 x 10-6 m2/s α = 7.71 x 10-2 m2/hk = 0.0259 W/mK
703.0/Pr
Example 3
4107.23uD
Re
m571.0)8.04.0(2)8.04.0(4
PA4
Dh
38.398PrRe023.0Nu 4.05/4D
Heat Leakage per unit length per unit temperature difference:
Convection 118
NuD = 398.38 k = 0.0259 W/mKD = 0.571 m
Example 3
Cm/W07.18D/)k.Nu(h 2DD
CW/m 37.432.1207.18hPQ
CALCULATE1. Heat transfer coefficient2. Total amount of heat transferred
Convection
Questions
Water flowing through a heated tube
Douter = 1.5 cmL = 3 m
WATERTinitial = 50 °CTfinal = 64 °Cu = 1 m/s
Constant Wall TemperatureTs = 90 °C
1
CALCULATEHeat transfer coefficient of air
Convection
Questions
Air flowing through annulus
ID = 3.125 cmOD = 5 cm
AIRTinitial = 16 °CTfinal = 32 °Cu = 30 m/s
Tsurface (of inner tube) = 50 °C
2
Part 24
CONVECTION
FREE CONVECTION
Free Convection Boundary LayerHeated Vertical Plate
Convection 119
TsTy
T∞
u(y)
δδt
VELOCITY BOUNDARY
LAYER
THERMAL BOUNDARY
LAYER
y, v
x, U
T∞, ρ∞,
g
Free Convection – Governing Equations
Continuity Equation
X Momentum Eqn.
Energy Equation
0yv
xu
Convection 120
2
2
yu
gxp1
yu
xu
u
2
2
p yT
ck
yT
xT
u
Free ConvectionX Momentum Equation
u0, ρρ∞ (density outside boundary layer)
Convection 121
2
2
yu
gxp1
yu
xu
u
gxp
2
2
yu
)(g
yu
xu
u
Free ConvectionX Momentum Equation
Convection 122
2
2
yu
)(g
yu
xu
u
pTp1
)TT(
2
2
yu
)TT(gyu
xu
u
T
Free ConvectionVolumetric Coefficient of thermal expansion, β
Convection 123
pT1
T1
RTp1
2
2p RT
pT1
TRp
T
RT/p
Free Convection
Convection 124
2
2
yu
)TT(gyu
xu
u
0yv
xu
2
2
yT
yT
xT
u
Summarizing the governing equations,
Free Convection
Convection 125
2*
*2
L
*20
s*
**
*
**
y
uRe
1T
uL)TT(g
yu
xu
u
Lx
x*
2*
*2
L*
**
*
**
y
TPrRe
1yT
xT
u
Identification of Dimensionless Groups
Ly
y* *0
*
uu
u
TTTT
T*
*
0yv
xu
*
*
*
*
Free Convection
2L
L2
0
2s
3
20
s
ReGr
)/Lu(/)TT(Lg
uL)TT(g
Rearranging
Where,
2s
3 )TT(Lg
Gr number,Grashof L
“ratio of buoyancy force to the viscous force in fluid” This number plays similar role in free convection as
does the Reynolds number in forced convection
Free Convection in External FlowsVertical SurfacesLaminar (Gr.Pr < 109)Constant Wall Temperature
Constant Heat Flux
Turbulent (Gr.Pr > 109)
25.0x
25.05.0x Gr.Pr)952.0(Pr 508.0Nu
Convection 127
11xx
52.0xxx 10NuGr10forPr).Nu.Gr(6.0Nu
333.0x Pr)Gr( 1.0Nu
Free Convection in External FlowsHorizontal SurfacesCharacteristic Length
Constant Wall Temperature
Constant Heat Flux
plate theof Perimeterplate theof area Surface
L
Convection 128
6425.0 108PrGr102forPr)Gr(54.0Nu 1163/1 10PrGr108forPr)Gr(15.0Nu
8x
333.0 102PrGrforPr)Gr(13.0Nu 1162.0 10PrGr10forPr)Gr(16.0Nu
Combined Free & Forced Convection
When air is flowing over heated surface at a low velocity, the effect of free and forced convections
are equally important
)negligible convection (free convection Forced1)Re/Gr( 2
Convection 129
)negligible convection (forced convection Free1)Re/Gr( 2
forced)and (free convection Mixed1)Re/Gr( 2
Combined Free & Forced ConvectionExternal Flow
Internal Flow (LAMINAR)
Graetz numberConvection 130
A)Re/(GrPrRe332.0Nu 2xx
3/12/1x if
A)Re/(GrGr)Pr952.0(Pr508.0Nu 2xx
4/1x
4/1-2/1 if
333.0333.1333.014.0
w)Gr.Gz(012.0Gz75.1Nu
)L/D.(Pr.ReGz D
Combined Free & Forced ConvectionInternal Flow (TURBULENT)
Applicable for ReD > 2000 and RaD (D/L) < 5000OrReD > 800 and RaD (D/L) > 2x 104
Convection 131
36.007.0D
21.027.0D (D/L)GrPrRe69.4Nu
Minimum spacing (L) to avoid interference of free convection boundary layers
Convection 132
Example – Convection between Vertical Plates
TSurface = 80 °C
?
δδ
L
3.5 cm
Twater = 20 °C
WATER
PLATE
Properties of water at Film Temperature
< 1 x 109 (LAMINAR)
Convection 133
Let, δ be the boundary layer thickness at trailing edgeMinimum spacing required = L = 2δFilm temperature = t∞ = (80 + 20) / 2 = 50 °C
Pr = 3.54 β = 0.48x10-3 K-1 ν = 0.567 x 10-6 m2/s
7
2
3
1076.3
Tgx
Grx
910894.0Pr.Gr
Example – Convection between Vertical Plates
Convection 134
Boundary layer thickness (δ)
Minimum Space to avoid interference
039.0.Pr)952.0(Pr93.3 25.025.05.0
x
x Grx
Example – Convection between Vertical Plates
Pr = 3.54 x = 0.035 m Gr = 0.2526x109
mm36.1035.0038.0
mm72.22
Part 24
Questions
1. Draw the free convection boundary layer on a heated vertical plate.
2. Write the governing equations for free convection3. What is the significance of Grashof number ?4. Explain the situations under which combined free
and forced convection should be considered.
Part 25
CONVECTION
FREE CONVECTIONProblems
Heat lost by pipe / metre length
Convection 135
Example 1 – Vertical PipeDouter = 10 cm
AIR (ambient)T = 20 °C
TSurface = 100 °C
Vertical pipe kept in a room
L = 30 cm
?
Properties of air at Film Temperature
Convection 136
Film Temperature
Pr = 0.696 β = 0.003003 K-1
ν = 18.97 x 10-6 m2/s k = 0.02896 W/m°C
K293C602
20100Tf
Example 1 – Vertical Pipe
103
s3
LL 1025.12Pr)TT(Lg
Pr.GrRa
L = 3 m T∞ = 100°C TS = 20°C
Checking
Then,
Convection 137
4/1L )Gr(35
LD
Example 1 – Vertical Pipe
RaL = 12.25 x 1010 k = 0.02896 W/m°CL = 3 m T∞ = 100°C TS = 20°C
true. is 4/110)106.17(35
1.03
488Ra1.0Nu 3/1LL
K71.4L/)k.Nu(h L2W/m
m/W37.118)TT).(D.(hQ s
Heat gained by duct / metre length
Convection 138
Example 2 – Horizontal Duct60 cm
Horizontal un-insulated Air Conditioning Duct
AIR (ambient)T = 25 °C
TSurface = 15 °C 30 cm
?
Properties of air at Film Temperature
Rate of Heat Gained per unit length of duct
Convection 139
Film Temperature
Pr = 0.705 β = 0.00341 K-1 ρ = 1.205 kg/m3
ν = 15.06 x 10-6 m2/s k = 0.02593 W/m°C
K293C202
2515Tf
bottomtopside QQQ2Q
Example 2 – Horizontal Duct
)TT(L2)hh()Lh2(Q scharbottomtopcharside
392
3
1001.0Pr)(
Pr. LTTLg
GrRa sLL
Heat gained from vertical wall (sides)
Laminar 45.13Ra59.0Nu 4/1
LL
539 107.23.0 1001.0 LRa
Example 2 – Horizontal Duct
K162.1L/)k.Nu(h L2W/m
m/W97.6)TT(Lh2Q svv
β = 0.00341 K-1 ρ = 1.205 kg/m3
ν = 15.06 x 10-6 m2/s k = 0.02593 W/m°CT∞ = 25°C TS = 15°C Pr = 0.705
Heat gained from top & bottom surfacesCharacteristic Length
Laminar
Similarly for bottom surface,
31.12Ra54.0Nu 4/1LL
539 107.23.0 1001.0 LRa
Example 2 – Horizontal Duct
K063.1L/)k.Nu(h Lt2W/m
m/W726.9)TT(L2)hh(Q sbtbt
β = 0.00341 K-1 ρ = 1.205 kg/m3
ν = 15.06 x 10-6 m2/s k = 0.02593 W/m°CT∞ = 25°C TS = 15°C Pr = 0.705
m3.02/wP/AL
TOP
SURF
ACE
4/1Lb )Ra(27.0)L/k(h
Convection 142
Rate of Heat GainedExample 2 – Horizontal Duct
bottomtopside QQQ2Q
m/W7.16Q
Qside = 6.97 W/m Q top + bottom = 9.73 W/m
Calculate the heat transfer coefficient
Convection 143
Combined Free & Forced Convection with Air
TSurface = 140 °C
0.4m
25 mmAIR
Tair = 27 °Cu = 30 cm/s
TUBEAIR
Air flowing through a horizontal tube3
Convection 144
Film Temperature
Properties of air at Film Temperature
Reynolds Number
Pr = 0.695 β = 2.805x10-3 K-1 ρ = 0.99 kg/m3
µ bulk = 2.1 x 10-5 kg/m.s k = 0.0305 W/m°Cµw = 2.337 x 10-5 kg/m.s
K5.356C5.832
27140Tf
Combined Free & Forced Convection with Air
53.3uD
Re
D = 0.025 mu = 0.3 m/s
Convection 145
Pr = 0.695 β = 2.805x10-3 K-1 ρ = 0.99 kg/m3
µf = 2.1 x 10-5 kg/m.s k = 0.0305 W/m°Cµw = 2.337 x 10-5 kg/m.s
Combined Free & Forced Convection with Air
353Re m
uD
5
2
310007.1
TgxGr
4677Ld
.Pr.Gr 33.15Ld
.Pr.ReGz
Convection 146
k = 0.0305 W/m°C µw = 2.337 x 10-5 kg/m.s µ = 1.8462 x 10-5 kg/m.s Gz = 15.33Gr = 1.007 x 109 d = 0.025 m
Combined Free & Forced Convection with Air
333.0333.1333.014.0
w)Gr.Gz(012.0Gz75.1Nu
7.7Nu
K4.9d/)k.Nu(h 2W/m
Panel : 0.915 m x 0.915 mOne side insulated, other side at 65.6 °CAmbient is at 10 °C
Questions
1
INSU
LATED
SURFA
CE
HOT
SURFA
CE
INSULATEDSURFACE
HOTSURFACE
HOTSURFACE
INSULATEDSURFACE
Calculate the mean heat transfer coefficient due to free convection
Estimate the heat gained by the duct.
Convection
Questions
30 X 20 cmDuct Surface at 5 °CAIR
T = 25 °C
2
Air flow through Rectangular Duct
Calculate the heat transferred considering combined free and forced convection
Convection
Questions
Air flowing through a tube
D = 20 mmL = 1 m
AIRT = 27 °C
u = 30 cm/s
Horizontal TubeTsurface = 127 °C
3