Convective Heat Transfer Coefficients of Automatic Transmission ...
Convective Heat Transfer - Rajiv Gandhi College of Engineering … YEAR/HEAT AND MASS... ·...
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Convective Heat Transfer
2.1. DIMENTIONAL ANALYSIS
Dimensional analysis is a mathematical methods which makes use of the study of the
dimensions for solving several engineering problems. This methods can be applied to all types
of fluid resistances, heat flow problems and many other problems in fluid mechanics and
thermodynamics.
2.1.1. Dimensions
In dimensional analysis, the various physical quantities used in fluid phenomenon can be
expressed in terms of fundamental quantities. These fundamental quantities are mass (M),
length (L), time (T), and temperature (θ)
The dimensions of commonly used quantities in heat transfer analysis is listed in
Table2.1 with reference to MLθT where
M = Mass,
L = Length,
Theta = Temperature,
T = Time.
For example,
Velocity V = Distance
Time =
L
T = LT -1
1
2.1.2. Buckingham π Theorem
A more general situation in which dimensional analysis may be profitably employed is
one in which there is no governing differential equation clearly applies. In such a situation, a
more general procedure is required which is known as Buckingham π theorem.
2
Buckingham π theorem states as follows.
“If there are n variables in a dimensionally homogeneous equation and if these contain
m fundamental dimensions, then the variables are arranged into (n - m) dimensionless terms.
These dimensionless term are called π terms.”
2.1.3. Advantages of Dimensional Analysis
1. If expresses the functional relationship between the variables in dimensionless
terms.
2. It enables getting up a theoretical solution in a simplified dimensionless form.
3. Design curves, by the use of dimensional analysis, can be developed from the
experimental data or direct solution of the problem.
4. The result of one series of tests can be applied to a large number of other similar
problems with the help of dimensional analysis.
2.1.4. Limitations of Dimensional Analysis
1. The complete information is not provided by dimensional analysis. It only indicates
that there is some relationship between the parameters.
2. No information is given about the internal mechanism of the physical phenomenon.
3. Dimensional analysis does not give any clue regarding the selection of variables.
2.2. DIMENSIONLESS NUMBERS AND THEIR PHYSICAL SIGNIFICANCE
2.2.1. Reynolds Number (Re)
It is defined as the ratio of inertia force to viscous force.
Re = 𝐼𝑛𝑒𝑟𝑡𝑖𝑎 𝐹𝑜𝑟𝑐𝑒
𝑉𝑖𝑠𝑐𝑜𝑢𝑠 𝐹𝑜𝑟𝑐𝑒
= ρU2L2
µ UL
Re = UL
v …(2.1)
Where U - Velocity, m/s,
L - Length, m,
3
V = µ
𝜌 - Kinematic Viscosity, m2/s.
Reynolds number, is therefore, a measure of relative magnitude of the inertia force to
the viscous force occurring in the flow.
2.2.2. Prandtl Number (Pr)
It is the ratio of the momentum diffusivity to the thermal diffusivity.
Pr = Momentum diffusivity
Thermal diffusivity
Pr = µ 𝐶𝑝
𝑘 =
V
𝛼 …(2.2)
Where V = Kinematic Viscosity, m2/s,
α = Thermal diffusivity, m2/s.
Prandtl number provides a measure of the relative effectiveness of the momentum and
energy transport by diffusion.
2.2.3. Nusselt Number (Nu)
It is defined as the ratio of the heat flow by convection process under an unit temperature
gradient to the heat flow rate by conduction under an unit temperature gradient through a stationary
thickness of L metre.
Nusselt Number (Nu) = 𝑞𝑐𝑜𝑛𝑣
𝑞𝑐𝑜𝑛𝑣 =
ℎ 𝐴 ∆𝑇
𝑘 𝐴 ∆𝑇
𝐿
= ℎ𝑘
𝐿
(Nu) = ℎ𝐿
𝑘 …(2.3)
Where h - Heat transfer coefficient W/m2k,
L - Length,m,
K - Thermal conductivity, W/mK.
The Nusselt number is a convenient measure of the convective heat transfer coefficient.
For a given value of the Nusselt Number, the convective heat transfer coefficient is directly
proportional to thermal conductivity of the fluid and inversely proportional to the significant
length.
4
2.2.4. Grashof Number (Gr)
It is defined as the ratio of product of inertia force and buoyancy force to the square of
viscous force.
Gr = 𝐼𝑛𝑒𝑟𝑡𝑖𝑎 𝑓𝑜𝑟𝑐𝑒 𝑥 𝐵𝑢𝑜𝑦𝑎𝑛𝑐𝑦 𝑓𝑜𝑟𝑐𝑒
(𝑉𝑖𝑠𝑐𝑜𝑢𝑠 𝑓𝑜𝑟𝑐𝑒 )2
Gr = ρ U2 L2 x ρβg∆TL3
(µUL )2
= g ρ2βL3∆T
µ2
Gr = g x β x L3 x ∆T
v2 …(2.4)
Where β - Coefficient of expansion, K-1
L - Length, m,
V - kinematic viscosity, m2/s,
∆𝑇 - Temperature difference, K.
Grashof Number has a role in free convection similar to that played by Reynolds number
in forced convection.
2.2.5. Stanton Number (St)
It is the ratio of Nusselt Number to the product of Reynolds number and Prandtl
Number.
5
2.2.6. Newtonion and Non-Newtonion fluids
The fluids which obey the Newton’s law of viscosity are called the Newtonion fluids and
those which do not obey are called non-newtonion fluids.
2.2.7. Laminar Flow
Laminar flow is sometimes also called stream line flow. In this type of flow, the fluid
moves in layers and each fluid particles follows a smooth and continuous path. The fluid
particles in each layer remain in an orderly sequence without making with each other.
2.2.8. Turbulent Flow
In addition to the laminar type of flow, a distinct irregular flow I frequently observed in
nature. This type of flow is called turbulent flow. The path of any individual particles is Zig-Zag
and irregular. Fig 2.1 shows the instantaneous velocity in laminar and turbulent flow.
2.3. Boundary layer concept
The concept of a boundary layer as proposed by Prandtl forms the starting point for the
simplification of the equation of motion and energy.
When a real fluid i.e., Viscous fluid, flows along a stationary solid boundary, a layer of
fluid which comes in contact with the boundary surfaces. Thus the layer of fluid which cannot
slip away the boundary surfaces and undergoes retardation. This retarded layer further causes
retardation for the adjacent layer of the fluid. So, small region is developed in the immediate
vicinity of the boundary surfaces in which the velocity of the flowing fluid increases rapidly from
zero at boundary surfaces and approaches the velocity of main stream.
6
The layer adjacent to the boundary is known as boundary layer. Boundary layer is
formed whenever there is relative motion between the boundary and the fluid.
In this concept, the flow over a body is divided into two regions.
1. A thin region near the body called the boundary layer, where the velocity and
temperature gradients are large.
2. The region outside the boundary layer where velocity and temperature gradients are
very nearly equal to their free stream values.
The thickness of the boundary layer has been defined as the distance from the surface
at which the local velocity or temperature reaches 99% of the external velocity or temperature.
Fig. 2.2. Boundary layer on flat plate
2.3.1. Types of Boundary Layer
1. Hydrodynamic boundary layer (or) Velocity boundary layer
2. Thermal boundary layer
2.3.2. Hydrodynamic Boundary Layer
In hydrodynamic boundary layer, velocity of the fluid is less than 99% of free stream
velocity.
2.3.3. Thermal Boundary Layer
In thermal boundary layer, temperature of the fluid is less than 99% of free stream temperature.
7
2.4. CONVECTION
Convection is a process of heat transfer that will occur between a solid surface and a
fluid medium when they are at different temperatures.
2.4.1. Newton’s Law of Convection
Heat transfer from the moving fluid to solid surfaces is given by the equation,
Q = h A (TW - T∞)
This equation is referred to as Newton’s Law of cooling.
Where, h = Local heat transfer coefficient in W/m2K,
A = Surface area in m2,
TW = Surface (or) Wall temperature in K,
T∞ = Temperature of the fluid in K.
2.4.2. Types of Convection
1. Free Convection, 2. Forced convection.
2.4.3. Free (or) Natural Convection
If the fluid motion is produced due to change in density resulting from temperature
gradients, the mode of heat transfer is said to be free or natural convection.
2.4.4. Forced Convection
If the fluid motion is artificially created by means of an external force like a blower or
fan, that type of heat transfer is known as forced convection.
2.5. THE LOCAL AND AVERAGE HEAT TRANSFER COEFFICIENTS FOR FLAT PLATE –
LAMINAR FLOW
At the surface of the flat plate, heat flow may be written as
…(2.6)
8
We know that,
9
We know that,
The average heat transfer coefficient, h is given by
10
… (2.9)
We know that,
Average Nusselt Number, Nu = ℎL
k
2.6. THE LOCAL AND AVERAGE HEAT TRANSFER COEFFICIENTS FOR FLAT PLATE –
TURBULENT FLOW
The heat transfer coefficient for turbulent flow can be derived by using Colburn analogy,
For colburn analogy, we know that,
11
…(2.11)
We know that,
…(2.12)
The average heat transfer coefficient, h is given by
12
We know that,
2.6.1. Heat Transfer coefficient for combination of Laminar and Turbulent Flow
Heat transfer coefficient for laminar – turbulent combined flow is given by
13
14
Transition occurs at critical Reynolds number, Rec = 5 x 105, i.e., Flow is laminar upto
Re = 5 x 105, after that flow is turbulent.
Substitute Rec = Rex = 5 x 105
… (2.15)
We know that,
Average Nusselt Number, Nu = ℎL
k
… (2.16)
15
2.7. BOUNDARY LAYER THICKNESS, SHEAR STRESS AND SKIN FRICTION COEFFICIENT
FOR TURBULENT FLOW
We know that, Von Karman momentum equation for boundary layer flow is
16
…(2.17)
We know that,
…(2.18)
Equating equation (2.17) and (2.18),
17
Assuming boundary layer is turbulent over the entire length of the plate.
… (2.19)
18
Local Skin Friction Coefficient, Cfx :
We know that,
19
Equating both equation,
…(2.21)
Average friction coefficient,
We know that,
20
21
1. Air at 20˚C at atmospheric pressure flows over a flat plate at a velocity of 3 m/s. If the plate
is 1 m wide and 80˚C, calculate the following at x = 300 mm.
1. Hydrodynamic boundary layer thickness.
2. Thermal boundary layer thickness,
3. Local friction coefficient,
4. Average friction coefficient,
5. Local heat transfer coefficient,
6. Average heat transfer coefficient,
7. Heat transfer.
Given:
Fluid temperature, T
= 20˚C
Velocity, U = 3 m/s
Wide, W = 1 m
Surface temperature, Tw = 80˚C
Distance, x = 300 m = 0.3 m
To find:
1. Hydrodynamic boundary layer thickness,
2. Thermal boundary layer thickness,
3. Local friction coefficient,
4. Average friction coefficient,
5. Local heat transfer coefficient,
6. Average heat transfer coefficient,
7. Heat transfer.
Solution:
We know
Film temperature Tf = 2
2080
2
TTw
Tf = 50˚C
22
Properties of air at 50˚C:
[From HMT data Book, Page No. 33]
Density, = 1.093 kg/m3
Kinematic viscosity, v = 17.95 10-6
m2/s
Prandt 1 Number = Pr = 0.698
Thermal conductivity, k = 0.02826 W/mK
We know that,
Reynolds Number, Re = v
UL
61095.17
3.03
mLx 3.0
< 5105
Since Re < 5105, flow is laminar
For Flat plate, laminar flow,
[Refer HMT data Book, Page No. 112]
1. Hydrodynamic boundary layer thickness:
5.04
5.0
1001.53.05
Re5
xhx
2. Thermal boundary layer thickness:
333.0698.0107.6 3
333.0
hx
hxTx Rr
Re = 5.01104
mhx
3107.6
mhx
3105.7
23
3. Local Friction coefficient:
5.04
5.0
1001.5664.0
Re664.0
fxC
4. Average friction coefficient:
3
5.04
5.0
109.5
1001.5328.1
Re328.1
fLC
5. Local heat transfer coefficient (h x ):
Local Nusselt Number
Nu x = 0.332 (Re) 0.5
(Pr) 0.333
= 0.332 (5.01104)0.5
(0.698)0.333
We know,
Nu x = k
Lhx
02826.0
3.09.65
xh
mLx 3.0
6. Average heat transfer coefficient (h):
h = 2 h x
= 2 6.20
31096.2 fxC
3109.5 fLC
h x = 6.20 W/m2K
Nu x = 9.65
h = 12.41 W/m2 K
24
7. Heat transfer:
We know that,
20803.0141.12
TThAQ w
Result:
1. ,107.6 3 mhx
2. ,105.7 3 mTx
3. ,1096.2 3fxC
4. ,109.5 3fLC
5. KmWhx
2/20.6
6. h = 12.41 W/m2K,
7. Q = 223.38 W.
2. Air at 30˚C flow over a flat plate at a velocity of 2 m/s. The plate is 2 m long and 1.5 m wide.
Calculate the following:
1. Hydrodynamic and thermal boundary layer thickness at the trailing edge of the
plate,
2. Total drag force,
3. Total mass flow rate through the boundary layer between 40x cm and 85x cm.
Given:
Fluid temperature, T = 30˚C
Velocity, U = 2 m/s
Length, L = 2 m
Wide, W = 1.5 m
To find:
1. Hydrodynamic and thermal boundary layer thickness.
2. Total drag force.
3. Total mass flow rate through the boundary layer between 40x cm and 85x cm.
Q = 223.38
Watts
25
Solution:
Properties of air at 30˚C
[From HMT data Book, Page No. 33]
= 1.165 kg/m3
v = 1610-6
m2/s
Pr = 0.701
K = 0.02675 W/mK
We know that,
Reynolds Number, Re = v
UL
= 61016
22
< 5 105
Since Re < 5 105, flow is laminar
For flat plate, laminar flow,
[From HMT data Book, Page No. 112]
Hydrodynamic boundary layer thickness
5.05
5.0
105.225
Re5
xTx mLx 2
Thermal boundary layer thickness,
333.0
333.0
701.002.0
Pr
hxTx
Re = 2.5105
mTx 02.0
Tx = 0.02
m
26
Average friction coefficient,
5.05
5.0
105.2328.1
Re328.1
fLC
fLC = 1.328 (Re) -0.5
= 1.328 (2.5105) -0.5
2
2165.11065.2
2
2
3
2
UC fL
==>
Drag force = Area Average shear stress
= 2 1.5 6.1 10-3
Drag force on two sides of the plate
= 0.018 2 0.036 N
Total mass flow rate between 40x cm and 50x
408
5 hxhxUm …… (1)
Hydrodynamic boundary layer thickness
Average shear stress, = 6.110-3
N/m2
Drag force = 0.018
N
Drag force, FD = 0.036 N
27
6
5.0
5.0
1016
85.0285.05
85.05
Re585.0
v
xU
xhx
mcmx 85.085
5.0
6
05.0
1016
40.0240.05
Re540.0
xhx
mhx
3
40.0109.8
3109.80130.02165.18
5)1( m
skgm /1097.5 3
Result:
1. Hydrodynamic boundary layer thickness, ,02.0 mhx
Thermal boundary layer thickness, ,0225.0 mTx
2. Drag force, FD = 0.036 N,
3. Total mass flow rate, m = 5.97 10-3
kg/s
3. Air at 30˚C flows over a flat plate at a velocity of 4 m/s. The plate is maintained at 90˚C. The
plate dimension is 90 30 cm2. Calculate the heat transfer for the following condition
1. First half of the plate,
2. Full plate,
3. Next half of the plate.
mhx 0130.085.0
28
Given:
Fluid temperature, T
= 30˚C
Velocity, U = 4 m/s
Plate surface temperature, T w = 90˚C
Plate dimension = 9030 cm2
= 0.90 030 m2
To fine:
Heat transfer for
1. First half of the plate, i.e., x = 0.45 m,
2. Full plate, i. e., x = 0.90 m,
3. Next half of the plate.
Solution:
We know that,
Film temperature, Tf = 2
TTw
2
3090
Properties of air at 60˚C:
[From HMT data Book, Page No. 33]
2/060.1 mkg
v = 18.97 10-6
m2/s
Pr = 0.696
K = 0.02896 W/mK
Reynolds number, Re = v
LU
Tf = 60˚C
29
61097.18
45.04
Since Re < 5105, flow is laminar
[From HMT data Book, Page No. 112]
Local Nusselt Number, Nu x = k
Lhx
02896.0
45.021.90
xh
==>
Average heat transfer coefficient
xhh 2
Heat transfer, TTAhQ w1
309030.045.061.11
TTWLh w
mWmLx 30.0;45.0
Case (ii):
For full pate x = L = 0.9 m
Re = 9.4 104
< 5105
xh = 5.80 W/m2K
Local heat transfer coefficient, h x = 5. 80 W/ m
2K
h = 11.61 W/m2K
Q1 = 94.04 W
30
Reynolds Number, Re = v
LU
61097.18
90.04
Since Re < 5 105, flow is laminar
For flat plate, laminar flow,
Local Nusselt Number, Nu x = 0.332(Re) 0.5
(Pr) 0.333
= 0.332 (1.89105) 0.5 (0.696)
0.333
We know, Nu x = k
Lhx
02896.0
90.018.128
xh
Local heat transfer coefficient
KmWhx
2/12.4
Average heat transfer coefficient
12.422 xhh
Heat transfer for entire plate
TTAhQ w2
309030.090.024.8
Re = 1.89105
< 5105
Nu x = 128.18
h = 8.24 W/m2K
Q2 = 133.48 W
31
Case (iii):
Heat lost from the next half of the plate
Q3 = Q2-Q1
= 133.48 – 94.04
Result:
1. Heat lost for first half of the plate Q1 = 94.04 W
2. Heat lost for entire plate Q2 = 133.48 W
3. Heat lost for next half of the plate Q3 = 39.44 W
4. Air at 40˚C flows over a flat plate, 0.8 m long at a velocity of 50 m/s. The plate surface is
maintained at 300˚C. Determine the heat transferred from the entire plate length to air taking
into consideration both laminar and turbulent portion of the boundary layer. Also calculate
the percentage error if the boundary layer is assumed to be turbulent nature from the very
leading edge of the plate.
Given:
Fluid temperature, T
= 40˚C
Length, L = 0.8 m
Velocity, U = 50 m/s
Plate surface temperature, Tw = 300˚C
To find:
1. Heat transferred for:
(i) Entire plate is considered as combination of both laminar and turbulent flow.
(ii) Entire plate is considered as turbulent flow.
Q3 = 39.44 W
32
2. Percentage error.
Solution:
We know that,
Film temperature, Tf = 2
TTw
K4432
40300
Properties of air at 170˚C:
= 0.790 kg/m3
v = 31.1010-6
m2/s
Pr = 0.6815
k = 0.037 W/mK
We know
Reynolds Number, Re = v
UL
6
61026.1
1010.31
8.050
Re > 5 105, so this is turbulent flow.
Case (i):
Laminar-turbulent combined. [It means, flow is laminar upto Reynolds number value is 5
105, after that flow is turbine.]
Average Nusselt Number Nu = (Pr) 0.333
[0.037 (Re) 0.8
– 871]
[From HMT data Book, Page No. 114]
Nu = (0.6815) 0.333
[0.037(1.286106)
0.8 -871]
Tf = 170˚C
Re = 1.286 106 > 5 10
5
Average Nusselt Number Nu = 1746.09
33
We know
Nu = k
hL
037.0
8.009.1746
h
Average heat transfer coefficient h = 80.75 W/m2K
Heat transfer, Q1 = TAAh w
4030018.075.80
TTWLh
TTWLh
w
w
Case (ii):
Entire plate is turbulent flow:
Local Nusselt Number Nu x = 0.029 (Re) 0.8 (Pr)
0.33
[From HMT data Book, Page No. 113]
Nu x = 0.029 (1.286106)
0.8 (0.6815) 0.33
We know
Nu x = k
Lhx
037.0
8.015.2010
xh
xh = 92.96 W/m2K
h = 80.75 W/m2K
Q1 = 16796
W
Nu x = 2010.15
34
Local heat transfer coefficient, xh = 92.96 W/m2 K
Average heat transfer coefficient (for fully turbulent flow) h = 1.25h x
= 1.2592.96
Average heat transfer coefficient h = 116.20 W/m2K
Heat transfer, Q2 = TTWAh w
4030018.020.116
TTWLh w
Percentage error = 1
12
Q
90.43
10016796
1679660.24169
Result:
1. Heat transfer (Laminar – Turbulent combined) Q1 = 16796 W
2. Heat transfer (Fully turbulent) Q2 = 24169.60 W
3. Percentage error 43.90
5. Air at 15˚C, 30 Km/h flows over a cylinder of 400 mm diameter and 1500 mm height with
surface temperature of 45˚C. Calculate the heat loss.
Given:
Fluid temperature,
T
= 15˚C
Velocity, U = 30 km/h
U = s
m
3600
1030 3
Q2 = 24169.60 W
35
U = 8.33 m/s
Diameter, D = 400 mm = 0.4 m
Length, L = 1500 mm = 1.5 m
Plate surface temperature, Tw = 45˚C
To fine:
Heat loss
Solution:
We know that,
Film temperature, Tf = 2
1545
2
TTw
Properties of air at 30˚C:
[From HMT data Book, Page No. 33]
Density, = 1.165 kg/m3
Kinematic viscosity, v = 16 10-6
m2/s
Prandtl Number, Pr = 0.701
Thermal conductivity, k = 0.02675 W/mK
We know,
Reynolds Number, Re = v
UD
61016
4.033.8
Nusselt Number, Nu = C (Re) in
(Pr) 0.333
Tf = 30˚C
ReD = 2.08105
36
[From HMT data Book, Page No. 115]
ReD value is 2.08105, corresponding C value is 0.0266 and m value is 0.805.
==> Nu = 0.0266 (2.08105)
0.805 (0.701)
0.333
We known that,
Nusselt number, Nu = k
hD
==> 451.3 = 02675.0
4.0h
==> h = 30.18 W/m2K
Heat transfer, Q = TThA w
= TTLDh w
DLA
= 30.18 0.41.5 (45 – i5)
Result:
Heat loss, Q = 1706.6 W
6. Air at 40˚C flows over a tube with a velocity of 30 m/s. The tube surface temperature is
120˚C; calculate the heat transfer coefficient for the following cases.
1. Tube could be square with a side of 6 cm.
2. Tube is circular cylinder of diameter 6 cm.
Given:
Fluid temperature, T
= 40˚C
Velocity, U = 30 m/s 6 cm
Nu = 451.3
Heat transfer coefficient, h = 30.18
W/m2K
Q = 1706.6 W
37
Tube surface temperature, Tw = 120˚C
To find:
Heat transfer coefficient, (h)
Solution:
We know that,
Film, temperature, Tf = 2
TTw
2
40120
Properties of air at 80˚C
[From HMT data Book, Page No. 33]
= 1kg/m3
v = 21.09 10-6
m2/s
Pr = 0.692
k = 0.03047 W/mK
Case (i):
Tube is considered as square of side 6 cm.
i.e., L = 6 cm = 0.06 m
Reynolds Number, Re = v
UD
61009.21
06.030
Tf = 80˚C
Re = 0.853105
38
Nusselt Number, Nu = C (Re) n
(Pr) 0.333
For square, n = 0.675
C = 0.092
[From HMT data Book, Page No. 118]
==> Nu = 0.092 (0.853 105)
0.675 (0.692)
0.333
==>
We know that,
Nu = k
hL
173.3 = 03047.0
06.0h
Case (ii):
Tube diameter, D = 6 cm = 0.06 m
Reynolds Number, Re = v
UD
61009.21
06.030
Nusselt Number, Nu = C (Re) m
(Pr) 0.333
[From HMT data Book, Page No. 115]
ReD value is 0.853105, corresponding C and m values are 0.0266 and 0.805 respectively.
Nu = 173.3
Heat transfer coefficient, h = 88
W/m2K
ReD = 0.853105
39
==> Nu = 0.0266 (0.853105)
0.805 (0.692)
0.333
We known that,
Nusselt number, Nu = k
hD
==> 219.3 = 03047.0
06.0h
==> h = 111.3 W/m2K
Result:
1. Heat transfer coefficient for square tube h = 88 W/m2K
2. Heat transfer coefficient for circular tube h = 111.3 W/m2K
7. in a surface condenser, water flows through staggered tubes while the air is passed in cross
flow over the tubes. The temperature and velocity of air are 30˚C and 8 m/s respectively. The
longitudinal and transverse pitches are 22 mm and 20 mm respectively. The tube outside
diameter is 18 mm and tube surface temperature is 90˚C. Calculate the heat transfer
coefficient.
Given:
Fluid temperature, T
= 30˚C
Velocity, U = 8 m/s
Longitudinal pitch, St = 22 mm = 0.022 m
Transverse pitch, St = 20 mm = 0.020 m
Diameter, D = 18 mm = 0.018 m
Tube surface temperature, Tw = 90˚C
To find:
1. Heat transfer coefficient
Nu = 219.3
Heat transfer coefficient, h = 111.3 W/m2K
W/ W/m2K
40
Solution:
We know that,
Film, temperature, Tf = 2
TTw
2
3090
Properties of air at 60˚C
[From HMT data Book, Page No. 33]
= 1.060kg/m3
v = 18.97 10-6
m2/s
Pr = 0.692
k = 0.02896 W/mK
We know that,
Maximum Velocity, DS
SUU
t
t
max
018.0020.0
020.08max
U
Reynolds Number, Re = v
DU max
61097.18
18.080
Tf = 60˚C
Re = 7.5104
maxU = 80 m/s
41
Nusselt Number, Nu = C (Re) n
(Pr) 0.333
11.1018.0
020.0
D
S t
22.1018.0
022.0
D
S t
,22.1,11.1 D
S
D
S lt Corresponding C, n values are 0.518 and 0.556 respectively.
[From HMT data Book, Page No. 112]
We know that,
Nusselt Number, Nu = 1.13 (Pr) 0.333
[C (Re) n]
[From HMT data Book, Page No. 112]
==> Nu = 1.13 (0.693) 0.333
105) [0.518 (7.510
4)
0.556
173.3 = 03047.0
06.0h
==> Nu = 0.0266 (0.853105)
0.805 (0.692)
0.333
C= 0.518
Nu = 219.3
11.1D
S t
22.1D
S t
n = 0.556
Nu = 266.3
42
Nusselt number, Nu = k
hD
==> 266.3 = 31096.28
018.0
h
Result:
Heat transfer coefficient h = 4238.6 W/m2K
8. Water at 50˚C enters 50 mm diameter and 4 m long tube with a velocity of 0.8 m/s. The tube
wall is maintained at a constant temperature of 90˚C. Determine the heat transfer coefficient
and the total amount of heat transferred if exit water temperature is 70˚C.
Given:
Inner temperature of water, Tmi= 50˚C
Diameter, D = 50 mm = 0.05 m
Length, L = 4 m
Velocity, U = 0.8 m/s
Tube wall temperature, Tw = 90˚C
Exit temperature of water Tmo = 70˚C
To find:
1. Heat transfer coefficient, (h)
2. Heat transfer, (Q)
Solution:
Bulk mean temperature, Tm= 2
momi TT
2
7050
Heat transfer coefficient, h = 428.6 W/m2K
W/ W/m2K
Tm = 60˚C
43
Properties of air at 60˚C
[From HMT data Book, Page No. 21]
= 985 kg/m3
v = 0.478 10-6
m2/s
Pr = 3.020
k = 0.6513 W/mK
let us first determine the type of flow:
v
UDRe
610478.0
05.08.0
Since Re > 2300, flow is turbulent.
160Pr6.0020.3Pr
000,101036.8Re
6080
8005.0
4
4
D
L
D
L
D
L Ratio is greater than 60. Re value is greater than 10,000 and Pr value is in between
0.6 and 160. So,
Nusselt Number, Nu = 0.023 (Re) 0.8
(Pr) n
[From HMT data Book, Page No. 125]
[Inlet temperature 50˚C, Exit temperature 70˚C Heating Process, So, n = 0.4]
Nu = 0.023 (8.36104)
0.8 (3.020)
0.4
Re = 8.36104
Nu = 310
44
We know that, k
hDNu
6513.0
05.0310
h
Heat transfer, Q = TThA w
= TTLDh w
= 4093.3 D L TTw
Result:
1. Heat transfer coefficient h = 4039.3W/m2K
2. Heat transfer, Q = 76139 W.
9. Air at 30˚C, 6 m/s flows in a rectangular section of size 300 800 mm. calculate the heat
leakage per metre length per unit temperature difference.
Given:
Air temperature, Tm= 30˚C
Velocity, U = 6 m/s
Area, A = 300 800 mm2
= 0.30.8 m2
To find:
1. Heat leakage per metre length per unit temperature difference.
C= 76139 W
Heat transfer coefficient, h = 4039.3
W/m2K
45
Solution:
Properties of air at 30˚C:
= 1.165 kg/m3
v = 16 10-6
m2/s
Pr = 0.701
k = 0.02675 W/mK
Equivalent diameter for 300800 mm2 cross section is given by
8.03.02
8.03.044
P
ADe
Where P – Perimeter = 2 (L + W)
We know that,
Reynolds Number, Re = v
UDe
61016
436.06
Since Re > 2300, flow is turbulent.
For turbulent flow general equation is (Re > 10000),
Nu = 0.023(Re) 0.8
(Pr) n
[From HMT data Book, Page No. 125]
Assuming the pipe wall temperature to be higher than air temperature. So, heating
process ==> n = 0.4.
==> Nu = 0.023 (16.3104)
0.8 (0.701)
0.4
De = 0.436 m
Re = 16.3 104
Nu = 294.96
46
We know,
Nusselt Number, Nu = k
hDe
31075.26
436.096.294
h
Heat leakage per unit length per unit temperature difference
Q = hP
= ]8.03.02[09.18
Result:
Heat leakage, Q = 39.79 W.
10. Air at 2 bar pressure and 60˚C is heated as it flows through a tube of diameter 25 mm at a
velocity of 15 m/s. If the wall temperature is maintained at 100˚C, find the heat transfer per
unit length of the tube. How much would be the bulk temperature increase over one metre
length of the tube.
Given:
Pressure, P = 2 bar = 2 105 N/m
2
Inlet temperature of air, Tmi = 60˚C
Diameter of tube D = 25 mm2
= 0.05 m
Velocity, U = 15 m/s
Tube wall temperature, Tw = 100˚C
Legth, L = 1 m
Q = 39.79 W
Heat transfer coefficient, h = 18.09 W/m2K
W/m2K
47
To find:
1. Heat transfer per unit length of the tube, Q.
2. Rise in bulk temperature of air, mimo TT
Solution:
Properties of air at 60˚C:
[From HMT data Book, Page No. 33]
= 1.060 kg/m3
v = 18.97 10-6
m2/s
Pr = 0.696
k = 0.02896 W/mK
Note:
Given pressure is above atmospheric pressure. So, kinematic viscosity, v and density, p
will vary with pressure. Pr. k, Cp are same for all pressures.
Kinematic viscosity, given
atm
atmP
Pvv
= 18.9710-6
bar
bar
2
1
[ Atmospheric pressure = 1 bar]
5
56
102
1011097.18
Density, RT
P
27360287
102 5
v = 9.48510-6
m2/s
P = 2.092 kg/m2
48
We know that,
Reynolds Number, Re = v
UD
610485.9
025.015
2300
Since Re > 2300, flow is turbulent.
For turbulent flow general equation is (Re > 10000),
Nu = 0.023(Re) 0.8
(Pr) n
[From HMT data Book, Page No. 125]
This is heating process. So, n = 0.4.
==> Nu = 0.023 (39.53103)
0.8 (0.696)
0.4
We know,
Nu = k
hD
02896.0
025.070.94
h
==>
Mass flow rate, m = AU
15025.04
092.2
4
2
2
UD
Heat transfer coefficient, h = 18.09 W/m2K
W/m2K
Re = 39.53 103
Nu = 94.70
h = 109.70 W/m2K
m = 0.015 kg/s
49
We know that,
Heat transfer, Q = mimop TTmC
= 601005015.0 moT
[For air Cp = 1005 J/kgK]
Q = Heat leakage per unit length per unit temperature difference
Q = hP 60moT …… (1)
We know that,
Heater transfer, Q = mimop TTmC
m
mw
T
TTDLh
1001025.070.109
…… (2)
Equating (1) and (2),
mmo TT 100615.860075.15
CT
T
TT
TT
TT
TTT
TT
mo
mo
mo
mo
mo
mo
mo
mo
momi
mo
mmo
78.77
94.174249.2
94.104301002
749.1
23010094.104749.1
2
60100104749.1
210060749.1
10060749.1
Outlet temperature of air, moT = 77.78˚C
50
Rise in bulk temperature of air, mimo TTT
= 77.78 – 60
Heat transfer, Q = mimop TTmC
= C78.171005015.0
Result:
1. Q = 268.03 W
2. CTTT mimo
78.17
11. A vertical plate of 0.75 m height is at 170˚C and is exposed to air at a temperature of 105˚C
and one atmosphere. Calculate:
1. Mean heat transfer coefficient,
2. Rate of heat transfer per unit width of the plate.
Given:
Legth, L = 0.75 m
Wall temperature, Tw = 170˚C
Fluid temperature, T
= 105˚C
To find:
1. Heat transfer coefficient, (h)
2. Heat transfer (Q) per unit width.
Solution:
Velocity (U) is not given. So this is natural convection type problem.
Q = 268.03 W
CT 78.17
51
Film temperature, Tf = 2
TTw
2
105170
Properties of air at Tf = 137.5˚C 140˚C
[From HMT data Book, Page No. 33]
Density, = 0.854 kg/m3
Kinematic viscosity, v = 27.80 10-6
m2/s
Prandtl Numbe Pr = 0.684
Thermal conductivity, k = 0.03489 W/mK
We know that,
Coefficient of thermal expansion,inKT f
1
==>
5.410
1
2735.137
1
==>
We know that,
Grashof Number, Gr = 2
3
v
TLg
[From HMT data Book, Page No. 134]
==>
26
33
1080.27
10517075.0104.281.9
Gr
Tf= 137.5˚C
= 2.4 10-3
K-1
52
==>
==> Gr Pr = 8.35 108 0.684
Since Gr Pr > 109, flow is laminar.
Gr Pr value is in between 104 and 109 i.e., 10
4 < Gr Pr < 10
9
So, Nusselt Number
Nu = 0.59 (Gr Pr) 0.25
[From HMT data Book, Page No. 134]
==> = 0.59 (5.71 108) 0.25
We know that,
Nusselt Number, Ne = k
hL
KmWh
h
2/24.4
03489.0
75.021.91
We know,
Heat transfer, Q = TThA w
10517075.0124.4
TTLWh w
[W = 1 m]
Heat transfer coefficient, h = 4.24 W/m2K
Q = 206.8 W
Gr = 8.35108
Gr Pr = 5.71 108
Nu = 91.21
53
Result:
1. Heat transfer coefficient, h = 4.24 W/m2K
2. Heat transfer, Q = 206.8 W
12. A vertical plate of 0.7 m wide and 1.2 m height maintained at a temperature of 90˚C in a
room at 30˚C. Calculate the convective heat loss.
Given:
Wide, W = 0.7 m
Height (or) Length, L = 1.2 m
Wall temperature, T w = 90˚C
Room temperature, T = 30˚C
To fine:
Convective heat loss (Q)
Solution:
Velocity (U) is not given. So, this is natural convection type problem.
We know that,
Film temperature, Tf = 2
TTw
2
3090
Properties of air at 60˚C:
[From HMT data Book, Page No. 33]
3/060.1 mkg
v = 18.97 10-6
m2/s
Pr = 0.696
Tf = 60˚C
54
K = 0.02896 W/mK
We know,
Coefficient of thermal expansion = inKT
L
f
1
1310327360
1
K
Grashor Number, Gr = 2
3
v
TLg
[From HMT data Book, Page No. 134]
26
33
1097.18
30902.110381.9
Gr Pr = 8.4109 0.696
Since Gr Pr > 109, flow is turbulent.
For turbulent flow,
NUSSELT Number, Nu = 0.10 (Gr Pr) 0.333
[From HMT data Book, Page No. 135]
Nu = 0.10 [5.9109]0.333
We know that,
Local Nusselt Number, Nu = k
hL
02896.0
2.13.179
h
13103 K
Gr = 8.4 109
Gr Pr = 5.9 109
Nu = 179.3
55
Heat loss, ThAQ
30902.17.032.4
TTLWh w
Result:
Convective heat loss, Q = 218.16 W
13. A horizontal plate of 800 mm long, 70 mm wide is maintained at a temperature of 140˚C in
a large tank of full of water at 60˚C. Determine the total heat loss from the plate.
Given:
Horizontal plate length, L = 800 mm = 0.8 m
Wide, W = 70 mm = 0.070 m
Plate temperature, Tw = 140˚C
Fluid temperature, T
= 60˚C
To find:
Solution:
Film temperature, Tf = 2
TTw
2
60140
Q = 218.16 W
Convective heat transfer coefficient h = 4.32 W/m2K
Tf = 100˚C
56
Properties of air at 100˚C:
[From HMT data Book, Page No. 21]
3/961 mkg
v = 0.293 10-6
m2/s
Pr = 1.740
K = 0.6804 W/mK
(water) = 0.7610-3
K-1
[From HMT data Book, Page No. 21]
Grahhof Number, Gr = 2
3
v
TLg c
……… (1)
For horizontal plate,
Lc = Characteristic length = 2
W
Lc = 2
070.0
= 0.035 m
(1) ==> Gr
26
33
10293.0
60140035.01076.081.9
Gr Pr = 0.297109 1.740
Gr Pr value is in between 8 106 and 10
11,
i.e., 8 106 < Gr Pr < 10
11
so, for horizontal plate, upper surface heated,
Lc = 0.035 m
Gr = 0.297 109
Gr Pr = 0.518 109
57
NUSSELT Number, Nu = 0.15 (Gr Pr) 0.333
[From HMT data Book, Page No. 135]
Nu = 0.15 [0.518109]0.333
We know that,
Local Nusselt Number, Nu = k
hL
6804.0
035.066.119
uh
For horizontal plate,
Lower surface heated,
[From HMT data Book, Page No. 136]
Nusselt Number, Nu = 0.27 [Gr Pr] 0.25
Nu – 0.27 [0.518 109]0.25
We know that,
Nusselt Number, Nu = k
Lh c1
6804.0
035.073.40 chl
Heat transfer coefficient for lower surface heated, hl = 791.79 W/m2K
Total heat transfer, TAhhQ lu
Nu = 119.66
Heat transfer coefficient for upper surface heated, hu = 2326.19 W/m2K
Nu = 40.73
58
30902.17.032.4
TTLWhh wlu
601408.0070.079.79119.2326 Q
Result:
Total heat loss, Q = 13, 968.55 W.
14. A steam pipe 80 mm in diameter is covered with 30 mm thick layer of insulation which has
a surface emissivity of 0.94. The insulation surface temperature is 85˚C and the pipe is placed
in atmospheric air at 15˚C. If the heat is lost both by radiation and free convection, find the
following:
1. The heat loss from 5 m length of the pipe.
2. The overall heat transfer coefficient.
3. Heat transfer coefficient due to radiation.
Given:
Diameter of pipe = 80 mm
= 0.080 m
Insulation thickness = 30 mm = 0.030 m
Actual diameter of the pipe, D = 0.080 + 2 0.030
= 0.14 m
Emissivity, = 0.94
Tube surface temperature, Tw = 85˚C
Air temperature, T
= 15˚C
Q = 13,968.55 W
59
To find:
1. Heat loss from 5 m length of the pipe, Q
2. Overall heat transfer coefficient, ht
3. Heat transfer coefficient due to radiation, hr
Solution:
Film temperature, Tf = 2
TTw
2
1585
Properties of air at 50˚C:
[From HMT data Book, Page No. 33]
3/093.1 mkg
v = 17.95 10-6
m2/s
Pr = 0.698
K = 0.02826 W/mK
Coefficient of thermal expansion , = inKT f
1
27350
1
We know that,
Grashof number, Gr = 2
3
v
TDg
[From HMT data Book, Page No. 134]
Tf = 50˚C
= 3.095 10-3
K-1
60
26
33
1095.17
158514.010095.381.9
Gr Pr = 18.10106 0.698
For horizontal cylinder,
Nusselt Number, Nu = C [Gr Pr] m
[From HMT data Book, Page No. 137]
Gr Pr = 1.263107
Corresponding C = 0.125, and m = 0.333
Nu = 0.125 [1.263107]0.333
We know that,
Nu = k
hD
02826.0
14.0952.28
h
h = 5.84 W/m2K
Gr Pr = 1.263 107
Nu = 28.952
Convective heat transfer coefficient, hc = 5.84 W/m2K
Gr = 18.10 106
61
Heat lost by convection,
1585514.084.5
TTLDh
ThAQ
w
conv
Heat lost by radiation,
Qrad = 44
TTA w
Where, = Emissivity
A = Area – m2
= Stefen Boltzmann constant
= 5.67 10-8
W/m2 K4
Tw = Surface temperature, K
T
= Fluid temperature, K
Tw = 85 + 273 T
= 15 + 273
==> Qrad = 44
TTDL w
448 2888358514.01067.594.0
Total heat transfer, radconvt QQQ
90.11199.898
Total heat transfer, Qt = TAht
Qconv = 898.99 W
Tw = 358 K
T
= 288
K
Qrad = 1118.90 W
Qt = 2017.89 W
62
1585514.089.2017
t
wt
h
TTDLh
==> ht = 13.108 W/m2K
Radiative heat transfer coefficient,
84.5108.13
ctr hhh
Result:
1. Heat loss from 5 m length of pipe
(i) By convection, Qc = 898.99 W
(ii) By radiation, Qr = 1118.90 W
2. Overall heat transfer coefficient, ht = 13.108 W/m2K
3. Radiative heat transfer coefficient, hr = 7.268 W/m2K
15. A vertical plate of 40 cm long is maintained at 80˚C and is exposed to air at 22˚C.
Calculate the following:
1. Boundary layer thickness at the tailing edge of the plate.
2. The same plate is placed in a wind tunnel and air is blown over it at a velocity
of 5 m/s. Calculate boundary layer thickness.
3. Average heat transfer coefficient for natural and forced convection for the
above mentioned data.
Overall heat transfer coefficient, ht = 13.108 W/m2K
hr = 7.268 W/m2K
63
Given:
Length, L = 40 cm = 0.40 m
Plate temperature, Tw = 80˚C
Fluid temperature, T∞ = 22˚C
To find: Case (i)
(i) Boundary layer thickness (Natural convection).
Case (ii)
(i) Boundary layer thickness at velocity U = 5 m/s (Forced convection).
(ii) Average heat transfer coefficient for forced convection,
Solution:
We know that,
Film temperature, Tf =
2
TTw
2
2280
Properties of air at 51˚C: 50˚C:
[From HMT data Book, Page No. 33]
3/093.1 mkg
v = 17.95 10-6
m2/s
Pr = 0.698
K = 0.02826 W/mK
= inKT f
1
Tf = 51˚C
64
= 27351
1
Case (i): for free convection,
Gr = 2
3
v
TLg
[From HMT data Book, Page No. 134]
26
33
1095.17
228014.010086.381.9
Gr Pr = 3.48108 0.698
Since Gr < 109, flow is laminar,
For free convection, laminar flow:
Boundary layer thickness, xGrx 25.025.05.0
Pr952.0Pr93.3 =
[From HMT data Book, Page No. 134]
==> 4.01048.3698.0952.0698.093.325.0825.05.0
x
mLx 40.0
==>
Case (ii): For forced convection,
Gr Pr = 2.43 106
Gr = 3.48 108
= 3.086 10-3
K-1
mx 0156.0
65
Reynolds number, Re = v
UL
61095.17
40.05
…… (1)
Since Re < 5105 flow is laminar.
For forced convection, laminar flow:
Boundary layer thickness 5Re5
xor hxx
[From HMT data Book, Page No. 134]
551011.140.05
x
mLx 40.0
mx
310003.6 ……. (2)
From equation (1) and (2), we know that, boundary layer thickness in forced convection
is less than that in free convection.
Case (iii): Average heat transfer coefficient for natural convection, h:
For free convection, laminar flow, vertical plate:
[From HMT data Book, Page No. 135]
Nusselt number, Nu = 0.59(Gr Pr) 0.25
= 0.59 (2.43106)
0.25
We know that,
Nu = k
hL
02826.0
4.029.23
h
Re = 1.11 105
Nu = 23.29
66
…….. (3)
Average heat transfer coefficient for forced convection, h:
For forced convection, laminar flow, flat plate:
Local Nusselt number,
Nu x = k
Lhx
02826.0
4.013.98 xh
Local heat transfer coefficient xh = 6.932 W/m2 K
Average heat transfer coefficient h = 26.932
…… (4)
From equation (3) and (4) we know that heat transfer coefficient in forced convection is
much larger than that in free convection.
Result:
Case (i): 1. x (Natural convection) = 0.0156 m
Case (ii): 1. x (Forced convection) = 6.003 10-3
m
Case (iii): 1. h (Natural convection) = 1.645 W/m2K
2. h = (Forced convection) = 13.86 W/m2K
h = 1.645 W/m2K
h = 13.86 W/m2K
67
16. Water is to be boiled at atmospheric pressure in a polished copper pan by means of an
electric heater. The diameter of the pan is 0.38 m and is kept at 115˚C. Calculate the following
1. Power required to boil the water
2. Rate of evaporation
3. Critical heat flux.
Given:
Diameter, d = 0.38 m;
Surface temperature, Tw = 115˚C
To find:
1. Power required, (P)
2. Rate of evaporation, (m)
3. Critical heat flux,
A
Q
Solution:
We know that, saturation temperature of water is 100˚C.
i.e.
Tsat = 100˚C
68
Density, l = 961akg/m3
Kinematic viscosity, v = 0.29310-6 m2/s
Prandtl Number, Pr = 1.740
Specific heat, Cpl = 4216 J/kg K
Dynamic viscosity, vl = 9610.29310-6
= 281.5710-6 Ns/m2
From Steam Table
At 100˚C
Enthalpy of evaporation, hfg = 2256.9 kJ/kg
hfg = 2256.9103 J/kg
Specific volume of vapour, gv = 1.673 m3/kg
Density of vapour, v = gv
1
673.1
1
T Excess temperature = satw TT = 115˚C – 100 = 15˚C
<50˚C. So, this is nucleate pool boiling process.
1. Power required to boil the water
For Nucleate pool boiling
Heat flux,
35.0
n
rfgfs
plvl
fglPhC
TCppgh
A
Q
…… (1)
Where = surface tension for liquid vapour interface
v = 0.597 kg/m3
T 15˚C
69
At 100˚C
n = 1 for water
Substitute
nhCTCpvph fgsfpllfgl ,,,,,,,,, And rP values in Equn (1)
3
1
5.0
36
74.11039.2256013.0
154216
0588.0
597.096181.9109.22561057.281)1(
A
Q
Heat flux, 25 /1083.4 mWA
Q
==> Heat transfer, AQ 51083.4
PQ
WQ
d
3
3
25
25
107.54
107.54
38.04
1083.4
41083.4
==>
2. Rate of evaporation, (m)
We know that,
Heat transferred, fghmQ
fgh
Qm
= 0.0588 N/m
For water – copper ==> Csf = surface fluid constant = 0.013
Power = 54.7103 W
70
3
3
109.2256
107.54
3. Critical heat flux
For Nucleate pool boiling, critical heat flux,
26
2
3
25.0
2
/1052.1
597.0
597.096181.90588.0597.0109.225618.0
18.0
mWA
Q
p
pgph
A
Q
v
vl
vfg
Result:
1. P = 54.7 103 W
2. M = 0.024 kg/s
3. ./10652.1 2mWqA
Q
17. Water is boiling on a horizontal tube whose wall temperature is maintained at 15˚C above
the saturation temperature of water. Calculate the nucleate boiling heat transfer co-efficient.
Assume the water to be at a pressure of 10 atm. And also find the change in value of heat
transfer co-efficient when
1. The temperature difference is increased to 30˚C at a pressure of 10 atm.
2. The pressure is raised to 20 atm at .15 CT
m = 0.024 kg/s
Critical Heat flux, 26 /1052.1 mWA
71
Given:
Wall temperature is maintained at 15˚C above the saturation temperature.
CTw 115 CTwCTsat 11515100;100
P = 10 atm = 10 bar
Case (i):
T
= 30˚C; p = 10 atm = 10 bar
Case (ii):
P = 20 atm = 20 bar; T
= 15˚C
Solution:
We know that, for horizontal surface, heat transfer co-efficient
3
3
3
10011556.5
56.5
56.5
sat
w TTh
Th
Heat transfer co-efficient other than atmospheric pressure
4.0
4.0
1018765
hphp
Case (i)
CTbarp 30;10
h = 150
103 W/m
2K
h = 18765 W/m2K
Heat transfer co-efficient, KmWhp 23 /1013.47
72
Heat transfer co-efficient other than atmospheric pressure
4.03
4.0
1010150
hphp
Case (ii)
CTbarp 15;20
Heat transfer co-efficient, h= 5.56 33
1556.5T
Heat transfer co-efficient other than atmospheric pressure
4.0
4.0
10150
hphp
Result:
Nucleate boiling heat transfer co-efficient
ph = 47.13103 W/m
2K
Case (i)
ph = 377 103 W/m
2K
Case (ii)
ph = 62.19 103 W/m
2K
ph = 377 103 W/m
2K
h = 18765 W/m2K
ph = 62.19 103 W/m
2K
73
18. A vertical flat plate in the form of fin is 500 mm in height and is exposed to steam at
atmospheric pressure. If surface of the plate is maintained at 60˚C, calculate the following
The film thickness at the trailing edge
Overall heat transfer co-efficient
Heat transfer rate
The condensate mass flow rate.
Given:
Height (or) Length, L = 500 mm 0.5 m
Surface temperature, Tw = 60˚C
To find:
1. x
2. h
3. Q
4. m
Solution:
We know that, saturation temperature of water is 100˚C
i.e.,
hfg = 2256.9 kJ/kg
hfg = 2256.9 103 J/kg
We know that,
Film temperature, Tf = 2
TTw
2
10060
Properties of air at 80˚C
Tsat = 100˚C
Tf = 80˚C
74
3/974 mkg
v = 0.364 10-6
m2/s
k = 0.6687 W/mK
610364.0974 vp
1. Film thickness x
We know for vertical plate,
Film thickness,
25.0
2
4
phg
TTkx
fg
wsat
x
Where
mLx 5.0
23
6
974109.225681.9
601005.06687.01053.3544
x
2. Average heat transfer co-efficient, (h)
For vertical surface, Laminar flow
25.023
943.0
wsat
fg
TTL
hgpkh
The factor 0.943 may be replaced by 1.13 for more accurate result as suggested by Mc Adams
Since Gr < 109, flow is laminar,
25.0
6
323
601005.01053.354
109.22569819746687.013.1
h
x =1.7310-4
m
26 /1053.354 mNs
75
3. Heat transfer rate, (Q)
Heat transfer, Q = wsat TThA
6010015.03.6164
wsat TTWLh
4. Condensate mass flow rate, (m)
We know that,
3109.2256
286,23,1
m
h
Qm
hmQ
fg
fg
Result:
1. x = 1.73
10-4
2. h = 6164.3 W/m2K
3. Q = 123286 W
4. m = 0.054 kg/s
19. Steam at 0.080 bar is arranged to condense over a 50 cm square vertical plate. The surface
temperature is maintained at 20˚C. Calculate the following.
a. Film thickness at a distance of 25 cm from the top of the plate.
b. Local heat transfer co-efficient at a distance of 25 cm from the top of the plate.
c. Average heat transfer co-efficient
d. Total heat transfer
e. Total steam condensation rate
h = 6164.3 W/m2/K
Q = 1, 23,286 W
m = 0.054 kg/s
76
f. What would be the heat transfer co-efficient if the plate is inclined at 30˚C with
horizontal plane.
Given:
Pressure, p = 0.080 bar
Area, A = 50 cm 0.50 = 0.25 m2
Surface temperature, Tw = 20˚C
Distance, x = 25 cm 0.25 m
To find:
a) x
b) xh
c) h
d) Q
e) m
f) h at 30˚C
Solution:
Properties of steam at 0.080 bar
Tsat = 41.53˚C
hfg = 2403.2 kJ/kg = 2403.2 103 J/kg
We know that,
Film temperature, Tf = 2
TTw
2
53.4120
Properties of air at 30.76˚C: 30˚C:
3/997 mkg
v = 0.83 10-6
m2/s
Tf = 30.76˚C
77
k = 0.612 W/mK
26
6
/1057.827
1083.0997
mNs
vp
a) Film thickness
We know, for vertical surfaces
=
23
6
25.0
2
997102.240381.9
2053.4125.0612.01051.8274
4
x
fg
wsat
xphg
TTkx
b) Local heat transfer co-efficient (h x )
x
x
kh
41046.1
612.0
xh
c) Average heat transfer co-efficient (h)
25.023
943.0
wsat
fg
TTL
hgpkh
The factor 0.943 may be replaced by 1.13 for more accurate result as suggested by Mc
Adams
25.023
13.1
wsat
fg
TTL
ghpkh
x = 1.46 10-4
m
xh = 4.191 W/m2K
78
Where L = 50 cm 0.5 m
25.0
6
323
2053.4151051.827
102.240381.9997612.013.1
h
d) Heat transfer (Q)
We know that,
2053.4125.06.5599
wsat
wsat
TTAh
TThAQ
e) Total steam condensation rate (m)
Heat transfer, fghmQ
3102.2403
8.139,30
m
h
Qm
fg
m = 0.0125 kg/s
f) If the plate is inclined at
h inclined = hvertical (sin ) ¼
hinclined = hvertical (sin 30) ¼
hinclined = 5599.6 (1/2)1/4
Let us check the assumption of laminar film condensation
We know that,
Q = 30,139.8 W
h = 5599.6 W/m2K
hinclined = 4,708.6 W/m2K
79
Reynolds Number, Re = W
m4
Where
W = width of the plate = 50 cm = 0.50 m
61051.82750.0
0125.04
eR
< 1800
So our assumption (laminar flow) is correct.
Result:
a) x = 1.46 10-4
b) xh = 4191 W/m2K
c) h = 5599.6 W/m2K
d) Q = 30,139.8 W
e) m = .0125 kg/s
f) h inclined = 4708.6 W/m2K
20. Saturated steam at tsat = 100˚C condenses on the outer surface of a 1.4 m long, 2 m outer
diameter vertical tube maintained at a uniform temperature Tw = 60˚C. Assuming film
condensation, find the following.
a) Local heat transfer co-efficient at the bottom of the tube.
b) Average heat transfer co-efficient over the entire length of the tube.
Given:
Saturation temperature, Tsat = 100˚C
Length, L = 1.4 m
Outer diameter, D = 2 m
Surface temperature, Tw = 60˚C
Re = 120.8
80
To find:
1. Local heat transfer co-efficient xh
2. Average heat transfer co-efficient, h
Solution:
Properties of steam at 100˚C
Enthalpy of evaporation,
kgJ
kgkJh fg
/109.2256
/9.2256
3
We know that,
Film temperature, Tf = 2
TTw
2
10060
Properties of air at 80˚C:
3/974 mkg
v = 0.364 10-6
m2/s
k = 0.6687 W/mK
26
6
/10354
10364.0974
mNs
vp
Assuming that the condensate film is laminar
For vertical surfaces, laminar flow,
Film thickness =
25.0
2
4
phg
TTkx
fg
wsat
x
Tf = 80˚C
81
25.0
23
6
974109.225681.9
601004.16687.01053.3544
mLx 4.1
Local heat transfer co-efficient (h x )
x
x
kh
41024.2
6687.0
xh
Average heat transfer co-efficient (h)
25.023
943.0
wsat
fg
TTL
hgpkh
The factor 0.943 may be replaced by 1.13 for more accurate result as suggested by Mc
Adams
25.023
13.1
wsat
fg
TTL
ghpkh
25.0
6
323
601004.11053.354
109.225681.99746687.013.1
h
Let us check the assumption of laminar film condensation
x = 1.46 10-4
m
xh = 2985.26 W/m2K
h = 4765.58W/m2K
82
We know that,
Reynolds Number, Re = P
m4
……. (1)
Heat transfer ThAQ
601004.1258.4765
wsat TTDLh
We know that,
36 109.22561067.1
m
mQ hfg
….. (2)
Perimeter, DP
2
…… (3)
Substitute P, m, values in equation (1)
61053.354283.6
739.04Re)1(
So our assumption (laminar flow) is correct
Result:
1. Local heat transfer co-efficient, xh = 2985.26 W/m2K
2. Average heat transfer co-efficient, h = 4765.58 W/m2K
Q= 1.67106
m = 0.739 kg/s
P = 6.283 m
Re 1327.04
83
TWO MARK QUESTIONS AND ANSWERS
1. What is dimensional analysis?
Dimensional analysis is a mathematical method which makes use of the study of the
dimensions for solving several engineering problems. This method can be applied to all types of
fluid resistances, heat flow problems in fluid mechanics and thermodynamics.
2. State Buckingham theorem.
Buckingham theorem states as follows: “If there are n variables in a dimensionally
homogeneous equation and if these contain m fundamental dimensions, then the variables are
arranged into (n – m) dimensionless terms. These dimensionless terms are called terms.
3. What are all the advantages of dimensional analysis?
It expresses the functional relationship between the variables in dimensional terms.
It enables getting up a theoretical solution in a simplified dimensionless form.
The results of one series of tests can be applied to a large number of other similar
problems with the help of dimensional analysis.
4. What are all the limitations of dimensional analysis?
The complete information is not provided by dimensional analysis. It only indicates that
there is some relationship between the parameters.
No information is given about the internal mechanism of physical phenomenon.
Dimensional analysis does not give any clue regarding the selection of variables.
5. Define Reynolds number (Re).
It is defined as the ratio of inertia force to viscous force.
ceViscousfor
ceInertiaforRe
6. Define Prandtl number (Pr)
It is the ratio of the momentum diffusivity to the thermal diffusivity.
fusivityThermaldif
ffusivityMomentumdiPr
84
7. Define Nusselt Number (Nu).
It is defined as the ratio of the heat flow by convection process under an unit temperature
gradient to the heat flow rate by conduction under an unit temperature gradient through a
stationary thickness (L) of metre.
Nusselt Number cond
conv
q
qNu )(
8. Define Grashof number (Gr)
It is defined as the ratio of product of inertia force and buoyancy force to the square of
viscous force.
2
ceViscousfor
ceBuoancyforceInertiaforGr
9. Define Stanton number (St).
It is the ratio of Nusselt number to the product of Reynolds number and Prandtl number.
PrRe
Nu
St
10. What is meant by laminar flow and turbulent flow?
Laminar flow: Laminar flow is sometimes called stream line flow. In this type of flow, the fluid
moves in layers and each fluid particle follows a smooth continuous path. The fluid particles in
each layer remain in an orderly sequence without mixing with each other.
Turbulent flow: In addition to the laminar type of flow, a distinct irregular flow is frequently
observed in nature. This type of flow is called turbulent flow. The path of any individual particle
is zig-zag and irregular. Fig. shows the instantaneous velocity in laminar and turbulent flow.
85
11. Define convection.
Convection is a process of heat transfer that will occur between a solid surface and a fluid
medium when they are at different temperatures.
12. State Newton’s law of convection.
Heat transfer from the moving fluid to solid surface is given by the equation
TThAQ w
This equation is referred to as Newton’s law of cooling.
Where
h - Local heat transfer coefficient in W/m2K
A - Surface area in m2
wT - Surface (or) Wall temperature in K
T - Temperature of fluid in K
13. What is meant by free or natural convection?
If the fluid motion is produced due to change in density resulting from temperature
gradients, the mode of heat transfer is said to be free or natural convection.
14. What is forced convection?
If the fluid motion is artificially created by means of an external forced like a blower or
fan, that type of heat transfer is known as forced convection.
15. What is the form of equation used to calculate heat transfer for flow through cylindrical
pipes?
Nu = 0.023 (Re) 0.8
(Pr) n
n = 0.4 for heating of fluids.
n = 0.3 for cooling of fluids.
16. What are the dimensionless parameters used in forced convection?
Reynolds number (Re).
Nusselt number (Nu).
Prandtl number (Pr).
86
17. Define boundary layer thickness.
The thickness of the boundary layer has been defined as the distance from the surface at
which the local velocity or temperature reaches 99% of the external velocity or temperature.
18. Indicate the concept or significance of boundary layer.
In the boundary layer concept the flow field over a body is divided into two regions;
A thin region near the body called the boundary layer where the velocity and the
temperature gradients are large.
The region outside the boundary layer where the velocity and the temperature gradients
are very nearly equal to their free stream values.
19. Write down the momentum equation for a steady, two dimensional flow of an
incompressible, constant property Newtonian fluid in the rectangular coordinate system and
mention the physical significance of each term.
Momentum equation,
2
2
2
2
y
u
x
uPF
y
uv
x
uuP
x
x
Where,
y
uv
x
uuP = Inertia forces.
F x = Body force.
x
P
= Pressure force.
2
2
2
2
y
u
x
u
= Viscous forces.
20. Sketch the boundary development of a flow.
87
21. Define displacement thickness.
The displacement thickness is the distance, measured perpendicular to the boundary, by
which the free stream is displaced on account of formation of formation of boundary layer.
22. Define momentum thickness.
The momentum thickness is defined as the distance through which the total loss of
momentum per second be equal to if it were passing a stationary plate.
23. Define energy thickness.
The energy thickness can be defined as the distance, measured perpendicular to the
boundary of the solid body, by which the boundary should be displaced to compensate for the
reduction in kinetic energy of the flowing fluid on account of boundary layer formation.
88