Convective Heat Transfer - Rajiv Gandhi College of Engineering … YEAR/HEAT AND MASS... ·...

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

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

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


= 1.165 kg/m3

v = 1610-6

m2/s

Pr = 0.701

K = 0.02675 W/mK

We know that,


UL

= 61016

22

< 5 105

Since Re < 5 105, flow is laminar

For flat plate, laminar flow,


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:


= 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



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


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


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:


= 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,


TTw

K4432

40300


= 0.790 kg/m3

v = 31.1010-6

m2/s

Pr = 0.6815

k = 0.037 W/mK

We know


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]


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


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

QQ

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,


1545

2

TTw




Kinematic viscosity, v = 16 10-6

m2/s

Prandtl Number, Pr = 0.701


We know,


UD

61016

4.033.8

Nusselt Number, Nu = C (Re) in

(Pr) 0.333

Tf = 30˚C

ReD = 2.08105

36


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:


= 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



= 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


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


==> 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


UD

61009.21

06.030

Nusselt Number, Nu = C (Re) m

(Pr) 0.333


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,


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:


= 30˚C

Velocity, U = 8 m/s

Longitudinal pitch, St = 22 mm = 0.022 m

Transverse pitch, St = 20 mm = 0.020 m



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



= 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


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.


We know that,

Nusselt Number, Nu = 1.13 (Pr) 0.333

[C (Re) n]


==> 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


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


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


W/ W/m2K

Tm = 60˚C

43



= 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


[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


= 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:


= 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,


UDe

61016

436.06


For turbulent flow general equation is (Re > 10000),

Nu = 0.023(Re) 0.8

(Pr) n


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


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:



= 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,


UD

610485.9

025.015

2300


For turbulent flow general equation is (Re > 10000),

Nu = 0.023(Re) 0.8

(Pr) n


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


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


= 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


TTw

2

105170

Properties of air at Tf = 137.5˚C 140˚C



Kinematic viscosity, v = 27.80 10-6

m2/s

Prandtl Numbe Pr = 0.684


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


==>

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


==> = 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,


10517075.0124.4

TTLWh w

[W = 1 m]


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,


TTw

2

3090



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


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


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


= 60˚C

To find:

Solution:


TTw

2

60140

Q = 218.16 W

Convective heat transfer coefficient h = 4.32 W/m2K

Tf = 100˚C

56



3/961 mkg

v = 0.293 10-6

m2/s

Pr = 1.740

K = 0.6804 W/mK

(water) = 0.7610-3

K-1


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


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,


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


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:


TTw

2

1585



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


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


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:


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


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 =


==> 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


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:


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;


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

Qq

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


4.03

4.0

1010150

hphp

Case (ii)

CTbarp 15;20

Heat transfer co-efficient, h= 5.56 33

1556.5T


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


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,


TTw

2

10060


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


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,


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


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,


TTw

2

10060


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

Convective Heat Transfer - Rajiv Gandhi College of Engineering … YEAR/HEAT AND MASS... ·...

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