chapter 5 principles of force convection · 2018. 9. 25. · 5.2 viscous flow . 5.2.1 flow on a...

Chapter 5

Principles of Force Convection

5.1 Introduction The preceding chapters have considered the mechanism and calculation of conduction

heat transfer. Convection was considered only insofar as it related to the boundary

conditions imposed on a conduction problem. We now wish to examine the methods

of calculating convection heat transfer and, in particular, the ways of predicting the

value of the convection heat-transfer coefficient h. Our development in this chapter is primarily analytical in character and is concerned

only with forced-convection flow systems. Subsequent chapters will present empirical

relations for calculating forced-convection heat transfer and will also treat the subjects

of natural convection.

5.2 Viscous Flow 5.2.1 Flow on a flat plate (external flow)

Consider the flow over a flat plate as shown in Figures 5.1 and 5.2. Beginning at the

leading edge of the plate, a region develops where the influence of viscous forces is

felt. These viscous forces are described in terms of a shear stress 𝜏 between the fluid

layers. If this stress is assumed to be proportional to the normal velocity gradient, we

have the defining equation for the viscosity,

𝜏 = 𝜇 𝑑𝑢𝑑𝑦

5.1

The region of flow that develops from the leading edge of the plate in which the

effects of viscosity are observed is called the boundary layer. Some arbitrary point is

used to designate the y position where the boundary layer ends; this point is usually

chosen as the y coordinate where the velocity becomes 99 percent of the free-stream

value. Initially, the boundary-layer development is laminar, but at some critical distance

from the leading edge, depending on the flow field and fluid properties, small

disturbances in the flow begin to become amplified, and a transition process takes

place until the flow becomes turbulent. The turbulent-flow region may be pictured as

a random churning action with chunks of fluid moving to and fro in all directions.

The transition from laminar to turbulent flow occurs when

𝑢∞𝑥𝜐

= 𝜌𝑢∞𝑥𝜇

= 𝑅𝑒 > 5 ∗ 105 (transiant)

𝑅𝑒 ≤ 5 ∗ 105 (laminar)

Where

𝑢∞= free stream velocity, m/s.

𝑥 = distance from leading edge, m.

𝜐 = 𝜇𝜌 =kinematic viscosity, m2/s.

𝑅𝑒𝑥 = 𝑢∞𝑥𝜐

5.2

Figure 5.1: Sketch showing different boundary-layer flow regimes on a flat plate.

Figure 5.2: Laminar velocity profile on a flat plate.

5.3 Laminar Boundary Layer on a Flat Plate

Consider the elemental control volume shown in Figure 5.3. We derive the equation

of motion for the boundary layer by making a force-and-momentum balance on this

element.

To simplify the analysis we assume:

1. The fluid is incompressible and the flow is steady.

2. There are no pressure variations in the direction perpendicular to the plate.

3. The viscosity is constant.

4. Viscous-shear forces in the y direction are negligible.

Figure 5.4 Elemental control volume for force balance on laminar boundary layer. The mass continuity equation for the boundary layer. 𝜕𝑢𝜕𝑥

+ 𝜕𝑣𝜕𝑦

= 0 5.3 The momentum equation of the laminar boundary layer with constant properties. 𝜌 �𝑢 𝜕𝑢

𝜕𝑥+ 𝑣 𝜕𝑢

𝜕𝑦� = 𝜇 𝜕2𝑢

𝜕𝑦2− 𝜕𝑝

𝜕𝑥 5.4

Figure 5.5: Elemental control volume for integral momentum analysis of laminar boundary layer.

The mass flow through plane 1 in Figure 5.5 is

�̇� = ∫ 𝜌𝑢 𝑑𝑦𝐻0

Von Kármán approximate solution of equations 5.3 and 5.4 gives

𝛿 = 4.64𝑥

𝑅𝑒𝑥12�

The exact solution of the boundary-layer equations

𝛿 = 5𝑥

𝑅𝑒𝑥12� 5.5

The velosity profile of the stream in x-direction within the baundary layer is given by:

𝑢𝑢∞

= 32𝑦𝛿− 1

2�𝑦𝛿�3 5.6

And the energy equation of the laminar boundary layer is:

𝑢 𝜕𝑇𝜕𝑥

+ 𝑣 𝜕𝑇𝜕𝑦

= 𝛼 𝜕2𝑇𝜕𝑦2

+ 𝜇𝜌𝑐𝑝

�𝜕𝑢𝜕𝑦�2 5.7

For low velocity incompressible flow, we have

𝑢 𝜕𝑇𝜕𝑥

+ 𝑣 𝜕𝑇𝜕𝑦

= 𝛼 𝜕2𝑇𝜕𝑦2

5.8

There is a striking similarity between equation 5.8 and the momentum equation for costant pressure,

𝑢 𝜕𝑢𝜕𝑥

+ 𝑣 𝜕𝑢𝜕𝑦

= 𝛼 𝜕2𝑢𝜕𝑦2

5.9

Example 5.1: Mass Flow and Boundary-Layer Thickness -3

Air at 27◦C and 1 atm flows over a flat plate at a speed of 2 m/s. Calculate the

boundary-layer thickness at distances of 20 cm and 40 cm from the leading edge of

the plate. Calculate the mass flow that enters the boundary layer between 𝑥 = 20 cm

and 𝑥 = 40 cm. The viscosity of air at 27◦C is 1.85×10−5 kg/m· s. Assume unit depth

in the z direction. Solution The density of air is calculated from

𝜌 = 𝑝𝑅𝑇

= 1.0132∗105

(287)(27+273)= 1.177 kg/m3


At 𝑥 = 20 𝑐𝑚: 𝑅𝑒𝑥 = 𝑢∞𝑥𝜐

= 𝜌𝑢∞𝑥𝜇

= 1.177∗2∗0.21.85×10−5

= 25,448

At 𝑥 = 40 𝑐𝑚: 𝑅𝑒𝑥 = 𝜌𝑢∞𝑥𝜇

= 1.177∗2∗0.41.85×10−5

= 50,896

The boundary layer thickness is calculated from:

𝛿 = 4.64𝑥

𝑅𝑒𝑥12�

At 𝑥 = 20 𝑐𝑚, 𝛿 = 4.64∗0.2

(25488)1 2�= 0.00582 m.

At 𝑥 = 40 𝑐𝑚, 𝛿 = 4.64∗0.4

(50896)1 2�= 0.00823 m.

To calculate the mass flow that enters the boundary layer from the free stream

between 𝑥 = 20 cm and 𝑥 = 40 cm, we simply take the difference between the mass

flow in the boundary layer at these two x positions. At any x position the mass flow in

the boundary layer is given by the integral

�̇� = ∫ 𝜌𝑢 𝑑𝑦𝛿0

𝑢 = 𝑢∞32𝑦𝛿− 1

2�𝑦𝛿�3

�̇� = ∫ 𝜌𝑢∞3

2

𝑦

𝛿−

1

2�𝑦𝛿�

3 𝑑𝑦𝛿

0 =5

8𝜌𝑢∞𝛿

∆�̇� = 58𝜌𝑢∞(𝛿40 − 𝛿20) = 5

8(1.177 ∗ 2(0.00823 − 0.00582) = 0.00354 kg/s

5.4 Thermal boundary layer

Just as the hydrodynamic boundary layer was defined as that region of the flow where

viscous forces are felt, a thermal boundary layer may be defined as that region where

temperature gradients are present in the flow. These temperature gradients would

result from a heat-exchange process between the fluid and the wall.

Consider the system shown in Figure 5.6. The temperature of the wall is 𝑇𝑤, the

temperature of the fluid outside the thermal boundary layer is 𝑇∞, and the thickness of

the thermal boundary layer is designated as 𝛿𝑡 . At the wall, the velocity is zero, and

the heat transfer into the fluid takes place by conduction. Thus the local heat flux per

unit area is: 𝑞𝐴

= −𝑘 𝜕𝑇𝜕𝑦�𝑤𝑎𝑙𝑙

5.10

From Newton’s law of cooling

𝑞𝐴

= ℎ(𝑇𝑤 − 𝑇∞) 5.11

where h is the convection heat-transfer coefficient. Combining these equations, we have

ℎ =−𝑘𝜕𝑇𝜕𝑦�𝑤𝑎𝑙𝑙(𝑇𝑤−𝑇∞)

5.12

Figure 5.6: Temperature profile in the thermal boundary layer.

Then we need only find the temperature gradiant at the wall to evaluate ℎ. Therefore, the temperature distribution is:

𝜃𝜃∞

= 𝑇−𝑇𝑤𝑇∞−𝑇𝑤

= 32𝑦𝛿𝑡− 1

2�𝑦𝛿𝑡�3 5.13

The thermal boundary layer can calculated from the equation below:

𝛿𝑡𝛿

= 11.026

𝑃𝑟−1 3� �1 − �𝑥𝑜𝑥�34� �13�

5.14

When the plate is heated over the entire length, 𝑥𝑜 = 0, and

𝛿𝑡𝛿

= 11.026

𝑃𝑟−1 3� 5.15

Where the Prandtl number is dimensionless when a consistent set of units is used:

𝑃𝑟 = 𝜐𝛼

=𝜇𝜌�

𝑘 𝜌𝑐𝑝�= 𝑐𝑝𝜇

𝑘 5.16

The local convective heat transfer coefficient is calculated from the equation as below:

ℎ𝑥 = 0.332𝑘 𝑃𝑟1 3� �𝑢∞𝜐𝑥�12� �1 − �𝑥𝑜

𝑥�34� �−1

3�

5.17

The equation may be nondimensionalized by multiplying both sides by 𝑥𝑘

, producing the dimensionless group on the left side,

𝑁𝑢𝑥 = ℎ𝑥𝑥𝑘

5.18

called the Nusselt number afterWilhelm Nusselt, who made significant contributions to the theory of convection heat transfer. Finally,

𝑁𝑢𝑥 = 0.332 𝑃𝑟1 3� 𝑅𝑒𝑥12� �1 − �𝑥𝑜

𝑥�34� �−1

3�

5.19

or, for the plate heated over its entire length, 𝑥𝑜 = 0 and

𝑁𝑢𝑥 = 0.332 𝑃𝑟1 3� 𝑅𝑒𝑥12� 0.6 < 𝑃𝑟 > 50 5.20 a

Equation (5.20 a) is applicable to fluids having Prandtl numbers between about 0.6

and 50. It would not apply to fluids with very low Prandtl numbers like liquid metals

or to high- Prandtl-number fluids like heavy oils or silicones. For a very wide range of

Prandtl numbers, Churchill and Ozoe have correlated a large amount of data to give

the following relation for laminar flow on an isothermal flat plate:

𝑁𝑢𝑥 = 0.3387𝑅𝑒𝑥12� 𝑃𝑟

13�

�1+�0.0468𝑃𝑟 �

23� �

14� 𝑅𝑒𝑥𝑃𝑟 > 100 5.20 b

Equations (5.17), (5.19), and (5.20 a) express the local values of the heat-transfer

coefficient in terms of the distance from the leading edge of the plate and the fluid

properties. For the case where 𝑥𝑜 = 0 the average heat-transfer coefficient and

Nusselt number may be obtained by integrating over the length of the plate:

ℎ� = ∫ ℎ𝑥𝑑𝑥𝐿0

∫ 𝑑𝑥𝐿0

= 2ℎ𝑥=𝐿 5.21

assuming the heated section is at the constant temperature 𝑇𝑤. For the plate heated

over the entire length,

𝑁𝑢��𝐿 = ℎ�𝐿𝑘

= 2𝑁𝑢𝑥=𝐿 5.22

Or


= 0.664𝑅𝑒𝐿12� 𝑃𝑟1 3� 5.23

Where:

𝑅𝑒𝐿 = 𝜌𝑢∞𝐿𝜇

5.24

For a plate where heating starts at 𝑥 = 𝑥𝑜, it can be shown that the average heat

transfer coefficient can be expressed as

ℎ�𝑥𝑜−𝐿 = ℎ𝑥=𝐿 �2𝐿1−�𝑥𝑜 𝐿� �

34�

𝐿−𝑥𝑜�

In this case, the total heat transfer for the plate would be 𝑞𝑡𝑜𝑡𝑎𝑙 = ℎ�𝑥𝑜−𝐿(𝐿 − 𝑥𝑜)(𝑇𝑤 − 𝑇∞)

The foregoing analysis was based on the assumption that the fluid properties were

constant throughout the flow. When there is an appreciable variation between wall

and free-stream conditions, it is recommended that the properties be evaluated at the

so-called film temperature 𝑇𝑓 , defined as the arithmetic mean between the wall and

free-stream temperature,

𝑇𝑓 = 𝑇𝑤+𝑇∞2

5.25

Constant Heat Flux

The above analysis has considered the laminar heat transfer from an isothermal

surface. In many practical problems the surface heat flux is essentially constant, and

the objective is to find the distribution of the plate-surface temperature for given

fluid-flow conditions. For the constant-heat-flux case it can be shown that the local

Nusselt number is given by

𝑁𝑢𝑥 = 0.453 𝑅𝑒𝑥12� 𝑃𝑟1 3� 5.26

which may be expressed in terms of the wall heat flux and temperature difference as

𝑁𝑢𝑥 = 𝑞𝑤𝑥𝑘(𝑇𝑤−𝑇∞)

5.27

Where: 𝑞𝑤 : heat flux, W/m2

Note that the heat flux 𝑞𝑤 = 𝑞𝐴 is assumed constant over the entire plate surface.

𝑇𝑤 − 𝑇∞�� = 1𝐿 ∫ (𝑇𝑤 − 𝑇∞

𝐿0 )𝑑𝑥 = 1

𝐿 ∫𝑞𝑤𝑥𝑘𝑁𝑢𝑥

𝐿0 𝑑𝑥

𝑇𝑤 − 𝑇∞�� =𝑞𝑤𝐿 𝑘�

0.6795𝑅𝑒𝐿12� 𝑃𝑟

13� 5.28

𝑞𝑤 = 3

2ℎ𝑥=𝐿(𝑇𝑤 −𝑇∞)

For constant heat flux case and the properties evaluated at the film temperature:

𝑁𝑢𝑥 = 0.4637𝑅𝑒𝑥12� 𝑃𝑟

13�

�1+�0.0207𝑃𝑟 �

23� �

14� 𝑅𝑒𝑥𝑃𝑟 > 100 5.29

Example 5.1: -3

Air at 27◦C and 1 atm flows over a flat plate at a speed of 2 m/s. Calculate the

boundary-layer thickness at distances of 20 cm and 40 cm from the leading edge of

the plate. Calculate the mass flow that enters the boundary layer between 𝑥 = 20 cm

and 𝑥 = 40 cm. The viscosity of air at 27◦C is 1.85×10−5 kg/m· s. Assume unit depth

in the z direction.

Example 5.2:Isothermal Flat Plate Heated Over Entire Length XAMPLE 5-4

For the flow system in Example 5.1 assume that the plate is heated over its entire

length to a temperature of 60◦C. Calculate the heat transferred in (a) the first 20 cm of

the plate and (b) the first 40 cm of the plate.

Solution


, 𝑇∞ = 27˚𝐶, 𝑢∞ = 2 m/s, 𝑇𝑤=60˚C

𝑇𝑓 = 60+272

= 43.5 + 273 = 316.5 K

We find the properties of air at film temperature.

𝜐 = 17.36 ∗ 10−6 m2/s, 𝑃𝑟 = 0.7 , 𝑘 = 0.02749 W/m.˚C.

𝑐𝑝 = 1.006 Kj/kg. ˚C.

At 𝒙 = 𝟐𝟎 𝒄𝒎


= 2∗0.217.36∗10−6

= 23041


𝑁𝑢𝑥 = 0.332 𝑃𝑟1 3� 𝑅𝑒𝑥12� = 0.332 (0.7)1 3� ∗ (23041)1 2� = 44.74

ℎ𝑥 = 𝑁𝑢𝑥𝑘𝑥

= 44.74 ∗ 0.027490.2

= 6.15 W/m2. ˚C

The average value of the heat-transfer coefficient is twice this value, or

ℎ� = 2ℎ𝑥 = 2 ∗ 6.15 = 12.30

𝑞 = ℎ�𝐴(𝑇𝑤 − 𝑇∞) = 12.3 ∗ (0.2)(60 − 27) = 81.18 W

At 𝒙 = 𝟒𝟎 𝒄𝒎


= 2∗0.417.36∗10−6

= 46082


𝑁𝑢𝑥 = 0.332 𝑃𝑟1 3� 𝑅𝑒𝑥12� = 0.332 (0.7)1 3� ∗ (46082)1 2� = 63.28


= 63.28 ∗ 0.027490.4

= 4.349 W/m2. ˚C

The average value of the heat-transfer coefficient is twice this value, or

ℎ� = 2ℎ𝑥 = 2 ∗ 4.349 = 8.698

𝑞 = ℎ�𝐴(𝑇𝑤 − 𝑇∞) = 8.698 ∗ (0.4)(60 − 27) = 114.8 W

EXAMPLE 5.3: Flat Plate with Constant Heat Flux

A 1.0-kW heater is constructed of a glass plate with an electrically conducting film

that produces a constant heat flux. The plate is 60 cm by 60 cm and placed in an

airstream at 27◦C, 1 atm with 𝑢∞ =5 m/s. Calculate the average temperature

difference along the plate.

Solution

Properties should be evaluated at the film temperature, but we do not know the plate

temperature. for an initial calculation, we take the properties at the free-stream

conditions of

At 𝑇∞ = 27 ˚C we find the properties of the fluid

𝜐 = 15.96 ∗ 10−6 m2/s, 𝑃𝑟 = 0.708 , 𝑘 = 0.02624 W/m.˚C.

𝑅𝑒𝐿 = 𝑢∞𝐿𝜐

= 5∗0.615.96∗10−6

= 1.88 ∗ 105 therefore the flow is laminar

𝑇𝑤 − 𝑇∞�� =𝑞𝑤𝐿 𝑘�

0.6795𝑅𝑒𝐿12� 𝑃𝑟

13�

𝑇𝑤 − 𝑇∞�� =�10000.62

��0.60.02624� �

0.6795(1.88∗105)1 2� (0.708)13�

= 241.85 ˚C

𝑇𝑤 = 241.85 + 27 = 268.85 ˚C

Now we find the properties at the film temperature


= 268.85+272

= (147.927 + 273) = 421 𝐾

𝜐 = 28.22 ∗ 10−6 m2/s, 𝑃𝑟 = 0.687 , 𝑘 = 0.035 W/m.˚C.

𝑅𝑒𝐿 = 5∗0.628.22∗10−6

= 1.06 ∗ 105

𝑇𝑤 − 𝑇∞�� =�10000.62

��0.60.035 � �

0.6795(1.06∗105)1 2� (0.687)13�

= 243.6 ˚C

𝑁𝑢𝑥 = 𝑞𝑤𝑥𝑘(𝑇𝑤−𝑇∞)

𝑇𝑤 − 𝑇∞ = 𝑞𝑤𝑥𝑘𝑁𝑢𝑥

𝑇𝑤 − 𝑇∞ = 𝑞𝑤𝑥𝑘𝑁𝑢𝑥

𝑁𝑢𝑥 = 0.453 𝑅𝑒𝑥12� 𝑃𝑟1 3�

𝑇𝑤 − 𝑇∞ = 𝑞𝑤𝑥

(𝑘)0.453 𝑅𝑒𝑥12� 𝑃𝑟

13�

At 𝑥 = 𝐿

𝑇𝑤 − 𝑇∞ = 𝑞𝑤𝐿

(𝑘)0.453 𝑅𝑒𝐿12� 𝑃𝑟

13�

𝑇𝑤 − 𝑇∞ =�10000.62

��0.60.035 � �

0.453 (1.06∗105)1 2� (0.687)1 3�= 365.9 ˚C

EXAMPLE 5.4:Plate with Unheated Starting Length

Air at 1 atm and 300 K flows across a 20-cm-square plate at a free-stream velocity of

20 m/s. The last half of the plate is heated to a constant temperature of 350 K.

Calculate the heat lost by the plate.

Solution

First we evaluate the air properties at the film temperature


= 350+3002

= 325 𝐾

𝜐 = 18.23 ∗ 10−6 m2/s, 𝑘 = 0.02814 W/m.˚C, 𝑃𝑟 = 0.7

At 𝑥 = 𝐿


= 20∗0.218.23∗10−6

= 2.194 ∗ 105

𝑅𝑒𝐿 < 5 ∗ 105

Therefore the flow is laminar

ℎ𝑥 = 0.332𝑘 𝑃𝑟1 3� �𝑢∞𝜐𝑥�12� �1 − �𝑥𝑜

𝑥�34� �−1

3�

ℎ𝐿 = 0.332 ∗ 0.02814( 0.7)1 3� � 2018.23∗10−6∗0.2

�12� �1 − �0.1

0.2�34� �−1

3�

ℎ𝐿 = 26.253 W/m2.˚C.

ℎ�𝑥𝑜−𝐿 = ℎ𝑥=𝐿 �2𝐿1−�𝑥𝑜 𝐿� �

34�

𝐿−𝑥𝑜�

ℎ�𝑥𝑜−𝐿 = 26.253 �2(0.2) 1−�0.1

0.2� �34�

0.2−0.1�

ℎ�𝑥𝑜−𝐿 = 42.566 W/m2.˚C

𝑞 = ℎ�𝑥𝑜−𝐿(𝐿 − 𝑥𝑜) ∗ 𝑤(𝑇𝑤 − 𝑇∞)

𝑞 = 42.566(0.2− 0.1) ∗ 0.2(350 − 300) = 42.566 W

EXAMPLE 5.5: Oil Flow Over Heated Flat Plate

Engine oil at 20◦C is forced over a 20-cm-square plate at a velocity of 1.2 m/s. The

plate is heated to a uniform temperature of 60◦C. Calculate the heat lost by the plate.

Solution

We first evaluate the film temperature:


= 60+202

= 40 + 273 = 313 K

𝜐 = 0.00024 m2/s, 𝑘 = 0.144 W/m.˚C, 𝑃𝑟 = 2870, 𝜌 = 876 kg/m3


= 1.2∗0.20.00024

= 1000

𝑁𝑢𝑥 = 0.3387𝑅𝑒𝑥12� 𝑃𝑟

13�

�1+�0.0468𝑃𝑟 �

23� �

14�

= 0.3387(1000)1 2� (2870)13�

�1+�0.04682870 �

23� �

14�

= 152.2


= 152.2 ∗ 0.1440.2

= 109.6 W/m2. ˚C

ℎ� = 2ℎ𝑥 = 2 ∗ 109.6 = 219.2 W/m2. ˚C

𝑞 = ℎ�𝐴(𝑇𝑤 − 𝑇∞) = 219.2 ∗ (0.2)2(60 − 20) = 350.6 W

5.5 The Relation Between Fluid Friction and Heat Transfer

We have already seen that the temperature and flow fields are related. Now we seek

an expression whereby the frictional resistance may be directly related to heat

transfer.

The shear stress at the wall may be expressed in terms of a friction coefficient 𝐶𝑓 :

𝜏𝑤 = 𝐶𝑓𝜌𝑢∞2

2 5.30

The exact solution of the boundary-layer equations yields

𝐶𝑓𝑥2

= 0.332𝑅𝑒𝑥−1

2� 5.31

𝑁𝑢𝑥 = 0.332 𝑃𝑟1 3� 𝑅𝑒𝑥12� 5.20 a

Equation (5.20 a) may be rewritten in the following form:

𝑁𝑢𝑥𝑅𝑒𝑥𝑃𝑟

= ℎ𝑥𝜌𝑐𝑝𝑢∞

= 0.332 𝑃𝑟−2 3� 𝑅𝑒𝑥−1

2�

The group on the left is called the Stanton number,

𝑆𝑡𝑥 = ℎ𝑥𝜌𝑐𝑝𝑢∞

So that

𝑆𝑡𝑥𝑃𝑟23� = 0.332𝑅𝑒𝑥

12� 5.32

Upon comparing Equations (5.31) and (5.32), we note that the right sides are alike

except for a difference of about 3 percent in the constant, which is the result of the

approximate nature of the integral boundary-layer analysis. We recognize this

approximation

And write

𝑆𝑡𝑥𝑃𝑟23� = 𝐶𝑓𝑥

2 0.6 < 𝑝𝑟 < 60 5.33

Equation (5.33), called the Reynolds-Colburn analogy, expresses the relation between

fluid friction and heat transfer for laminar flow on a flat plate. The heat-transfer

coefficient thus could be determined by making measurements of the frictional drag

on a plate under conditions in which no heat transfer is involved.

It turns out that Equation (5.33) can also be applied to turbulent flow over a flat plate

and in a modified way to turbulent flow in a tube. It does not apply to laminar tube

flow.

EXAMPLE 5.3: A 1.0-kW heater is constructed of a glass plate with an electrically

conducting film that produces a constant heat flux. The plate is 60 cm by 60 cm and

placed in an airstream at 27◦C, 1 atm with 𝑢∞ =5 m/s. Calculate the average

temperature difference along the plate.

EXAMPLE 5.6 Drag Force on a Flat Plate

For the flow system in Example 5.2 compute the drag force exerted on the first 40 cm

of the plate using the analogy between fluid friction and heat transfer.

Solution

We use Equation (5.33) to compute the friction coefficient and then calculate the drag

force. An average friction coefficient is desired, so

𝑆𝑡𝑥𝑃𝑟23� = 𝐶𝑓𝑥

2

From example 5.2


, 𝑇∞ = 27˚𝐶, 𝑢∞ = 2 m/s, 𝑇𝑤=60˚C, 𝑐𝑝 = 1.006 Kj/kg. ˚C.

𝑇𝑓 = 60+272

= 43.5 + 273 = 316.5 K

𝜌 = 𝑝𝑅𝑇

= 1.0132∗105

(287)(316.5)= 1.115 kg/m3

For 40 cm length

From example 5.2

ℎ� = 8.698

𝑆𝑡� = ℎ�

𝜌𝑐𝑝𝑢∞= 8.698

1.115∗1006∗2= 3.88 ∗ 10−3

𝐶𝑓��

2= 𝑆𝑡� 𝑃𝑟2 3� = 3.88 ∗ 10−3(0.7)2 3� = 3.06 ∗ 10−3

𝜏𝑤�� = 𝐶𝑓��𝜌𝑢∞2

2= 3.06 ∗ 10−3 ∗ 1.115 ∗ (2)2 = 0.0136 N/m2

𝐷 = 𝜏𝑤��𝐿 = 0.0136 ∗ 0.4 = 5.44 N.m

5.6 Turbulent-Boundary-Layer Heat Transfer

Schlichting has surveyed experimental measurements of friction coefficients for

turbulent flow on flat plates. We present the results of that survey so that they may be

employed in the calculation of turbulent heat transfer with the fluid-friction–heat-

transfer analogy. The local skin-friction coefficient is given by

𝐶𝑓𝑥 = 0.0592𝑅𝑒𝑥−15 5 ∗ 105 < 𝑅𝑒𝑥 < 107 5.34

𝐶𝑓𝑥 = 0.370(log𝑅𝑒𝑥)−2.584 107 < 𝑅𝑒𝑥 < 109 5.35

The average-friction coefficient for a flat plate with a laminar boundary layer up to

𝑅𝑒crit and turbulent thereafter can be calculated from

𝐶�̅� = 0.455(log𝑅𝑒𝐿)2.584 −

𝐴𝑅𝑒𝐿

𝑅𝑒𝐿 < 109 for laminar and turbulent 5.36

where the constant A depends on 𝑅𝑒crit in accordance with Table 5.1. A somewhat

simpler formula can be obtained for lower Reynolds numbers as

𝐶�̅� = 0.074

𝑅𝑒𝐿15− 𝐴

𝑅𝑒𝐿 𝑅𝑒𝐿 < 107 5.37

Applying the fluid-friction analogy

𝑆𝑡 𝑃𝑟2/3 = 𝐶𝑓 2

we obtain the local turbulent heat transfer as:

𝑆𝑡𝑥 𝑃𝑟2/3 = 0.0296 𝑅𝑒𝑥−15 5 ∗ 105 < 𝑅𝑒𝑥 < 107 5.38

Or

𝑆𝑡𝑥 𝑃𝑟2/3 = 0.185(log𝑅𝑒𝑥)−2.584 107 < 𝑅𝑒𝑥 < 109 5.39 The average heat transfer over the entire laminar-turbulent boundary layer is

𝑆𝑡� 𝑃𝑟2/3 = 𝐶𝑓 2

5.40

For 𝑅𝑒𝑐𝑟𝑖𝑡 = 5 × 105 and 𝑅𝑒𝐿 < 107, Equation (5.37) can be used to obtain

𝑆𝑡� 𝑃𝑟2/3 = 0.037𝑅𝑒𝐿−1 5� − 871𝑅𝑒𝐿−1 5.41

𝑆𝑡� = 𝑁𝑢��

(𝑅𝑒𝐿Pr)

𝑁𝑢��𝐿 = ℎ𝐿𝑘

= 𝑃𝑟13(0.037𝑅𝑒𝐿0.8 − 871) 5.42

For higher Reynolds numbers the friction coefficient from Equation (5.36) may be used, so that For 107 < 𝑅𝑒𝐿 < 109 and 𝑅𝑒𝑐𝑟𝑖𝑡 = 5 × 105

𝑁𝑢𝐿 = ℎ�𝐿𝑘

= [0.228𝑅𝑒𝐿(log𝑅𝑒𝐿)−2.584 − 871]𝑃𝑟13 5.43

If 𝑅𝑒𝑐𝑟𝑖𝑡 differs from 5 × 105, An alternative equation is suggested by Whitaker that

may give better results with some liquids because of the viscosity-ratio term:

𝑁𝑢��𝐿 = 0.036𝑃𝑟0.43(𝑅𝑒𝐿0.8 − 9200) �𝜇∞𝜇𝑤�14 5.44

0.7 < 𝑃𝑟 < 380

2 ∗ 105 < 𝑅𝑒𝐿 < 5.5 ∗ 106

0.26 < 𝜇∞𝜇𝑤

< 3.5

𝜇∞ : the viscosity evaluated at 𝑇∞.

𝜇𝑤 : the viscosity evaluated at 𝑇𝑤.

For the gases the viscosity ratio is dropped and the properties are evaluated at 𝑇𝑓.

Costant Heat Flux For constant-wall-heat flux in turbulent flow that the local Nusselt number is only

about 4 percent higher than for the isothermal surface; that is,

𝑁𝑢𝑥 = 1.04𝑁𝑢𝑥]𝑇𝑤=𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 5.45

Example 5.7:Turbulent Heat Transfer from Isothermal Flat Plate

Air at 20◦C and 1 atm flows over a flat plate at 35 m/s. The plate is 75 cm long and is

maintained at 60◦C. Assuming unit depth in the z direction, calculate the heat transfer

from the plate.

Solution

We evaluate properties at the film temperature:


= 60+202

= 40 + 273 = 313 K

𝜌 = 𝑝𝑅𝑇

= 1.0132∗105

(287)(313)= 1.128 kg/m3.

𝜇 = 1.906 ∗ 10−5 kg/m.s, 𝑘 = 0.02723 W/m.˚C, 𝑃𝑟 = 0.7, 𝐶𝑝 = 1.007 kJ/kg.˚C.


= 𝜌𝑢∞𝐿𝜇

= 1.128∗35∗0.751.906∗10−5

= 1.553 ∗ 106 > 5 ∗ 105

then the flow is turbulent

therefore we use equation 5.42


= 𝑃𝑟13(0.037𝑅𝑒𝐿0.8 − 871)

𝑁𝑢��𝐿 = (0.7)13[(0.037)(1.553 ∗ 106)0.8 − 871] = 2180

ℎ� = 𝑁𝑢��𝐿𝑘𝐿

= (2180)(0.02723)0.75

= 79.1 W/m2.˚C.

𝑞 = ℎ�𝐴(𝑇𝑤 − 𝑇∞) = (79.1)(0.75)(60 − 20) = 2373 W.

5.7 Turbulent-Boundary-Layer Thickness

The turbulent boundary layer thichness is calculated from the equation

below:

1. The boundary layer is fully turbulent from the leading edge of the

plate.

𝛿 = 0.381𝑥

𝑅𝑒𝑥15

5.46

2. The boundary layer follows a laminor growth pattern up to

𝑅𝑒𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 = 5 ∗ 105 and a turbulent growth thereafter.

𝛿 = �0.381𝑅𝑒𝑥−15 − 10,256𝑅𝑒𝑥−1�𝑥 5 ∗ 105 < 𝑅𝑒𝑥 < 107 5.47

Example 5.8: Turbulent-Boundary-Layer Thickness

Calculate the turbulent-boundary-layer thickness at the end of the plate for Example

5.7, assuming that it develops (a) from the leading edge of the plate and (b) from the

transition point at 𝑅𝑒𝑐𝑟𝑖𝑡 = 5 ∗ 105.

Solution

Since we have already calculated the Reynolds number as 𝑅𝑒𝐿 = 1.553 × 106, it is a

simple matter

to insert this value in Equations (5.46) and (5.47) along with 𝑥 = 𝐿 = 0.75 m to give

(a) 𝛿 = 0.381𝑥

𝑅𝑒𝑥15

=(0.381)(0.75)

(1.553∗106)15

= 0.0165 m= 16.5 mm

(b) 𝛿 = �0.381𝑅𝑒𝑥−15 − 10,256𝑅𝑒𝑥−1�𝑥

𝛿 = �(0.381)(1.553 ∗ 106)−15 − (10,256)(1.553 ∗ 106)−1� ∗ 0.75 = 9.9 mm

5.8 Internal Forced Convection

5.8.1 Heat Transfer in Laminar Tube Flow

Consider the flow in a tube as shown in Figure 5.7. A boundary layer develops at the

entrance, as shown. Eventually the boundary layer fills the entire tube, and the flow is

said to be fully developed. If the flow is laminar, a parabolic velocity profile is

experienced, as shown in Figure 5.7a. When the flow is turbulent, a somewhat blunter

profile is observed, as in Figure 5.7b. In a tube, the Reynolds number is again used as

a criterion for laminar and turbulent flow. For

𝑅𝑒𝐷ℎ = 𝑢𝑚𝐷ℎ𝑣

> 2300 5.48

Where

𝐷ℎ = 4𝐴𝑐𝑃

the flow is usually observed to be turbulent 𝐷 is the tube diameter.

Again, a range of Reynolds numbers for transition may be observed, depending on the

pipe roughness and smoothness of the flow. The generally accepted range for

transition is

2000 < 𝑅𝑒𝐷 < 4000

Figure 5.7 Velocity profile for (a) laminar flow in a tube and (b) turbulent tube flow.

Mean velocity The value of the mean velocity 𝑢𝑚 in a tube is determined from

�̇� = 𝜌𝑢𝑚𝐴𝑐 = ∫𝜌 𝑢(𝑟, 𝑥)𝑑𝐴𝑐 5.49

𝑢𝑚 = ∫ 𝜌𝑢(𝑟,𝑥)2𝜋𝑟𝑑𝑟𝑟𝑜0

𝜌𝜋𝑟𝑜2= 2

𝑟𝑜2∫ 𝑢(𝑟, 𝑥)𝑟𝑑𝑟𝑟𝑜0 5.50

The velocity distribution for the fully developed flow in tube may be written 𝑢𝑢𝑜

= 1 − 𝑟2

𝑟𝑜2 5.51

The temperature distribution in the tube can calculated from the equation below:

𝑇 − 𝑇𝑐 = 1𝛼𝜕𝑇𝜕𝑥

𝑢𝑜𝑟𝑜2

4�� 𝑟𝑟𝑜�2− 1

4� 𝑟𝑟𝑜�4� 5.52

Where:

𝑇𝑐 : center temperature.

The Bulk Temperature

In tube flow the convection heat-transfer coefficient is usually defined by

𝑞𝐴

= ℎ(𝑇𝑤 − 𝑇𝑏) (local heat flux) 5.53

Where

𝑇𝑤 :is the wall temperature and

𝑇𝑏 :is the so-called bulk temperature, or energy-average fluid temperature across the

tube, which may be calculated from

𝑇𝑏 = 𝑇� = ∫ 𝜌2𝜋𝑟𝑑𝑟𝑢𝑐𝑝𝑇𝑟00∫ 𝜌2𝜋𝑟𝑑𝑟𝑢𝑐𝑝𝑟00

5.54

𝑇𝑏 = 𝑇𝑐 + 796

𝑢𝑜𝑟𝑜2

𝛼𝜕𝑇𝜕𝑥

5.55

and for the wall temperature

𝑇𝑤 = 𝑇𝑐 + 316

𝑢𝑜𝑟𝑜2

𝛼𝜕𝑇𝜕𝑥

5.56

The heat-transfer coefficient is calculated from

𝑞 = ℎ𝐴(𝑇𝑤 − 𝑇𝑏) = 𝑘𝐴 �𝜕𝑇𝜕𝑟�𝑟=𝑟𝑜

5.57

ℎ =𝑘�𝜕𝑇𝜕𝑟�𝑟=𝑟𝑜

(𝑇𝑤−𝑇𝑏) 5.58

The temperature gradient is given by

𝜕𝑇𝜕𝑟�𝑟=𝑟𝑜

= 𝑢𝑜𝛼𝜕𝑇𝜕𝑥�𝑟2− 𝑟3

4𝑟𝑜2�𝑟=𝑟𝑜

= 𝑢𝑜𝑟𝑜4𝛼

𝜕𝑇𝜕𝑥

5.59

Substituting Equations (5.55), (5.56), and (5.59) in Equation (5.58) gives

ℎ = 2411

𝑘𝑟𝑜

= 4811

𝑘𝐷

5.60

Expressed in terms of the Nusselt number, the result is

𝑁𝑢𝐷 = ℎ𝑑𝑘

= 4.364 5.61

ℎ = 4.364 ∗ 𝑘𝑑

5.8.2 Turbulent Flow in a Tube

The developed velocity profile for turbulent flow in a tube will appear as shown in

Figure 5.8.

Figure 5.8 Velocity profile in turbulent tube flow.

𝑆𝑡 = ℎ𝜌𝑐𝑝𝑢𝑚

= 𝑁𝑢𝐷𝑅𝑒𝐷𝑃𝑟

= 𝑓8 𝑃𝑟 ≈ 1 5.62

𝑓 = 0.316

𝑅𝑒𝐷

14

for 4 ∗ 103 < 𝑅𝑒𝐷 ≤ 2 ∗ 105 and 5.63

𝑁𝑢𝐷𝑅𝑒𝐷𝑃𝑟

= 0.0395𝑅𝑒𝐷−1

4

𝑁𝑢𝐷 = 0.0395𝑅𝑒𝐷34 for 4 ∗ 103 < 𝑅𝑒𝐷 ≤ 2 ∗ 105 and 𝑃𝑟 ≈ 1 5.64

𝑆𝑡𝑃𝑟23 = 𝑓

8 𝑃𝑟 ≠ 1 5.65

𝑁𝑢𝐷 = 0.0395𝑅𝑒𝐷34 𝑃𝑟

13 for 4 ∗ 103 < 𝑅𝑒𝐷 ≤ 2 ∗ 105 and 𝑃𝑟 ≠ 1 5.66

5.8.3 Empirical Relation for Flow In Tube

Thus, for the tube flow depicted in Figure 5.9 the total energy added can be expressed

in terms of abulk-temperature difference by

𝑞 = �̇�𝑐𝑝(𝑇𝑏2 − 𝑇𝑏1) = ℎ𝐴(𝑇𝑤 − 𝑇𝑏)𝑎𝑣 5.67

𝑇𝑏 = 𝑇𝑏1+𝑇𝑏2

2

Figure 5.9 Total heat transfer in terms of bulk-temperature difference.

• Empirical Relation In Laminar Flow In Tube

Constant Heat Flux

Fully developed laminar flow in circular tube

𝑁𝑢𝐷 = ℎ𝐷𝑘

= 4.36 5.68

Constant Surface Temperature

Fully developed laminar flow in circular tube

𝑁𝑢𝐷 = ℎ𝐷𝑘

= 3.66 5.69

Laminar flow in the Entrance Region

𝑁𝑢��𝐷 = ℎ𝐷𝑘

= 3.66 +0.065�𝐷𝐿�𝑅𝑒 𝑃𝑟

1+0.04��𝐷𝐿�𝑅𝑒 𝑃𝑟�23 5.70

When the difference between the surface and the fluid temperatures is large, it may be

necessary to account for the variation of viscosity with temperature. The average

Nusselt number for developing laminar flow in a circular tube in that case can be

determined from [Sieder and Tate (1936),

𝑁𝑢��𝐷 = 1.86 �𝑅𝑒𝐷 Pr𝐷𝐿

�13 �𝜇𝑏

𝜇𝑤�0.14

𝑅𝑒𝐷 Pr𝐷𝐿

> 10 5.71

All properties are evaluated at the bulk mean fluid temperature, except for 𝜇𝑤, which is evaluated at the𝑇𝑤.

Entry Length

𝐿𝑡,𝑙𝑎𝑚𝑖𝑛𝑎𝑟 ≈ 0.05𝑅𝑒𝐷𝑃𝑟𝐷

page 439 in the book.

• Imperical Relations for Turbulent Flow in Tube

For fully developed, smooth surface and moderate temperature differences:

𝑁𝑢𝐷 = 0.023𝑅𝑒𝐷0.8𝑃𝑟𝑛 0.6 < 𝑃𝑟 < 100 5.72

The properties in this equation are evaluated at the average fluid bulk temperature, and the exponent 𝑛 has the following values:

𝑛 = �0.4 for heating of the fluid0.3 for cooling of the fluid

More recent information by Gnielinski suggests that better results for turbulent flow

in smooth tubes may be obtained from the following:

𝑁𝑢𝐷 = 0.0214(𝑅𝑒𝐷0.8 − 100)𝑃𝑟0.4 � 0.5 < 𝑃𝑟 < 1.5104 < 𝑅𝑒𝐷 < 5 ∗ 106 5.73

Or

𝑁𝑢𝐷 = 0.012(𝑅𝑒𝐷0.87 − 280)𝑃𝑟0.4 � 1.5 < 𝑃𝑟 < 500103 < 𝑅𝑒𝐷 < 106 5.74

To take into account the property variations, Sieder and Tate recommend the

following relation:

𝑁𝑢𝐷 = 0.027𝑅𝑒𝐷0.8𝑃𝑟13 �𝜇𝑏

𝜇𝑤�0.14

5.75

For entrance region

The above equations are apply to fully developed turbulent flow in tubes. In the

entrance region the flow is not developed, and Nusselt recommended the following

equation:

𝑁𝑢𝐷 = 0.036𝑅𝑒𝐷0.8𝑃𝑟13 �𝐷

𝐿�0.055

For 10 < 𝐿𝐷

< 400 5.76

Entry Length

𝐿𝑡,𝑡𝑢𝑟𝑏𝑢𝑙𝑒𝑛𝑡 ≈ 10𝐷

The above equations offer simplicity in computation, but uncertainties on the order of

±25 percent are not uncommon. Petukhov has developed a more accurate, although

more complicated, expression for fully developed turbulent flow in smooth tubes:

𝑁𝑢𝐷 =�𝑓8�𝑅𝑒𝐷𝑃𝑟

1.07+12.7�𝑓8�0.5�𝑃𝑟

23−1�

�𝜇𝑏𝜇𝑤�𝑛

5.77

Where:

𝑛 = 0.11 for 𝑇𝑤 > 𝑇𝑏,

𝑛 = 0.25 for 𝑇𝑤 < 𝑇𝑏, and

𝑛 = 0 for constant heat flux or for gases.

All properties are evaluated at

𝑇𝑓 = (𝑇𝑤 +𝑇𝑏)2

except for 𝜇𝑏 and 𝜇𝑤.

The friction factor may be obtained either from the following for smooth tubes:

𝑓 = (1.82 log10 𝑅𝑒𝐷 − 1.64 )−2 5.78

Equation (5.77) is applicable for the following ranges:

0.5 < 𝑃𝑟 < 200 for 6 percent accuracy

0.5 < 𝑃𝑟 < 2000 for 10 percent accuracy

104 > 𝑅𝑒𝐷 > 5 ∗ 106

0.8 < 𝜇𝑏/𝜇𝑤 < 40

All above equations are for smooth pipes. For rough pipe we use the equation below:

𝑆𝑡𝑏𝑃𝑟𝑓23 = 𝑓

8

𝑁𝑢𝐷𝑅𝑒𝐷𝑃𝑟

𝑃𝑟𝑓23 = 𝑓

8

The friction coefficient f is defined by

∆𝑝 = 𝑓 𝐿𝐷𝜌 𝑢𝑚2

2

An empirical relation for the friction factor for rough tubes is given as

𝑓 = 1.325

�ln� 𝜀3.7𝐷�+

5.74𝑅𝑒𝐷

0.9�2

For 10−6 < 𝜀

𝐷< 10−3 , and 5000 < 𝑅𝑒𝐷 < 108

• Isothermal Parallel Plates

The average Nusselt number for the thermal entrance region of flow between

isothermal parallel plates of length L is expressed as (Edwards et al., 1979)

𝑁𝑢𝐷ℎ = ℎ𝐷ℎ𝑘

= 7.45 +0.03�

𝐷ℎ𝐿 �𝑅𝑒𝐷ℎ 𝑃𝑟

1+0.016��𝐷ℎ𝐿 �𝑅𝑒𝐷ℎ 𝑃𝑟�

23 entry region, laminar flow 𝑅𝑒 ≤ 2800

𝐷ℎ = 2𝑆

Where

𝑆 : is the space between the two plates.

Example 5.9: Turbulent Heat Transfer in a Tube

Air at 2 atm and 200◦C is heated as it flows through a tube with a diameter of 1 in

(2.54 cm) at a velocity of 10 m/s. Calculate the heat transfer per unit length of tube if

a constant-heat-flux condition is maintained at the wall and the wall temperature is

20◦C above the air temperature, all along the length of the tube. How much would the

bulk temperature increase over a 3-m length of the tube?

Solution

We first calculate the Reynolds number to determine if the flow is laminar or

turbulent, and then select the appropriate empirical correlation to calculate the heat

transfer. The properties of air at a bulk temperature of 200◦C are

𝜌 = 𝑝𝑅𝑇

= (2)(1.0132∗105)(287)(473)

= 1.493 kg/m3

𝑃𝑟 = 0.681 , 𝜇 = 2.57 ∗ 10−5 kg/m.s, 𝑘 = 0.0386 W/m.˚C, 𝑐𝑝 = 1.025 kj/kg. ˚C.

𝑅𝑒𝐷 = 𝜌𝑢𝑚𝐷𝜇

= (1.493)(10)(0.0254)2.57∗10−5

= 14756

So that the flow is turbulent.

Check the entry length

𝐿𝑡,𝑡𝑢𝑟𝑏𝑢𝑙𝑒𝑛𝑡 ≈ 10𝐷 = 10(0.0254) = 0.254 m

Its too shorter than the length of the tube. Therefore the flow assumed fully

developed. We therefore use Equation (5.72) to calculate the heat-transfer coefficient.

𝑁𝑢𝐷 = 0.023𝑅𝑒𝐷0.8𝑃𝑟𝑛

For heating the fluid n=0.4

𝑁𝑢𝐷 = 0.023(14756)0.8(0.681)0.4 = 42.67

ℎ = 𝑁𝑢𝐷𝑘𝐷

= (42.67)(0.0386)0.0254

= 64.85 W/m2.˚C. 𝑞𝐿

= ℎ𝜋𝐷(𝑇𝑤 − 𝑇𝑏) = (64.85)(𝜋)(0.0254)(20) = 103.5 W/m.

𝑞 = �̇�𝑐𝑝∆𝑇𝑏 = 𝐿 �𝑞𝐿�

�̇� = 𝜌𝑢𝑚𝜋𝐷2

4= (1.493)(10)(𝜋) (0.0254)2

4= 7.565 ∗ 10−3 kg/s.

∆𝑇𝑏 =𝐿�𝑞𝐿�

�̇�𝑐𝑝= (3)(103.5)

(7.565∗10−3)(1025)= 40.04 ˚C.

EXAMPLE 5.10: Heating ofWater in Laminar Tube Flow

Water at 60◦C enters a tube of 1-in (2.54-cm) diameter at a mean flow velocity of 2

cm/s. Calculate the exit water temperature if the tube is 3.0 m long and the wall

temperature is constant at 80◦C.

Solution

We first evaluate the Reynolds number at the inlet bulk temperature to determine the

flow regime.

The properties of water at 60◦C are

𝜌 = 985 kg/m3



= (985)(0.02)(0.0254)4.71∗10−4

= 1062

So the flow is laminar.

Check for enrty length

𝐿𝑡,𝑙𝑎𝑚𝑖𝑛𝑎𝑟 ≈ 0.05𝑅𝑒𝐷𝑃𝑟𝐷 = (0.05)(1062)(3.02)(0.0254) = 4 m

Therefore the flow not fully developed (i.e the flow in the interence region)

Calculating the additional parameter, we have 𝑅𝑒𝐷 Pr𝐷

𝐿= (1062)(3.02)(0.0254)

3= 27.15 > 10

So Equation (5.71) or Equation (5.70) is applicable. We do not yet know the mean

bulk temperature to evaluate properties so we first make the calculation on the basis

of 60◦C,

At the wall temperature of 80◦C we have

𝜇𝑤 = 3.55 ∗ 10−4 kg/m.s.


�13 �𝜇𝑏

𝜇𝑤�0.14

𝑁𝑢��𝐷 = 1.86 �(1062)(3.02)(0.0254)3

�13 �4.41

3.55�0.14

= 5.816


= (5.816)(0.651)0.0254

= 149.1 W/m2.˚C.

�̇� = 𝜌𝑢𝑚𝜋𝐷2

4= (985)(0.02)(𝜋) (0.0254)2

4= 9.982 ∗ 10−3 kg/s.

𝑞 = ℎ𝜋𝐷𝐿 �𝑇𝑤 −𝑇𝑏1+𝑇𝑏2

2� = �̇�𝑐𝑝(𝑇𝑏2 − 𝑇𝑏1)

(149.1)(𝜋)(0.0254)(3) �80 − 60+𝑇𝑏22

� = (9.982 ∗ 10−3)(4180)(𝑇𝑏2 − 60)

𝑇𝑏2 = 71.98 ˚C

𝑇𝑏,𝑎𝑣𝑔 = 𝑇𝑏1+𝑇𝑏22

= 60+71.982

= 66 ˚C

𝜌 = 982 kg/m3



= (982)(0.02)(0.0254)4.63∗10−4

= 1147

𝑅𝑒𝐷 Pr𝐷𝐿

= (1147)(2.78)(0.0254)3

= 27.00 > 10


�13 �𝜇𝑏

𝜇𝑤�0.14

𝑁𝑢��𝐷 = 1.86(27)13 �4.36

3.55�0.14

= 5.743


= (5.743)(0.656)0.0254

= 148.3 W/m2.˚C.

(148.3)(𝜋)(0.0254)(3) �80 − 60+𝑇𝑏22

� = (9.982 ∗ 10−3)(4185)(𝑇𝑏2 − 60)

𝑇𝑏2 = 71.88 ˚C

5.9 Flow Across Cylinders And Spheres As the flow progresses along the front side of the cylinder, the pressure would

decrease and then increase along the back side of the cylinder, resulting in an increase

in free-stream velocity on the front side of the cylinder and a decrease on the back

side. The transverse velocity (that velocity parallel to the surface) would decrease

from a value of 𝑢∞ at the outer edge of the boundary layer to zero at the surface. As

the flow proceeds to the back side of the cylinder, the pressure increase causes a

reduction in velocity in the free stream and throughout the boundary layer.

Figure 5.10 Cylinder in cross flow.

Figure 5.11 Velocity distributions indicating flow separation on a cylinder in cross flow.

The pressure increase and reduction in velocity are related through the Bernoulli

equation written along a streamline: 𝑑𝑝𝜌

= −𝑑 � 𝑢2

2𝑔𝑐�

When the velocity gradient at the surface becomes zero, the flow is said to have reached a separation point:

seperation point at 𝜕𝑢𝜕𝑦�𝑦=0

= 0

The drag coefficient for bluff bodies is defined by

Drage force = 𝐹𝐷 = 𝐶𝐷𝐴𝜌𝑢∞2

2𝑔𝑐 5.79

where 𝐶𝐷 is the drag coefficient and

A is the frontal area of the body exposed to the flow, which, for a cylinder, is the

product of diameter and length. 𝐴 = 𝐿𝐷 The values of the drag coefficient for cylinders and spheres are given as a function of

the Reynolds number in Figures 5.12 and 5.13.

Figure 5.12: Drag coefficient for circular cylinders as a function of the Reynolds number.

Figure 5.13: Drag coefficient for spheres as a function of the Reynolds number.

5.9.1 Cylinder

The resulting correlation for average heat-transfer coefficients in cross flow over

circular cylinders is

𝑁𝑢𝐷𝑓 = ℎ𝐷𝑘𝑓

= 𝐶 �𝑢∞𝐷𝑣𝑓�𝑛𝑃𝑟𝑓

13 ( flow of gas) 5.80

where the constants 𝐶 and 𝑛 are tabulated in Table 5.1. Properties for use with

Equation (5.80) are evaluated at the film temperature as indicated by the subscript 𝑓 . Table 5.1 Constants for use with Equation (5.80), 𝑅𝑒𝐷𝑓 𝐶 𝑛

0.4–4 0.989 0.330

4–40 0.911 0.385

40–4000 0.683 0.466

4000–40,000 0.193 0.618

40,000–400,000 0.0266 0.805

For non-circular cylinder table 5.2 used to evaluate the constants in equation 5.80.

Table 5.2 Constants for heat transfer from noncircular cylinders for use with Equation (5.80).

Fand has shown that the heat-transfer coefficients from liquids to cylinders in cross

flow may be better represented by the relation

𝑁𝑢𝐷𝑓 = (0.35 + 0.56𝑅𝑒𝐷𝑓0.52)𝑃𝑟𝑓0.3 10−1 > 𝑅𝑒𝐷𝑓 < 105 flow of liquid 5.81

For equation 5.81 the properties at 𝑇𝑓 Still a more comprehensive relation is given by Churchill and Bernstein that is

applicable over the complete range of available data:

For the flow of air, water, and liquid sodium

𝑁𝑢𝐷 = 0.3 + 0.62𝑅𝑒𝐷

12𝑃𝑟

13

�1+�0.4𝑃𝑟�

23�

14�1 + � 𝑅𝑒𝐷

282000�58�

45

5.82

For 102 < 𝑅𝑒𝐷 < 107, 𝑃𝑒𝐷 > 0.2

Where:

𝑃𝑒𝐷 = 𝑅𝑒𝐷𝑃𝑟

𝑁𝑢𝐷 = �0.8237 − ln�𝑃𝑒𝐷12 ��

−1

𝑃𝑒𝐷 < 0.2 5.83

5.9.2 Spheres

McAdams recommends the following relation for heat transfer from spheres to a

flowing gas:

• Flow of gas

ℎ𝐷𝑘

= 𝑁𝑢𝐷 = 0.37 �𝑢∞𝐷𝑣�0.6

17 < 𝑅𝑒𝐷 < 7 ∗ 104 5.84

All properties of fluid at film temperature 𝑇𝑓.

Achenbach has obtained relations applicable over a still wider range of Reynolds

numbers for air with Pr =0.71:

𝑁𝑢𝐷 = 2 + (0.25𝑅𝑒𝐷 + 3 ∗ 10−4𝑅𝑒𝐷1.6)12 �100 < 𝑅𝑒𝐷 < 3 ∗ 105

𝑃𝑟 = 0.71 5.84

𝑁𝑢𝐷 = 430 + 𝑎 𝑅𝑒𝐷 + 𝑏 𝑅𝑒𝐷2 + 𝑐 𝑅𝑒𝐷3 �3 ∗ 105 < 𝑅𝑒𝐷 < 5 ∗ 106𝑃𝑟 = 0.71

5.85

All properties of fluid at film temperature 𝑇𝑓.

• Flow of liquid

For flow of liquids past spheres, the data of Kramers may be used to obtain the

Correlation

𝑁𝑢𝐷𝑃𝑟−0.3 = 0.97 + 0.68 �𝑢∞𝐷𝑣�0.5

1 < 𝑅𝑒𝐷 < 2000 5.86

All properties at film temperature 𝑇𝑓.

heat transfer from spheres to oil and water over a more extended range of Reynolds

numbers from 1 to 200,000:

𝑁𝑢𝐷𝑃𝑟−0.3 �𝜇𝑤𝜇∞�0.25

= 1.2 + 0.53𝑅𝑒𝐷0.54 1 < 𝑅𝑒𝐷 < 2 ∗ 105 5.87

where all properties are evaluated at free-stream conditions 𝑇∞, except 𝜇𝑤, which is

evaluated at the surface temperature of the sphere 𝑇𝑤.

• Liquid and Gases

All the above data have been brought together by Whitaker to develop a single

equation for gases and liquids flowing past spheres:

𝑁𝑢𝐷 = 2 + �0.4𝑅𝑒𝐷12 + 0.06𝑅𝑒𝐷

23�𝑃𝑟0.4 �𝜇∞

𝜇𝑤�14 �3.5 < 𝑅𝑒𝐷 < 8 ∗ 104

0.7 < 𝑃𝑟 < 380 5.88

Properties in Equation (5.88) are evaluated at the free-stream temperature 𝑇∞.

Example 5.11: Airflow Across Isothermal Cylinder

Air at 1 atm and 35◦C flows across a 5.0-cm-diameter cylinder at a velocity of 50 m/s.

The cylinder surface is maintained at a temperature of 150◦C. Calculate the heat loss

per unit length of the cylinder.

Solution:

We first determine the Reynolds number and then find the applicable constants from

Table 5.1 for use with Equation (5.80). The properties of air are evaluated at the film

temperature:


= 150+352

= 92.5 + 273 = 365.5 K

𝜌 = 𝑝𝑅𝑇

= (1.0132∗105)(287)(365.5)

= 0.966 kg/m3

𝑃𝑟 = 0.695 , 𝜇 = 2.14 ∗ 10−5 kg/m.s, 𝑘 = 0.0312 W/m.˚C.

𝑅𝑒𝐷 = 𝜌𝑢∞𝐷𝜇

= (0.966)(50)(0.05)2.14∗10−5

= 1.129 ∗ 105

From table 5.1 𝐶 = 0.0266 , 𝑛 = 0.805



13

ℎ𝐷𝑘𝑓

= (0.0266)(1.129 ∗ 105)0.805(0.695)13 = 275.1

ℎ = (275.1)(0.0312)0.05

= 171.7 W/m2.˚C. 𝑞𝐿

= ℎ𝜋𝐷(𝑇𝑤 − 𝑇∞) 𝑞𝐿

= (171.7)(𝜋)(0.05)(150 − 35) = 3100 W/m

H.W

A fine wire having a diameter of 3.94*10−5 m is placed in a 1-atm airstream at 25◦C

having a flow velocity of 50 m/s perpendicular to the wire. An electric current is

passed through the wire, raising its surface temperature to 50◦C. Calculate the heat

loss per unit length.

1. Use equation 5.80 2. Use equation 5.82

Example 5.12: Heat Transfer from Sphere

Air at 1 atm and 27◦C blows across a 12-mm-diameter sphere at a free-stream

velocity of 4 m/s. Asmall heater inside the sphere maintains the surface temperature at

77◦C. Calculate the heat lost by the sphere. Solution Consulting Equation (5.88) we find that the Reynolds number is evaluated at the free-

stream temperature.We therefore need the following

properties: at 𝑇∞ =27◦C=300 K,

𝑃𝑟 = 0.708 , 𝜇∞ = 2.14 ∗ 10−5 kg/m.s, 𝑘 = 0.02624 W/m.˚C,

𝑣 = 15.96 ∗ 10−6 m2/s

At 𝑇𝑤 = 77˚C = 350 K, 𝜇𝑤 = 2.075 ∗ 10−5 kg/m.s

𝑅𝑒𝐷 = 𝑢∞𝐷𝑣

= (4)(0.012)15.96∗10−6

= 3059

𝑁𝑢𝐷 = 2 + �0.4𝑅𝑒𝐷12 + 0.06𝑅𝑒𝐷

23�𝑃𝑟0.4 �𝜇∞

𝜇𝑤�14

𝑁𝑢𝐷 = 2 + �(0.4)(3059)12 + (0.06)(3059)

23� (0.708)0.4 �1.8462

2.075�14 = 31.4

ℎ = (𝑁𝑢𝐷)(𝑘)𝐷

= (31.4)(0.02624)0.012

= 68.66 W/m2.˚C.

𝑞 = ℎ4𝜋𝐷2(𝑇𝑤 − 𝑇∞)

𝑞 = (68.66)(4𝜋)(0.006)2(77 − 27) = 1.553 W

5.10 Flow Across Tube Banks

Cross-flow over tube banks is commonly encountered in practice in heat transfer

equipment such as the condensers and evaporators of power plants, refrigerators, and

air conditioners. In such equipment, one fluid moves through the tubes while the other

moves over the tubes in a perpendicular direction.

In a heat exchanger that involves a tube bank, the tubes are usually placed in a shell

(and thus the name shell-and-tube heat exchanger), especially when the fluid is a

liquid, and the fluid flows through the space between the tubes and the shell. There

are numerous types of shell-and-tube heat exchangers.

Because many heat-exchanger arrangements involve multiple rows of tubes, the

heattransfer characteristics for tube banks are of important practical interest. The heat-

transfer characteristics of staggered and in-line tube banks were studied by Grimson.

Determination of Maximum Flow Velocity

For flows normal to in-line tube banks the maximum flow velocity will occur through

the minimum frontal area (𝑆𝑛 − 𝑑) presented to the incoming free stream velocity

𝑢∞. Thus,

𝑢𝑚𝑎𝑥 = 𝑢∞ �𝑠𝑛

(𝑠𝑛−𝑑)� (in-line arrangement) 5.89

For staggered

If ��𝑆𝑛2�2

+ 𝑆𝑝2�12− 𝑑� ∗ 2 < (𝑆𝑛 − 𝑑)

Then

𝑢𝑚𝑎𝑥 =12𝑠𝑛𝑢∞

��𝑆𝑛2 �2+𝑆𝑝2�

12−𝑑

(staggered arrangement) 5.90

If ��𝑆𝑛2�2

+ 𝑆𝑝2�12− 𝑑� ∗ 2 > (𝑆𝑛 − 𝑑)

𝑢𝑚𝑎𝑥 = 𝑢∞ �𝑠𝑛

(𝑠𝑛−𝑑)�

The nomenclature for use with Table 5.3 is shown in Figure 5.14. The data of Table

5.3 pertain to tube banks having 10 or more rows of tubes in the direction of flow. For

fewer rows the ratio of h for N rows deep to that for 10 rows is given in Table 5.4.

𝑁𝑢𝑑𝑓 = ℎ𝑑𝑘𝑓

= 𝐶 �𝑢∞𝑑𝑣𝑓�𝑛𝑃𝑟𝑓

13 (5.80)

Table 5.3 Modified correlation of Grimson for heat transfer in tube banks of 10 rows or more, , for use with Equation (5.80).

Figure 5.14 Nomenclature for use with Table 5.3: (a) in-line tube rows; (b) staggered tube rows.

Table 5.4 Ratio of h for N rows deep to that for 10 rows deep, for use with Equation (5.80).

Zukauskas has presented additional information for tube bundles that takes into

account wide ranges of Reynolds numbers and property variations. The correlating

equation takes the form

𝑁𝑢 = ℎ�𝑑𝑘

= 𝐶𝑅𝑒𝑑,𝑚𝑎𝑥𝑛 𝑃𝑟0.36 � 𝑃𝑟

𝑃𝑟𝑤�14 �10 < 𝑅𝑒𝑑,𝑚𝑎𝑥 < 106

0.7 < 𝑃𝑟 < 500 5.91

where all properties except 𝑃𝑟𝑤 are evaluated at 𝑇∞ and the values of the constants are

given in Table 5.5 for greater than 20 rows of tubes. For gases the Prandtl number

ratio has little influence and is dropped. Once again, note that the Reynolds number is

based on the maximum velocity in the tube bundle. For less than 20 rows in the

direction of flow the correction factor in Table 5.6 should be applied. It is essentially

the same as for the Grimson correlation.

Table 5.5 Constants for Zukauskas correlation [Equation (5.91)] for heat transfer in tube banks of 20 rows or more.

Table 5.6 Ratio of h for N rows deep to that for 20 rows deep for use with Equation (5.91).

Example 5.13: Heating of Air with In-Line Tube Bank

Air at 1 atm and 10◦C flows across a bank of tubes 15 rows high and 5 rows deep at a

velocity of 7 m/s measured at a point in the flow before the air enters the tube bank.

The surfaces of the tubes are maintained at 65◦C. The diameter of the tubes is 1 in

[2.54 cm]; they are arranged in an in-line manner so that the spacing in both the

normal and parallel directions to the flow is 1.5 in [3.81 cm]. Calculate the total heat

transfer per unit length for the tube bank and the exit air temperature.

Solution

The constants for use with Equation (5.80) may be obtained from Table 5.3, using

The properties of air are evaluated at the film temperature, which at entrance to the

tube bank is



13 (5.80)

Because there are only 5 rows deep, this value must be multiplied by the factor 0.92,

as determined from Table 5.4.

The total surface area for heat transfer, considering unit length of tubes, is

chapter 5 principles of force convection · 2018. 9. 25. · 5.2 viscous flow . 5.2.1 flow on a...

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