Chapter 6 Temperature at which extreme values of average convection

PROBLEM 6.17

KNOWN: Velocity and temperature of water flowing over a flat plate. Length of plate. Variation of

local convection coefficient with x for laminar and turbulent flow.

FIND: Minimum and maximum average convection coefficient for roughness applied over the range

xr ≤ x ≤ L. Temperature at which extreme values of average convection coefficient occur and

corresponding values of xr.

SCHEMATIC:

ASSUMPTIONS: (1) Steady–state conditions, (2) Constant properties, (3) Transition occurs at a

critical Reynolds number of 5 × 105 for the smooth plate, (4) Incompressible flow.

PROPERTIES: Table A.6, Liquid water (T = 300 K):

ρ

= vf

-1 = 997 kg/m3,

m

= 855 × 10-6 N⋅s/m2;

Liquid water (T = 350 K):

ρ

= vf

-1 = 974 kg/m3,

m

= 3655 × 10-6 N⋅s/m2.

ANALYSIS: The smooth plate transition location, xc, was found in Example 6.4 to be 0.43 m and

0.19 m, for T = 300 K and 350 K, respectively. For roughness applied over the range 0 ≤ xr ≤ xc,

transition occurs at xr. From Eq. 6.14,

The critical locations xr,c that give rise to minimum or maximum values of

h

can be found by

differentiating Eq. (1) and setting the result to zero:

Evaluating xr,c for the two temperatures yields:

PROBLEM 6.17 (Cont.)

These values are both less than their respective turbulence transition locations, xc, therefore Eq. (1)

holds at these locations. Substituting these values into Eq. (1), the average heat transfer coefficients for

roughness applied at these locations can be calculated.

For T = 300 K,



The subscript max is used above to indicate that these are the maximum values, since they are larger

than the respective values for xr > xc, namely:

COMMENTS: (1) The maximum value of

h

exists when transition occurs very close to the leading

edge of the plate. It does not occur exactly at the leading edge because the laminar heat transfer

coefficient equation yields a slightly higher value than the turbulent heat transfer coefficient equation

very near x = 0. (2) Turbulent heat transfer coefficients are usually (but not always) larger than laminar

heat transfer coefficients. Therefore, tripping the transition to turbulence at or near the leading edge

results in enhanced heat transfer. (3) The conclusion that the laminar heat transfer coefficient is

slightly higher than the turbulent heat transfer coefficient very near x = 0 may not be accurate.

PROBLEM 6.18

KNOWN: Transition Reynolds number. Velocity and temperature of atmospheric air, engine

oil, and mercury flow over a flat plate.

FIND: Distance from leading edge at which transition occurs for each fluid.

SCHEMATIC:

ASSUMPTIONS: Transition Reynolds number is

Re .

x,c= ×5 105

PROPERTIES: For the fluids at T = 300 K and 350 K:

ν

(m2/s)

Fluid Table T = 300 K T = 350 K

ANALYSIS: The point of transition is

Substituting appropriate viscosities, find

xc(m)

Fluid T = 300 K T = 350 K <

COMMENTS: (1) Note the great disparity in transition length for the different fluids. Due to

the effect which viscous forces have on attenuating the instabilities which bring about

transition, the distance required to achieve transition increases with increasing

ν

. (2) Note the

temperature-dependence of the transition length, in particular for engine oil. (3) As shown in

Example 6.4, the variation of the transition location can have a significant effect on the

average heat transfer coefficient associated with convection to or from the plate.

PROBLEM 6.19

KNOWN: Pressure dependence of the dynamic viscosity, thermal conductivity and specific heat.

FIND: (a) Variation of the kinematic viscosity and thermal diffusivity with pressure for an

ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Transition at Rex,c = 5

× 105, (4) Ideal gas behavior.

PROPERTIES: Table A.4, air (350 K): m = 208.2 × 10-7 N⋅s/m2, k = 0.030 W/m⋅K, cp = 1009

J/kg⋅K, ρ = 0.995 kg/m3.

ANALYSIS:

(a) For an ideal gas

ν = μ/ρ

or ν is independent of pressure. <

The thermal diffusivity is

(b) For T = 350 K, p = 1 atm, the thermal diffusivity of air is

PROBLEM 6.19 ( Cont.)

(c) For transition over a flat plate,

Using Equation 4, at p = 5 atm

-6 2 -6 2

ν = 20.92 × 10 m /s 5 = 4.18 × 10 m /s

At p = 10 atm,

-6 2 -6 2

ν = 20.92 × 10 m /s 10 = 2.09 × 10 m /s

COMMENT: Note the strong dependence of the transition length upon the pressure for the gas

(the transition length is independent of pressure for the incompressible liquid).

PROBLEM 6.20

KNOWN: Nondimensional form of the thermal boundary layer equation and boundary conditions,

expressions for x*, y*, u*, v* and T*. Laminar, incompressible flow with negligible viscous

dissipation.

FIND: Expressions for (a) the thermal boundary conditions and (b) energy equation in dimensional

form.

ASSUMPTIONS: (1) Steady–state conditions, (2) Constant properties.

ANALYSIS: (a) From Equation 6.39, the thermal boundary conditions in nondimensional form are

*

s

TT

∞

−

=−

*x

xL

=

Substituting

*

s

TT

TTT

∞

−

=−

from Equation 6.33 and

*x

xL

=

from Equation 6.31 into Equation (2)

(b) Note that

( )

*

* (/)

s

TT

T LT

x xL T T x

∞



−

∂

−

∂∂



= =

∂ ∂ −∂

. Likewise,

( )

*

*/

s

TT

T LT

y yL T T y

∞



−

∂

−

∂∂



= =

∂ ∂ −∂

.

PROBLEM 6.20 (Cont.)

COMMENTS: (1) Equations 6.36 and 6.39 are nondimensional forms of Equations 6.29 and the

boundary conditions. When converted to their nondimensional forms, the resulting equations explicitly

illustrate the importance of the Reynolds and Prandtl numbers in describing the thermal boundary

layer. (2) For a flat plate subject to parallel flow, the Reynolds number is usually expressed as ReL =

ρ

u∞L/

m

, or u∞L/

ν

since u∞ = V.

PROBLEM 6.21

KNOWN: Laminar boundary layer flow over a flat plate. Atmospheric pressure. Type of fluid, and/or

temperature, and/or

δ

/

δ

t.

FIND: Missing information for type of fluid, and/or temperature, and or

δ

/

δ

t.

SCHEMATIC:

x

ASSUMPTIONS: (1) Steady-state, (2) Laminar boundary layer flow, (3) Properties are constant at the

specified temperature.

PROPERTIES: Table A-4, Air, (T = 300 K): Pr = 0.707. Table A-6, Water, (T = 350 K): Pr = 2.29.

ANALYSIS: For air at 300 K and water at 350 K, the Prandtl number can be found in Appendix A, and

the ratio of boundary layer thicknesses is

For engine oil, the Prandtl number can be found from the given boundary layer thickness ratio:

COMMENTS: Most gases have Pr < 1, and therefore

δ

<

δ

t. Most liquids have Pr > 1, and therefore

δ

>

δ

t. Liquid metals are an exception to this trend. Their Prandtl numbers are very small, and consequently

δ

«

δ

t for liquid metals.

PROBLEM 6.22

KNOWN: Critical Reynolds number for a cylinder in cross flow. Critical Mach number.

FIND: Critical cylinder diameter below which, if the flow of air at atmospheric pressure and

temperature is turbulent, compressibility effects may be important.

SCHEMATIC:

ASSUMPTIONS: (1) Steady–state conditions. (2) Air behaves as an ideal gas.

PROPERTIES: Table A.4, air (T = 300 K): M = 28.97 kg/kmol, cp = 1.007 kJ/kg∙K,

m

= 184.6

× 10-7 N∙s/m2.

ANALYSIS: The density of an ideal gas may be found from the equation of state,

Substituting the preceding equations into the definition of the Reynolds number yields

Before evaluating the critical cylinder diameter, we note that the gas constant for air is

Continued…

PROBLEM 6.22 (Cont.)

and the specific heat at constant volume, cv, is

Therefore, the ratio of specific heats for air is

For the conditions of the problem, the critical cylinder diameter is

COMMENTS: (1) The expression for the critical Reynolds number (ReD,c = 2 × 105) is plotted using

log-log scales in the figure below. Laminar flow occurs to the left of the sloped line, while turbulent flow

occurs to the right of the sloped line. The velocity associated with the critical Mach number is identified

(2) The value of the critical Reynolds number is geometry-dependent. Care must be taken to

apply the correct value of the critical Reynolds number in any calculation involving convection

heat transfer.

PROBLEM 6.23

KNOWN: Characteristic length, surface temperature and average heat flux for an object

placed in an airstream of prescribed temperature and velocity.

FIND: Average convection coefficient if characteristic length of object is increased by a

factor of five and air velocity is decreased by a factor of five.

ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties.

ANALYSIS: For a particular geometry,

( )

LL

Nu f Re , Pr .=

The Reynolds numbers for each case are

2

For Case 1, using the rate equation, the convection coefficient is

Hence, it follows that for Case 2

COMMENTS: If ReL,2 were not equal to ReL,1, it would be necessary to know the specific

form of f(ReL, Pr) before

h2

could be determined.

PROBLEM 6.24

KNOWN: Heat transfer rate from a turbine blade for prescribed operating conditions.

FIND: Heat transfer rate from a larger blade operating under different conditions.

SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Surface area A is

directly proportional to characteristic length L, (4) Negligible radiation, (5) Blade shapes are

geometrically similar.

ANALYSIS: For a prescribed geometry,

Hence, with constant properties,

L,1 L,2

Re Re .=

Also,

Pr Pr .

1 2

=

Therefore,

Hence, the heat rate for the second blade is

COMMENTS: The slight variation of ν from Case 1 to Case 2 would cause ReL,2 to differ

from ReL,1. However, for the prescribed conditions, this non-constant property effect is

small.

PROBLEM 6.25

KNOWN: Experimental measurements of the heat transfer coefficient for a square bar in

cross flow.

FIND: (a)

h

for the condition when L = 1m and V = 15m/s, (b)

h

for the condition when L

= 1m and V = 30m/s, (c) Effect of defining a side as the characteristic length.

SCHEMATIC:

ASSUMPTIONS: (1) Functional form

mn

Nu CRe Pr=

applies with C, m, n being

constants, (2) Constant properties.

ANALYSIS: (a) For the experiments and the condition L = 1m and V = 15m/s, it follows

that Pr as well as C, m, and n are constants. Hence

Using the experimental results, find m. Substituting values

giving m = 0.782. It follows then for L = 1m and V = 15m/s,

(b) For the condition L = 1m and V = 30m/s, find

COMMENTS: The foregoing Nusselt number relation is used frequently in heat transfer

analysis, providing appropriate scaling for the effects of length, velocity, and fluid properties

on the heat transfer coefficient.

PROBLEM 6.26

KNOWN: Form of the Nusselt number correlation for forced convection and fluid properties.

FIND: Expression for figure of merit FF and values for air, water and a dielectric liquid.

PROPERTIES: Prescribed. Air: k = 0.026 W/m⋅K, ν = 1.6 × 10-5 m2/s, Pr = 0.71. Water: k =

0.600 W/m⋅K, ν = 10-6 m2/s, Pr = 5.0. Dielectric liquid: k = 0.064 W/m⋅K, ν = 10-6 m2/s, Pr = 25

ANALYSIS: With

mn

LL

Nu ~ Re Pr ,

the convection coefficient may be expressed as

The figure of merit is therefore

and for the three fluids, with m = 0.80 and n = 0.33,

Water is clearly the superior heat transfer fluid, while air is the least effective.

COMMENTS: The figure of merit indicates that heat transfer is enhanced by fluids of large k, large

Pr and small ν.

PROBLEM 6.27

KNOWN: Form of the Nusselt number correlation for forced convection and fluid properties.

Properties of xenon and He-Xe mixture. Temperature and pressure. Expression for specific heat for

monatomic gases.

FIND: Figures of merit for air, pure helium, pure xenon, and He-Xe mixture containing 0.75 mole

fraction of helium.

PROPERTIES: Table A-4, Air (300 K): k = 0.0263 W/m⋅K,

ν

= 15.89 × 10-6 m2/s, Pr = 0.707.

Table A-4, Helium (300 K): k = 0.152 W/m⋅K,

ν

= 122 × 10-6 m2/s, Pr = 0.680. Pure xenon (given):

k = 0.006 W/m⋅K, m = 24.14 × 10-6 N∙s/m2. He-Xe mixture (given): k = 0.0713 W/m·K, m = 25.95 ×

10-6 N·s/m2.

ANALYSIS: With

mn

LL

Nu ~ Re Pr ,

the convection coefficient may be expressed as

ν

For xenon and the He-Xe mixture, we must find the density and specific heat. Proceeding for pure

xenon:

Thus

ν

= m/ρ = 24.14 × 10-6 N∙s/m2/5.33 kg/m3 = 4.53 × 10-6 m2/s and Pr = mcp/k = 24.14 × 10-6

N∙s/m2 × 158 J/kg/0.006 W/m·K = 0.636.

Finally, for the four fluids, with m = 0.85 and n = 0.33, we can calculate the figure of merit from

Equation (1):

FF (W·s0.85/m2.7·K) <

COMMENTS: The effectiveness of the He–Xe mixture is much higher than that of pure He, pure

Xe, or air. By blending He and Xe, the high thermal conductivity of helium and the high density of

xenon are both exploited in a manner that leads to a high figure of merit.

PROBLEM 6.28

KNOWN: Base fluid (water) and nanofluid properties. Fixed surface and ambient temperatures, fixed

characteristic velocity. Fixed geometry. Form of Nusselt number correlation.

FIND: (a) Prandtl numbers of the base fluid and nanofluid. (b) Ratio of Reynolds numbers of the two

fluids and ratio of Nusselt numbers necessary to provide the same convection heat transfer

coefficients. (c) Whether the base fluid can provide greater convection heat transfer rates than the

nanofluid.

SCHEMATIC:

ASSUMPTIONS: (1) Steady–state conditions, (2) Constant properties, (3) Negligible viscous

dissipation.

PROPERTIES: Table A.6 (T = 300 K): Water; kbf = 0.613 W/m⋅K,

ρ

bf = 997 kg/m3, cp,bf = 4.179

kJ/kg⋅K,

m

bf = 855 × 10-6 N⋅s/m2. Example 2.2: Nanofluid, knf = 0.705 W/m⋅K,

ρ

nf = 1146 kg/m3, cp,nf =

3.587 kJ/kg⋅K,

m

nf = 962 × 10-6 N⋅s/m2.

ANALYSIS: (a) The Prandtl numbers for the water and nanofluid are:

(b) For a given velocity and characteristic length,

For the same convection heat transfer coefficients,

(c) For the base fluid to have a greater convection heat transfer rate would require

bf nf

hh>

, or

PROBLEM 6.28 (Cont.)

COMMENTS: (1) The conclusion regarding the relative efficacy of the nanofluid to the base fluid is

valid only for situations involving unconstrained boundary layers. (2) For internal flow situations,

such as will be discussed in Chapter 8, one cannot draw a general conclusion that the nanofluid would

outperform the base fluid. In fact, in common instances, the base fluid would outperform the

nanofluid.

PROBLEM 6.29

KNOWN: Air, water, engine oil or mercury at 300K in laminar, parallel flow over a flat plate.

FIND: Sketch of velocity and thermal boundary layer thickness.

ASSUMPTIONS: (1) Laminar flow.

PROPERTIES: For the fluids at 300K:

Fluid Table Pr

ANALYSIS: For laminar, boundary layer flow over a flat plate.

n

t

~ Pr

δ

where n > 0. Hence, the boundary layers appear as shown below.

COMMENTS: Although Pr strongly influences relative boundary layer development in laminar

flow, its influence is weak for turbulent flow.

PROBLEM 6.30

KNOWN: Flow over a flat plate. Velocity and temperature of two fluids. Variation of boundary

layer thickness with x for laminar flow.

FIND: (a) Location where transition to turbulence occurs for each fluid, (b) Plot of velocity boundary

layer thickness for 0 ≤ x ≤ xc for each fluid, (c) Plot of thermal boundary layer thickness over the same

range. Which fluid has the largest local temperature gradient at the surface, Nusselt number, and heat

transfer coefficient.

ASSUMPTIONS: (1) Steady–state conditions, (2) Incompressible flow, (3) Transition occurs at a

critical Reynolds number of 5 × 105.

PROPERTIES: Table A.4, Air (T = 300 K):

ν

= 15.89 × 10-6 m2/s, k = 0.0263 W/m⋅K, Pr = 0.707.

Table A.5, Engine Oil (T = 380 K):

ν

= 16.9 × 10-6 m2/s, k = 0.136 W/m⋅K, Pr = 233.

ANALYSIS: (a) Transition occurs at Rex,c = u∞x/

ν

= 5 × 105. Therefore, for air

(b) The velocity boundary layer thickness is given by

5

x

xRe

δ

=

. Thus, for air

x

PROBLEM 6.30 (Cont.)

0.03

(c) From Eq. 6.55 with n = 1/3,

δ

t =

δ⋅

Pr–1/3. Thus, for air

Engine oil

0.03

Engine oil

Air