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2* Analysis of Laminar Forced Convection in a Long Tube

2* Analysis of Laminar Forced Convection in a Long Tube

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6.2 Analysis of Laminar Forced Convection In a Long Tube



τ (2πr dx) = –µ



u(r)



361



du

(2πr dx)

dr



r

x



( p + dp)πr 2



pπr 2



rs



dx



FIGURE 6.7 Force balance on a cylindrical fluid element inside a tube of

radius rs.

From this relation, we obtain

du =



1 dp

a

br dr

2m dx



where dp> dx is the axial pressure gradient. The radial distribution of the axial

velocity is then

u(r) =



1 dp 2

a

br + C

4m dx



where C is a constant of integration whose value is determined by the boundary condition that u = 0 at r = rs. Using this condition to evaluate C gives the velocity distribution

u(r) =



r2 - r2s dp

4m

dx



(6.9)



The maximum velocity umax at the center (r = 0) is

umax = -



r2s dp

4m dx



(6.10)



so that the velocity distribution can be written in dimensionless form as

u

umax



= 1 - a



r 2

b

rs



(6.11)



The above relation shows that the velocity distribution in fully developed laminar

flow is parabolic.

In addition to the heat transfer characteristics, engineering design requires consideration of the pressure loss and pumping power required to sustain the convection flow

through the conduit. The pressure loss in a tube of length L is obtained from a force

balance on the fluid element inside the tube between x = 0 and x = L (see Fig. 6.7):

¢ppr2s = 2prstsL



where



¢p = p1 - p2 = pressure drop in length L(¢ p = - (dp> dx)L) and

ts = wall shear stress (ts = - m(du> dr)|r = rs)



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(6.12)



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Chapter 6 Forced Convection Inside Tubes and Ducts

The pressure drop also can be related to a so-called Darcy friction factor f according to

¢p = f



q2

L rU

D 2gc



(6.13)



where Uq is the average velocity in the tube.

It is important to note that f, the friction factor in Eq. (6.13), is not the same

quantity as the friction coefficient Cf, which was defined in Chapter 4 as

Cf =



ts

q

rU 2/2gc



(6.14)



Cf is often referred to as the Fanning friction coefficient. Since ts = - m(du>dr)r = r

it is apparent from Eqs. (6.12), (6.13), and (6.14) that

Cf =



f

4



For flow through a pipe the mass flow rate is obtained from Eq. (6.9)

#

m = r



rs



L0



u2pr dr =



rx

¢ppr

¢ppr4s r

(r2 - r2s )r dr = 2Lm L0

8Lm



(6.15)



and the average velocity Uq is

Uq =



#

¢pr2s

m

=

8Lm

rpr2s



(6.16)



equal to one-half of the maximum velocity in the center. Equation (6.13) can be

rearranged into the form

p1 - p2 = ¢p =



q2

64Lm U

64 L rUq 2

=

ReD D 2gc

rUq 2D 2



(6.17)



Comparing Eq. (6.17) with Eq. (6.13), we see that for fully developed laminar flow

in a tube the friction factor in a pipe is a simple function of Reynolds number

f =



64

ReD



(6.18)



The pumping power, Pp, is equal

to the product of the pressure drop and the volu#

metric flow rate of the fluid, Q, divided by the pump efficiency, ␩p, or

#

Pp = ¢pQ> hp

(6.19)

The analysis above is limited to laminar flow with a parabolic velocity distribution in pipes or circular tubes, known as Poiseuille flow, but the approach taken to

derive this relation is more general. If we know the shear stress as a function of the

velocity and its derivative, the friction factor also could be obtained for turbulent

flow. However, for turbulent flow, the relationship between the shear and the average

velocity is not well understood. Moreover, while in laminar flow, the friction factor

is independent of surface roughness; in turbulent flow, the quality of the pipe surface

influences the pressure loss. Therefore, friction factors for turbulent flow cannot be

derived analytically but must be measured and correlated empirically.

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6.2 Analysis of Laminar Forced Convection In a Long Tube



363



6.2.1 Uniform Heat Flux

For the energy analysis, consider the control volume shown in Fig. 6.8. In laminar

flow, heat is transferred by conduction into and out of the element in a radial direction, whereas in the axial direction, the energy transport is by convection. Thus, the

rate of heat conduction into the element is

dqk,r = - k2pr dx



0T

0r



while the rate of heat conduction out of the element is

dqk,r + dr = - k2p(r + dr)dxc



0T

0 2T

+ 2 dr d

0r

0r



The net rate of convection out of the element is

dqc = 2pr dr rcpu(r)



0T

dx

0x



Writing a net energy balance in the form

net rate of conduction

net rate of convection

=

into the element

out of the element

we get, neglecting second-order terms,

ka



0T

0 2T

0T

+ r 2 bdx dr = r rcpu

dx dr

0r

0x

0r



which can be recast in the form

rcp 0T

0T

1 0

ar

b =

ur 0r

0r

k 0x



dqc,in = (2πr dr)ρcpu(r)T(x)



(6.20)



dqc,out = (2πr dr)ρcpu(τ) T(x) + ∂T dx

∂x



dqr+dr



dqr

r

dr

Tube r = rs



dx



FIGURE 6.8 Schematic sketch of control volume for energy analysis in flow

through a pipe.

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Chapter 6 Forced Convection Inside Tubes and Ducts

The fluid temperature must increase linearly with distance x since the heat flux over

the surface is specified to be uniform, so

0T

= constant

0x



(6.21)



When the axial temperature gradient 0T> 0x is constant, Eq. (6.20) reduces from a

partial to an ordinary differential equation with r as the only space coordinate.

The symmetry and boundary conditions for the temperature distribution in

Eq. (6.20) are

0T

= 0

0r



`k



at r = 0



0T

`

= q–s = constant at r = rs

dr r = rs



To solve Eq. (6.20), we substitute the velocity distribution from Eq. (6.11).

Assuming that the temperature gradient does not affect the velocity profile, that is,

the properties do not change with temperature, we get

0

0T

1 0T

r2

ar

b =

u max a1 - 2 br

a 0x

0r

0r

rs



(6.22)



The first integration with respect to r gives

r



0T

1 0T umax r2

r2

=

a1 - 2 b + C1

a 0x

0r

2

2rs



(6.23)



A second integration with respect to r gives

T(r, x) =



1 0T umax 2

r2

r a1 - 2 b + C1 ln r + C2

a 0x 4

4rs



(6.24)



But note that C1 = 0 since (0T> 0r)r = 0 = 0 and that the second boundary condition

is satisfied by the requirement that the axial temperature gradient 0T> 0x is constant.

If we let the temperature at the center (r ϭ 0) be Tc, then C2 = Tc and the temperature distribution becomes

T - Tc =



1 0T u max r2s

r 2

1 r 4

ca b - a b d

rs

a 0x

4

4 rs



(6.25)



The average bulk temperature Tb that was used in defining the heat transfer coefficient can be calculated from

rs



rs



Tb =



30



(pucpT)(2pr dr)



30

=



rs



30



(pucp)2pr dr



(pucpT )2pr dr

#

cpm



(6.26)



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6.2 Analysis of Laminar Forced Convection In a Long Tube



365



Since the heat flux from the tube wall is uniform, the enthalpy of the fluid in the tube

must increase linearly with x, and thus 0Tb> 0x = constant. We can calculate the bulk

temperature by substituting Eqs. (6.25) and (6.11) for T and u, respectively, in Eq.

(6.26). This yields

Tb - Tc =



7 umax r2s 0T

a

96

0x



(6.27)



3 umax r2s 0T

a

16

0x



(6.28)



while the wall temperature is

Ts - Tc =



In deriving the temperature distributions, we used a parabolic velocity distribution,

which exists in fully developed flow in a long tube. Hence, with ѨT> Ѩx equal to a

constant, the average heat transfer coefficient is

hqc =



k(0T/0r) r = rs

qc

=

A(Ts - Tb)

Ts - Tb



(6.29)



Evaluating the radial temperature gradient at r ϭ rs from Eq. (6.23) and substituting

it with Eqs. (6.27) and (6.28) in the above definition yields

qhc = 24k = 48k

11rs

11D



(6.30)



or

NuD =



EXAMPLE 6.1



hqcD

= 4.364 for q–s = constant

k



(6.31)



Water entering at 10°C is to be heated to 40°C in a tube of 0.02-m-ID at a

mass flow rate of 0.01 kg/s. The outside of the tube is wrapped with an

insulated electric-heating element (see Fig. 6.9) that produces a uniform flux

of 15,000 W > m 2 over the surface. Neglecting any entrance effects, determine



Insulation

L=?

Water in

10°C

0.01 kg/s



Heater

Tube

Water out

40°C



Electric power supply



FIGURE 6.9 Schematic diagram of water flowing through electrically

heated tube, Example 6.1.



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Chapter 6 Forced Convection Inside Tubes and Ducts

(a)

(b)

(c)

(d)

(e)

(f)

(g)



SOLUTION



the Reynolds number

the heat transfer coefficient

the length of pipe needed for a 30°C increase in average temperature

the inner tube surface temperature at the outlet

the friction factor

the pressure drop in the pipe

the pumping power required if the pump is 50% efficient.



From Table 13 in Appendix 2, the appropriate properties of water at an average temperature between inlet and outlet of 25°C are obtained by interpolation:

r = 997 kg/m3

cp = 4180 J/kg K

k = 0.608 W/m K

m = 910 * 10-6 N s/m2

(a) The Reynolds number is

#

rUqD

(4)(0.01 kg/s)

4m

=

ReD =

=

= 699

m

pDm

(p)(0.02 m)(910 * 10-6 N s/m2)

This establishes that the flow is laminar.

(b) Since the thermal-boundary condition is one of uniform heat flux, NuD ϭ 4.36

from Eq. (6.31) and

qhc = 4.36 k = 4.36 0.608 W/m K = 132 W/m2 K

D

0.02 m

(c) The length of pipe needed for a 30°C temperature rise is obtained from a heat

balance

#

q–pDL = mcp(Tout - Tin)

Solving for L when Tout - Tin = 30 K gives

#

mcp ¢T

(0.01 kg/s)(4180 J/kg K)(30 K)

L =

=

= 1.33 m

pDq–

(p)(0.02 m)(15,000 W/m2)

Since L> D = 66.5 and 0.05 ReD = 33.5, entrance effects are negligible according to

Eq. (6.7). Note that if L> D had been significantly less than 33.5, the calculations

would have to be repeated with entrance effects taken into account, using relations

to be presented.

(d) From Eq. (6.1)

qc

q– =

= qhc(Ts - Tb)

A

and

qc

15,000 W/m2

Ts =

+ Tb =

+ 40°C = 154°C

Ahqc

132 W/m2 °C



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6.2 Analysis of Laminar Forced Convection In a Long Tube



367



(e) The friction factor is found from Eq. (6.18):

f =



64

64

=

= 0.0915

ReD

699



(f) The pressure drop in the pipe is, from Eq. (6.17),

p1 - p2 = ¢p = f a



rUq 2

L

ba

b

D

2gc



Since

#

4m

q

U =

=

rpD2



4a0.01

a997



kg

m3



kg

b

s



b(p)(0.02 m)2



= 0.032



m

s



we have



¢p = (0.0915)(66.5)



a997



b10.032



kg

m3

2a1



kg m

N s2



m 2

2

2



b



= 3.1



N

m2



(g) The pumping power Pp is obtained from Eq. 6.19 or

(0.01 kg/s)(3.1 N/m2)

# ¢p

Pp = m

=

= 6.2 * 10-5 W

rhp

(997 kg/m3)(0.5)



6.2.2* Uniform Surface Temperature

When the tube surface temperature rather than the heat flux is uniform, the analysis

is more complicated because the temperature difference between the wall and bulk

varies along the tube, that is, 0Tb> 0x = f (x). Equation (6.20) can be solved subject

to the second boundary condition that at r = rs, T(x, rs) = constant, but an iterative

procedure is necessary. The result is not a simple algebraic expression, but the

Nusselt number is found (for example, see Kays and Perkins [11]) to be a constant:

Nu D =



qhc D

= 3.66

k



(Ts = constant)



(6.32)



In addition to the value of the Nusselt number, the constant-temperature

boundary condition also requires a different temperature to evaluate the rate of

heat transfer to or from a fluid flowing through a duct. Except for the entrance

region, in which the boundary layer develops and the heat transfer coefficient

decreases, the temperature difference between the surface of the duct and the bulk

remains constant along the duct when the heat flux is uniform. This is apparent

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Chapter 6 Forced Convection Inside Tubes and Ducts

T



Entrance

region



Fully

developed

region



T



Ts(x)



Surface

temperature, Ts(x)

(Ts – Tb)



(Ts – Tb)



Tb(x)



ΔTin

Bulk

temperature, Tb(x)

0



x



0



Distance from entrance

(a)



x

Distance from entrance

(b)



FIGURE 6.10 Variation of average bulk temperature with constant heat flux

and constant wall temperature: (a) constant heat flux, qs(x) ϭ constant;

(b) constant surface temperature, Ts(x) ϭ constant.



from an examination of Eq. (6.20) and is illustrated graphically in Fig. 6.10. For a

constant wall temperature, on the other hand, only the bulk temperature increases

along the duct and the temperature potential decreases (see Fig. 6.10). We first

write the heat balance equation

#

dqc = mcp dTb = q–s P dx

where P is the perimeter of the duct and qsЉ is the surface heat flux. From the preceding

we can obtain a relation for the bulk temperature gradient in the x-direction

dTb

q–s P

P

= #

= # hc(Ts - Tb)

dx

mcp

mcp



(6.33)



Since dTb> dx = d(Tb - Ts)> dx for a constant surface temperature, after separating

variables, we have

¢Tout



3¢Tin



L

d(¢T )

P

= - #

hc dx

¢T

mcp 30



(6.34)



where ¢T = Ts - Tb and the subscripts “in” and “out” denote conditions at the inlet

(x ϭ 0) and the outlet (x ϭ L) of the duct, respectively. Integrating Eq. (6.34) yields

ln a



¢Tout

PL

b = - # qhc

¢Tin

mcp



(6.35)



where

L

1

hqc =

h dx

L 30 c



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6.2 Analysis of Laminar Forced Convection In a Long Tube



369



Rearranging Eq. (6.35) gives

¢Tout

- hqcPL

= exp a #

b

¢Tin

mcp



(6.36)



The rate of heat transfer by convection to or from a fluid flowing through a duct with

Ts ϭ constant can be expressed in the form

#

#

qc = mcp[(Ts - Tb,in) - (Ts - Tb,out)] = mcp(¢Tin - ¢Tout)

#

and substituting mcp from Eq. (6.35), we get

qc = qhc As c



¢Tout - ¢Tin

d

ln(¢Tout/¢Tin)



(6.37)



The expression in the square bracket is called the log mean temperature difference

(LMTD).



EXAMPLE 6.2



Used engine oil can be recycled by a patented reprocessing system. Suppose that

such a system includes a process during which engine oil flows through a 1-cm-ID,

0.02-cm-wall copper tube at the rate of 0.05 kg/s. The oil enters at 35°C and is to be

heated to 45°C by atmospheric-pressure steam condensing on the outside, as shown

in Fig. 6.11. Calculate the length of the tube required.



SOLUTION



We shall assume that the tube is long and that its temperature is uniform at 100°C.

The first approximation must be checked; the second assumption is an engineering

approximation justified by the high thermal conductivity of copper and the large heat

transfer coefficient for a condensing vapor (see Table 1.4). From Table 16 in

Appendix 2, we get the following properties for oil at 40°C:

cp

r

k

m

Pr



=

=

=

=

=



1964 J/kg K

876 kg/m3

0.144 W/m K

0.210 N s/m2

2870



Condensing steam

Oil in

35°C

0.05 kg/s



Oil out

45°C

1 cm



0.02 cm



Copper tube

L=?



FIGURE 6.11 Schematic diagram for Example 6.2.

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Chapter 6 Forced Convection Inside Tubes and Ducts

The Reynolds number is

ReD =



#

(4)(0.05 kg/s)

4m

=

= 30.3

mpD

(p)(0.210 N s/m2)(0.01 m)



The flow is therefore laminar, and the Nusselt number for a constant surface temperature is 3.66. The average heat transfer coefficient is

qhc = NuD k = 3.66 0.144 W/m K = 52.7 W/m2 K

D

0.01 m

The rate of heat transfer is

#

qc = cpm(Tb,out - Tb,in)

= (1964 J/kg K)(0.05 kg/s)(45 - 35) K = 982 W



Recalling that ln(1> x) = - ln x, we find the LMTD is

LMTD =



¢Tout - ¢Tin

55 - 65

10

=

=

= 59.9 K

>

>

0.167

ln(¢Tout ¢Tin)

ln(55 65)



Substituting the preceding information in Eq. (6.37), where As = LpDi, gives

L =



qc

pDihqcLMTD



=



982 W

= 9.91 m

(p)(0.01 m)(52.7 W/m2 K)(59.9 K)



Checking our first assumption, we find L> D ϳ 1000, justifying neglect of entrance

effects. Note also that LMTD is very nearly equal to the difference between the surface temperature and the average bulk fluid temperature halfway between the inlet

and outlet. The required length is not suitable for a practical design with a straight

pipe. To achieve the desired thermal performance in a more convenient shape, one

could route the tube back and forth several times or use a coiled tube. The first

approach will be discussed in Chapter 8 on heat exchanger design, and the coiledtube design is illustrated in an example in the next section.



6.3



Correlations for Laminar Forced Convection

This section presents empirical correlations and analytic results that can be used in

thermal design of heat transfer systems composed of tubes and ducts containing

gaseous or liquid fluids in laminar flow. Although heat transfer coefficients in laminar flow are considerably smaller than in turbulent flow, in the design of heat

exchange equipment for viscous liquids, it is often necessary to accept a smaller heat

transfer coefficient in order to reduce the pumping power requirements. Laminar gas

flow is encountered in high-temperature, compact heat exchangers, where tube

diameters are very small and gas densities low. Other applications of laminar-flow

forced convection occur in chemical processes and in the food industry, in electronic

cooling as well as in solar and nuclear power plants, where liquid metals are used as

heat transfer media. Since liquid metals have a high thermal conductivity, their heat

transfer coefficients are relatively large, even in laminar flow.

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6.3 Correlations for Laminar Forced Convection



371



6.3.1 Short Circular and Rectangular Ducts

The details of the mathematical solutions for laminar flow in short ducts with

entrance effects are beyond the scope of this text. References listed at the end of

this chapter, especially [4] and [11], contain the mathematical background for the

engineering equations and graphs that are presented and discussed in this section.

For engineering applications, it is most convenient to present the results of analytic and experimental investigations in terms of a Nusselt number defined in the

conventional manner as hcD> k. However, the heat transfer coefficient hc can vary

along the tube, and for practical applications, the average value of the heat transfer

coefficient is most important. Consequently, for the equations and charts presented

in this section, we shall use a mean Nusselt number, NuD = hqcD> k, averaged with

respect to the circumference and length of the duct L:

LD

qhcD

1

hc (x) dx =

L 3

k

k

0



NuD =



where the subscript x refers to local conditions at x. This Nusselt number is often

called the log mean Nusselt number, because it can be used directly in the log mean

rate equations presented in the preceding section and can be applied to heat exchangers (see Chapter 8).

Mean Nusselt numbers for laminar flow in tubes at a uniform wall temperature have

been calculated analytically by various investigators. Their results are shown in Fig. 6.12



0.2



0.5



1.0



2.0



5.0



10



20



50



100



100

Very “long” tubes



Very “short” tubes



50

Uniform velocity

Boundary-layer analysis

modified for tube



NuD



20



Short duct

approximation



10



Noris and streid interpolation



5



Parabolic velocity

Region of interest in



2 gas flow heat exchangers

0.1



0.2



0.5



1.0



2.0



5.0



10



20



50



100



ReD PrD

× 10–2

L



FIGURE 6.12 Analytic solutions and empirical correlations for heat

transfer in laminar flow through circular tubes at constant wall

temperature, NuD versus ReDPrD/L. The dots represent Eq. (6.38).

Source: Courtesy of W. M. Kays, “Numerical Solution for Laminar Flow Heat Transfer in

Circular Tubes,” Trans. ASME, vol. 77, pp. 1265–1274, 1955.



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