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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