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Unformatted text preview: t for a specified flow rate, the pressure drop and thus the required pumping power is proportional to the length of the tube and the viscosity of the fluid, but it is
inversely proportional to the fourth power of the radius (or diameter) of the
tube. Therefore, the pumping power requirement for a piping system can be
reduced by a factor of 16 by doubling the tube diameter (Fig. 8–19). Of course
the benefits of the reduction in the energy costs must be weighed against the
increased cost of construction due to using a larger diameter tube.
The pressure drop is caused by viscosity, and it is directly related to the wall
shear stress. For the ideal inviscid flow, the pressure drop is zero since there
are no viscous effects. Again, Eq. 8–47 is valid for both laminar and turbulent
flows in circular and noncircular tubes. ⋅
Wpump = 16 hp
D ⋅
Wpump = 1 hp 2D FIGURE 8–19
The pumping power requirement for
a laminar flow piping system can
be reduced by a factor of 16 by
doubling the pipe diameter. dx mCpTx R4 P
8L PR2
R2
8L ave Ac dx r
dr Temperature Profile and the Nusselt Number
In the analysis above, we have obtained the velocity profile for fully developed flow in a circular tube from a momentum balance applied on a volume
element, determined the friction factor and the pressure drop. Below we obtain the energy equation by applying the energy balance to a differential volume element, and solve it to obtain the temperature profile for the constant
surface temperature and the constant surface heat flux cases.
Reconsider steady laminar flow of a fluid in a circular tube of radius R. The
fluid properties , k, and Cp are constant, and the work done by viscous
stresses is negligible. The fluid flows along the xaxis with velocity . The
flow is fully developed so that is independent of x and thus
(r). Noting that energy is transferred by mass in the xdirection, and by conduction in
the rdirection (heat conduction in the xdirection is assumed to be negligible),
the steadyflow energy balance for a cylindrical shell element of thickness dr
and length dx can be expressed as (Fig. 8–20) Qr ·
m CpTx mCpTx
Qr FIGURE 8–20
The differential volume element
used in the derivation of
energy balance relation. dr ·
where m
Ac
after rearranging, ·
m CpTx dx ·
Qr ·
Qr dr 0 (849) (2 rdr). Substituting and dividing by 2 rdrdx gives, Cp Tx dx dx Tx ·
·
1 Q r dr Q r
2 rdx
dr (850) cen58933_ch08.qxd 9/4/2002 11:29 AM Page 435 435
CHAPTER 8 or
·
Q
2 Cp rdx r T
x 1 (851) where we used the definition of derivative. But
·
Q
r k2 rdx r Substituting and using T
r 2 kdx r r T
r (852) k/ Cp gives
α
T
r dr r r T
x (853) which states that the rate of net energy transfer to the control volume by mass
flow is equal to the net rate of heat conduction in the radial direction. Constant Surface Heat Flux
For fully developed flow in a circular pipe subjected to constant surface heat
flux, we have, from Eq. 8–24,
dTs
dx T
x ·
2qs
mCpR dTm
dx constant (854) If heat conduction in the xdirection were considered in the derivation of
Eq. 8–53, it would give an additional term 2T/ x2, which would be equal to
zero since T/ x constant and thus T T(r). Therefore, the assumption that
there is no axial heat conduction is satisfied exactly in this case.
Substituting Eq. 8–54 and the relation for velocity profile (Eq. 8–41) into
Eq. 8–53 gives
·
4qs
1
kR r2
R2 dT
1d
r dr r dr (855) which is a secondorder ordinary differential equation. Its general solution is
obtained by separating the variables and integrating twice to be
T ·
qs 2
r
kR r2
4R2 C1 r C2 (856) The desired solution to the problem is obtained by applying the boundary
conditions T/ x 0 at r 0 (because of symmetry) and T Ts at r R.
We get
T Ts ·
qs R 3
k4 r2
R2 r4
4R4 (857) cen58933_ch08.qxd 9/4/2002 11:29 AM Page 436 436
HEAT TRANSFER The bulk mean temperature Tm is determined by substituting the velocity and
temperature profile relations (Eqs. 8–41 and 8–57) into Eq. 8–4 and performing the integration. It gives
Tm ·
Combining this relation with qs
h Ts ·
11 qs R
24 k h(Ts Tm) gives 48 k
11 D 24 k
11 R (858) 4.36 k
D (859) or
Circular tube, laminar (q·x constant): Nu hD
k 4.36 (860) Therefore, for fully developed laminar flow in a circular tube subjected to
constant surface heat flux, the Nusselt number is a constant. There is no dependence on the Reynolds or the Prandtl numbers. Constant Surface Temperature
A similar analysis can be performed for fully developed laminar flow in a circular tube for the case of constant surface temperature Ts. The solution procedure in this case is more complex as it requires iterations, but the Nusselt
number relation obtained is equally simple (Fig. 8–21): Ts = constant
64
f = –––
Re D Nu = 3.66 max
m Fully developed
laminar flow FIGURE 8–21
In laminar flow in a tube with
constant surface temperature,
both the friction factor and
the heat transfer coefficient
remain constant in the fully
developed region. Circular tube, laminar (Ts constant): Nu hD
k 3.66 (861) The thermal conductivity k for use in th...
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This note was uploaded on 01/28/2010 for the course HEAT ENG taught by Professor Ghaz during the Spring '10 term at University of Guelph.
 Spring '10
 Ghaz

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