Chapter 8: Internal Forced ConvectionYoav PelesDepartment of Mechanical, Aerospace and Nuclear Engineering Rensselaer Polytechnic Institute

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ObjectivesWhen you finish studying this chapter, you should be able to: Obtain average velocity from a knowledge of velocity profile, and average temperature from a knowledge of temperature profile in internal flow, Have a visual understanding of different flow regions in internal flow, such as the entry and the fully developed flow regions, and calculate hydrodynamic and thermal entry lengths, Analyze heating and cooling of a fluid flowing in a tube under constant surface temperature and constant surface heat flux conditions, and work with the logarithmic mean temperature difference, Obtain analytic relations for the velocity profile, pressure drop, friction factor, and Nusselt number in fully developed laminar flow, and Determine the friction factor and Nusselt number in fully developed turbulent flow using empirical relations, and calculate the pressure drop and heat transfer rate.

Introduction Pipe circular cross section. Duct noncircular cross section. Tubes small-diameter pipes. The fluid velocity changes from zero at the surface (no-slip) to a maximum at the pipe center. It is convenient to work with an average velocity, which remains constant in incompressible flow when the cross-sectional area is constant.

Average Velocity The value of the average velocity is determined from the conservation of mass principle

= Vavg AC = m

Ac

u ( r ) dAC

(8-1)

For incompressible flow in a circular pipe of radius R

Vavg =

Ac

u ( r ) dA AC

C

=

R

0

u ( r ) 2 rdr R 2

2 = 2 u ( r ) rdr R 0(8-2)

R

Average Temperature It is convenient to define the value of the mean temperature Tm from the conservation of energy principle. The energy transported by the fluid through a cross section in actual flow must be equal to the energy that would be transported through the same cross section if the fluid were at a constant temperature Tm E fluid = mc pTm = c pT ( r ) m = mAc

c T ( r ) u ( r )VdAp

c

(8-3)

For incompressible flow in a circular pipe of radius R

Tm =

m

c pT ( r ) m p mc

=

Ac

c T ( r ) u ( r ) 2 rdrp

Vavg ( R 2 ) c pR

(8-4)

2 = T ( r ) u ( r ) rdr 2 Vavg R 0 The mean temperature Tm of a fluid changes during heating or cooling.

Idealized

Actual

Laminar and Turbulent Flow in Tubes For flow in a circular tube, the Reynolds number is defined as Vavg D Vavg D (8-5) Re = =

For flow through noncircular tubes D is replaced by the hydraulic diameter Dh.

4 Ac Dh = P

(8-6)

laminar flow: Re10,000.

The Entrance Region A fluid entering a circular pipe at a uniform velocity. The no-slip condition - the flow in a pipe is divided into two regions: (i) boundary layer region, (ii) irrotational (core) flow region

The thickness of this boundary layer increases in the flow direction until it reaches the pipe center.Irrotational flow Boundary layer

Hydrodynamic entrance region the region from the pipe inlet to the point at which the boundary layer merges at the centerline. Hydrodynamically fully developed region the region beyond the entrance region in which the velocity profile is fully developed and remains unchanged. The velocity profile in the fully developed region is parabolic in laminar flow, and somewhat flatter or fuller in turbulent flow.

Thermal Entrance Region Consider a fluid at a uniform temperature entering a circular tube whose surface is maintained at a different temperature. Thermal boundary layer along the tube is developing. The thickness of this boundary layer increases in the flow direction until the boundary layer reaches the tube center. Thermal entrance region. Thermally fully developed region the region beyond the thermal entrance region in which the dimensionless temperature profile expressed as (Ts-T)/(Ts-Tm) remains unchanged.

Hydrodynamically fully-developed:

u ( r , x ) x

= 0 u = u (r )

(8-7)

Thermally fully-developed:

Ts ( x ) T ( r , x ) =0 x Ts ( x ) Tm ( x )

(8-8)

( T r ) r = R Ts T = f ( x ) (8-9) Ts Tm r Ts Tm r = R

Surface heat flux can be expressed as k ( T r ) r = R T s = hx (Ts Tm ) = k q hx = (8-10) r T T r=R s m h For thermally fully developed region From (Eq. (8-9))

( T

r ) r = R

Ts Tm

f ( x)Fully developed flow

hx f ( x )

hx = constant Fully developed flow

Heat Transfer coefficient and Friction factor

Developing region

Fully developed region

Lh Lt ?

Entry Lengths ()Laminar flow Hydrodynamic

Lh ,laminar 0.05 Re D

(8-11)

Thermal Lt ,laminar 0.05 Re Pr D = Pr Lh ,laminar (8-12)

Turbulent flow Hydrodynamic Thermal (approximate)

Lh ,turbulent = 1.359 D Re

14

(8-13)

Lh ,turbulent Lt ,turbulent 10 D

(8-14)

Turbulent flow Nusselt Number The Nusselt numbers are much higher in the entrance region. The Nusselt number reaches a constant value at a distance of less than 10 diameters. The Nusselt numbers for the uniform surface temperature and uniform surface heat flux conditions are identical in the fully developed regions, and nearly identical in the entrance regions. Nusselt number is insensitive to the type of thermal boundary condition.

General Thermal Analysis In the absence of any work interactions, the conservation of energy equation for the steady flow of a fluid in a tube

= mc p (Te Ti ) Q

(W)

(8-15)

The thermal conditions at the surface can usually be approximated as: constant surface temperature, or constant surface heat flux.

The mean fluid temperature Tm must change during heating or cooling. Either Ts= constant or qs = constant at the surface of a tube, but not both.

Constant Surface Heat Flux In the case of constant heat flux, the rate of heat transfer can also be expressed as =q (8-17) s As = mc p (Te Ti ) (W) Q Then the mean fluid temperature at the tube exit becomes s As q (8-18) Te = Ti + p mc The surface temperature in the case of constant surface heat flux can be determined from s q (8-19) s = h (Ts Tm ) Ts = Tm + q hh

In the fully developed region, the surface temperature Ts will also increase linearly in the flow direction Applying the steady-flow energy balance to a tube slice of thickness dx, the slope of the mean fluid temperature Tm can be determined s p dTm q p dTm = q s ( pdx ) mc = = constant p dx mc Noting that both the heat flux and h (for fully developed flow) are constants (8-19) :

(8-20)

dTm dTs = dx dx

(8-21)

In the fully developed region (Ts-Tm=constant) Ts T x Ts Tm T dTs 1 Ts T =0 = =0 Ts Tm x x x dx (8-22)

Combining Eqs. 820, 821, and 822 givess p T dTs dTm q = = = = constant p x dx dx mc(8-23)

For a circular tube ( p = 2R = D )s 2q T dTs dTm = = = = constant (8-24) x dx dx Vavg c p R

Constant Surface Temperature The energy balance on a differential control volume

= mc p dTm = h (Ts Tm ) dAs Q

(8-27)

Since the mean temperature of the fluid Tm increases in the flow direction, the heat flux decays with x. The surface temperature is constant (dTm=-d(Ts-Tm)) and dAs=pdx, therefore,

d (Ts Tm ) Ts Tm

hp = dx p mc

(8-28)

Integrating Eq. 6-28 from x=0 (tube inlet where Tm=Ti) to x=L (tube exit where Tm=Te) givesTs Te hAs ln = p Ts Ti mc(8-29)

Taking the exponential of both sides and solving for Te p) Te = Ts (Ts Ti ) exp ( hpL mc(8-30)

or

p) Tm ( x ) = Ts (Ts Ti ) exp ( hpx mc x

The temperature difference between the fluid and the surface decays exponentially in the flow direction, and the rate of decay depends on the magnitude of the exponent

p hAs mc This dimensionless parameter is called the number of transfer units (NTU). Large NTU value increasing tube length marginally increases heat transfer rate. Small NTU value heat transfer increases significantly with increasing tube length.

Solving Eq. 829 for mcp gives p= mc hAs ln (Ts Te ) (Ts Ti )

(8-31)

Substituting this into Eq. 815

= mc p (Te Ti ) Q

(W)(8-32)

= hA T Q s lnwhere

Ti Te Te Ti Tln = = ln (Ts Te ) (Ts Ti ) ln [ Te Ti ]

(8-33)

ln is the logarithmic mean temperature difference.

Laminar Flow in TubesAssumptions: steady laminar flow, incompressible fluid, constant properties, fully developed region, and straight circular tube.

The velocity profile u(r) remains unchanged in the flow direction. no motion in the radial direction. no acceleration.

Consider a ring-shaped differential volume element. A force balance on the volume element in the flow direction gives

( 2 r dr P ) x ( 2 r dr P ) x+ dx + ( 2 r dx )r ( 2 r dx )r + dr = 0(8-34)

Dividing by 2 dr dx and rearranging

Px + dx Px ( r )r + dr ( r )r r + =0 dx dr

(8-35)

Taking the limit as dr, dx 0 givesdP d ( r ) r + =0 dx dr(8-36)

Substituting = (du/dr) gives d du dP r = r dr dr dx(8-37)

Rearranging and integrating it twice to give1 dP 2 u (r ) = r + C1 ln r + C2 4 dx Boundary Conditions:(8-38)

symmetry about the centerline : u/r = 0 at r = 0, no-slip condition: u = 0 at r = R.

Eq. 6-38 with the boundary conditionsR 2 dP r2 u (r ) = 1 2 4 dx R (8-39)

Substituting Eq. 839 into Eq. 82, and performing the integration gives the average velocityVavgR R 2 2 R 2 dP r2 = 2 u ( r ) rdr = 2 1 2 rdr R 0 R 0 4 dx R

R 2 dP = 8 dx

(8-40)

Combining the last two equations, the velocity profile is rewritten as

r2 u ( r ) = 2Vavg 1 2 R

;

umax = 2Vavg

(8-41)

Pressure Drop One implication from Eq. 8-37 is that the pressure drop gradient (dP/dx) must be constant (the left side is a function only of r, and the right side is a function only of x). Integrating from x=x1 where the pressure is P1 to x=x1=L where the pressure is P2 gives ()P2 P dP 1 = constant = dx L

(8-43)

Substituting Eq. 843 into the Vavg expression in Eq. 840P = P 1P 2 = 8 LVavg R2

=

32 LVavg D2

(8-44)

A pressure drop due to viscous effects represents an irreversible pressure loss. It is convenient to express the pressure loss for all types of fully developed internal flows in terms of the dynamic pressure and the friction factordynamic pressure friction factor

PL = ( P 1P 2) =

P f

L D

P 2 Vavg 2

(8-45)

Setting Eqs. 844 and 845 equal to each other and solving for f gives Circular tube, laminar:

64 64 f = = DVavg Re

(8-46)

Temperature Profile and the Nusselt Number Energy is transferred by mass in the x-direction, and by conduction in the r-direction. The steady flow energy balance for a cylindrical shell element can be expressed as:

Q pTx mc pTx + dx + Q mc r r + dr = 0

(8-49)

Substituting

= uAc = u ( 2 rdr ) = m

and dividing by 2r dr dx gives, after rearranging

Tx + dx Tx Q 1 r + dr Qr ) = ( )( ) c pu ( 2 rdx dx dr

(8-50)

Or

T 1 Q u = ( ) 2 c p rdx r x

(8-51)

Since Fouriers Law () Q T = k (2 rdx) r r r T = 2 kdx r r r (8-52)

Eq 8-51 becomes ( Energy Equation)T T = u r x r dr r ; k = cp(8-53)

Constant Surface Heat Flux Laminar Fully Developed flow Substituting Eqs. 8-24 and 8-41 into Eq. 8.53 r2 u ( r ) = 2Vavg 1 2 R (8-41)

s 2q T = = constant x Vavg c p R(8-24) (8-53)

T T u = r x r dr r

s 4q kR

r 2 1 d dT (8-55) 1 2 = r R r dr dr

Separating the variables and integrating twices 2 r 4 q T= r 2 + C1r + C2 kR 4R (8-56)

Boundary conditions Symmetry at r = 0: At r = R:

T ( r = 0 ) rT(r=R) = Ts

=0

C1=0 C2

s R 3 r 2 q r4 T = Ts 2+ 4 k 4 R 4R

(8-57)

Substituting the velocity and temperature profile relations (Eqs. 841 and 857) into Eq. 84 and performing the integrationTm = = m

c T ( r ) u ( r ) 2 rdr c T ( r ) mp p

p mcR

=

Ac

Vavg ( R 2 ) c p(8-4)

2 = T ( r ) u ( r ) rdr 2 Vavg R 0

(8-58)

s R 11 q Tm = Ts 24 k

s h (Ts Tm ) q(8-59)

Constant heat flux (circular tube, laminar)

24 k 48 k k h= = = 4.36 11 R 11 D D

hD Nu = = 4.36 = k

(8-60)

Constant Surface temperature (circular tube, laminar)

hD Nu = = 3.66 k()

(8-61)

Laminar Flow in Noncircular Tubes The friction factor ( f ) and the Nusselt number relations are given in Table 81 for fully developed laminar flow in tubes of various cross sections.

Laminar flow Hydrodynamic Thermal

Lh ,laminar 0.05 Re D

Lt ,laminar 0.05 Re Pr D = Pr Lh ,laminar

Developing Laminar Flow in the Entrance Region For a circular tube of length L subjected to constant surface temperature, the average Nusselt number for the thermal entrance region (hydrodynamically developed flow)

Nu = 3.66 +

0.065 ( D L ) Re Pr 1 + 0.04 ( D L ) Re Pr 0.03 ( Dh L ) Re Pr23

(8-62)

For flow between isothermal parallel plates

Nu = 7.54 +

1 + 0.016 ( Dh L ) Re Pr

23

(8-64)

Turbulent flow in Tubes Most correlations for the friction and heat transfer coefficients in turbulent flow are based on experimental studies. For smooth tubes, the friction factor in turbulent flow can be determined from the explicit first Petukhov equation

f = ( 0.79 ln Re 1.64 )

2

3000