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Qconv

 

 

 

 

 

 

 

 

T0 ]

= mcp [ Ti

From Eq. (5.241)

 

 

 

 

 

 

 

 

 

 

 

 

 

T

 

 

hpL

ln

 

 

o

 

= −

 

 

 

 

 

 

 

m cp

 

 

Ti

 

 

 

mc

 

= −

hpL

 

p

 

T

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ln

 

o

 

 

 

 

 

 

 

 

 

Ti

Substituting Eq. (5.248) in Eq. (5.246)

 

 

 

 

Q =

 

 

L p( T0 Ti )

h

 

 

 

 

 

 

T

conv

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ln

 

0

 

 

 

 

 

 

 

 

Ti

 

Qconv = hAs (LMTD)

Tlm = LMTD =

( To

 

Ti )

T

 

 

ln

 

o

 

 

 

Ti

 

 

 

5.6 Internal flows 161

(5.246)

(5.247)

(5.248)

(5.249)

(5.250)

(5.251)

Given h, the area As and the temperatures Qconv can be determined. In Eq. (5.251), LMTD is the logarithmic mean temperature difference, which physically is the mean

temperature difference that is driving the convection heat transfer. Since from Eq. (5.242) it is seen that the temperature distribution is exponential, the mean temperature difference is logarithmic. A mean of the arithmetic temperature difference at the outlet and the inlet is crude and is against the physics of the problem.

Analytical solution for Nusselt number for a fully developed flow

The boundary layer energy equation for flow inside a duct or tube can be derived by modifying the energy equation given in Chapter 4 and employing boundary layer simplifications presented earlier in this chapter. The final equation turns out to be

 

∂T

+ v

∂T

=

α

∂T

 

u

∂x

∂r

 

 

r

 

(5.252)

 

 

 

 

 

r ∂r

∂r

 

For fully developed flow, v = 0, ux = 0.

For constant surface heat flux and fully developed flow, we know the following relation:

u

 

 

r

2

 

 

 

= 2

1

 

 

 

(5.253)

u

 

r

m

 

 

 

 

 

 

 

 

 

0

 

 

162 CHAPTER 5  Forced convection

 

 

 

 

 

 

∂T

=

dTm

 

 

 

 

 

(5.254)

 

 

 

 

 

 

∂x

dx

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1 d

 

dT

 

 

2u

m

 

r

 

2 dT

 

 

 

 

=

 

1

 

 

 

m

 

 

 

r

 

 

 

 

 

 

 

(5.255)

 

 

 

α

 

 

 

 

r dr

dr

 

 

 

r0

 

 

dx

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Eq. (5.255) can be integrated twice in order to obtain an expression for temperature distribution as follows.

T =

2u

m

 

dT r2

r4

 

+ A lnr + B

 

 

 

m

 

 

 

 

(5.256)

α

 

 

16r02

 

 

dx

4

 

 

 

 

The two boundary conditions are

 

 

 

 

 

 

 

 

 

At r = r0 , T = Ts

 

(5.257)

 

 

 

At r = 0, T is finite

(5.258)

Substituting the above boundary conditions in Eq. (5.256), we get

A = 0

 

 

 

 

 

 

B = Ts

2u

m

dT

 

3r2

(5.259)

 

m

 

0

 

α

 

dx

16

 

Substituting for A and B in Eq. (5.256), we have

T (r) = T

 

2um

 

dTm

 

3r02

+

r4

r2

 

(5.260)

 

 

 

 

2

 

s

 

α

 

 

 

 

 

 

 

 

 

 

 

 

dx 16 16r0

 

4

 

We know that the bulk mean temperature is given by

 

2

r

(5.261)

Tm =

 

0 u T r dr

2

 

 

um r0

 

 

Substituting for T (r) from Eq. (5.260) in Eq. (5.261) and completing the integration, we get the following expression for Tm

Tm = Ts

11

 

2um dTm r02

(5.262)

 

96

 

α dx

 

From Newton’s law of cooling,

q = h(Ts Tm ) = h ×

11

×

2um

 

dTm

r02

(5.263)

96

 

 

 

 

α dx

 

By working on eqn 5.263 with the understanding that dTm /dx can be related to q through eqn.5.232, we can show that

Tm Ts = − 11 qD

48 k

= 48 k h

11 d

(5.264)

(5.265)

5.6 Internal flows 163

NuD =

hD

=

48

= 4.36

(5.266)

k

11

 

 

 

 

The above relationship is valid for constant heat flux.

For the constant wall temperature, the analysis is more difficult and the Nusselt number turns out to be NuD = 3.66. The analytical solution can be found in advanced texts on convection.

Correlation for turbulent flow inside tubes and ducts

For fully developed (both hydrodynamically and thermally)

Colburn equation:

Recall the relation fD = 0.184ReD–0.2 , ReD 2 ×104 (Eq. 5.214).

Combining this with the modified Reynolds analogy (Eq. 5.178), we get an expression for the average Nusselt number as

cf

=

fD

= StPr2/3 ReD > 104

 

(5.267)

 

 

2

 

8

 

 

 

 

 

 

 

From the definition of Stanton number, St, from Eq. (5.267) the expression for average Nusselt number turns out to be

NuD

= 0.023Re0.8 Pr1/3

(5.268)

Dittus-Boelter equation

Dittus and Boelter worked out a correlation for both heating and cooling, which is given as

 

= 0.023Re0.8 Prn

(5.269)

NuD

In Eq. (5.269), the exponent n = 0.4 for heating, and n = 0.3 for cooling of the fluid flowing inside the tube or duct. This equation is slightly more accurate than Eq. (5.268). Dittus-Boelter equation is the workhorse of convective heat transfer calculations in internal flows that are turbulent. These correlations can have errors of the order of ±25%. The Dittus-Boelter equation can be used for both constant wall flux and constant wall temperature conditions and can be used for turbulent flow in an annulus too. Current state of the art with three-dimensional turbulent flow and heat transfer correlations can give us Nusselt numbers far closer to truth than the above.

Example 5.5: Consider the flow of water in a tube of diameter 25 mm. The flow rate of water in the tube is 0.16 kg/s. The tube surface is maintained at 120 °C (by, say, a vapor, like steam, condensing at its saturation pressure). Water enters the tube at 30 °C and leaves the tube at 90 °C. If the convective heat transfer coefficient associated with the flow of water inside the tube is 2250 W/m2k, determine the length of the tube required to achieve the objective.

164 CHAPTER 5  Forced convection

Solution:

Water properties are to be evaluated at a bulk mean temperature

Tm,i + Tm,o = 30 + 90 = 60 C

2 2

ρ = 985 kg/m3 , cp = 4.18 kJ/kgK, k = 0.65 W/mK, ν = 5.1 × 107m2 /s and Pr = 3.3

The problem details are sketched in Fig. 5.12. Let us now draw a temperature length diagram for the heat transfer to the water in the tube. Fig. 5.13

The heat transfer rate Q in the problem under consideration is given by the following set of relations.

Q = mc

T

T

= LMTD

 

p m,o

m,i

 

FIGURE 5.12

Flow of water in a tube.

FIGURE 5.13

Temperature variation of water passing through the tube in Example 5.5.

5.6 Internal flows 165

Where LMTD is the logarithmic mean temperature difference given by

LMTD =

(Ts Tm,o )(Ts Tm,i )

 

T

T

 

 

 

 

 

ln

s

m,o

 

 

 

 

 

 

 

Ts Tm,i

LMTD =

30 − 90

= 54.61 C

 

 

30

 

 

 

 

ln

 

 

 

 

 

 

90

 

 

 

Q = 0.16 × 4.184 × 103 × (90 30) = 40166W = 40.166 kW Q = h A LMTD

40.166 × 103 = 2250 × π × 0.025 × L × LMTD

40.166 × 103 = 2250 × π × 0.025 × L × 54.61 L = 4.164 m

Hence, the length of the tube required to accomplish the heat transfer is 4.164 m. In this problem, there is no need for us to use an analytical result/correlation to determine the heat transfer coefficient “h,” as it is already specified in the problem. However, this has been specified only for the purpose of demonstrating internal convection calculations through an example. In engineering practice, invariably what we seek is the heat transfer coefficient, which can be determined from the data given together with additional details if the flow is fully developed. Let us now solve one more example directly related to Example 5.5 to see if the heat transfer coefficient

2250 W/m2K given in the problem is reasonable.

Example 5.6: Revisit Example 5.5. All the information given to us in Example 5.5 is available except that the heat transfer coefficient is as yet unknown. However, a key additional piece of information, namely the fact that the flow is fully developed from the inlet, is given to us. The unknown again is the length of the tube required to accomplish the heat transfer.

Solution:

The Reynolds number of the flow is

ReD = uνmd

Where the mean velocity of the water in the tube is

 

 

 

.

 

 

 

u

 

=

4 m

 

 

 

m

πρd2

 

 

 

 

um =

 

4 × 0.16

= 0.33m/s

3.14

× 0.0252 × 985

Re

 

=

0.33

× 0.025 = 16176

 

D

 

5.1 ×107