5.4 Approximate solution to the boundary layer equations 137

N uL = 0.662Re1/2 Pr1/3

N uL = 0.662 × (2.66 ×105 )1/2 (0.711)1/3

N uL = 304.77

N uL = hL = 304.77 kf

h = 304.77 × 0.027

1

h = 8.23 W/m2 K

The rate of heat transfer is given by

q = hA T

q = 494.1 W/m2

Flow over a cylinder

Flow over a cylinder is commonly encountered in several heat transfer applications like evaporator, condenser, recuperator, automative radiator, and so on. The special difficulty associated with flow over a cylinder is the curvature and its effect on the boundary layer development. Consider a flow over a cylinder, as shown in Fig. 5.3.

The key difference between the flat plate and this flow is that U∞ is no longer a constant and varies with x, as indicated in the figure. For θ = 0 to 90 , u∞ (x) increases

FIGURE 5.3

Development and separation of boundary layer over a cylinder.

138 CHAPTER 5  Forced convection

with x and this results in a favorable pressure gradient, whereas for 90 ≤ θ ≤ 180 , the fluid decelerates consequent upon which there is an adverse pressure gradient, and separation of the boundary layer occurs from the wall the condition for which is mathematically given by ∂u/∂y = 0. When the boundary layer detaches at the surface, a wake is formed in the downstream. Please refer to Incropera et al. (2013) for a more comprehensive discussion on this special flow. If ReD ≤ 2 ×105 , the flow stays laminar and separation occurs at θ = 80 . However, beyond a ReD of 2 ×105 , transition to turbulence (more on this in Section 5.5) occurs and separation takes place at θ = 140 .

For flow over a cylinder, the correlation for Nusselt number (NuD ) proposed by Zukauskas (1972) is of the form

The constants C, m are reported in Table 5.1.

This correlation is valid for 0.7 ≤ Pr ≤ 500 and 1 ≤ ReD ≤ 106. If Pr ≤ 10 then n = 0.37 and if Pr >10, then n = 0.36. Churchill and Bernstein (1977) have proposed a single comprehensive equation as follows

The above equation is recommended for all the values of ReD Pr ≥ 0.2

Flow over a sphere

Whitaker (1972) proposed a correlation for the average Nusselt number for flow over a sphere as follows.

5.4 Approximate solution to the boundary layer equations 139

Heat transfer in flows across a bank of tubes

Flow and heat transfer over a bank of tubes or rod bundles frequently occur in industrial practice, as, for example, in heat exchangers, regenerators (for example, in an application where exhaust gases preheat the incoming fluid, say air or water), and in nuclear fuel rods. A variety of configurations are possible. Two frequently encountered ones are shown below in Figs. 5.4A and 5.4B.

In the above two figures, there are three rows of tubes. The difference between the two is that in (a) the arrangement is aligned and is frequently referred to as “in-line arrangement” while in (b) the arrangement of the tubes is staggered. There are two key geometrical parameters that enter the problem. These are the transverse pitch ST and the longitudinal pitch SL. For such flows, it is intuitive to expect that the average Nusselt number is a function of the pertinent parameters as follows.

FIGURE 5.4

Arrangement of (A) aligned and (B) staggered rows of tubes in tube/rod bundle.

140 CHAPTER 5  Forced convection

Following the development presented in the earlier portions of this chapter, the Nusselt number should vary as

Professor Zukauskas from Lithuania did extensive work and has presented reviews of heat transfer in these geometries. In fact, he synthesized the work of all the investigators before him and proposed a generic expression for NuD as

where n = 0 for gases and 0.25 for liquids. All properties need to be evaluated at the bulk fluid temperature, while Prw is evaluated at the wall temperature Tw .

For 102 ≤ ReD ≤ 103

Aligned rows : Z = 0.52ReD0.5

Staggered rows : Z = 0.71ReD0.5

For 103 ≤ ReD ≤ 2 ×105

Aligned rows : Z = 0.27 ReD0.63 , ST /SL ≥ 0.7

Staggered rows : Z = 0.35(ST /SL )0.2 ReD0.6 , ST /SL ≤ 2

For even higher Reynolds numbers and for liquid metals, please refer to Lienhard and Lienhard (2020), who have presented a comprehensive account of heat transfer across tube bundles.

Example 5.2: Consider a long pin fin of diameter 8 mm. The fin is made of aluminum (k = 205 W /mK ) and is maintained at a base temperature of 100 °C. Air at 30 °C flows over the fin with a velocity of 8 m/s.

a.Determine the heat transfer coefficient from the fin.

b.Determine the net heat transfer from the fin.

c.If the diameter of the fin is changed to 14 mm, with all other conditions being the same, determine the heat transfer coefficient and the heat transfer from the fin.

Now, we obtain the air properties at 338 K from Table 5.2 given on the next page.

5.4 Approximate solution to the boundary layer equations 141

Table 5.2  Thermophysical properties of air at atmospheric pressure (Kadoya et al., 1985; Jacobsen, R. T., et al., 1992).

5.4 Approximate solution to the boundary layer equations 143

The key takeaway from this problem is that if we had tried to solve this problem with knowledge of only fin heat transfer but without any idea of how the heat transfer coefficient will change for a change in diameter, our solution would have been

Q2 = 2.311

Q1

The actual ratio of Q2 /Q1 We obtained is 2.04 (i.e., 22.37 W/10.94 W), which leads to a significantly different result by about 15%.

Example 5.3: A heated sphere made of aluminum (k = 205 W/mK) of 8 mm diameter at 100 °C cools in a current of air at 30 °C, with a free stream velocity of 8 m/s. The density and specific heat of aluminum are 2700 kg/m3 and 900 J /kgK respectively.

a.Determine the heat transfer coefficient from the sphere.

b.Determine the time constant of the sphere.

c.How much time will it take for the sphere to cool to 50 °C.

Neglect radiation from the sphere.

Solution:

a. The Reynolds number ReD , is given by

144 CHAPTER 5  Forced convection

b. Time constant

τ= 27.33 s

c.Let us now calculate the Biot number.

Bi = hr = 118.58 × 4 ×10−3 3k 3× 205

Bi = 0.00077

Since Bi < 0.1, the lumped capacitance method is valid. Recalling the solution for a spatially lumped object undergoing cooling (from Chapter 3),

((T − T∞ )) = e−t /T

Ti − T∞

(50 − 30) = e−t /27.33

(100 − 30)

0.2857 = e−t /27.33t = 34.24 s

5.5  Turbulent flow

Consider flow and heat transfer over a flat plate. Up to ReL = 5 ×105 , the flow stays laminar and beyond this there is transition to turbulence. The boundary layer grows thicker and eddies, and consequent mixing will be in the transition region and in the fully turbulent region beyond the transition. A depiction of the boundary layer flow over a flat plate is shown in Fig. 5.5.

FIGURE 5.5

Forced convection boundary layer development over a horizontal flat plate depicting the laminar, transition, and turbulent regions of flow.