# Some Analytical Considerations

For air temperatures up to 1000 K, the specific heats are essentially constant, thus justifying the assumption of a calorically perfect gas for this range. Moreover, the

temperature variations of /і and к over this range are virtually identical. As a result, the Prandtl number, ficp/ k, is essentially constant up to temperatures on the order of 1000 K. This is shown in Figure 16.11, obtained from Schetz (Reference 53). Note that Pr ~ 0.71 for air; this is the value that was used in Example 16.1.

Question: How high a Mach number can exist before we would expect to en­counter temperatures in the flow above 1000 K? Answer: An approximate answer is to calculate that Mach number at which the total temperature is 1000 K. Assuming a static temperature T = 288 K, from Equation (8.40),

Hence, for most aeronautical applications involving flight at a Mach number of 3.5 or less, the temperature within the viscous portions of the flow field will not exceed 1000 K. A Mach number of 3.5 or less encompasses virtually all operational aircraft today, with the exception of a few hypersonic test vehicles.

In light of the above, many viscous flow solutions are carried out making the justi­fiable assumption of a constant Prandtl number. For the case of compressible Couette flow, the assumption of Pr = constant allows the following analysis. Consider the energy equation, Equation (16.3), repeated below:

Inserting b and c into Equation (16.79) and simplifying, we obtain

и2 и Pr

h – b Pr — =hw—– (he – hw) + —(uue)

2 ue 2

Assume the lower wall is adiabatic; that is, (Bh/Sy)w = 0. Differentiating Equa­tion (16.80) with respect to y, we have

Recall that the condition for an adiabatic wall is that (dh/dy)w = 0. Applying Equation (16.81) at у = 0 for an adiabatic wall, where и = 0 and by definition hw = haw, we have

Since (du/dy)w is finite, then

This is identical to Equation (16.39) obtained for incompressible flow. Hence, we have just proven that the recovery factor for compressible Couette flow, assuming constant Prandtl number, is also

[16.83]

Since the recovery factors for the incompressible and compressible cases are the same (as long as Pr = constant), what can we say about Reynolds analogy? Does Equation (16.59) hold for the compressible case? Let us examine this question. Return to Equation (16.3), repeated below:

Recalling that, from the definitions,

then Equation (16.3) can be written as

Integrating Equation (16.86) with respect to y, we have

where a is a constant of integration. Evaluating Equation (16.87) at у = 0, where и = 0 and q = qw, we find that

Hence, Equation (16.87) is

q + xu = qw

Inserting Equations (16.84) and (16.85) into (16.88), we have

 . dT du 4w=k~ + їли — dy dy [16.89] or qw к dT du x x dy dy [16.90] Recall that the shear stress is constant throughout the flow; hence, du r = ~ r>" dy or ?W a — ——– du/dy [16.91] Also, к _ Cp fl Pr [16.92]

Inserting Equation (16.91) into the left-hand side of Equation (16.90), and Equa­tion (16.92) into the right-hand side of Equation (16.90), we have

qw du cp dT d{u2/2)

rw dy Pr 3у dy

Integrate Equation (16.93) between the two plates, keeping in mind that qw, ru

and Pr are all fixed values:

or which yields

CtLUe^%Te~Tw)+1^ Tw Pr 2

Rearranging Equation (16.94), and recalling that h = cpT, we have

Equation (16.98) is Reynolds analogy—a relation between heat transfer and skin friction coefficients. Moreover, it is precisely the same result as obtained in Equa­tion (16.59) for incompressible flow. Hence, for a constant Prandtl number, we have shown that Reynolds analogy is precisely the same form for incompressible and com­pressible flow.

Consider the geometry given in Figure 16.2. The two plates are separated by a distance of 0.01 in (the same as in Example 16.1). The temperature of the two plates is equal, at a value of 288 К (standard sea level temperature). The air pressure is constant throughout the flow and equal to 1 atm. The upper plate is moving at Mach 3. The shear stress at the lower wall is 72 N/m2. (This is about 1.5 lb/ft2—a much larger value than that associated with the low-speed case treated in Example 16.1.) Calculate the heat transfer to either plate. (Since the shear stress is constant throughout the flow, and the plates are at equal temperature, the heat transfer to the upper and lower plates is the same.)

Solution

The velocity of the upper plate is

ue = Meae = Mey/yRTe = 3v7(1 -4)(288)(287) = 1020 m/s The air density at both plates is (noting that 1 atm =1.01 x 105 N/m2)

1.22 kg/m3

Hence, the skin friction coefficient is

(1020)2 ,

haw = (1004.5)(288) + (0.71)- – – – = 6.59 x 105 J/kg

[Note: This gives Taw = huw/cp = (6.59 x 105)/1004.5 = 656 K. In the adiabatic case, the wall would be quite warm.] Hence, from the definition of the Stanton number [Equa­tion (16.55)], and noting that hw = cpT„, = (1004.6)(288) = 2.89 x 105 J/kg,

qw = peue(hau,-hw)CH = ( 1.22)(1020)[(6.59- 2.89) x 105](8 x КГ5)