Question 12.3: We neglected the viscous forces and the base pressure in est......
We neglected the viscous forces and the base pressure in estimating the forces acting on the sharp cone of Example 12.2. Consider a sharp cone \left(\theta_{c}=10^{\circ}\right) \mathrm{ex}- posed to the Mach 8 flow of Tunnel B at the Arnold Engineering Development Center (AEDC), as shown in Fig. 12.18. For this problem, the subscript t 1 designates the flow conditions in the nozzle reservoir, the subscript 1 designates those in the test section, the subscript e designates those at the edge of the boundary layer of the sharp cone (and which are constant along the entire length of the cone), and the subscript w designates conditions at the wall of the cone. If T_{t 1}=1350^{\circ} \mathrm{R}, p_{t 1}=850 \mathrm{psia}, T_{w}=600^{\circ} \mathrm{R}, and \alpha=0^{\circ}, develop expressions for the forebody pressure coefficient \left(C_{p, e}\right), the base pressure coefficient \left(C_{p, b}\right), and the skin friction. What is the total drag acting on the vehicle?
Let us use Figs. 8.15 \mathrm{~b} and \mathrm{c} to obtain values for the pressure coefficient and for the Mach number of the inviscid flow at the edge of the boundary layer for a 10^{\circ}-half-angle cone in a Mach 8 stream. Inherent in this use of these figures is the assumption that the boundary layer is thin and that there are no significant viscous/inviscid interactions which perturb the pressure field. Thus, M_{e}=6 and C_{p, e}=0.07. Since q_{1}=\frac{1}{2} \rho_{1} U_{1}^{2} =(\gamma / 2) p_{1} M_{1}^{2},
p_{e}=p_{1}\left(1+\frac{\gamma}{2} M_{1}^{2} C_{p, e}\right)
We can use Table 8.1 to calculate the free-stream static pressure \left(p_{1}\right) in the Mach 8 flow of the wind-tunnel test section:
p_{1}=\frac{p_{1}}{p_{t 1}} p_{t 1}=\left(102 \times 10^{-6}\right)(850)=0.867 \text { psia }
Thus,
p_{e}=0.0867[1.0+(0.7)(64)(0.07)]=0.3586 \mathrm{psia}
Since M_{e}=6, we can use Table 8.1 and the fact that the edge flow is that for an adiabatic flow of a perfect gas, so that T_{t 1}=T_{t e}. Thus,
T_{e}=\frac{T_{e}}{T_{t e}} T_{t e}=(0.12195)(1350)=164.63^{\circ} \mathrm{R}
To aid in our decision as to whether the boundary layer is laminar or turbulent, let us calculate the Reynolds number for the flow at the boundarylayer edge:
\operatorname{Re}_{l}=\frac{\rho_{e} U_{e} l}{\mu_{e}}=\left[\frac{p_{e}}{R T_{e}}\right]\left[M_{e} \sqrt{\gamma R T_{e}}\right]\left[\frac{T_{e}+198.6}{2.27 \times 10^{-8} T_{e}^{1.5}}\right] l
where, as shown in Fig. 12.18, l is the wetted distance along a conical generator. Thus,
\begin{aligned} \operatorname{Re}_{l} & =\left(0.0001828 \frac{\mathrm{lbf} \cdot \mathrm{s}^{2}}{\mathrm{ft}^{4}}\right)\left(3773.80 \frac{\mathrm{ft}}{\mathrm{s}}\right)\left(7.575 \times 10^{6} \frac{\mathrm{ft}^{2}}{\mathrm{lbf} \cdot \mathrm{s}}\right) l \\ & =5.226 \times 10^{6} l \end{aligned}
Based on the wetted length of a conical ray \left(l_{c}=17.276 ~\mathrm{in}.\right),
\operatorname{Re}_{l c}=7.523 \times 10^{6}
It should be noted that, although hypersonic boundary-layers are very stable, boundary-layer transition should occur for this flow, probably within the first one-half of the cone length. Thus, we can use the correlation presented in Fig. 12.19 to determine the base pressure:
\frac{p_{b}}{p_{e}}=0.02so that p_{b}=0.00717~ \mathrm{psi}.
Mach number for turbulent flow. [From Cassanto (1973).]
Note that since p_{b}=0.02 p_{e},
\begin{aligned} C_{p, b} & =\frac{p_{b}-p_{1}}{q_{1}}=\left(\frac{p_{b}}{p_{1}}-1\right) \frac{2}{\gamma M_{1}^{2}} \\ & =\left(\frac{p_{b}}{p_{e}} \frac{p_{e}}{p_{1}}-1\right) \frac{2}{\gamma M_{1}^{2}}=-0.0205 \end{aligned}
Recall that the assumption for Newtonian flow is C_{p, b}=0.0.
To calculate the skin friction contribution, let us use Reynolds analogy (see Chapter 4). For supersonic flow past a sharp cone, the heat-transfer rate (or, equivalently, the Stanton number) can be calculated using the relatively simple approximation obtained through the Eckert reference temperature approach [Eckert (1955)]. In this approach, the heat-transfer rates are calculated using the incompressible relations [e.g., equation (4.103)]
\mathrm{St} \equiv C_h=\frac{\dot{q}}{\rho u_e c_p\left(T_e-T_w\right)} (4.103)
with the temperature-related parameters evaluated at Eckert’s reference temperature \left(T^{*}\right). According to Eckert (1955),
T^{*}=0.5\left(T_{e}+T_{w}\right)+0.22 r\left(T_{i e}-T_{e}\right)
where r is the recovery factor. For turbulent flow, the recovery factor is \sqrt[3]{P r}, which is equal to 0.888 . Thus,
\begin{aligned}T^* & =0.5(164.63+600)+0.22(0.888)(1350-164.63) \\& =613.89^{\circ} R\end{aligned}
Thus,
\begin{aligned} \mathrm{Re}_{l}^{*} & =\left[\frac{p_{e}}{R T^{*}}\right]\left[U_{e}\right]\left[\frac{T^{*}+198.6}{2.27 \times 10^{-8} T^{* 1.5}}\right] l \\ & =\left(0.00004901 \frac{\mathrm{lbf} * \mathrm{~s}^{2}}{\mathrm{ft}^{4}}\right)\left(3773.80 \frac{\mathrm{ft}}{\mathrm{s}}\right)\left(2.353 \times 10^{6} \frac{\mathrm{ft}^{2}}{\mathrm{lbf} \cdot \mathrm{s}}\right) l \\ & =4.353 \times 10^{5} l \end{aligned}
Thus, using Reynolds’ analogy, we have
C_{f}=\frac{0.0583}{\left(\operatorname{Re}_{l}^{*}\right)^{0.2}}=\frac{0.004344}{l^{0.2}}
Thus,
\tau=C_{f}\left(\frac{\gamma}{2} p_{e} M_{e}^{2}\right)=\frac{0.03926}{l^{0.2}} \frac{\mathrm{lbf}}{\mathrm{in}^{2}}
To calculate the total drag,
D=\int p_{e} 2 \pi r d l \sin \theta_{c}+\int \tau_{w} 2 \pi r d l \cos \theta_{c}-p_{b} \pi R_{b}^{2}
Noting that p_{e} is constant along the length of the conical generator, d l \sin \theta_{c}=d r, and r=l \sin \theta_{c},
\begin{aligned} D & =p_{e} \pi R_{b}^{2}+0.03926(2 \pi)\left(\cos \theta_{c} \sin \theta_{c}\right) \int_{0}^{l_{c}} l^{0.8} d l-p_{b} \pi R_{b}^{2} \\ & =(0.3586)(9 \pi)+(0.04218) \frac{l_{c}^{1.8}}{1.8}-(0.00717)(9 \pi) \\ & =10.139+3.956-0.203=13.892 \mathrm{lbf} \end{aligned}
