A typical World War I biplane fighter (such as the French SPAD shown in Figure 3.50) has a number of vertical interwing struts and diagonal bracing wires. Assume for a given airplane that the total length for the vertical struts (summed together) is 25ft, and that the struts are cylindrical with a diameter of 2 in. Calculate the drag (in pounds) contributed by the struts and bracing wires when the airplane is flying at \(120 \;mi/h\) at standard sea level.

Assume for a given airplane that the total length for the vertical struts (summed together) is \(25\; ft\), and that the struts are cylindrical with a diameter of \(2\; in\). Assume also that the total length of the bracing wires is \(80\; ft\), with a cylindrical diameter of \(3/32\;in\). Calculate the drag (in pounds) contributed by these struts and bracing wires when the airplane is flying at \(120 \;mi/h\) at standard sea level. Compare this component of drag with the total zero-lift drag for the airplane, for which the total wing area is \(230\; ft^{2}\) and the zero-lift drag coefficient is \(0.036\).

Worldtech Asked on 5th November 2019 in Aerodynamics.
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    At a standard sea level conditions  \[\rho _{\infty}=0.002377\;slug/ft^{3}
    \\\mu _{\infty}=3.737\times 10^{-7}\;slug/ft\cdot sec\]

    \[V=120\;mph=120\left ( \frac{88}{60} \right )ft/sec=176\;ft/sec
    \\q_{\infty}=\frac{1}{2}\rho _{\infty}V_{\infty}^{2}=\frac{1}{2}\times0.002377\times176^{2}=36.8\;lb/ft^{2}\]

    \[D=2\;inch=0.167\;ft\]

    Reynolds number \[Re=\frac{\rho V D}{\mu}=\frac{0.002377\times187.7\times0.167}{3.737\times10^{-7}}
    \\=199,382\]

    Total frontal surface area of the struts is \[25\left ( 0.167 \right )=4.175\;ft^{2}\]

    Drag due to struts , here \(C_{D}=1\) \[D_{s}=q_{\infty}SC_{D}=36.8\times4.175\times 1 =153\;lb\]

    For the bracing wires \[D=\frac{3}{32}\;inch=0.0078\;ft\]

    \[Re=199382\left ( \frac{0.0078}{0.167} \right )=9312\]

    For the wires \(C_{D}=1\).Total frontal surface area of wires is \[80\times 0.0078=0.624\;ft^{2}\]

    Therefore drag due to wires is \[D_{w}=q_{\infty}SC_{D}=36.8\times0.624\times1=23\;lb\]

    Total drag due to struts and wires = \(D_{s}+D_{w}=153+23=176\;lb\)

    Total zero-lift drag for the airplane (including struts and wires ) is \[D=q_{\infty}SC_{D,0}=36.8\times230\times0.036=304.8\;lb\]

     

    Worldtech Answered on 5th November 2019.
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