Chapter 5, Problem 5.9P

Consider the Superrnarine Spitfire shown in Figure 5.19. The first version of the Spitfire was the Mk I. which first flew in 1936. Its maximum velocity is 362 mi/h at an altitude of 18,500 ft. Its weight is 5820 lb, wing area is $242{\text{ft}}^{\text{2}}$, and wing span is 36.1 ft. It is powered by a supercharged Merlin engine, which produced 1050 horsepower at 18,500 ft. (a) Calculate the induced drag coefficient of the Spitfire at the flight condition of $V_{\max }$ at 18,500 ft. (b) What percentage of the total drag coefficient is the induced drag coefficient? Note: To calculate the total drag. we note that in steady, level flight of the airplane, $T=D$, where T is the thrust from the propeller. In turn, the thrust is related to the power by the basic mechanical relation $T{V}_{\infty }=P$, where P is the power supplied by the propeller-engine combination. Because of aerodynamic losses experienced by the propeller, P is less than the shaft power provided by the engine by a ratio $\eta$, defined as the propeller efficiency. That is, if HP is the shaft horsepower provided by the engine, and since $550\text{ft}\cdot \text{lb/s}$ equals I horsepower, then the power provided by the engine-propeller combination in foot-pounds per second is $P=550\eta HP$. See Chapter 6 of Reference 2 for more details. For this problem, assume the propeller efficiency for the Spitfire is 0.9.

To determine

The induced drag coefficient of the spitfire at the flight condition of $V_{\max }$ at $18000\text{ft}$ .

Answer to Problem 5.9P

The induced drag coefficient of the spitfire at the flight condition of $V_{\max }$ at $18000\text{ft}$ is $9\cdot 67\times {10}^{-4}$ .

Explanation of Solution

Given:

The maximum velocity of a spitfire at an altitude of $18000\text{ft}$ is $362\text{mi}/\text{h}$ , the weight of super-marine spitfire is $5820\text{lb}$ , the power generated by the Merlin engine is $1050\text{hp}$ , the area of the wing is $242{\text{ft}}^{\text{2}}$ , and the wingspan is $36\cdot 1\text{ft}$ .

The expression for the coefficient of drag is given by,

  $C_L=\frac{W}{\frac{1}{2}\rho V^{2}S}$ ..... (1)

Here,

  $W$ is the weight of the super-marine spitfire.

  $\rho$ is the density of the air.

  $S$ is the area of the wing.

  $V$ is the velocity of a spitfire at an altitude of $18000\text{ft}$ .

Substitute $362\text{mi}/\text{h}$ for $V$ , $5820\text{lb}$ for $W$ , $242{\text{ft}}^{\text{2}}$ for $S$ and $0\cdot 00133\text{slug}/{\text{ft}}^{\text{3}}$ for $V$ in equation (1).

  $\begin{array}{c}{C}_L=\frac{\left(5820\text{lb}\right)}{\frac{1}{2}\left(0\cdot 00133\text{slug}/{\text{ft}}^{\text{3}}\right){\left(362\text{mi}/\text{h}\right)}^2\left(242{\text{ft}}^{\text{2}}\right)}\\ =\frac{\left(5820\text{lb}\right)}{\frac{1}{2}\left(0\cdot 00133\text{slug}/{\text{ft}}^{\text{3}}\right){\left(362\text{mi}/\text{h}\left(\frac{5280\text{ft}/\text{mi}}{3600\text{s}/\text{h}}\right)\right)}^2\left(242{\text{ft}}^{\text{2}}\right)}\\ =0\cdot 128\end{array}$

The formula for aspect ratio is given by,

  $AR=\frac{b^2}{S}$ ..... (2)

Here,

  $b$ is the wingspan.

Substitute $36\cdot 1\text{ft}$ for $b$ and $242{\text{ft}}^{\text{2}}$ for $S$ in equation (2).

  $\begin{array}{c}AR=\frac{{\left(36\cdot 1\text{ft}\right)}^2}{\left(242{\text{ft}}^{\text{2}}\right)}\\ =5\cdot 39\end{array}$

The expression for induced drag coefficient is given by,

  $C_{d,i}=\frac{C_L{}^2}{\pi eAR}$ .......(3)

Here,

  $e$ is the span efficiency factor.

For an elliptic plan-form, the span efficiency factor is $1$ .

Substitute $1$ for $e$ , $5\cdot 39$ for $AR$ and $0\cdot 128$ for $C_L$ in equation (3).

  $\begin{array}{c}{C}_{d,i}=\frac{{\left(0\cdot 128\right)}^2}{\pi \left(1\right)\left(5\cdot 39\right)}\\ =9\cdot 67\times {10}^{-4}\end{array}$

Hence, The induced drag coefficient of the spitfire at the flight condition of $V_{\max }$ at $18000\text{ft}$ is $9\cdot 67\times {10}^{-4}$ .

To determine

The percentage of total drag coefficient as compared to induced drag coefficient.

Answer to Problem 5.9P

The percentage of total drag coefficient as compared to induced drag coefficient is $4\cdot 49$ percent.

Explanation of Solution

Given:

The propeller efficiency of the Merlin engine is $0\cdot 9$ .

The expression for the power generated by the engine is given by

  $P_A=550\eta \left(\text{HP}\right)$

Here,

  $\eta$ is the propeller efficiency of the Merlin engine.

  $HP$ is the power generated by the Merlin engine.

Substitute $0\cdot 9$ for $\eta$ and $1050\text{hp}$ for $\text{HP}$ in the above equation.

  $\begin{array}{c}{P}_A=550\left(0\cdot 9\right)\left(1050\text{hp}\right)\\ =519\cdot 75\times {10}^3\text{lb}\cdot \text{ft}/\text{s}\end{array}$

For the steady and level flight, the drag force is equal in magnitude to the thrust produced by impeller,

  $D=T$

Here,

  $D$ is the drag force.

  $T$ is the thrust produced by the impeller.

The expression for the drag force or thrust for steady flight is given by,

  $D=\frac{P_A}{V}$ .......(4)

Substitute $519\cdot 75\times {10}^3\text{lb}\cdot \text{ft}/\text{s}$ for $P_A$ and $362\text{mi}/\text{h}$ for $V$ in equation (4).

  $\begin{array}{c}D=\frac{519\cdot 75\times {10}^3\text{lb}\cdot \text{ft}/\text{s}}{\left(362\text{mi}/\text{h}\right)}\\ =\frac{519\cdot 75\times {10}^3\text{lb}\cdot \text{ft}/\text{s}}{\left(362\text{mi}/\text{h}\left(\frac{5280\text{ft}/\text{mi}}{3600\text{s}/\text{h}}\right)\right)}\\ =978\cdot 94\text{lb}\end{array}$

The expression for the total drag coefficient is given by,

  $C_D=\frac{D}{\frac{1}{2}\rho V^{2}S}$

Substitute $362\text{mi}/\text{h}$ for $V$ , $978\cdot 94\text{lb}$ for $D$ , $242{\text{ft}}^{\text{2}}$ for $S$ and $0\cdot 00133\text{slug}/{\text{ft}}^{\text{3}}$ for $V$ in equation (1).

  $\begin{array}{c}{C}_L=\frac{\left(978\cdot 94\text{lb}\right)}{\frac{1}{2}\left(0\cdot 00133\text{slug}/{\text{ft}}^{\text{3}}\right){\left(362\text{mi}/\text{h}\right)}^2\left(242{\text{ft}}^{\text{2}}\right)}\\ =\frac{\left(978\cdot 94\text{lb}\right)}{\frac{1}{2}\left(0\cdot 00133\text{slug}/{\text{ft}}^{\text{3}}\right){\left(362\text{mi}/\text{h}\left(\frac{5280\text{ft}/\text{mi}}{3600\text{s}/\text{h}}\right)\right)}^2\left(242{\text{ft}}^{\text{2}}\right)}\\ =0\cdot 0215\end{array}$

The expression for the percentage of total drag coefficient as compared to induced drag coefficient is given by

  $r=\left(\frac{C_{d,i}}{C_D}\right)\times 100$

Substitute $0\cdot 0215$ for $C_D$ and $9\cdot 67\times {10}^{-4}$ for $C_{d,i}$ in above equation.

  $\begin{array}{c}r=\left(\frac{9\cdot 67\times {10}^{-4}}{0\cdot 0215}\right)\times 100\\ =4\cdot 49\end{array}$

Hence, the percentage of the total drag coefficient as compared to the induced drag coefficient is $4\cdot 49$ percent.