Consider the single-engine turboprop utility aircraft Cessna 208 Caravan with the following features: mTO = 3995 kg, S = 25.9 m², P = 647 kW, b = 15.88, Vs = 61 knot, Vmax = 185 knot. Assume: CDo = 0.038, ηP = 0.85, e = 0.9. 1. Determine the corner speed of the aircraft. Then, determine the

Question

Consider the single-engine turboprop utility aircraft Cessna 208 Caravan with the following features:

[latex]m_{TO}[/latex] = 3995 kg, S = 25.9 m², P = 647 kW, b = 15.88, [latex]V_s[/latex] = 61 knot, [latex]V_{\max}[/latex] = 185 knot.
Assume: [latex]C_{Do}[/latex] = 0.038, [latex]η_P[/latex] = 0.85, e = 0.9.

1. Determine the corner speed of the aircraft. Then, determine the corresponding load factor, turn rate, and turn radius.
2. What is the maximum producible load factor in a level turn? Then, determine its corresponding airspeed and bank angle.
3. Assume the aircraft is turning with an airspeed from corner speed to maximum speed. Plot the variations of load factor, bank angle, turn rate, and turn radius.
Assume sea level, constant prop efficiency, and constant zero-lift drag coefficient.

Accepted Answer

We first need to find two parameters:

[latex]AR=\frac{b^2}{S}=\frac{15.88^2}{25.9}=9.74 \quad \quad \quad \quad (3.9)\ \space \ K=\frac{1}{\pi eAR}=\frac{1}{3.14 imes 0.9 imes 9.74}=0.036 \quad \quad \quad \quad (3.8)\ \space \ \Big(\frac{L}{D}\Big)_{\max}=\frac{1}{2\sqrt{KC_{D_o}}}=\frac{1}{2\sqrt{0.036 imes 0.038}}=13.5 \quad \quad \quad \quad (5.28)[/latex]

1. Corner speed and its corresponding load factor, turn rate, and turn radius

[latex]C_{L_{\max}}=\frac{2mg}{ ho SV^2_s}=\frac{2 imes 3995 imes 9.81}{1.225 imes 25.9 imes (61 imes 0.514)}=2.51 \quad \quad \quad \quad (2.49)[/latex]

• Corner speed

[latex]V^{\ast}=\Big[\frac{2P_{\max}\eta_P}{ ho S(C_{D_{o}}+KC_{L_{\max}}^2)}\Big]^{\frac{1}{3}}=\Big[\frac{2 imes 647,000 imes 0.85}{1.225 imes 25.9 imes (0.038+0.036 imes (2.51)^2)}\Big]^{\frac{1}{3}} \quad \quad \quad \quad (9.90)\ \space \ \Rightarrow V^{\ast}=50.7\space m/s=98.5 \space knot.[/latex]

• Corresponding load factor

[latex]n_{\max_C}=\frac{ ho(V^{\ast})^2SC_{L_{\max}}}{2W}=\frac{1.225 imes (50.7)^2 imes 25.9 imes 2.51}{2 imes 3995 imes 9.81}=2.61 \quad \quad \quad \quad (9.91)[/latex]

• Corresponding turn rate

[latex]\omega_{\max}=\frac{g\sqrt{n^2_{\max_C}-1}}{V^{\ast}}=\frac{9.81\sqrt{2.61^2-1}}{50.7}=0.466 \space rad/s=26.7 \space deg/s \quad \quad \quad \quad (9.92)[/latex]

• Corresponding turn radius

[latex]R_{\min}=\frac{V^{\ast^2}}{g\sqrt{n_{\max_C}^2}-1}=\frac{50.7^2}{9.81\sqrt{2.61^2-1}}=108.7 \space m \quad \quad \quad \quad (9.93)[/latex]

2. Maximum producible load and its corresponding airspeed and bank angle

[latex]n_{\max_{\max}}=0.687\Big[\frac{ ho \eta_P^2P_{\max}^2S(L/D)_{\max}}{KW^3}\Big]^{\frac{1}{3}}=0.687\Big[\frac{1.225 imes 0.85^2 imes 647,000^2 imes 25.9 imes 13.5}{0.036 imes (3,995 imes 9.81)^3}\Big]^{\frac{1}{3}}\quad \quad \quad \quad (9.84) \ \space \ n_{\max_{\max}}=2.68 \ \space \ V_{\max_{n}}=\Big[\frac{P_{\max}\eta_P}{2 ho SC_{D_o}}\Big]^{\frac{1}{3}}=\Big[\frac{647,000 imes 0.85}{2 imes 1.225 imes 25.9 imes 0.038}\Big]^{\frac{1}{3}}=61.3 \space m/s=119.3 \space knot \quad \quad \quad \quad (9.82) \ \space \ \phi_{\max_{\max}}=\cos^{-1}\Big(\frac{1}{n_{\max_{\max}}}\Big)=\cos^{-1}\Big(\frac{1}{2.68}\Big)=68.1^{\circ}\quad \quad \quad \quad (9.85)[/latex]

3. Plot (variations of load factor, bank angle, turn rate, and turn radius)
For each airspeed, first Equation 9.80 is used to determine its corresponding maximum load factor

[latex]n_{\max}=\frac{S}{W}\sqrt{\frac{ ho VP_{\max}\eta_P}{2KS}-\frac{ ho^2V^4C_{D_o}}{4K}}\quad \quad \quad \quad (9.80)[/latex]

Then, Equation 9.13 is utilized to determine its corresponding bank angle.

[latex]\phi_{\max}=\cos^{-1}\Big(\frac{1}{n_{\max}}\Big)\quad \quad \quad \quad (9.13)[/latex]

Then, Equations 9.32 and 9.24 are employed to determine its corresponding turn rate and turn radius.

[latex]\omega=\frac{g\sqrt{n_{\max}^2-1}}{V} \quad \quad \quad \quad (9.32) \ \space \ R=\frac{V^2}{g\sqrt{n_{\max}^2-1}}\quad \quad \quad \quad (9.24)[/latex]

Figure 9.12 and Table 9.2 demonstrate the variations of load factor, turn rate, bank angle, and turn radius versus airspeed. As expected, the maximum turn rate and minimum turn radius occur at a corner speed of 98.5 knot. Furthermore, at the maximum speed, the load factor is unity, the turn rate is zero, and the turn radius is infinity. The maximum achievable bank angle is 68.1°.

(3.9): [latex]AR=\frac{b}{C}=\frac{bb}{Cb}=\frac{b^2}{S}[/latex]

(5.28): [latex]\Big(\frac{C_L}{C_D}\Big)_{\max}=\frac{1}{2\sqrt{KC_{D_o}}}[/latex]

(2.49): [latex]V_s=\sqrt{\frac{2W}{ ho SC_{L_{\max}}}}[/latex]

(9.13): [latex]n_{\max}=\frac{1}{\cos\phi_{\max}}[/latex]

(9.32): [latex]\omega=\frac{g\sqrt{n^2-1}}{V}[/latex]

(9.24): [latex]R=\frac{V^2}{g\sqrt{n^2-1}}[/latex]

Table 9.2 Turn performance for the aircraft in Example 9.7

No.	V (knot)	[latex]n_{\max}[/latex]	ϕ (deg)	ω (deg/s)	R (m)	Remarks
1.	98.5	2.61	67.47	26.73	108.6	Corner speed, Max turn rate, Max turn radius
2.	100	1.62	67.56	26.4	111.4
3.	110	2.668	67.99	24.5	132
4.	119.3	2.683	68.11	22.8	154.3	Max load factor, Max bank angle
5.	120	2.682	68.1	22.6	156
6.	130	2.66	67.92	20.7	185
7.	140	2.591	67.23	18.6	221
8.	150	2.464	66.05	16.4	269.6
9.	160	2.26	63.7	13.8	340.9
10.	170	1.944	59.04	10.7	467.8
11.	180	1.428	45.5	6.18	857.6
12.	185	1	0	0	∞	Max speed

Question 9.7: Consider the single-engine turboprop utility aircraft Cessna...

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