Question 10.26: A hot-air balloon with a diameter of 15 ft, height of 13 ft,...

(a) In order to determine the lift force, Equation 10.28 \sum\limits_{z}{F_{z}} = F_{L} + F_{B} – W – F_{D} = ma_{z} = 0 (Newton’s second law of motion) is applied in the vertical direction. The frontal area is used to compute the drag force for the hot-air balloon, and the drag force is determined by applying Equation 10.12 F_{D} = C_{D} \frac{1}{2} \rho v^{2}A as follows:

D_{b}: = 15 ft                               A_{frontb}: = \frac{\pi . D_{b}^{2}}{4} = 176.715 ft^{2}                               h_{b}: = 13 ft                               C_{Db}: = 1.3

V: = 25 \frac{ft}{sec}                               b_{c}: = 4 ft                               d_{c}: = 4 ft                               h_{c}: = 4 ft                               \gamma _{air}: = 0.056424 \frac{lb}{ft^{3}}

W_{b}: = 200 lb                               W_{c}: = 300 lb                               W_{bc}: = W_{b} + W_{c} = 500 lb

Guess value:                               F_{B}: = 100 lb                               F_{D}: = 100 lb                               F_{L}: = 100 lb

Given

F_{L} + F_{B} – W_{bc} – F_{D} = 0                               F_{B} = \gamma _{air} . V_{bc}                        F_{D} = C_{Db} \frac{1}{2} \rho _{air} .V^{2} . A_{frontb}
\left ( \begin{matrix} F_{B} \\ F_{D} \\ F_{L} \end{matrix} \right ) : =Find ( F_{B},F_{D}, F_{L})

F_{B} = 109.022 lb                               F_{D} = 170.631 lb                               F_{L} = 561.609 lb

(b) The planform area of the hot-air balloon is used to model the lift force for the hotair balloon, and the lift coefficient is determined by applying the lift force equation, Equation 10.21 F_{L} = C_{L} \frac{1}{2} \rho.V^{2} . A, as follows:

slug: = 1 lb \frac{sec^{2}}{ft}                               \rho _{air} : = 0.0017555 \frac{slug}{ft^{3}}                               \mu _{air}: = 0.35343 \times 10^{-6} lb \frac{sec}{ft^{2}}

Guess value:                               R : = 10000                               C_{L} : = 1.0

F_{L} = C_{L} \frac{1}{2} \rho _{air} .V^{2} . Aplanc                               R = \frac{\rho _{air} .V .D_{b}}{\mu _{air}}
\left ( \begin{matrix} R \\ C_{L} \end{matrix} \right ) : = Find (R, C_{L})
R = 1.863 \times 10^{6}                              C_{L} = 5.793

Since R = 1.863 \times 10^{6} , this confirms the assumption that R > 10,000 (see Table 10.6). Furthermore, it is important to note that because such simplistic assumptions (for instance, the planform area of the hot-air balloon in the lift equation) have been made in the estimation and modeling of the lift force, F_{L} , the estimated value of the lift coefficient, C_{L} is a rough estimate at best.

The Drag Coefficient, C_{D} for Three-Dimensional Bodies at Given Orientation to the Direction of Flow, v for R > 10,000
			C_{D} = 1.3			ν (m/s)	C_{D}
						10	1.0 – 1.4
						20	0.8 – 1.2
						30	0.5 – 0.9
						40	0.3 – 0.7
Parachute, A = \pi D^{2}/4				Tree, A= frontal area
			C_{D} = 1.0 – 1.4			Position	C_{D} A ( ft^{2} )
						Standing	9
						Sitting	6
						Crouching	2.5
High-rise buildings, A = frontal area				Average person
	Position	A ( ft^{2} )	C_{D}				C_{D} = 0.4
	Upright	5.5	1.1
	Racing	3.9	0.88
	Drafting	3.9	0.5
	Streamlined	5.0	0.12
Bikes				Large birds, A = frontal area
			Without deflector C_{D} = 0.96				Minivan: C_{D} = 0.4
			With deflector: C_{D} = 0.76				Passengercar or sports car: C_{D} = 0.3
Tractor-trailer truck, A= frontal area				Automotive, A = frontal area