Question

A hovercraft, shown below, operates by using a fan to fill a skirt of high hovers. Consider a round hovercraft weighing 4000 lbm with a diameter of skirt of the hovercraft, and an equal amount of air must be supplied by skirt and the ground is 0.05 inches. Take p = 0.07518 lbm/ft. = 1.227x (a) What is the air pressure under the hovercraft in psig? (b) What is the volumetric flow rate through the (c) If the fan is 80% efficient, what is the power requirement of fan by using a fan to fill a skirt of high pressure air on which the craft bm with a diameter of 8 ft Air escapes under the equal amount of air must be supplied by the fan. The gap between the P = 0.07518 lbm/it'. = 1.227 x 10 Ibm/fts. metric flow rate through the skirt in ft/s? You will need to find v first!) cient, what is the power requirement of the fan in hp? skirt ground

Answer 1

given ground. 4000 lbm The hovercraft weighs the hovercraft = 8ft The diameter of in psig. under hovercraft (2) Pressure egn. (Am) Force balance air pressure to hovercraft = force applied by to lift it weig mg = PA 2 (m= mass of hovercraft (1 bm) g - acceleration due to gravity upthrust weight of due to pressure A area of skire hovercraft Wow As I (da). (84)50,24 12

. .. p- mg 4000 lbmx 32.17 ft S2 50.24 ft? 50.24842 = 17.78 lbl pe 2561.30 lbf IE2 or|P> 17.78 porn or P= 17.78 psig! through the air flow of b) Find volumetric the okirt in ft3/s. Now P. = // PV₂²

lbl Ibm x ve = 0.07518 0.0751 / 2561.30 3 E d=2 f should be the units correct now to (Inlug = 32.17 Ibm) in alugs/ft3 HE3, Ps 0.07518 lbm - 0.00233 ning ..VD 1280]" v>: [2x 2510 me 2 x 2561. 30 0.00233 slug /tt3 ve 1482.74 ft/s

(6) : The volumchic flowrate Q = AV where A = area between the skirto the ground. A: h = TDXL gap between skirt and ground - diameter of hovercraft A T inich (88t) (0.05 inches) x 10.0833 ft) A = 0.1046 Jt2. 1482.74 It Iln Q = Av = 0.1046 &t ² x Sa. 155.13 NS power requirement. c) Fan is 80% efficient. Find eqn or An) using extended the mechanical energy Bernoulli ean between points 1,2

pe + V, 2 +93 in + Wohaft - loss a +982 l 29 pe pressure at respetive points v= velocity of flow e = density of fluid z = elevation added to air by an Wshaft - lon = amt of work Now P, chmo» puri Pчиш •o V=0, 3₂-3, =0, P2 : 2561.30 lb L2 ft/s. V = 1482.74 Find Wrhaft - lon. ² : 2561.30 lb it + P2 + 1482.74 stje Wnhaft-lom = 0.00233 slug fts 21.98 XL tta sa (5)

ww (lbs) (pt) Wshaft -lon now efficiency na wa wow me ? I wonaft P: power of fan and Wohaft Wihaft mi= mass flow rate Wshaft - lon Wnhaft = 21.98 xior J6²/2. 2018 Xv0*120) 0.80 Wanuft ; 27.48 x1072 and m= PAV = la 155.13 It 0.00233 nlug x Ats min 0.3614 plug

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