280 Solutions Manual x Fluid Mechanics, Fifth Edition 3.160 The air-cushion vehicle in Fig. P3.160 brings in sea-level standard air through a fan and discharges it at high velocity through an annular skirt of 3-cm clearance. If the vehicle weighs 50 kN, estimate (a) the required airflow rate and (b) the fan power in kW. Solution: The air inside at section 1 is nearly stagnant (V | 0) and supports the weight and also drives the flow out of the interior into the atmosphere: p1 | po1: po1 patm weight area 50,000 N S (3 m)2 Solve for Vexit | 54.2 m/s, whence Q e Then the power required by the fan is P Qe'p Fig. P3.160 1 2 U Vexit 2 A e Ve 1 2 | 1768 Pa (1.205)Vexit 2 m3 S (6)(0.03)(54.2) 30.6 s (30.6)(1768) | 54000 W Ans. 3.161 A necked-down section in a pipe flow, called a venturi, develops a low throat pressure which can aspirate fluid upward from a reservoir, as in Fig. P3.161. Using Bernoulli’s equation with no losses, derive an expression for the velocity V1 which is just sufficient to bring reservoir fluid into the throat. Fig. P3.161 Solution: Water will begin to aspirate into the throat when pa p1 Volume flow: V1 V2 (D2 /D1 )2 ; Bernoulli ('z 0): p1 Ugh. Hence: 1 1 U V12 | patm U V22 2 2 Chapter 3 x Integral Relations for a Control Volume Solve for pa p1 U 2 D2 2gh , or: V2 t D1 D4 1 (D 4 1)V 22 t U gh, D 2gh 1 (D1 /D 2 )4 D 2V2, min Similarly, V1, min 281 Ans. Ans. 3.162 Suppose you are designing a 3 u 6-ft air-hockey table, with 1/16-inch-diameter holes spaced every inch in a rectangular pattern (2592 holes total), the required jet speed from each hole is 50 ft/s. You must select an appropriate blower. Estimate the volumetric flow rate (in ft3/min) and pressure rise (in psi) required. Hint: Assume the air is stagnant in the large manifold under the table surface, and neglect frictional losses. Solution: Assume an air density of about sea-level, 0.00235 slug/ft3. Apply Bernoulli’s equation through any single hole, as in the figure: p1 'prequired p1 pa U 2 V 2jet U 2 V12 pa U 2 2 V jet , or: 0.00235 (50)2 2 2.94 lbf ft 2 0.0204 lbf in 2 Ans. The total volume flow required is 2 ft · S § 1/16 · § Q VA1hole (# of holes) ¨ 50 ¸ ¨ ft ¸ (2592 holes) s ¹ 4 © 12 © ¹ ft 3 2.76 s ft 3 166 min It wasn’t asked, but the power required would be P 8.1 ft·lbf/s, or about 11 watts. Q 'p Ans. (2.76 ft3/s)(2.94 lbf/ft2)

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# P3.160 brings in sea-level standard air