SOLUTIONS MANUAL TO ACCOMPANY
INTRODUCTION TO FLIGHT 7th Edition
By John D. Anderson, Jr.
Chapter 2
2.1 5 2 3 /(1.2)(1.0110)/(287)(300) 1.41kg/m 1/1/1.410.71m/kg ρ ρ ρ × = =pRT= v===
2.2 Mean kinetic energy of each atom 23 20 33 (1.3810)(500)1.03510J 22 =×× kT==
One kg-mole, which has a mass of 4 kg, has 6.02 × 1026 atoms. Hence 1 kg has 26 26 1 (6.0210)1.50510 4 = ×× atoms. 20 26 6
Totalinternalenergy(energyperatom)(numberofatoms) (1.03510)(1.50510)1.55810J= =´´=´
2.3 3 2116 slug 0.00237 (1716)(46059)ft ρ == = + p RT 3 Volumeoftheroom(20)(15)(8)2400ft
Total mass in the room(2400)(0.00237)5.688slug Weight(5.688)(32.2)183lb ==
2.4 3 2116 slug 0.00274 (1716)(46010) ft p RT ρ == = -
Since the volume of the room is the same, we can simply compare densities between the two problems. 3 slug 0.002740.002370.00037 ft Δρ =-= 0.00037 %change(100)15.6%increase 0.00237
2.5 First, calculate the density from the known mass and volume, 3 m 15009001.67lbft ρ //== In consistent units, 3 1.67/32.20.052slugft. ρ / == Also, T = 70 F = 70 + 460 = 530 R. Hence, (0.52)(1716)(530) ρ pRT== 2 47,290lb/ft p = or 47,290/211622.3atm p ==
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2.6 ρ p RT = npnpnRnT =++
Differentiating with respect to time, 111 ρ ρ =+ dpddT p dtdtTdt
or, T ρ ρ =+ dppdpdT dtdtdt
or, ρ ρ =+ dpddT RTR dtdtdt (1)
At the instant there is 1000 lbm of air in the tank, the density is 3 m 1000/9001.11lb/ft
Also, in consistent units, is given that T = 50 + 460 = 510 R and that 1/min1/min0.016/sec === dT FRR dt
From the given pumping rate, and the fact that the volume of the tank is 900 ft3, we also have
m m 3 0.5lb/sec 0.000556lb/(ft)(sec) 900ft ρ == d dt 53 0.000556 1.7310slug/(ft)(sec) 32.2 ρ ==× d dt
Thus, from equation (1) above, 5 2 (1716)(510)(1.7310)(0.0345)(1716)(0.0167) 16.1 15.10.9916.1lb/(ft)(sec) 2116 0.0076atm/sec ρ =×+ =+== = d dt
2.7 In consistent units, 10273263K=−+= T
Thus,
pRT
2.8 53 3 /0.510/(287)(240)0.726kg/m 1/1/0.7261.38m/kg ρ ρ ==×= === pRT v
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23 0 Forceduetopressure (211610) [21165]6303lb perpendicular to wall.
to shear stress= τ (9) [180(9)]623.554083.5lbtangentialtowall.
Magnitude of the resultant aerodynamic force = 22 (6303)(835)6303.6lb
2.10 3 sin 2 θ ∞VV =
Minimum velocity occurs when sin θ = 0, i.e., when θ = 0° and 180°. Vmin = 0 at θ = 0° and 180°, i.e., at its most forward and rearward points.
Maximum velocity occurs when sin θ = 1, i.e., when θ = 90°. Hence, max 3 (85)(1)127.5mphat90, 2 θ ° V= == i.e., the entire rim of the sphere in a plane perpendicular to the freestream direction.
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2.11 The mass of air displaced is
The weight of this air is 3 air (5.2310)(32.2)0.168lb W
This is the lifting force on the balloon due to the outside air. However, the helium inside the balloon has weight, acting in the downward direction. The weight of the helium is less than that of air by the ratio of the molecular weights 4 (0.168)0.0233lb. 28.8 c WH
Hence, the maximum weight that can be lifted by the balloon is 0.168 0.0233 = 0.145 lb.
2.12 Let p3, ρ3, and T3 denote the conditions at the beginning of combustion, and p4, ρ4, and T4 denote conditions at the end of combustion. Since the volume is constant, and the mass of the gas is constant, then p4 = ρ3 = 11.3 kg/m3. Thus, from the equation of state, 72 444 (11.3)(287)(4000)1.310N/m ρ pRT===´ or, 7 4 5 1.310
2.13 The area of the piston face, where the diameter is 9 cm = 0.09 m, is
(a) The pressure of the gas mixture at the beginning of combustion is
The
the piston
the piston
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2.14 Let p3 and T3 denote conditions at the inlet to the combustor, and T4 denote the temperature at the exit. Note: 62 34 410N/m pp==´
(a) 6 3 3 3 3 410 =15.49kg/m (287)(900) ρ p RT ´ ==
(b) 6 3 4 4 4 410 =9.29kg/m (287)(1500) ρ p RT ´ ==
2.15 1 mile = 5280 ft, and 1 hour = 3600 sec. So: miles5280ft1hour 6088 ft/sec. hourmile3600 sec
A very useful conversion to remember is that 60mph88ft/sec = also, 1 ft = 0.3048 m ft0.3048mm 8826.82 sec1ftsec
Thus ftm 8826.82 secsec =
2.16 miles88 ft/sec 6921015 ft/sec hour60 mph
miles26.82 m/sec 692309.3 m/sec hour60 mph
2.17 On the front face
On the back face
55 (1.071510)(2)2.14310 N ff FpA==×=×
55 (1.0110)(2)2.0210 N bb FpA==×=×
The net force on the plate is
55 (2.1432.02)100.12310 N fb FFF=−=−×=×
From Appendix C, 1 4.448 N. f lb =
So, 0.123105 2765 lb 4.448 F × ==
This force acts in the same direction as the flow (i.e., it is aerodynamic drag.)
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
2.18
2.21 From Fig. 2.16, length
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.