Solution Manual For Gas Dynamics, 3rd Edition

From Gas Dynamics, Third Edition, by James E. John and Theo G. Keith. ISBN 0-13-120668-0. © 2006 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No Portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Chapter Two WAVE PROPAGATION IN COMPRESSILBE MEDIA Problem 1. – Using the expansion wave and control volume depicted in Figs. 2.8 and 2.9 along with the continuity and momentum equations, rederive Eq. (2.4). moving wave dV p – dp ρ – dρ dV a gas at rest dV moving wave p – dp ρ – dρ dV a gas at rest Continuity equation (ρ − dρ)(a + dV )A − ρaA = 0 Expand, neglect products of derivatives and simplify to get ρdV − adρ = 0 (1) Momentum equation pA − (p − dp )A = ρaA[(a + dV ) − a ] or dp = ρadV (2) Combining Eqs. (1) and (2) gives dp = a 2 dρ Since the process is reversible and adiabatic, i.e., isentropic, this can be written as: ⎛ ∂p ⎞ a = ⎜⎜ ⎟⎟ ⎝ ∂ρ ⎠ s 17 From Gas Dynamics, Third Edition, by James E. John and Theo G. Keith. ISBN 0-13-120668-0. © 2006 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No Portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 2. – (a) Derive an expression for ks, for a perfect gas, substitute the result into Eq. (2.10), and thereby demonstrate Eq. (2.7); (b) Derive an expression for kT, for a perfect gas, substitute the result into Eq. (2.11), and thereby demonstrate Eq. (2.7) and finally; (c) Derive an expression for βs, for a perfect gas, substitute your result into Eq. (2.14), and thereby demonstrate Eq. (2.7). (a) ks = 1 ⎛ ∂ρ ⎞ ⎜ ⎟ ρ ⎜⎝ ∂p ⎟⎠ s An isotropic process involving a perfect gas is described by P = cρ γ ∴ dp γcρ γ γ p = = γcρ γ −1 = dρ ρ ρ Hence, ⎛ ∂ρ ⎞ ρ ⎜⎜ ⎟⎟ = ⎝ ∂p ⎠ s γ p ks = 1 ⎛ ∂ρ ⎞ 1 1 ⎜⎜ ⎟⎟ = = ρ ⎝ ∂p ⎠ s γ p γ ρ RT So, a= (b) kT = ρ= 1 = γRT ρk s 1 ⎛ ∂ρ ⎞ ⎜ ⎟ ρ ⎜⎝ ∂p ⎟⎠ T p RT ⎛ ∂ρ ⎞ 1 ⎜⎜ ⎟⎟ = ⎝ ∂p ⎠ T RT kT = 1 ρRT So, 18 From Gas Dynamics, Third Edition, by James E. John and Theo G. Keith. ISBN 0-13-120668-0. © 2006 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No Portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. a= (c) γ = γRT ρk T ⎛ γp⎞ ⎛ ∂p ⎞ βs = ρ⎜⎜ ⎟⎟ = ρ⎜⎜ ⎟⎟ = γ p = γρRT ⎝ ρ ⎠ ⎝ ∂ρ ⎠ s a= βs = γRT ρ Problem 3. – Use dimensional analysis to develop an expression for the speed of sound in terms of the isentropic compressibility, the density and gc. a = f (k s , ρ, g e ) a~ L , T ks ~ ML L2 M , gc ~ , ρ~ 3 F FT 2 L a b c L ⎛⎜ L2 ⎞⎟ ⎛ M ⎞ ⎛ ML ⎞ ⎜⎜ ⎟⎟ ⎜⎜ ⎟⎟ = F −a −c M b+c L2a −3b+c T −2c = 3 2 ⎟ ⎜ T ⎝ F ⎠ ⎝ L ⎠ ⎝ FT ⎠ F : −a − c = 0 M:b+c = 0 L : 2a − 3b + c = 1 T : −2c = 1 1 1 a=− Hence, c = 2 2 b=− 1 2 So, a= gc ρk s Problem 4. – Using the data provided in Tables 2-1, 2-2 and 2-3, i.e., the density, and the isentropic compressibility or the bulk modulus, calculate the velocity of sound at 20°C and one atmosphere pressure in (a) helium, (b) turpentine, and (c) lead. (a) Helium: ρ = 0.16 kg m 3 , k s = 5,919 1 GPa 19 From Gas Dynamics, Third Edition, by James E. John and Theo G. Keith. ISBN 0-13-120668-0. © 2006 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No Portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. a= (b) Turpentine: ρ = 870 kg m 3 , k s = 0.736 a= (c) Lead: ρ = 11,300 kg m3 1 109 = = 1027.6 m / s (0.16)(5,919) ρk s 1 GPa 1 = ρk s 10 9 = 1249.7 m / s (870)(0.736) , β s = 16.27 GPa a= βs = ρ (16.27) 10 9 = 1199.9 m / s 11,300 Problem 5. – In Example Problem 2.3 the speed of sound of superheated steam was determined by using a finite difference representation of the compressibility and steam table data (Table 2-4). Using the same steam table data, determine the speed of sound of superheated steam for the same pressure and temperature, i.e., at p = 500 kPa and T = 300˚C. However, use the following finite differences to obtain two estimates for the speed of sound: γ γ a2 = = ⎛ ∂ρ ⎞ ⎡ ∂ (1 v ) ⎤ ⎜⎜ ⎟⎟ ⎢ ⎥ ⎝ ∂p ⎠ T ⎣ ∂p ⎦ T a2 = a2 = 1 1 = ⎛ ∂ρ ⎞ ⎡ ∂(1 v )⎤ ⎜⎜ ⎟⎟ ⎢ ⎥ ⎝ ∂p ⎠ s ⎣ ∂p ⎦ s γ γ 2γ∆p = = 1 1 1 1 ⎡ ∂ (1 v ) ⎤ − − ⎢ ∂p ⎥ v(p + ∆p, T ) v(p − ∆p, T ) v(p + ∆p, T ) v(p − ∆p, T ) ⎣ ⎦T 2∆p From Example 2.3 v(p + ∆p1T ) = 0.4344 M3 , kg v(p − ∆p1T ) = 0.6548 M3 , and ∆p = 100,000 Pa kg 20 From Gas Dynamics, Third Edition, by James E. John and Theo G. Keith. ISBN 0-13-120668-0. © 2006 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No Portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. (2)(1.327 )(100,000) = 342,521.5 m 2 a2 = 1 1 − 0.4344 0.6548 s2 a = 585.3 m / s a2 = 1 2∆p = 1 1 ⎡ ∂ (1 v ) ⎤ − ⎢ ∂p ⎥ ⎣ ⎦ s v(p + ∆p, s ) v(p − ∆p, s ) From Example 2.3 v(p + ∆p, s ) = 0.4544 M3 M3 , v(p − ∆p, s ) = 0.6209 and ∆p = 100,000 Pa kg kg a2 = (2)(100,000) 1 1 − 0.4544 0.6209 = 338,903.2 m2 s2 a = 582.2 m / s Problem 6. – Equation (2.16) provides a convenient expression for calculating the speed of sound in air: a = 20.05 T , where T is the absolute temperature in degrees Kelvin. Derive the following linear equation for the speed of sound in air: a = a 0 + 0.6t where a0 is the speed of sound in air at 0°C and t is °C. To accomplish this make use of Eq. (2-16) and the expansion (x + y )n = x n + nx n −1y + ….. a = γRT = [γR (273 + t )]1 / 2 1/ 2 t ⎞ ⎛ = γR (273) ⎜1 + ⎟ ⎝ 273 ⎠ ⎛ 1 t ⎞ = a o ⎜1 + + …..⎟ ⎝ 2 273 ⎠ 21 From Gas Dynamics, Third Edition, by James E. John and Theo G. Keith. ISBN 0-13-120668-0. © 2006 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No Portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. a o = γR (273) = 20.05 273 = 331 m , s ao 331 = = 0.6 (2)(273) 546 ∴ a = 331 + 0.6t Problem 7. – Rather than measure the bulk modulus directly it may be easier to measure the speed of sound as it propagates though a material and then use it to compute the bulk modulus. For a Lucite plastic of density 1,200 kg/m3, the speed of sound is measured as 2,327 m/s. Determine the bulk modulus. What is the corresponding isentropic compressibility? kg m , a = 2,327 Now ρ = 1,200 3 s m a= βs ρ 2 so, kg ⎞⎛ m ⎞ ⎛ N ⋅ s2 ⎞ ⎛ ⎟⎟ = 6.498 × 109 Pa = 6.498 GPa 2 , 327 β3 = ρa 2 = ⎜1,200 ⎟⎜ ⎟ ⎜⎜1 ms ⎠⎝ s ⎠ ⎝ kg ⋅ m ⎠ ⎝ ks = 1 1 = 0.1539 GPa βs Problem 8. – An object of diameter d (m) is rotated in air at a speed of N revolutions per minute. Draw a plot of the rotational speed required for the velocity at the outer edge of the object to just reach sonic velocity for a given diameter. Take the speed of sound of the air to be 331m/s. The highest speed will occur at R. 1 min m ⎛ rev ⎞⎛ 2πrad ⎞ V = N⎜ = a = 331 ⎟⎜ ⎟R (m ) 60 s s ⎝ min ⎠⎝ rev ⎠ π m = ND, 60 s 22 From Gas Dynamics, Third Edition, by James E. John and Theo G. Keith. ISBN 0-13-120668-0. © 2006 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No Portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. The following is a log base 10 plot of N = 6,321.6/D. Rotational Speed, RPM 5.6 5.4 5.2 5.0 4.8 4.6 4.4 4.2 4.0 3.8 3.6 Supersonic Region Subsonic Region 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 Diameter, m Problem 9. – (a) Newton assumed that the sound wave process was isothermal rather than isentropic. Determine the size of error made in computing the speed of sound by making this assumption. (b) A flash of lightening occurs in the distance. 20 seconds later the sound of thunder is heard. The temperature in the area is 23°C. How far away was the lightening strike? (a) as = 1 ρk s aT = ⎞ a T − a s ⎛⎜ 1 = − 1⎟ 100 ⎜ γ ⎟ as ⎝ ⎠ (b) 1 ρk T ∴ aT = as for γ = 1.4 ks kT aT − as = −15.5% as L = a∆t = (344.86 )(20 ) = 6,897.2 m Problem 10. – (a) The pressure increase across a compression pulse moving into still air at 1 atmosphere pressure and 30°C is 100 Pa. Determine the velocity following the pulse. (b) The velocity changes by 0.1 m/s across a pressure wave that moves into hydrogen gas that is at rest at a pressure of 100 kPa and temperature 300K. Determine the pressure behind the wave. Use Eq (2.2) and write the expression in difference form as 23 From Gas Dynamics, Third Edition, by James E. John and Theo G. Keith. ISBN 0-13-120668-0. © 2006 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No Portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. (a) air: ∆p , ∆p = 100 Pa ρa 101,000 kg ρ= = 1.1615 ⎛ 8314 ⎞ m3 ⎜ ⎟(303) ⎝ 28.97 ⎠ ∆V = a = 20.05 303 = 349.0 m/s Therefore, (b) ∆V = 100 = 0.247 m/s (1.1615)(349.0) ∆p = ρa∆V , hydrogen: ρ= a= Therefore, ∆V = 0.1 m/s p 100,000 kg = 0.0808 = RT ⎛ 8314 ⎞ m3 ⎜ ⎟(300 ) ⎝ 2.016 ⎠ (1.41)⎛⎜ 8314 ⎞⎟(300) = 1320.8 m/s ⎝ 1.016 ⎠ ∆p = (0.0808)(1320.8)(0.1) = 10.68 Pa Problem 11. – (a) Helium at 35°C is flowing at a Mach number of 1.5. Find the velocity and determine the local Mach angle. (b) Determine the velocity of air at 40°C to produce a Mach angle of 38° (a) helium: T = 35°C = 308K M = 1.5 V = aM a = γRT = (1.667)⎛⎜ 8,314 ⎞⎟(308) = 1,032.7 m/s ⎝ 4,003 ⎠ V = (1,032.7 )(1.5) = 1,549.0 m/s ⎛1⎞ µ = sin −1 ⎜⎜ ⎟⎟ = 41.8° ⎝µ⎠ (b) air: T = 40°C = 313K a = 20.05 223.3 = 299.6 m/s 24 From Gas Dynamics, Third Edition, by James E. John and Theo G. Keith. ISBN 0-13-120668-0. © 2006 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No Portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. ⎛1⎞ µ = sin −1 ⎜ ⎟ ⎝M⎠ M= 1 V = sinµ a V= a 354.6 = = 576.0 m/s sinµ sin (38) Problem 12. – (a) A jet plane is traveling at Mach 1.8 at an altitude of 10 km where the temperature is 223.3K. Determine the speed of the plane. (b) Air at 320 K flows in a supersonic wind tunnel over a 2-D wedge. From a photograph the Mach angle is measured to be 45°. Determine the flow velocity, the local speed of sound and the Mach number of the tunnel. (a) M = 1.8 , T = 223.3 K , a = 20.05 223.3 = 299.6 m/s V = aM = (299.6 )(1.8) = 539.3 m/s (b) air: T = 320 K, µ = 45° , a = 20.05 320 = 358.7 m/s V= a 358.7 = = 507.2 m/s sinµ sin (45) M= V 1 = = 1.414 a sin µ Problem 13. – A supersonic aircraft, flying horizontally a distance H above the earth, passes overhead. ∆t later the sound wave from the aircraft is heard. In this time increment, the plane has traveled a distance L. Show that the Mach number of the aircraft can be computed from: 2 2 ⎛ V∆t ⎞ ⎛L⎞ M = ⎜ ⎟ +1 = ⎜ ⎟ +1 ⎝H⎠ ⎝ H ⎠ Hint: first show that the Mach angle µ can be expressed as tan −1 ⎛⎜1 ⎝ connect the Mach angle, µ, to the geometric parameters H and L. 1 sin µ = µ M H M 2 − 1 ⎞⎟ and then ⎠ µ L 25 From Gas Dynamics, Third Edition, by James E. John and Theo G. Keith. ISBN 0-13-120668-0. © 2006 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No Portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 1 tan µ = M2 −1 = H L M 1 µ 2 ⎛L⎞ ∴ M = ⎜ ⎟ +1 ⎝H⎠ 2 M -1 but L = V∆t 2 ⎛ V∆t ⎞ = ⎜ ⎟ +1 ⎝ H ⎠ Problem 14. – Given speeds and temperatures, determine the corresponding Mach numbers of the following (note: 1 mile = 5,280 ft = 1,609.3 m; 1 mi/hr = 1.6093 km/hr = 0.447 m/s): (a) A cheetah running at top speed of 60 mi/hr; the local temperature is 40°C (b) A Peregrine falcon in a dive at 217 mi/hr; local temperature of 25°C (c) In June 1999 in Athens Greece, Maurice Greene became the world’s fastest human by running 100 m in 9.79 s; the temperature was 21°C (d) In June 1999, Alexander Popov became the world’s fastest swimmer by swimming 50 m in 21.64s; the temperature of the water was 20°C (a) a = 20.05 313 = 354.7 m/s mi m/s ( 60 ) (0.447 ) V hr mi/hr = 0.076 M= = (354.7 ) m/s a (b) a = 20.05 298 = 346.1 m/s V (217 )(0.447 ) = = 0.28 (346.1) a M= (c) a = 20.05 294 = 343.8 m/s M= mi ⎞ ⎛ 10.21 = 22.9 ⎟ ⎜= hr ⎠ ⎝ 0.447 50 = 2.31 m/s 21.64 mi ⎞ ⎛ 2.31 = 5.17 ⎟ ⎜= hr ⎠ ⎝ 0.447 10.21 = 0.03 343.8 (d) a = 1,481 m/s (from Table 2 – 2) V = M= 100 = 10.21 m/s 9.79 V= 2.31 = 0.00156 1,481 26 From Gas Dynamics, Third Edition, by James E. John and Theo G. Keith. ISBN 0-13-120668-0. © 2006 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No Portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 15. – Given speeds and Mach numbers, assuming air is a perfect gas, determine the corresponding local temperature (note: 1 mi/hr = 0.447 m/s) for the following: (a) A Boeing 747-400 at a cruise speed of 910 km/hr; M = 0.85. (b) Concorde at a cruise speed of 1,320 mi/hr; M = 2.0 (c) The fastest airplane, the Lockheed SR-71 Blackbird, flying at 2,200 mi/hr; M = 3.3 (d) The fastest boat, the Spirit of Australia, that averaged 317.6 mi/hr; M = 0.41 (e) The fastest car, the ThrustSSC, averaged 760.035 mi/hr; M = 0.97 (a) V= 910,000m m = 252.8 3600s s M = 0.85 2 a= V 252.8 m = = 297.4 M .85 s 2 ⎛ a ⎞ ⎛ 297.4 ⎞ T=⎜ ⎟ =⎜ ⎟ = 220°K = −53°C ⎝ 20.05 ⎠ ⎝ 20.05 ⎠ (b) V = (1320)(0.447 ) = 590.0 m s 2 M = 2 .0 a= V = 295.0 m/s M 2 ⎛ 295 ⎞ ⎛ a ⎞ T=⎜ ⎟ = 216.5K = −56.5°C ⎟ =⎜ ⎝ 20.05 ⎠ ⎝ 20.05 ⎠ (3) V = (2200)(0.447 ) = 983.4 m s 2 M = 3 .3 a= 983.4 = 298.0 m/s 3.3 2 ⎛ a ⎞ ⎛ 298.0 ⎞ T=⎜ ⎟ =⎜ ⎟ = 220.9K = −52.1°C ⎝ 20.05 ⎠ ⎝ 20.05 ⎠ (d) V = (317.6 )(0.447 ) = 142.0 2 m s M = 0.41 a= 142 = 346.3 m/s .41 2 ⎛ 346.3 ⎞ ⎛ a ⎞ T=⎜ ⎟ = 298.2K = 25.2°C ⎟ =⎜ ⎝ 20.05 ⎠ ⎝ 20.05 ⎠ (e) V = (760.035)(0.447 ) = 339.7 m/s 2 M = 0.97 a= 339.7 = 350.2 m/s .97 2 ⎛ 350.2 ⎞ ⎛ a ⎞ T=⎜ ⎟ = 305.1K = 32.1°C ⎟ =⎜ ⎝ 20.05 ⎠ ⎝ 20.05 ⎠ 27 From Gas Dynamics, Third Edition, by James E. John and Theo G. Keith. ISBN 0-13-120668-0. © 2006 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No Portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 16. – A baseball, which has a mass of 145 grams and a diameter of 3.66 cm, when dropped from a very tall building reaches high speeds. If the building is tall enough the speed will be controlled by the drag, as the baseball will reach terminal speed. At this state W = FD Where W (weight) = mg, g (acceleration of gravity) = 9.81 m/s2, FD (drag force) = CDρairAV2/2, CD (drag coefficient) = 0.5 and A (projected area of sphere) = πR2. Find the terminal speed of the baseball and determine the corresponding Mach number if the ambient air temperature is 23°C and the ambient air pressure is 101 kPa.. The density of the air is first determined: ρ air = p 101 = = 1.19kg / m 3 RT (0.287 )(296) Now W = mg = FD = C D Aρ air V 2 2 Hence, V= 2mg = C D ρ air A M= V = a 2(0.145)(9.81) = 33.76 m / s (0.5)(1.19)(0.0042) 33.76 (1.4)(287 )(296) = 0.098 Problem 17. – Derive the following equation for the speed of sound of a real gas from Berthelot’s equation of state: p= ρRT αρ 2 − 1 − βρ T ⎡ RT RTρβ 2αρ ⎤ − a = γ⎢ + ⎥ T ⎥⎦ ⎢⎣1 − βρ (1 − βρ )2 28 From Gas Dynamics, Third Edition, by James E. John and Theo G. Keith. ISBN 0-13-120668-0. © 2006 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No Portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Since T is treated as a constant, we may simply use information from Section 2.6 where ⎛ ∂p ⎞ a = γ⎜⎜ ⎟⎟ ⎝ ∂ρ ⎠ T ⎛ ∂p ⎞ RT RTρβ ⎜⎜ ⎟⎟ = + − 2αρ ⎝ ∂ρ ⎠ T 1 − βρ (1 − βρ )2 Now replace α with α/T. Thus, from Eq. (2.24) ⎡ RT RTρβ 2αρ ⎤ a = γ⎢ + − ⎥ T ⎥⎦ ⎢⎣1 − βρ (1 − βρ )2 Problem 18. – Using the speed of sound expression from the previous problem and the following constants for nitrogen R = 296.82 (N·m)/(kg·K) α = 21,972.68 N·m4/kg2 β = 0.001378 m3/kg γ = 1.4 determine the speed of sound for the two cases described in Example 2.4. Case (1) p 0.3 MPa and T = 300K Iteration 1 2 3 v 0.296823 0.297386 0.297384 f(v) df /dv v-f/(df/dv) -4.9286E-05 8.7530E-02 0.297386 1.8796E-07 8.8198E-02 0.297384 2.6975E-12 8.8195E-02 0.297384 ρ 3.3690 3.3626 3.3627 a 353.7517 353.7505 353.7505 The result differs from the experimental value 353.47 m/s by 0.08%. Case (2): p 30.0 MPa and T = 300K 29 From Gas Dynamics, Third Edition, by James E. John and Theo G. Keith. ISBN 0-13-120668-0. © 2006 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No Portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Iteration 1 2 3 4 5 6 7 v 0.002968 0.005658 0.004594 0.004088 0.003953 0.003944 0.003944 f(v) df /dv v-f/(df/dv) ρ -8.2594E-09 3.0708E-06 0.005658 336.9016 5.2436E-08 4.9296E-05 0.004594 176.7430 1.3084E-08 2.5826E-05 0.004088 217.6647 2.2920E-09 1.7035E-05 0.003953 244.6426 1.4088E-10 1.4959E-05 0.003944 252.9695 6.6552E-13 1.4817E-05 0.003944 253.5736 1.5099E-17 1.4817E-05 0.003944 253.5765 a 604.3973 426.1798 457.9898 483.2795 491.8702 492.5088 492.5118 The result differs from the experimental value 483.18 m/s by 1.9%. Problem 19. –Employ the finite difference method of Example 2.5 to determine the speed of sound in nitrogen using the Redlich-Kwong equation of state p= a oρ 2 RTρ − 1 − βρ (1 + βρ ) T where for nitrogen: R = 296.823 (N·m)/(kg·K) ao = 1979.453 (N·m4·√K )/(kg2) β = 0.0009557 m3/kg γ = 1.4 Compute the speed at a pressure of 30.1 MPa and a temperature of 300 K. Experimental values of the speed of sound of nitrogen may be found in Ref. (11). For the given conditions the measured value is 483.730 m/s. The Redlich-Kwong equation of state is: p = ao RT − . Rearrange to obtain: v − β v (v + β ) T ao ⎞ a β ⎛ RT ⎞ 2 ⎛ 2 RTβ ⎟v − o = 0 ⎟⎟ v − ⎜ β + f (v ) = v 3 − ⎜⎜ − ⎟ ⎜ p p T⎠ p T ⎝ p ⎠ ⎝ ⎛ ao ⎞ ⎛ RT ⎞ df RTβ ⎟ ⎟⎟ v − ⎜ β 2 + − = 3v 2 − 2⎜⎜ ⎟ ⎜ dv p p T ⎝ p ⎠ ⎠ ⎝ Use Newton-Raphson to find v = 0.003279 m3/kg. Thus, ρ = 304.9917 kg/m3. Use ∆ρ = 0.1 and compute 30 From Gas Dynamics, Third Edition, by James E. John and Theo G. Keith. ISBN 0-13-120668-0. © 2006 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No Portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. p(ρ+∆ρ,T) = p(305.0917,300) = 30,112,951.62 Pa p(ρ−∆ρ,T) = p(304.8917,300) = 30,087,052.10 Pa a= γ ∆p m = 425.79 ∆ρ s The result is 12% too small compared to the experimental value of 483.73m/s. However, if a more appropriate value of γ at this pressure and temperature is used, i.e., γ = 1.704, a = 469.75m/s, which is in error by only 2.9%. 31