Chapter9_SM - Chapter 9 Compressible Flow 9.1 An ideal gas...

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Chapter 9 Compressible Flow 9.1 An ideal gas flows adiabatically through a duct. At section 1, p 1 = 140 kPa, T 1 = 260 ° C, and V 1 = 75 m/s. Farther downstream, p 2 = 30 kPa and T 2 = 207 ° C. Calculate V 2 in m/s and s 2 s 1 in J/(kg K) if the gas is (a) air, k = 1.4, and (b) argon, k = 1.67. Fig. P9.1 Solution: (a) For air, take k = 1.40, R = 287 J/kg K, and cp = 1005 J/kg K. The adiabatic steady-flow energy equation (9.23) is used to compute the downstream velocity: += = + = + 22 2 p 2 11 1 c T V constant 1005(260) (75) 1005(207) V or . 2 Ans 2 m V 335 s 21 p 2 1 2 1 207 273 30 Meanwhile, s s c ln(T /T ) R ln(p /p ) 1005ln 287ln , 260 273 140 + ⎛⎞ −= = ⎜⎟ ⎝⎠ + or s2 s1 = 105 + 442 337 J/kg K Ans . (a) (b) For argon, take k = 1.67, R = 208 J/kg K, and cp = 518 J/kg K. Repeat part (a): 2 p2 1 c T V 518(260) (75) 518(207) V , solve . 2 Ans + = + 2 m V 246 s = + = −+ ≈ + 207 273 30 s s 518ln 208ln 54 320 . (b) 260 273 140 Ans 266 J/kg K 9.2 Solve Prob. 9.1 if the gas is steam. Use two approaches: (a) an ideal gas from Table A.4; and (b) real steam from the steam tables [15]. Solution: For steam, take k = 1.33, R = 461 J/kg K, and cp = 1858 J/kg K. Then 2 1 c T V 1858(260) (75) 1858(207) V , solve . (a) 2 Ans + = + 2 m V 450 s + = − + ≈ + 207 273 30 s s 1858ln 461ln 195 710 . (a) 260 273 140 Ans 515 J/kg K
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636 Solutions Manual Fluid Mechanics, Fifth Edition (b) For real steam, we look up each enthalpy and entropy in the Steam Tables: 1 J at 140 kPa and 260 C, read h 2.993E6 ; °= 2 kg J at 30 kPa and 207 C, h 2.893E6 kg += + = + 22 2 2 11 1 Then h V 2.993E6 (75) 2.893E6 V , solve . (b) 2 Ans 2 m V 453 s 12 JJ at 140 kPa and 260 C, read s 7915 , at 30 kPa and 207 C, s 8427 kg K kg K ° = −= 21 Thus s s 8427 7915 . (b) Ans 512 J/kg K These are within ± 1% of the ideal gas estimates (a). Steam is nearly ideal in this range. 9.3 If 8 kg of oxygen in a closed tank at 200 ° C and 300 kPa is heated until the pressure rises to 400 kPa, calculate (a) the new temperature; (b) the total heat transfer; and (c) the change in entropy. Solution: For oxygen, take k = 1.40, R = 260 J/kg K, and cv = 650 J/kg K. Then ρρ ⎛⎞ =∴ = =+ = ⎜⎟ ⎝⎠ 2 1 400 , T T (p /p ) (200 273) 631 K . (a) 300 Ans 358 C ° v Q mc T (8)(650)(358 200) . (b) Ans = 8.2E5 J + = + v 2 1 358 273 s s mc ln(T /T ) (8)(650)ln . (c) 200 273 Ans J 1500 K 9.4 Compressibility becomes important when the Mach number > 0.3. How fast can a two-dimensional cylinder travel in sea-level standard air before compressibility becomes important somewhere in its vicinity? Solution: For sea-level air, T = 288 K, a = [1.4(287)(288)] 1/2 = 340 m/s. Recall from Chap. 8 that incompressible theory predicts V max = 2 U on a cylinder. Thus === = = max max 2 0.3(340) 0.3 when . 340 2 V U M aU a mf t 51 167 ss A n s
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Chapter 9 Compressible Flow 637 9.5 Steam enters a nozzle at 377 ° C, 1.6 MPa, and a steady speed of 200 m/s and accelerates isentropically until it exits at saturation conditions. Estimate the exit velocity and temperature. Solution: At saturation conditions, steam is not ideal . Use the Steam Tables: At 377 ° C and 1.6 MPa, read h1 = 3.205E6 J/kg and s1 = 7153 J/kg K At saturation for s1 = s2 = 7153, read p2 = 185 kPa, T2 = 118 ° C , and h2 = 2.527E6 J/kg 22 2 2 11 1 Then h V 3.205E6 (200) 2.527E6 V , solve .
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Chapter9_SM - Chapter 9 Compressible Flow 9.1 An ideal gas...

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