Fluid Mechanics-Frank M White Solution Ch9

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Fluid Mechanics-Frank M White Solution Ch9에 대한 보고서 자료입니다.

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9.1 An ideal gas flows adiabatically through a duct. At section 1, p1 = 140 kPa, T1 = 260°C, and V1 = 75 m/s. Farther downstream, p2 = 30 kPa and T2 = 207°C. Calculate V2 in m/s and s2 − s1 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:


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):



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 EES or the steam tables [15].
Solution: For steam, take k = 1.33, R = 461 J/kg ⋅ K, and cp = 1858 J/kg ⋅ K. Then



2 (b) For real steam, we look up each enthalpy and entropy in EES or the Steam Tables:




These are within ±1.5% 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




P9.4 Consider steady adiabatic airflow in a duct. At section B, the pressure is 154 kPa and the density is 1.137 kg/m3. At section D, the pressure is 28.2 kPa and the temperature is 19°C. (a) Find the entropy change, if any. (b) Which way is the air flowing?
Solution: Convert TD = -19+273 = 254 K. We need the temperature at section B:


The entropy is higher at B. Therefore the (adiabatic) flow is from D to B. Ans.(b)

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