(c, d) Finally, the entropy change and mass flow follow from the properties known above: 900 460 + 79 æöæö s s -= 5167 - =ln+ 1130 ˜ ln 2368 266. (b) The steady-flow energy equation, with no shaft work, yields the heat transfer per mass:ġ1 22 q c=-+-= (T T )-+V-V 5167( 90022 400 ) p2 1 2 1 22 ft (a) The downstream pressure is computed from one-dimensional continuity: pp A V A= V, cancel, V V, cancel A R, 12 111 2 22 1 2 RT RT Solution: For carbon dioxide, take k =1.28, R =1130 ft Hint: This problem requires the continuity equation. Compute (a) p2, (b) the heat added between sections, (c) the entropy change between sections, and (d) the mass flow per unit area. Farther downstream the properties are V2 =1000 ft/s and T2 =900 ☏. (a) The density is unchanged because the container is constant volume. Solution: From Table A.4 for helium, k =1.66 and R =2077 m 2 /s2
Estimate (a) the new temperature, in ☌ and (b) the change in entropy, in J/(kg We are assu ming with this calculation that a (supersonic) shock wave does not form.ĩ.6Helium at 300 ☌ and 200 kPa, in a closed container, is cooled to a pressure of 100 kPa. This exit flow is supersonic, with a Mach number exceeding 2.0. Solution: At saturation conditions, steam is At 377 ☌ and 1.6 MPa, read h At saturation for sġ1 1 Then h V 3.205E6 += + = 222 (200) + 2.527E6 V, solve. Estimate the exit velocity and temperature. Thus 8ĩ.5 Steam enters a nozzle at 377 ☌, 1.6 MPa, and a steady speed of 200 m/s and accelerates isentropically until it exits at saturation conditions. 8 that incompressible theory predicts VU 2 0.3(340) 8 Ma U Ans=max =˜= 0.3 when. 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, Chap. (c) 21 v 21ĩ.4 Compressibility becomes important when the Mach number >0.3. Steam is nearly ideal in this range.ĩ.3 If 8 kg of oxygen in a closed tank at 200 ☌ 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. (b) For real steam, we look up each enthalpy and entropy in the Steam Tables: Jġ1 1 Then h V 2.993E6 += + = 222 (75) + 2.893E6 V, solve. Use two approaches: (a) an ideal gas from Table A.4 and (b) real steam from the steam tables. The adiabatic steady-flow energy equation (9.23) is used to compute the downstream velocity: 111 222 c T V+= constant = + = +1005(260) (75) 1005(207) V or. Solution: (a) For air, take k =1.40, R =287 J/kg Calculate V2 in m/s and s2 - s1 in J/(kg Farther downstream, p2 =30 kPa and T2 =207 ☌. 9.1 An ideal gas flows adiabatically through a duct.
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