Chapter 3 Thermodynamics solution set manual for Hipolito Sta maria, Cheat Sheet of Mathematics

Download Chapter 3 Thermodynamics solution set manual for Hipolito Sta maria and more Cheat Sheet Mathematics in PDF only on Docsity! Thermodynamics 1 Hipolito B. Sta. Maria Chapter 3 : The Ideal Gas Solution Manual 1. An automobile tire is inflated to 32psig pressure at 50 F. After being driven, the temperature rises to 75 F. Determine the final gage pressure assuming the volume remains constant. Ans. 34.29 psig (Electrical Engineering Board Exam Problem) Given: Required: P1 = 32 psig P2 = ? T1 = 50 F + 460 = 510 R T2 = 75 F + 460 = 535 R Solution: Since volume is constant, use Charles’ Law on constant volume processes. 𝑃1 𝑇1 = 𝑃2 𝑇2 Note: P and T should be absolute. 𝑃2 = 𝑃1𝑇2 𝑇1 𝑃2 = (32 𝑝𝑠𝑖𝑔 + 14.696 𝑝𝑠𝑖)( 535 R ) 510 R 𝑃2 = 48.9850 Converting back to gage pressure 𝑃2 = 48.9850 − 14.696 𝑃2 = 34.29 𝑝𝑠𝑖𝑔 2. If 100 ft3 of atmospheric air at zero Fahrenheit temperature are compressed to a volume of 1 ft3 at a temperature of 200 F, what will be the pressure of the air in psi? Ans. 2109 psia (Electrical Engineering Board Exam Problem) Given: Required: V1 = 100 ft3 V2 = 1 ft3 P2 = ? T1 = 0 F + 460 = 460 R T2 = 200 F + 460 = 660 R P1 = 14.696 psia Solution: 𝑃1𝑉1 𝑇1 = 𝑃2𝑉2 𝑇2 𝑃2 = 𝑃1𝑉1𝑇2 𝑇1𝑉2 𝑃2 = (14.696 psia )(100 ft3)(660 R) (460 R)(1 ft3) 𝑃2 = 2109 𝑝𝑠𝑖𝑎 7. An automobile tire contains 3730 cu in. of air at 32 psig and 80 F. (a) What mass of air is in the tire? (b) In operation, the air temperature increases to 145 F. If the tire is inflexible, what is the resulting percentage increase in gage pressure? (c) What mass of the 145 F air must be bled off to reduce the pressure back to its original value? Ans. (a) 0.5041 Ib; (b) 17.53%; (c) 0.0542 lb Given: Required: V1 = 3730 in3 T2 = 145 F + 460 = 605 R a. m = ? P 1 = 32 psig 𝑅 = 53.342 𝑙𝑏 ∙ 𝑓𝑡 𝑙𝑏 ∙𝑅 b. %P = ? T1 = 80 F + 460 = 540 R c. m = ? Solution: a. 𝑃𝑉 = 𝑚𝑅𝑇 𝑚1 = 𝑃𝑉 𝑅 𝑇1 = (32 𝑝𝑠𝑖𝑔 + 14.692 𝑝𝑠𝑖𝑎)(3730 in3) (53.342 𝑙𝑏 ∙ 𝑓𝑡 𝑙𝑏 ∙ 𝑅 ) (540 R)(12 𝑖𝑛 𝑓𝑡 ) 𝑚1 = 0.50386 𝑙𝑏 b. V1 = 3730 in3 = V2 𝑃1 𝑇1 = 𝑃2 𝑇2 𝑃2 = (32 𝑝𝑠𝑖𝑔 + 14.692 𝑝𝑠𝑖𝑎)(605 R) 540 R 𝑃2 = 52.6148 𝑝𝑠𝑖𝑎 − 14.692 𝑝𝑠𝑖𝑎 = 37.63 𝑝𝑠𝑖𝑔 %𝑃 = 37.63 𝑝𝑠𝑖𝑔 − 32 𝑝𝑠𝑖𝑔 32 𝑝𝑠𝑖𝑔 ×100 %𝑃 = 17.59 % c. 𝑃𝑉 = 𝑚𝑅𝑇 𝑚2 = 𝑃𝑉 𝑅 𝑇1 = (32 𝑝𝑠𝑖𝑔 + 14.692 𝑝𝑠𝑖𝑎)(3730 in3) (53.342 𝑙𝑏 ∙ 𝑓𝑡 𝑙𝑏 ∙ 𝑅 ) (605 R)(12 𝑖𝑛 𝑓𝑡 ) 𝑚2 = 0.45232 𝑙𝑏 𝑚3 = 𝑚1 − 𝑚2 = 0.50386 𝑙𝑏 − 0.45232 𝑙𝑏 𝑚3 = 0.05154 𝑙𝑏 8. A spherical balloon is 40 ft in diameter and surrounded by air at 60 F and 29.92in Hg abs. (a) If the balloon is filled with hydrogen at a temperature of 70 F and atmospheric pressure, what total load can it lift? (b) If it contains helium instead of hydrogen, other conditions remaining the same, what load can it lift? (c) Helium is nearly twice as heavy as hydrogen. Does it have half the lifting force? R for hydrogen is 766.54 and for helium is 386.04 ft.lb/lb. R. Ans. (a) 2381 lb; (b) 2209 lb Given: Required: d = 40 ft Thydrogen = 70 F + 460 = 530 R a. m = ? T = 60 F + 460 = 520 R 𝑅hydrogen = 766.54 𝑙𝑏 ∙ 𝑓𝑡 𝑙𝑏 ∙𝑅 b. m = ? P = 29.92 in Hg abs 𝑅helium = 386.04 𝑙𝑏 ∙ 𝑓𝑡 𝑙𝑏 ∙𝑅 c. Solution: 𝑉 = 4 3 𝜋(20𝑓𝑡)3 = 33510.32 𝑓𝑡3 𝑃𝑉 = 𝑚𝑅𝑇 𝑚𝑎𝑖𝑟 = 𝑃𝑉 𝑅𝑇 = (29.92 in Hg × 14.692 𝑝𝑠𝑖𝑎 29.92 in Hg )(33510.32 𝑓𝑡3)( 144𝑖𝑛2 𝑓𝑡2 ) (53.342 𝑙𝑏 ∙ 𝑓𝑡 𝑙𝑏 ∙𝑅 )(520 R) 𝑚𝑎𝑖𝑟 = 2557.4201 𝑙𝑏 a. 𝑃𝑉 = 𝑚𝑅𝑇 𝑚ℎ𝑦𝑑𝑟𝑜𝑔𝑒𝑛 = 𝑃𝑉 𝑅𝑇 = (14.692 𝑝𝑠𝑖𝑎)(33510.32 𝑓𝑡3)( 144𝑖𝑛2 𝑓𝑡2 ) (766.54 𝑙𝑏 ∙ 𝑓𝑡 𝑙𝑏 ∙𝑅 )(530 R) 𝑚ℎ𝑦𝑑𝑟𝑜𝑔𝑒𝑛 = 174.6014 𝑙𝑏 𝑚𝑙𝑖𝑓𝑡 = 𝑚𝑎𝑖𝑟 − 𝑚ℎ𝑦𝑑𝑟𝑜𝑔𝑒𝑛 = 2557.4201 𝑙𝑏 − 174.6014 𝑙𝑏 𝑚𝑙𝑖𝑓𝑡 = 2382.8187 𝑙𝑏 a. b. 𝑃𝑉 = 𝑚𝑅𝑇 𝑚ℎ𝑒𝑙𝑖𝑢𝑚 = 𝑃𝑉 𝑅𝑇 = (14.692 𝑝𝑠𝑖𝑎)(33510.32 𝑓𝑡3)( 144𝑖𝑛2 𝑓𝑡2 ) (386.04 𝑙𝑏 ∙ 𝑓𝑡 𝑙𝑏 ∙𝑅 )(530 R) 𝑚ℎ𝑒𝑙𝑖𝑢𝑚 = 346.6971 𝑙𝑏 𝑚𝑙𝑖𝑓𝑡 = 𝑚𝑎𝑖𝑟 − 𝑚ℎ𝑒𝑙𝑖𝑢𝑚 = 2557.4201 𝑙𝑏 − 346.6971 𝑙𝑏 𝑚𝑙𝑖𝑓𝑡 = 2210.723 𝑙𝑏 c. No 9. A reservoir contains 2.83 cu m of carbon monoxide at 6895 kPa and 23.6 C. An evacuated tank is filled from the reservoir to a pressure of 3497 kPa and a temperature of 12.4 C, while the pressure in the reservoir decreases to 6205 kPa and the temperature to 18.3 C. What is the volume of the tank? R for CO is 296.92 J/kg.K. Ans. 0.451 m Given: Required: Vco = 2.83 m3 𝑅co = 296.92 𝐽 𝑘𝑔 ∙𝐾 Vtank = ? P1 = 6895 kPa P2 = 3497 kPa Ptank = 6205 kPa T1 = 23.6 C + 273 = 296.6 K T2 = 12.4 C + 273 = 285.4 K Ttank = 18.3 C +273 = 291.3 K Solution: 𝑃𝑉 = 𝑚𝑅𝑇 𝑚1 = 𝑃1𝑉𝑐𝑜 𝑅 𝑇1 = (6895 kPa)(2.83m3) (0.29692 𝑘𝑁 ∙ 𝑚 𝑘𝑔 ∙ 𝐾 ) (296.6 K) 𝑚1 = 221.5696 𝑘𝑔 𝑚𝑡𝑎𝑛𝑘 = 𝑃𝑡𝑎𝑛𝑘𝑉𝑐𝑜 𝑅 𝑇𝑡𝑎𝑛𝑘 = (6205 kPa)(2.83m3) (0.29692 𝑘𝑁 ∙ 𝑚 𝑘𝑔 ∙ 𝐾 ) (291.3 K) 𝑚𝑡𝑎𝑛𝑘 = 203.024 𝑘𝑔 𝑚 = 𝑚1 − 𝑚𝑡𝑎𝑛𝑘 = 221.5696 𝑘𝑔 − 203.024 𝑘𝑔 𝑚 = 18.5451 𝑘𝑔 𝑃𝑉 = 𝑚𝑅𝑇 𝑉𝑡𝑎𝑛𝑘 = 𝑚𝑅𝑇2 𝑃2 = (18.5451 𝑘𝑔) (296.92 𝑁 ∙ 𝑚 𝑘𝑔 ∙ 𝐾 ) (285.4 K) (3497 kPa) 𝑉𝑡𝑎𝑛𝑘 = 0.449 𝑚3 10. A gas initially at 15 psia and 2 cu ft undergoes a process to 90 psia and 0.60 cu ft, during which the enthalpy increases by 15.5 Btu; cv = 2.44Btu/lb. R. Determine (a) ΔU, (b) cp, and (c) R. Ans. (a) 11.06 Btu; (b) 3.42 Btu/lb.R; (c) 762.4ft.lb/lbR Given: Required: P1 = 15 psia P2 = 90 psia ΔH = 15.5 BTU a. ΔU = ? V1 = 2 ft3 V2 = 0.60 ft3 𝑐𝑣 = 2.44 𝐵𝑇𝑈 𝑙𝑏 ∙ °𝑅 b. Cp = ? c. R Solution: ∆𝑊𝑓 = {(90psia×0.60 ft3) − (15psia×2 ft3) }× 144 𝑖𝑛2 𝐵𝑇𝑈 778 𝑓𝑡3 ∙ 𝑙𝑏 ∆𝑊𝑓 = 4.0720 𝐵𝑇𝑈 a. ΔH = 15.5 Btu = Qp 𝑄𝑝 = ∆𝑈 + ∆𝑊𝑓 ∆𝑈 = 𝑄𝑝 − ∆𝑊𝑓 ∆𝑈 = 15.5 𝐵𝑇𝑈 − 4.0720 𝐵𝑇𝑈 ∆𝑈 = 11.06 𝐵𝑇𝑈 b. ∆U = 11.06 BTU = Qv 𝑄𝑣 = 𝑚𝐶𝑣(𝑇2 − 𝑇1) 𝑚(𝑇2 − 𝑇1) = 𝑄𝑣 𝐶𝑣 = 11.06 𝐵𝑇𝑈 2.44 𝐵𝑇𝑈 𝑙𝑏 ∙ °𝑅 𝑚(𝑇2 − 𝑇1) = 4.53 𝑙𝑏 ∙ °𝑅 𝑄𝑝 = 𝑚𝐶𝑝(𝑇2 − 𝑇1) 𝐶𝑝 = 𝑄𝑝 𝑚(𝑇2 − 𝑇1) 𝐶𝑝 = 15.5 BTU 4.53 𝑙𝑏 ∙ °𝑅 𝐶𝑝 = 3.42 𝐵𝑇𝑈 𝑙𝑏 ∙ °𝑅 c. 𝐶𝑝 = 𝐶𝑣 + 𝑅 𝑅 = 𝐶𝑝 − 𝐶𝑣 𝑅 = 3.42 𝐵𝑇𝑈 𝑙𝑏 ∙ °𝑅 − 2.44 𝐵𝑇𝑈 𝑙𝑏 ∙ °𝑅 𝑅 = 0.98 𝐵𝑇𝑈 𝑙𝑏 ∙ °𝑅 × 778 𝑓𝑡 ∙ 𝑙𝑏 𝐵𝑇𝑈 𝑅 = 762.44 𝑓𝑡 ∙ 𝑙𝑏 𝑏 ∙ °𝑅 11. For a certain gas, R = 0.277 kJ/kg.K and k = 1.384. (a) What are the value of cp and cv? (b) What mass of this gas would occupy a volume of 0.425 cu m at 517.11 kPa and 26.7 C? (c) If 31.65 kJ are transferred to this gas at constant volume in (b), what are the resulting temperature and pressure? Ans. (a) 0.7214 and 0.994 kJ/kg.R; (b)2.647 kg; (c) 43.27 C, 545.75 kPa Given: Required: 𝑅 = 0.277 𝑘𝐽 𝑘𝑔∙𝐾 𝑉 = 0.425 𝑚 3 a. cp = ?, cv = ? k = 1.384 P = 517.11 kPa b. m = ? Qv = 31.65 kJ T = 26.7 C + 273 = 299.7 K c. T2 = ?, P2 = ? Solution: a. 𝐶𝑝 = 𝑘𝑅 𝑘−1 = (1.384)( 0.277 𝑘𝐽 𝑘𝑔∙𝐾 ) 1.384 − 1 𝐶𝑝 = 0.998 𝑘𝐽 𝑘𝑔 ∙ 𝐾 𝐶𝑣 = 𝑅 𝑘 − 1 = ( 0.277 𝑘𝐽 𝑘𝑔 ∙ 𝐾 ) 1.384 − 1 𝐶𝑣 = 0.7214 𝑘𝐽 𝑘𝑔 ∙ 𝐾 b. 𝑃𝑉 = 𝑚𝑅𝑇 𝑚 = 𝑃𝑉 𝑅𝑇 = (517.11 kPa)(0.425 𝑚 3) (0.277 𝑘𝐽 𝑘𝑔 ∙ 𝐾 ) (299.7 K) 𝑚 = 2.647 𝑘𝑔 c. 𝑄𝑣 = 𝑚𝐶𝑣(𝑇2 − 𝑇1) 𝑇2 = 𝑄𝑣 𝑚𝐶𝑣 + 𝑇1 𝑇2 = 31.65 𝑘𝐽 (2.647 𝑘𝑔) (0.7214 𝑘𝐽 𝑘𝑔 ∙ 𝐾 ) + 299.7 K 𝑇2 = 316.28 𝐾 𝑃1 𝑇1 = 𝑃2 𝑇2 𝑃2 = 𝑃1𝑇2 𝑇1 = (517.11 kPa)(316.28 𝐾) 299.7 K 𝑃2 = 545.72 𝑘𝑃𝑎