Question

2 Thermodynamics Equations (Cycles)

• Solve the following problems from the textbook (Thermodynamics: An Engineering Approach, by Yunus A. Cengel and Michael A. Boles, 8th Edition, SI Version, 2015): Gas Power Cycles: Obtain problem statements from the textbook Air-Standard Cycles (Otto Cycles) Problems No.: 9-32 An ideal Otto cyele has a compression ratio of 8. At the beginning of the compression process, air at 95 kPa and 27°C, and 750 kJ/kg of heat is transferred to air during the constant- volume heat addition process. Taking into account the variation of specific heats with temperature. a. Sketch the P-V and T-s diagrams for the cycle and determine: b. the maximum temperature and pressure of the cycle. c. the specific work output from the cycle. d. the thermal efficiency of the cycle. e. the mean effective pressure for the cycle. f. the specific entropy increased due to the heat addition. g. the specific heat rejected from the cycle. Problems No.: 9.35 A six-cylinder, four-sroke, spark ignition engine operating on the ideal Otto cycle takes in air at 95 kPa and 40°C, and is limited to a maximum cycle temperature of 1300°C. Each eyelinder has a bore of 8.9 em, and each piston has a stroke of 9.9 cm. The minimum enclosed Volume is 9.8 percent of the maximum enclosed volume. Use constant specific heats at (a) sketch the P.v and T-s diagrams for the cycle and determine: (b) the number of revolutions per eyelem @ the displaced volume of the engine. (d) the network output per cycle. (e) the power produced by the engine when it is operated at 2500 rpm the thermal efficiency of the engine.

Answer 1

The values from property table can varyfrom table to table slightly so there can be slight variation in answer.Just replace the values from the table you have. Procedure will be same.

a) - Given. 7=8 = P = 95 kPa T=270 300k a = 750 kJ/ieg Evom pторск о. R=0.987 Kolgok (6) As we can see that, Marimum pressure and temperature in otto cycle occurs at 3 Thus, we have to calculate Pressure and temperature at 3 It Now, from property table, # "At T = 3ook h = 214.07 kJ / 19 Ver, e Vere are specific rolume 621.2 Now, - YOV 2 300 # V = Vasen W XVI, = *21.2 = 97.65 Now, at vera = 77.65 To = 673.1k he = 491 2 kJ/1g Apply Gas quation at I and a pro Pers >> P W T 84673.4 x95 - 1705.187 kPa constant heat addition (2-3) a = hz-ho - Az = do the = 750 +491.2 = 1241. 2 KJ/14 from table, at n3 =124 1.2 kJ /icg Tg = 1539 K and Ve=V2 for constant heat adding: P-T2 ) P3 - T2 xPQ = 1539 x1705.187! : Pg = 3898.8 kPa Tg =1539k 673.1

( II Specific work output = din dout dout -ha-h we have to find ha for proces 3-4 (isentropic expansion) Vra = u larg = x 6.588 = 52.70 (Vorg = 6.588 at ha = 124 2.2 kuling) At Vora from table. TA = 774.5k ha = 572.69 kJ/1g (d) Efficiency. Whet ha 392.4 - 0.5232 750 so, Lout = 571.69 -214.07 - 357.62 kJ / 16 so, Wmet = 750 – 357.62 = 392.4 KJ / kg @ Mean effective pressure = Whet V-V2 V = RTA = 0.267*300 = R. T. 0.287 X 300 95 = 0.906 0.906 so, MEP: 392,4 5 495.0 kPa 0.906 - 0.906 ( 2-3 process specific entropy increased during AS = S3-S2 from property table Sa fet 1539 k) =4.88 kJ/kg-K Sg (at 673k) - 3.35 Killy-K AS = 4.28-3.35 = 0.93 KI/kg-K. 116) Specific heat rejected - Yout = ha-h, = 357.62 KJ/18