Question

1-9 please

We Isentropic 04/13/2020 By writing my name above. I understand that no programmable calculators are allowed. Cols phones should be used only for showing yourself on Zoom Video. No notes or solved problems allowed in text book, and any form of cheating is not allowed during this test. Any violation will result in F grade outright. Your video should be turned on throughout the test. Problem Statement for Test 2-Part 1 The P-v diagram of an ideal Otto cycle is shown in Figure 1. 1 T=1300K The beginning of compression is at state 3 and the pressure and temperature at the beginning of compression are P-102 kPa and Ty-290 K respectively. The compression ratio, V/v2-8. The highest temperature in the cycle, T-1300 K The temperature at the end of the isentropic expansion, T-565.9K. For the entire analysis, assume the air is an ideal gas with constant specific heats and that -1.005 kJ/kg-K -0.718 kJ/kg-K, and R-0.287 kJ/kg-K and the sentropic exponent 1.4. There is no need to read the gas tables. 2 T-565,9K 1. State 3: Looking at state 3 on the Figure, compute the specific volume, vs, of the air at state 3.1.5 minutes) Your 2. State 4: Looking at state 4 on the Figure, and the fact that rvs/V=8 3 Ty290 K the compression ratio, -vy8, computeva. [1.5 Figure 1 ideal Otto Cycle P-102 hPa minutes) TDC BDCU 3. Applying the isentropic equation for the isentropic compression process between states 3 and 4, compute the temperature Te at the end of compression. 12 minutes 1 U Jsentropic With your computation of T4 and the given temperatures T, T. and T shown on the Figure 4. Applying energy balance for the isentropic compression process between states 3 and 4 compute w, 12.5 minutes 5. Applying energy balance for the heat input process between states and I compute 2.5 minutes 6. Applying energy balance for the sentropic expansion process between state and 2. compute the work done during the sentropic expansion process, W -12.5 minutes] 7. Applying energy balance for the heat rejection proces between state 2 and 3. compule 2.5 minutes 8. Calculate the network, w [1.5 minutes) 9. Calculate the thermal efficiency of the cycle. [1.5 minutes]

Answer 1

T, = 1300k Wexp P3 = 102 kea T3 - 290k - V3 - 8 Va = 8 T2 = 565.9k TDC , BDO v p = 1.005 kJ lige r=0.218 kJ 1 kgk R = 0.287 kJ lige. K - 1.4. PV = RT3. TV - RT3 - 0.2870 290 malka 102 3) 1 - Vez V Verona 0,25 - 0,102 MB) ly 0.816 = 0.102 m3 1 kg HEER 0.4 T4 = 666.245 k 290 . We = Q - DU. Q =0 for Isentropic process We = -AU. = - [Puvu - P3 V₂ ] = P3 Vz - Pu va W. RT3-T4) -0.284x(666.245 - 290) =-269.956 KT) kg K- T.

-ve on indicate work done. the system. We =-269.956 kJ 1kg 5. for process 4-1 4-1 for isochronic or constant volm Q = sunt Wang I process Worl=0 Ein = vi-ua. - Cv (Ti-T4) = 0.7 18 x (1300- 666.245) Lin = 455.04 ks) kg For process 1-2. КІ 2 1.4-1 Tz 1300 = 80.4 T2 = 565.9k) Q1 2 = AU 1-2 + Wexp. (for Isentropic expon) Wexp = - DU,2 = - cv (T2-T) = G(T-T2) = 0.7188 (1300- 565.9) Wexp = 527.08 kJ 1 kg

7 for process 2-3 22.3 = (U3-U₂) & W2-3, W2-3.50 for const. volm process. - - U3-Uz . 22-3. = Cv (T3-T2) = 0.718x (290-565-9) = - 198.09 kilks - ve sign indicate heat rejection out = 198.09 kJ 1 kg 8) - Wnet - Wexs - Wcomp Wret = 527.08 - 269.956 Wwet = 257. 124 Kilkg awet = lin - Yout = 257 kJ kg I wnet -a net verify لملم لما 257.124 Qin 455.04 - 56.517
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