Question

A four-cylinder four-stroke engine is modelled using the air standard Otto cycle (two engine revolutions per cycle). Given the conditions at state 1, total volume (V1) of each cylinder, compression ratio (r), rate of heat addition (Q), and engine speed in RPM, determine the efficiency and other values listed below. The gas constant for air is R=0.287 kJ/kg-K.

--Given Values--

T1 (K) = 335

P1 (kPa) = 175

V1 (cm^3) = 600

r = 11

Q (kW) = 57

Speed (RPM) = 4400

Please answer only the missing answers!

(h-s) will rate!

a) Determine the total mass (kg) of air in the engine. Your Answer Correct! Exact Answer-4.3684E-03 +/- 9.3E-06 b) Determine the specific internal energy (kJ/kg) at state 1. Your Answer =293215 Incorrect. c) Determine the specific volume (m^3/kg) at state 1. Your Answer GSM Correct! Exact Answer=0.5494 +/- 3.4E-03 d) Determine the relative specific volume at state 1. Your Answer 133 Correct! Exact Answer-471.75 +/- 2.5E+00 e) Determine the relative specific volume at state 2. Your Answer = 20 Correct! Exact Answer=42.886 +/- 2.5E-01 1) Determine the temperature (K) at state 2. Your Answer 300 C orrect! Exact Answer=833.07 +/- 2.9E+00 g) Determine the pressure (kPa) at state 2. Your Answer - CT9404 Correct! Exact Answer-4787 +/- 9.7E+00 h) Determine the specific internal energy (kJ/kg) at state 2. Your Answer = i) Determine the clearance volume (cm 3) for a single cylinder. Your Answer = j) Determine the specific internal energy (kJ/kg) at state 3. Your Answer k) Determine the temperature (K) at state 3. Your Answer = 1) Determine the pressure (kPa) at state 3. Your Answer = m) Determine the relative specific volume at state 3. Your Answer = n) Determine the relative specific volume at state 4. Your Answer = o) Determine the temperature (K) at state 4. Your Answer = p) Determine the pressure (kPa) at state 4. Your Answer = 9) Determine the specific internal energy (kJ/kg) at state 4. Your Answer = r) Determine the net power output (kW). Your Answer s) Determine the efficiency (%) of the engine. Your Answer

Answer 1

Solution : Given values are T 335K P = 175 kla A V = 600 cm3 T P-V DIAGRAM OTTO CYCLE 3 +-- Q = 57kW Speed - 4400 RPM Also given T2 = 833.07K Massmair 4.368 X/0-3kg Pi Vi= MRT 175X103 x boux(10-2)3= mx0.287x335 m1.092 kg. Vz-V3 We know that! Compression ratio, r - - Total cylinder volume Clearance volume Vivu » V₂ = Vihr z 600/11 = 54.54 cm3 And clearance volume, Vc = V₂ = 54.54 cm3 In process 2-3 (constant volume heat addition) - Q - AU+W Here Woo. - AU = Cv (Tz-T2) Trat T2=8331 = 0.812(Iz - 833.07) = 0.812 kilig 7 57 = 0.812 Tz - 676.45 - T3 = 903.26 K. Specific internal energy at state 3 is givenas Au = Cultz-T)° = 57.268X1033 - - 13049 kJing.

In process 1-2 (Adiabatic compression) Дu - С Дт сут, -1 ) -0.717 (833.07 - 335) =357.11 - . Specific internal inergy at state 2: - Au - 357.) -32.7 kJ/kg m 11092 tho, in process 2-3: (Constant volume combustion) T & P₂ = P2 x Tz 9 4787 x zag 903.26 833.07 .P₂ - 5190 kPa. process 3-4 : (Adiabatic expansion) TV - =C In sy. & V3 = V2 = 54.54 cm3 & Vy= V = 600 cm3 Ty = T3 x (su 600 - 903:26 x 10.0909)" - 903,26 x 0.383 .. T4 = 345.95K Py Vy Pava And,

Pu= 54.54 345.951 600 9 03.26) 1 x 5190 Py = 0.0348 X 5190 si Py = 180.612 kla Also internal energy au at state 4 : - Au=-G AT =-Cv (Tu-T3) = -0.812 (345.95-903.26) ) au = 452.53 Now, specific internal energy at state 4. - - - A - 452.53 – 414.4 kilig m - 1.092 Relative specific volume at state 3 is from the table of properties of air at Tz = 903.26 V 34.31 - Relative specific volume at state 4 i.e. at Ty = 345.95k Ves - 438.15 - - Work done during process 1-2 : - W12=-Cr CT-T) E-0.812 ( 833:07 - 335) =-404.43 Klug Work done during process 3-4 Wzy - CvCT4-T3) = -0.812 (345.95-903:26) = 452.53 kJ kg . Net Work = Wiz t Way =-404.43 +452.53 = 48.09 kJ/kg

- Power = Whet x cycle per second • V8.09 x 900 . Net Power = 352606 KW. Efficiency of engine, n - When - 3526.6 57 n = 61.87%
Summary:Answers from (h-s )are: (h) 327Kj/kg.(i) Vc=54.54cm^3.(j) 13049kj/kg.(k)T3=903.26K.(l)P3=5190kPa.(m)Vr=34.31 (n)Vr=438.15 (O)T4=345.95K.(p)P4=180.612kPa.(q)414.4kj/kg.(r)P=3526.6KW.(s) Efficiency=61.87%.