Question

Q1 Otto Cycle Bookwork Question An air standard Otto cycle:- 1. Draws air from the environment at 100 kPa and 5°C. 2. The cycle has a compression ratio of 9.2: 1. 3. Heat (880 kJ/kg) is added to the compressed gases at constant volume. 4. The working fluid is air, a perfect gas with ratio of specific heats y = 1.4 and gas constant R = 287 J/kgk. Follow the instructions below: a) Describe each of the four processes from the thermodynamic point of view. [2 marks] b) Sketch the P-v and T-S plots for this cycle add labels starting as air intake at (1) [2 marks] Calculate the peak in cylinder pressure. [3 marks] Calculate the thermal efficiency of the cycle. [2 marks] Evaluate the Break Mean Effective Pressure. [1 mark] oo Unseen Question Q1 Otto Cycle f) Nehotto = 1 - 9) Above is the well-known expression for the thermal efficiency of an air standard Otto cycle in terms of its compression ratio ry and the ratio of specific heats y. Derive the equation. [5 marks] Use your answer to f) above to calculate the overall efficiency of the engine in the Bookwork part: compare it to your result. [1 mark] Using your result of f) state how the Otto thermal efficiency varies with the ratio of specific heats. [2 marks] If we try and increase the thermal efficiency by changing the compression ratio what unfortunate behaviour can occur and what effect might this have on the engine? [2 marks] h) i)

Answer 1

Otto cycle
It is also known as constant volume cycle because both hear addtion and rejection take place at constant volume .

1) Point 1 represent the position at which new charge take entry in a cylinder . Cylinder is full of air with volume V1 and pressure P1 and absolute temprature T1

2) Line 1-2 represents the adiabatic compression where piston moves BDC to TDC . As we are studing air standard cycle with all assummtions we know that reversible adiabatic cycle is isentropic cycle thus entropy remain same while pressure reaches to P2 volume reaches to V2 and temprature is increased upto T2

3) Line 2-3 shows the heat supply to the air at constant volume with the help of spark and combustion of fuel, pressure and temprature reaches to its maximum point of this cycle i.e P3 and T3

4) line 3-4 represents adiabatic expansion hear we get power from engine . Hear entropy remains same , volume get increased to V4= V1 while temprature and pressureare reduced to T4 and P4

5) line 4-1 shows heat rejection at constant volume hear all the exhaust gases escaped out and at thi point cycle completed , and all the thermodynamics properties comes to an intial stati V1 P1 T1

PAGE triv 5°C = 278K iven D = lookea Se 8.2 1 HA = 880 KI @ T4 PT Adiabatic Vol-Coretant 2 Heat Addito 2 5 Heat Rejection Vot-Coretant Adiabatic S

3) Peak Pressure ? · Adiabatic Compression (1-2) (S 12 Y- V 24 .५- (9.२ 278 0.4 12 = 228_x(5.२) T: 635.39 Also १ : P, भ १ 22 P वत

le loo = (9.2).4 22.35.13 kPa a 2 • Constant Volune (2-3) 0 Weat added Cal Tz - T2) 880 = HA Cp - Y Cj Cp - Cv = R Cve C. = =R Cv R Y-1 (Is-Te) = 880 0.282 (Iz=675.39) = 880 Y-1 1.4-1 Iz = 1226.48_7675.39

T3= 1901.87 V At Constant Voline Pz Iz B 2.235.13 675.39 1901287 Pg= 6294.03 kPa lascimum Cylinder Pressure 4) Thermal efficancy ܩܩܩܛ ܥܠܦܘܠܝܬܐ ܒܓܝܫܡܢܨܛC Heat Supplied lare to 2-1

PAGE Loto 1.4- Lotto - Porto 0.4116 58.83 % Motto 5) Break mean Effective Pressure Do are [lre 1) Grp. 1 (ोहन Where to Po - L Pe - Pu Pressure ratio Pe Spe6294.03 2235 013 Putting all 1ooxa. 2 [(s. 24-1) (2.815 -17 (1.4-1) (9. 2-1) P 2.815 Values

Pm = 920 x 1.429 X 1.815 0.4 x 8.2 Pm= 727.482 kPa

We can find efficancy by finding all tempretures but with the help of direct formula its very easy and quick to calculate
Similarly with mean effective pressure use of direct formula helps a lot in exams

As per HomeworkLib guideline i can answer 4 questions only but i still solved 5 to help you in better way .
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