Problem: An engine operating on a cold air standard Otto cycle takes air at 100 kPa and 25 °C and compresses it isentropically to 2.2 MPa. The work output from the cycle is 200 kJ / kg of air. Find the efficiency of the cycle, and the maximum temperature reached in the cycle. Find: Efficiency

Question

Problem: An engine operating on a cold air standard Otto cycle takes air at 100 kPa and 25 °C and compresses it isentropically to 2.2 MPa. The work output from the cycle is 200 kJ / kg of air. Find the efficiency of the cycle, and the maximum temperature reached in the cycle.

Find: Efficiency [latex] η_{th,Otto} [/latex] of the Otto cycle, maximum temperature [latex] T_3 [/latex] reached.

Known: Cold air standard Otto cycle, intake air pressure [latex] P_1 [/latex] = 100 kPa, intake temperature [latex] T_1 [/latex] = 25 °C, pressure after compression [latex] P_2 [/latex] = 2.2 MPa, work output [latex] w_{net} = 200 kJ / kg. [/latex]

Diagram: Figure E10.1.

Assumptions: Constant specific heats evaluated at [latex] T_1 [/latex] = 25 °C, air behaves as an ideal gas.

Governing Equations:

Otto cycle efficiency (isentropic, ideal gas) [latex] \eta _{th,Otto}=\frac{w_{net}}{q_H} =1-\frac{1}{r^{\gamma -1}} [/latex]

Heat added in cycle [latex] q_H=c_v (T_3-T_2) [/latex]

Properties: Specific heat ratio of air [latex] \gamma [/latex] = 1.400 kJ / kgK (Appendix 1), specific heat of air at constant volume [latex] c_v [/latex] = 0.717 kJ / kg (Appendix 1).

Accepted Answer

The compression ratio of the cycle can be found, since the process uses an ideal gas and is isentropic:

[latex] r=\frac{V_1}{V_2} =\left\lgroup\frac{P_2}{P_1} ight group ^{1/\gamma }=\left\lgroup\frac{2.2 \ MPa}{0.1 \ MPa} ight group ^{1/1.400}=9.096 \ 48. [/latex]

The cycle efficiency can then be found,

[latex] \eta _{th,Otto}=1-\frac{1}{r^{\gamma -1}} =1-\frac{1}{9.096 \ 48^{0.400}} =0.586 \ 524. [/latex]

The air temperature at the end of compression is

[latex] T_2=T_1\left\lgroup\frac{P_2}{P_1} ight group ^{(\gamma -1)/\gamma }=298.15 \ K imes \left\lgroup\frac{2.2 \ MPa}{0.1 \ MPa} ight group ^{0.400/1.400}=721.081 \ K. [/latex]

The heat added to the cycle, per unit mass of air is

[latex] q_H=\frac{w_{net}}{\eta _{th,Otto}} =\frac{200 \ kJ/kg}{0.586 \ 524} =340.992 \ kJ/kg. [/latex]

Then the maximum temperature in the cycle is

[latex] T_3=T_2+\frac{q_H}{c_v} =721.081 \ K+\frac{340.992 \ kJ/kg}{0.717 \ kJ/kgK} =1196.66 \ K. [/latex]

Answer: The efficiency of the cycle is 58.7% and the maximum air temperature in the cycle is 1197 K.

Question 10.1: Problem: An engine operating on a cold air standard Otto cyc...

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