Question

An ideal Otto cycle has a compression ratio of 9.2 and uses air as the working fluid. At the beginning of the compression process, air is at 98 kPa and 27°C. The pressure is doubled during the constant-volume heat-addition process. Accounting for the variation of specific heats with temperature, determine (a) the amount of heat transferred to the air (qout), (b) the net work output, (c) the thermal efficiency, and (d) the mean effective pressure for the cycle.

Answer 1

(a) The amount of heat transferred to the air,
`q_(out)` is 215.5077 kJ/kg

(b) The net work output,
`W_(net)`, is 308.07 kJ/kg

(c) The thermal efficiency is 58.8%

(d) The Mean Effective Pressure, MEP, is 393.209 kPa

Step-by-step explanation:

(a) The assumptions made are;

`c_p` = 1.005 kJ/(kg·K),
`c_v` = 0.718 kJ/(kg·K), R = 0.287 kJ/(kg·K),

Process 1 to 2 is isentropic compression, therefore;

`T_(2)= T_(1)\left ((v_(1))/(v_(2)) \right )^(k-1) = 300.15* 9.2^(0.4) = 729.21 \, K`

From;

`(p_(1)* v_(1))/(T_(1)) = (p_(2)* v_(2))/(T_(2) )`

We have;

`p_(2) = (p_(1)* v_(1)* T_(2))/(T_(1) * v_(2)) = (98* 9.2* 729.21)/(300.15 ) = 2190.43 \, kPa`

Process 2 to 3 is reversible constant volume heating, therefore;

`(p_3)/(T_3) =(p_2)/(T_2)`

p₃ = 2 × p₂ = 2 × 2190.43 = 4380.86 kPa

`T_3 = (p_3 * T_2)/(p_2) =(4380.86 * 729.21)/(2190.43) = 1458.42 \, K`

Process 3 to 4 is isentropic expansion, therefore;

`T_(3)= T_(4)\left ((v_(4))/(v_(3)) \right )^(k-1)`

`1458.42= T_(4) * \left (9.2 \right )^(0.4)`

`T_4 = (1458.42)/((9.2)^(0.4)) = 600.3 \, K`

`q_(out) = m * c_v * (T_4 - T_1) = 0.718 * (600.3 - 300.15) = 215.5077 \, kJ/kg`

The amount of heat transferred to the air,
`q_(out)` = 215.5077 kJ/kg

(b) The net work output,
`W_(net)`, is found as follows;

`W_(net) = q_(in) - q_(out)`

`q_(in) = m * c_v * (T_3 - T_2) = 0.718 * (1458.42 - 729.21) = 523.574 \, kJ/kg`

`\therefore W_(net) = 523.574 - 215.5077 = 308.07 \, kJ/kg`

(c) The thermal efficiency is given by the relation;

`\eta_(th) = (W_(net))/(q_(in)) * 100= (308.07)/(523.574) * 100= 58.8\%`

(d) From the general gas equation, we have;

`V_(1) = (m* R* T_(1))/(p_(1)) = (1* 0.287* 300.15)/(98) =0.897\, m^(3)/kg`

The Mean Effective Pressure, MEP, is given as follows;

`MEP =(W_(net))/(V_1 - V_2) = (W_(net))/(V_1 * (1- 1/r))= (308.07)/(0.897* (1- 1/9.2)) = 393.209 \, kPa`

The Mean Effective Pressure, MEP = 393.209 kPa.