Question

An ideal Otto cycle has a compression ratio of 8 and takes
in air at 95 kPa and 15oC, and the maximum cycle temperature is
1200oC. Determine the heat transferred to and rejected from this
cycle, as well as the cycle’s thermal efficiency.

Answer 1

Question #83234

An ideal Otto cycle has a compression ratio of 8 and takes in air at $95\mathrm{kPa}$ and $15^{\circ}\mathrm{C}$, and the maximum cycle temperature is $1200^{\circ}\mathrm{C}$. Determine the heat transferred to and rejected from this cycle, as well as the cycle's thermal efficiency.

Answer:

Figure 1 shows an ideal Otto cycle on pV-diagram.

FIGURE 1. An ideal Otto cycle

The thermal efficiency $\eta$ of an ideal Otto cycle is given by:

$$\eta = 1 - r ^ {1 - k},$$

where $r = 8$ – the compression ratio,

$k = 1.4$ – the adiabatic index for the ideal air.

Substitute into (1):

$$\eta = 1 - 8 ^ {1 - 1. 4} = 0. 5 6 5.$$

The heat $Q_{H}$ is transferred to the cycle in process 2-3:

$$Q _ {H} = c _ {v} \left(T _ {3} - T _ {2}\right),$$

where the heat capacity at constant volume for the ideal (diatomic) air is given by:

$$c _ {v} = \frac {5}{2} R,$$

$R = 287\mathrm{J.kg^{-1}K^{-1}}$ – the gas constant of air.

Now, let us determine the temperatures $T_{2}$ and $T_{3}$.

The maximum cycle temperature is achieved at state 3. So, $T_{3} = 1200^{\circ}\mathrm{C} = 1473\mathrm{K}$.

In the isentropic (adiabatic) process 1-2, we have:

$$T _ {2} = T _ {1} \left(\frac {V _ {1}}{V _ {2}}\right) ^ {k - 1} = T _ {1} r ^ {k - 1},$$

where $T_{1} = 15^{\circ}\mathrm{C} = 288\mathrm{K}$ – the initial absolute temperature of the air.

Substitute into (4), (3) and (2) to obtain the heat $Q_{H}$ transferred to the cycle:

$$T _ {2} = 2 8 8 \cdot 8 ^ {0. 4} = 6 6 1. 7 \mathrm {K},$$ $$c _ {v} = \frac {5}{2} \cdot 2 8 7 = 7 1 7. 5 \mathrm {J / kg . K} = 0. 7 1 7 5 \mathrm {k J / kg . K},$$ $$Q _ {H} = 0. 7 1 7 5 (1 4 7 3 - 6 6 1. 7) = 5 8 2. 1 \mathrm {k J / kg}.$$

The thermal efficiency of a cycle is given by:

$$\eta = 1 - \frac {Q _ {L}}{Q _ {H}},$$

we can find the heat $Q_{L}$ rejected from the cycle as follow:

$$Q _ {L} = Q _ {H} (1 - \eta),$$ $$Q _ {L} = 5 8 2. 1 (1 - 0. 5 6 5) = 2 5 3. 4 \mathrm {k J / kg}.$$

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