Question

At the beginning of the compression process of an air-standard Otto cycle, P1 = 1.0 bar,
T1 = 290 K, V1 = 400 cm3. The maximum temperature in the cycle is 2200 K and the
compression ratio is 8. Determine: a) the heat addition in kJ, b) the net work in kJ, c) the
thermal efficiency, and d) the mean effective pressure, in bar.

Answer 1

Concepts and Reason

The concept required to solve this problem is Otto cycle.

Otto cycle: It is an ideal cycle which is used to describe the processes taking place in a spark ignition internal combustion engine. It consists of two isentropic processes and two constant volume processes. The cycle is plotted on a P-V diagram as:

In the PV plot process 1 to 2 represents an isentropic compression process and the process 2 to 3 represents a constant volume heat addition process. The process 3 to 4 represents isentropic expansion process and process 4 to 1 represents constant volume heat rejection process.

Swept volume for an Otto cycle is .

Isentropic process: It is considered as a process for which there is no change in entropy of the system throughout the process.

Isochoric Process: Processes for which there is no change in the volume are called as isochoric/constant volume processes.

Initially, calculate the properties of air at different states by applying isentropic relations. Apply energy balance to the process to find the heat addition, net work, and efficiency of the cycle. Finally, use the expression of mean effective pressure to calculate its value.

Fundamentals

The expression for compression ratio is:

$r = \frac{{{V_1}}}{{{V_2}}} = \frac{{{V_4}}}{{{V_3}}}$

Here, $r$ is the compression ratio.

The following are the relations used in isentropic process:

$\frac{{{v_{r2}}}}{{{v_{r1}}}} = \frac{{{V_2}}}{{{V_1}}}$

The heat addition in an Otto cycle is,

${Q_{in}} = m\left( {{u_3} - {u_2}} \right)$

Here, the mass of air is m and u is specific internal energy.

Heat rejected in an Otto cycle is,

${Q_{out}} = m\left( {{u_4} - {u_1}} \right)$

The net work done during the cyclic process is,

${W_{net}} = {Q_{in}} - {Q_{out}}$

Thermal efficiency is given by,

$\eta = \frac{{{W_{net}}}}{{{Q_{in}}}}$

The mean effective pressure is given by,

$\begin{array}{c}\\{P_m} = \frac{{{W_{net}}}}{{{V_1} - {V_2}}}\\\\ = \frac{{{W_{net}}}}{{{V_1}\left( {1 - \frac{{{V_2}}}{{{V_1}}}} \right)}}\\\\ = \frac{{{W_{net}}}}{{{V_1}\left( {1 - \frac{1}{r}} \right)}}\\\end{array}$

Draw the $p - V$ diagram of the Otto cycle.

Calculate the mass of air using the ideal gas equation.

$m = \frac{{{P_1}{V_1}}}{{R{T_1}}}$

Here, the pressure, volume, and temperature at state 1 are ${P_1}$ , ${V_1}$ , and ${T_1}$ respectively. The specific gas constant is R.

Substitute 100 kPa for ${P_1}$ , 290 K for ${T_1}$ , $400 \times {10^{ - 6}}{\rm{ }}{{\rm{m}}^3}$ for ${V_1}$ , and $0.287{\rm{ kJ/kg}} \cdot {\rm{K}}$ for R.

$\begin{array}{c}\\m = \frac{{100 \times 400 \times {{10}^{ - 6}}}}{{0.287 \times 290}}\\\\ = 4.8059 \times {10^{ - 4}}{\rm{ kg}}\\\end{array}$

Obtain properties of air at compression inlet temperature ${T_1} = 290\,{\rm{K}}$ from ideal gas properties of air.

Specific internal energy, ${u_1} = 206.91\,{\rm{kJ/kg}}$

Relative volume, ${v_{r1}} = 676.1$

Similarly, Obtain properties of air at turbine inlet temperature ${T_3} = 2200\,{\rm{K}}$ from ideal gas properties of air.

Specific internal energy, ${u_3} = 1872.4\,{\rm{kJ/kg}}$

Relative volume, ${v_{r3}} = 2.012$

Apply isentropic relation to the process 1-2.

$\frac{{{v_{r2}}}}{{{v_{r1}}}} = \frac{{{V_2}}}{{{V_1}}}$

Substitute 8 for $\frac{{{V_1}}}{{{V_2}}}$ and 676.1 for ${v_{r1}}$ .

$\begin{array}{l}\\\frac{{{v_{r2}}}}{{676.1}} = \frac{1}{8}\\\\{v_{r2}} = 84.512\\\end{array}$

Find the internal energy at state 2 at the corresponding value of ${v_{r2}} = 84.512$ from ideal gas properties of air table.

$\begin{array}{c}\\{u_2} = 473.25 + \left( {481.01 - 473.25} \right)\left( {\frac{{84.512 - 85.35}}{{81.89 - 85.34}}} \right)\\\\ = 475.134\,{\rm{kJ/kg}}\\\end{array}$

Apply isentropic relation to the process 3-4.

$\frac{{{v_{r3}}}}{{{v_{r4}}}} = \frac{{{V_3}}}{{{V_4}}} = \frac{{{V_2}}}{{{V_1}}}$

Substitute 8 for $\frac{{{V_1}}}{{{V_2}}}$ and $2.012$ for ${v_{r3}}$ .

$\begin{array}{l}\\\frac{{2.012}}{{{v_{r4}}}} = \frac{1}{8}\\\\{v_{r4}} = 16.096\\\end{array}$

Find the internal energy at state 4 at the corresponding value of ${v_{r4}} = 16.096$ from ideal gas properties of air table.

$\begin{array}{c}\\{u_4} = 880.35 + \left( {897.91 - 880.35} \right)\left( {\frac{{16.096 - 16.946}}{{16.064 - 16.946}}} \right)\\\\ = 897.27\,{\rm{kJ/kg}}\\\end{array}$

(a)

Calculate the constant volume heat addition to the air in process 2-3 by applying energy equation.

${Q_{in}} = {W_{2 - 3}} + m\left( {{u_3} - {u_2}} \right)$

Since the process 2-3 is constant volume process, work done during the process is zero.

Substitute $4.8059 \times {10^{ - 4}}{\rm{ kg}}$ for m, $1872.4\,{\rm{kJ/kg}}$ for ${u_3}$ , and $475.134\,{\rm{kJ/kg}}$ for ${u_2}$

$\begin{array}{c}\\{Q_{in}} = 4.8059 \times {10^{ - 4}}\left( {1872.4 - 475.134} \right)\\\\ = 0.6715\,{\rm{kJ}}\\\end{array}$

(b)

Calculate the constant volume heat rejection from air in process 4-1 by applying energy equation.

${Q_{out}} = {W_{4 - 1}} + m\left( {{u_4} - {u_1}} \right)$

Since the process 4-1 is constant volume process, work done during the process is zero.

Substitute $4.8059 \times {10^{ - 4}}{\rm{ kg}}$ for m, $897.27\,{\rm{kJ/kg}}$ for ${u_4}$ , and $206.91\,{\rm{kJ/kg}}$ for ${u_1}$ .

$\begin{array}{c}\\{Q_{out}} = 0 + 4.8059 \times {10^{ - 4}} \times \left( {897.27 - 206.91} \right)\\\\ = 0.33178\,{\rm{kJ}}\\\end{array}$

Find the net work done by using the following equation:

$\begin{array}{c}\\{W_{net}} = {Q_{in}} - {Q_{out}}\\\\ = 0.6715 - 0.33178\\\\ = 0.33972\,{\rm{kJ}}\\\end{array}$

Calculate the thermal efficiency of the cycle.

$\begin{array}{c}\\\eta = \frac{{{W_{net}}}}{{{Q_{in}}}}\\\\ = \frac{{0.33972}}{{0.6715}}\\\\ = 0.5059\\\\ = 50.59\% \\\end{array}$

(d)

Determine the mean effective pressure.

${P_m} = \frac{{{W_{net}}}}{{{V_1}\left( {1 - \frac{1}{r}} \right)}}$

Substitute 0.33972 kJ for ${W_{net}}$ , 8 for r, and $400 \times {10^{ - 6}}{\rm{ }}{{\rm{m}}^3}$ for ${V_1}$ .

$\begin{array}{c}\\{P_m} = \frac{{0.33972}}{{400 \times {{10}^{ - 6}}\left( {1 - \frac{1}{8}} \right)}}\\\\ = 970.628{\rm{ kPa}}\\\\{\rm{ = 9}}{\rm{.706}}\,{\rm{bar}}\\\end{array}$

Ans: Part a

The heat addition is 0.6715 kJ.

Part b

The net work done is 0.33972 kJ.

Part c

The thermal efficiency of the cycle is 50.59%

Part d

The mean effective pressure is 9.706 bar.