Question

An ideal Otto cycle has a compression ratio of 7. At the beginning of the compression process, P1 = 90 kPa, T1 = 27°C, and V1 = 0.004 m3. The maximum cycle temperature is 1167°C. For each repetition of the cycle, calculate the heat rejection and the net work production. Also, calculate the thermal efficiency and mean effective pressure for this cycle. Use constant specific heats at room temperature. The properties of air at room temperature are cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.4.

The heat rejection is _____ kJ.

The net work production is _______ kJ.

The thermal efficiency is _________ .

The mean effective pressure ______ kPa

Answer 1

Answer:

heat rejection $1,28KJ$

net work  $-1,26KJ$ 

termal efficency $\eta =0,53$

mean effective pressure $MEP=367,5Kpa$

Explanation:

The cycle can be graphicated as the index image

We can determinate the mass of air contained in the system by (assuming that the air behaves as an ideal gas)

$m=\frac{P_{1} V_{1}}{R T_{1}}=\frac{90 Kpa \times 0,004 m^{3}}{0,287\frac{Kpa m^{3}}{Kg K} \times 300,15K}=4,18 \times 10^{-3} Kg$

The unknown temperatures for the steps 2 and 4 are (where k is the adiabatic constant for the air)

$T_{2}=T_{1}\times(\frac{V_{1}}{V_{2}})^{k-1}=300K\times(\frac{7}{1})^{1,4-1}=653,37K$

$T_{4}=T_{3}\times(\frac{V_{2}}{V_{1}})^{k-1}=1440K\times(\frac{1}{7})^{1,4-1}=661,19K$

Application of the first law of thermodynamics to the four steps of the otto cycle processes gives

$W_{1-2}=m \times C_{v}\times (T_{2}-T_{1})=4,18 \times 10^{-3} Kg \times 0,718\frac{KJ}{KgK}\times (653,37-300)K=1,06KJ$

$Q_{2-3}=m \times C_{v}\times (T_{3}-T_{2})=4,18 \times 10^{-3} Kg \times 0,718\frac{KJ}{KgK}\times (1440-653,37)K=2,36KJ$

$W_{3-4}=m \times C_{v}\times (T_{4}-T_{3})=4,18 \times 10^{-3} Kg \times 0,718\frac{KJ}{KgK}\times (661,19-1440)K=-2,32KJ$

$Q_{4-1}=m \times C_{v}\times (T_{1}-T_{4})=4,18 \times 10^{-3} Kg \times 0,718\frac{KJ}{KgK}\times (300-661,19)K=-1,08KJ$

The heat rejection then is  

$Q_{rejected}= Q_{2-3}+Q_{4-1}=1,28KJ$

The net work

$W_{net}= W_{1-2}+ W_{3-4}=-1,26KJ$  (it’s negative because it gets out of the system)

The termal efficency can be calculated by

$\eta =\frac{\left | W_{net} \right |}{Q_{2-3}}=\frac{1,26KJ}{2,36KJ}=0,53$

The mean effective pressure in the engine is

$MEP=\frac{\left | W_{net} \right |}{V_{1}-V_{2}}=\frac{1,26KJ}{0,004-\frac{0,004}{7}m^{3}}\times \frac{1KPa m^{3}}{KJ}=367,5Kpa$