Answer:
a. [tex]T_3=2027.1 K,P_3=5750.22 KPa[/tex]
b. [tex]\eta =0.564[/tex]
c. Work out put = 564 KJ/kg
d. [tex] P_{mean}=786.61 KPa[/tex]
Explanation:
Given that
Heat in put = 1000 KJ/kg
Compression ratio,r = 8
[tex]T_1=15 C[/tex]
[tex]P_1=100 KPa [/tex]
Process 1-2
[tex]\dfrac{T_2}{T_1}=r^{\gamma -1}[/tex]
[tex]\dfrac{T_2}{273+15}=8^{1.4 -1}[/tex]
[tex]T_2=661.65 K[/tex]
[tex]\dfrac{P_2}{P_1}=r^{\gamma}[/tex]
[tex]\dfrac{P_2}{100}=8^{1.4}[/tex]
[tex]P_2=1837.9 KPa [/tex]
Process 2-3
We know that for air
[tex]C_v=0.71\ \frac{KJ}{kg-K}[/tex]
[tex]Q=C_v(T_3-T_2)[/tex]
[tex]1000=0.71(T_3-661.5)[/tex]
[tex]T_3=2027.1 K[/tex]
[tex]\dfrac{P_3}{P_2}=\dfrac{T_3}{T_2}[/tex]
[tex]\dfrac{P_3}{1837.9}=\dfrac{2027.1}{661.5}[/tex]
[tex]P_3=5750.22 KPa [/tex]
We know that efficiency of otto cycle
[tex]\eta =1-\dfrac{1}{r^{\gamma -1}}[/tex]
[tex]\eta =1-\dfrac{1}{8^{1.4-1}}[/tex]
[tex]\eta =0.564[/tex]
[tex]\eta =\dfrac{Work\ out\ put}{Heat\ in\ put}[/tex]
[tex]0.564 =\dfrac{Work\ out\ put}{1000}[/tex]
Work out put = 564 KJ/kg
[tex]v_1=\dfrac{RT_1}{P_1}[/tex]
[tex]v_1=\dfrac{0.287\times 288}{100}[/tex]
[tex]v_1=0.82656\ \frac{m^3}{kg}[/tex]
So
[tex]v_2=\dfrac{0.82656}{8}\ \frac{m^3}{kg}[/tex]
[tex]v_2= 0.103\frac{m^3}{kg}[/tex]
[tex]Work\ out\ put\ = P_{mean}\times (v_1-v_2)[/tex]
[tex]564= P_{mean}\times (0.82-0.103)[/tex]
[tex] P_{mean}=786.61 KPa[/tex]