Answer:
[tex]\dot Q_{in} = 228.659\,kW[/tex], [tex]T_{3} = 1573.662\,K\,(1300.512\,^{\textdegree}C)[/tex]
Explanation:
The ideal efficiency of the Diesel cycle is given by this expression:
[tex]\eta_{th} = \left\{1 - \frac{1}{r^{k-1}} \cdot \left[\frac{r_{c}^{k}-1}{k\cdot (r_{c}-1)} \right]\right\}\times 100\%[/tex]
Where [tex]r[/tex] and [tex]r_{c}[/tex] are the compression and cutoff ratios, respectively.
[tex]\eta_{th} = \left\{1-\frac{1}{18^{0.4}}\cdot \left[\frac{1.5^{1.4}-1}{1.4\cdot (1.5-1)} \right] \right\}\times 100\%[/tex]
[tex]\eta_{th} = 65.648\,\%[/tex]
The heat addition to the cycle is:
[tex]\dot Q_{in} = \frac{\dot W}{\eta_{th}}[/tex]
[tex]\dot Q_{in} = \frac{150\,kW}{0.656}[/tex]
[tex]\dot Q_{in} = 228.659\,kW[/tex]
The temperature at state 2 is:
[tex]T_{2} = T_{1} \cdot r^{k-1}[/tex]
[tex]T_{2} = (330.15\,K)\cdot 18^{0.4}[/tex]
[tex]T_{2} = 1049.108\,K\,(775.958\,^{\textdegree}C)[/tex]
And the temperature at state 3 is:
[tex]T_{3} = T_{2}\cdot r_{c}[/tex]
[tex]T_{3} = (1.5)\cdot (1049.108\,K)[/tex]
[tex]T_{3} = 1573.662\,K\,(1300.512\,^{\textdegree}C)[/tex]
