Answer:600 kW
Explanation:
Given
Power delivered [tex]P=150 kW[/tex]
Fuel heating value [tex]q=35,000 kJ/kg[/tex]
thermal Efficiency [tex]\eta =\frac{Power}{Heat}[/tex]
[tex]0.2=\frac{150}{35,000\times \dot{m}}[/tex]
where, [tex]\dot{m} =mass\ flow\ rate\ of\ fuel[/tex]
[tex]\dot{m}=\frac{150}{35,000\times 0.2}[/tex]
[tex]\dot{m}=0.0214 kg/s[/tex]
For an Engine
[tex]q\dot{m}=P+Q_r , where Q_r=heat\ rejected[/tex]
[tex]Q_r=q\dot{m}-P[/tex]
[tex]Q_r=750-150=600 kW[/tex]
