Answer:
0.8 kilograms of fuel are consumed each second.
Step-by-step explanation:
As turbines are steady-state devices, the thermal efficiency of a turbine is equal to the percentage of the ratio of the output power to fluid power, that is:
![\eta_(th) = (\dot W)/(\dot E) * 100\,\%](https://img.qammunity.org/2021/formulas/engineering/college/3uqljr6o0fh8v6p0mie4349hw7mtg8de53.png)
The fluid power is:
![\dot E = (\dot W)/(\eta) * 100\,\%](https://img.qammunity.org/2021/formulas/engineering/college/pe6u3z0riu16hq4omhpehczfgbtio3s7v1.png)
![\dot E = (8\,MW)/(20\,\%)* 100\,\%](https://img.qammunity.org/2021/formulas/engineering/college/127txk0qfeqb9ueuqfft7bmaq06qy6r1nh.png)
![\dot E = 40\,MW](https://img.qammunity.org/2021/formulas/engineering/college/b8tclyqlh66bo2rw0h1e4sokwyg4cz82fw.png)
Which means that gas turbine consumes 40 megajoules of fluid energy each second, which is heated and pressurized with help of the fuel, whose amount of consumption per second is:
![50\,(MJ)/(kg) = (40\,MJ)/(m_(fuel))](https://img.qammunity.org/2021/formulas/engineering/college/x78zssglygwt5f9plss9cxof59wgoinbvo.png)
![m_(fuel) = 0.8\,kg](https://img.qammunity.org/2021/formulas/engineering/college/qrmfy7rsszxesif6jlr6x1vt8vtf9mn5oo.png)
0.8 kilograms of fuel are consumed each second.