Answer:
0.8 kilograms of fuel are consumed each second.
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} = \frac{\dot W}{\dot E} \times 100\,\%](https://tex.z-dn.net/?f=%5Ceta_%7Bth%7D%20%3D%20%5Cfrac%7B%5Cdot%20W%7D%7B%5Cdot%20E%7D%20%5Ctimes%20100%5C%2C%5C%25)
The fluid power is:
![\dot E = \frac{\dot W}{\eta} \times 100\,\%](https://tex.z-dn.net/?f=%5Cdot%20E%20%3D%20%5Cfrac%7B%5Cdot%20W%7D%7B%5Ceta%7D%20%5Ctimes%20100%5C%2C%5C%25)
![\dot E = \frac{8\,MW}{20\,\%}\times 100\,\%](https://tex.z-dn.net/?f=%5Cdot%20E%20%3D%20%5Cfrac%7B8%5C%2CMW%7D%7B20%5C%2C%5C%25%7D%5Ctimes%20100%5C%2C%5C%25)
![\dot E = 40\,MW](https://tex.z-dn.net/?f=%5Cdot%20E%20%3D%2040%5C%2CMW)
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\,\frac{MJ}{kg} = \frac{40\,MJ}{m_{fuel}}](https://tex.z-dn.net/?f=50%5C%2C%5Cfrac%7BMJ%7D%7Bkg%7D%20%3D%20%5Cfrac%7B40%5C%2CMJ%7D%7Bm_%7Bfuel%7D%7D)
![m_{fuel} = 0.8\,kg](https://tex.z-dn.net/?f=m_%7Bfuel%7D%20%3D%200.8%5C%2Ckg)
0.8 kilograms of fuel are consumed each second.
Answer:
<em>Solution:</em>
(A). For the evaluation of Type A uncertainty data, a single repeatability test containing mean, standard deviation and degrees of freedom
<em>Mean </em>
= (21.0 + 20.2 + 20.7 + 21.3 + 20.5) / 5
= 20.74 psig
<em>Standard Deviation</em>
= (0.26-0.54-0.04 + 0.56-0.24)2/5 = 0
<em>Degrees of Freedom </em>
= n-1 = 5-1
<em>= 4 </em>
(B). For the evaluation of Type B uncertainty data, because the standard deviation is zero, therefore Type B will be uncertain.
(C). Average standard
=20.74 x 5
=<em>103.70 psig</em>
Answer:
f = 0.05Hz
Explanation:
Look at the graph. You can see that the wave complete one cycle in 20 seconds, so we can say that the period is 20s.
T = 20
frequency is just the inverse the period so
f = 1/T = 1/20 = 0.05Hz