Respuesta :
Answer:
a) 6498.84 kW
b) 0.51
c) 0.379
Explanation:
See the attached picture below for the solution
![Ver imagen barackodam](https://us-static.z-dn.net/files/d20/1d22b1e081aec03283409e5b8c783358.jpg)
![Ver imagen barackodam](https://us-static.z-dn.net/files/d5f/d8a5805a774aff6585adb40c589bdde2.jpg)
Given-
Temperature at state 1 [tex]T_{1}[/tex] is 40 degree cel. This is equal to the 313 K.
Temperature at state 1 [tex]T_{2}[/tex] is 650 degree cel. This is equal to the 923 K.
To know the mass flow rate use the idol gas equation. Idol gas equation for a substance can be given as,
[tex]PV=nRT[/tex]
Rewrite this equation for [tex]n[/tex],
[tex]n=\dfrac{PV}{RT}[/tex]
Put the value of known variable from the question,
[tex]n=\dfrac{100\times\dfrac{850}{60} }{0.287\times 313}[/tex]
[tex]n=15.77[/tex]
Thus the value of n is 15.77 kg/sec.
Temperature at state 2 is,
In Bryon cycle we know that the processes 1-2 and 3-4 are isotropic. Hence,
[tex]T_{2} =T_{1}\left (1+ \dfrac{r_{p}^{\dfrac{\gamma-1}{\gamma}-1} }{c_{v} } \right )[/tex]
[tex]T_{2} =313\left (1+ \dfrac{16^{\dfrac{1.35-1}{1.35}-1} }{0.821 } \right )[/tex]
[tex]T_{2} =700[/tex]
Temperature at state 3 is,
[tex]T_{3} =\dfrac{T_4}{1+c_p(r_p^{\dfrac{\gamma-1}{\gamma}-1}-1)}[/tex]
Put the value in above equation we get,
[tex]T_3=1682[/tex]
- Net power output is,
[tex]W=mc_p(T_i{n}-T_{out})[/tex]
[tex]W=mc_p(T_3-T_2-T_4+T_{1})[/tex]
[tex]W=15.77\times1.109(1682-700-923+313)[/tex]
[tex]W=65[/tex]
Thus the net power output is 70 watt.
- The back work ratio,
[tex]r_b=\dfrac{T_2-T_1}{T_3-T_4}[/tex]
[tex]r_b=\dfrac{700-313}{1682-923}[/tex]
[tex]r_b=0.51[/tex]
Hence the back work ratio is 0.51.
- The thermal efficiency-
[tex]\eta=\dfrac{W}{Q_{in}}[/tex]
[tex]\eta=\dfrac{65}{mc_p(T_3-T_2)}[/tex]
[tex]\eta=0.379[/tex]
Hence, for the given problem the net power output is 70 watt,the back work ration is 0.51, and the thermal efficiency is 0.379.
For more about the Brayton cycle, follow the link below-
https://brainly.com/question/24696911