A gas-turbine power plant operates on the simple Brayton cycle between the pressure limits of 100 and 1600 kPa. The working fluid is air, which enters the compressor at 40∘C at a rate of 850m3/min and leaves the turbine at 650∘C. Assuming a compressor isentropic efficiency of 85 percent and a turbine isentropic efficiency of 88 percent, determine
(a) the net power output,
(b) the back work ratio, and
(c) the thermal efficiency.
Use constant specific heats with
cv=0.821kJ/kg?
K,cp=1.108kJ/kg?
K, and k=1.35.

Respuesta :

Answer:

a) 6498.84 kW

b) 0.51

c) 0.379

Explanation:

See the attached picture below for the solution

Ver imagen barackodam
Ver imagen barackodam

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

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