Respuesta :
Answer:
The value of temperature at compressor outlet = 543.43 K
The temperature at turbine outlet = 773.04 K
Back work ratio = 0.388
The value of efficiency of the Ideal Brayton cycle =0.448
Explanation:
Pressure ratio ([tex]r_{p}[/tex])= 8 And specific heat ratio ([tex]\gamma[/tex]) = 1.4
Compressor inlet temperature ([tex]T_{1}[/tex]) = 300 K
Turbine inlet temperature ([tex]T_{3}[/tex]) = 1400 K
(a). Temperature ratio inside the compressor is given by
[tex]\frac{T_{2} }{T_{1} }[/tex] = [tex]r_{p}\frac{\gamma - 1}{\gamma}[/tex]
Where [tex]T_{1}[/tex] & [tex]T_{2}[/tex] represents the compressor inlet and outlet temperature.
Put all the values in the given formula we get
⇒ [tex]\frac{T_{2} }{300}[/tex] = [tex]8^{\frac{1.4 - 1}{1.4} }[/tex]
⇒ [tex]\frac{T_{2} }{300}[/tex] = 1.811
⇒ [tex]T_{2}[/tex] = 543.43 K
This is the value of temperature at compressor outlet.
The temperature ratio inside the turbine is given by
[tex]\frac{T_{3} }{T_{4} }[/tex] = [tex]r_{p}\frac{\gamma - 1}{\gamma}[/tex]
Where [tex]T_{3}[/tex] & [tex]T_{4}[/tex] represents the turbine inlet & outlet temperatures.
Put all the values in the given formula we get
⇒ [tex]\frac{1400}{T_{4} }[/tex] = [tex]8^{\frac{1.4 - 1}{1.4} }[/tex]
⇒ [tex]T_{4}[/tex] = 773.05 K
This is the temperature at turbine outlet.
(b). The work done by the turbine is given by the formula ([tex]W_{T}[/tex]) = [tex]C_{p}[/tex] ([tex]T_{3}[/tex] - [tex]T_{4}[/tex] )
Put all the values in the above formula we get [tex]W_{T}[/tex] = 1.005 × (1400 - 773.05)
[tex]W_{T}[/tex] = 630.08 [tex]\frac{KJ}{Kg}[/tex]
And work done inside the pump is given by [tex]W_{c}[/tex] = [tex]C_{p}[/tex] ([tex]T_{2}[/tex] - [tex]T_{1}[/tex])
Put all the values in the above formula we get [tex]W_{c}[/tex] = 1.005 × (543.43 - 300)
[tex]W_{c}[/tex] = 244.65 [tex]\frac{KJ}{Kg}[/tex]
Back work ratio is given by [tex]r_{b}[/tex] = [tex]\frac{W_{C} }{W_{}T}[/tex]
Put the values of [tex]W_{c}[/tex] & [tex]W_{T}[/tex] in above formula we get
⇒ [tex]r_{b}[/tex] = [tex]\frac{244.65}{630.08}[/tex]
⇒ [tex]r_{b}[/tex] = 0.388
This is the value of back work ratio.
(C). Thermal Efficiency is given by E = 1 - [tex]\frac{1}{r_{p} ^{\frac{\gamma - 1}{\gamma} } }[/tex]
⇒ E = 1 - [tex]\frac{1}{8 ^{\frac{1.4 - 1}{1.4} } }[/tex]
⇒ E = 0.448
This is the value of efficiency of the Ideal Brayton cycle.
