Answer:
- the fraction of the turbine work output used to drive the compressor is 50%
- the thermal efficiency is 29.6%
Explanation:
given information:
= 300 k
= 700 kPa
= 580 K
Qin = 950 kj/kg
efficiency, η = 86% = 0.86
Assume, Ideal gas specific heat capacities of air,
= 1.005
k = 1.4
(a) the fraction of the turbine work output used to drive the compressor
Qin =
(
-
)
-
= Qin /
= Qin /
+
thus,
=
= 1525 K
=
\
=
so,
=
\
=
\
= 1525 K ![(\frac{100}{700} )^{1.4-1/1.4}](https://tex.z-dn.net/?f=%28%5Cfrac%7B100%7D%7B700%7D%20%29%5E%7B1.4-1%2F1.4%7D)
= 875 K
=
(
-
)
= 1.005 kj/kgK (580 K - 300 K)
= 281.4 kJ/kg
= η
(
-
)
= 0.86 (1.005 kj/kgK) (1525 k - 875 K)
= 562.2 kJ/kg
therefore,
the fraction of turbine = ![\frac{W_{in} }{W_{out} }](https://tex.z-dn.net/?f=%5Cfrac%7BW_%7Bin%7D%20%7D%7BW_%7Bout%7D%20%7D)
= ![\frac{281.4}{562.2}](https://tex.z-dn.net/?f=%5Cfrac%7B281.4%7D%7B562.2%7D)
= 50%
(b) the thermal efficiency
η = ΔW/![Q_{in}](https://tex.z-dn.net/?f=Q_%7Bin%7D)
= (
-
)/![Q_{in}](https://tex.z-dn.net/?f=Q_%7Bin%7D)
=
/950
= 29.6%