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

= 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 =

=

= 50%
(b) the thermal efficiency
η = ΔW/

= (
-
)/

=
/950
= 29.6%