Page 51 - Handbook of Electrical Engineering

30 HANDBOOK OF ELECTRICAL ENGINEERING

Therefore,
1223(1.0 − 0.51796)η c (0.86) − 298(1.93063 − 1.0)
η p = 0.32 =
1223η c − 298(1.93063 − 1.0 + η c )
Transposing for η c results in η c = 0.894. Hence the compressor efficiency would be 89.4%.

2.2.2 Maximum Work Done on the Generator

If the temperatures T 2e and T 4e are used in (2.11) to compensate for the efficiencies of the compressor
and turbine, then it is possible to determine the maximum power output that can be obtained as a
function of the pressure ratio r p .
The revised turbine work done U te is,

U te = C p (T 3 − T 4 )η t kJ/kg (2.21)

The revised compressor work done U ce is,

1
U ce = C p (T 2 − T 1 ) kJ/kg (2.22)
η c
The revised heat input from the fuel U fe is,

U fe = C p (T 3 − T 2e ) kJ/kg (2.23)
where,
β

r p − 1 + η c
T 2e = T 1
η c
From (2.19),
δ
T 4 = T 3 r p (2.24)
and
β
T 2 = T 1 r p (2.25)
Substituting for T 2 , T 2e and T 4 gives the resulting output work done U oute to be,

β

δ T 1 r p − T 1
U oute = U te − U ce = C p (T 3 − T 3 r p )η t − C p
η c

δ T 1 β
= C p T 3 (1 − r )ηt − (r p − 1) kJ/kg (2.26)
η c
To find the maximum value of U oute differentiate U oute with respect to γ p and equate the result
to zero. The optimum value of γ p to give the maximum value of U oute is,

d

T 1
r pmax = (2.27)
T 3 η c η t