Handbook of Electrical Engineering

30 HANDBOOK OF ELECTRICAL ENGINEERING

Therefore,

ηp= 0. 32 =

1223 ( 1. 0 − 0. 51796 )ηc( 0. 86 )− 298 ( 1. 93063 − 1. 0 )

1223 ηc− 298 ( 1. 93063 − 1. 0 +ηc)

Transposing forηcresults inηc= 0 .894. Hence the compressor efficiency would be 89.4%.

2.2.2 Maximum work done on the generator

If the temperaturesT 2 eandT 4 eare 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 ratiorp.

The revised turbine work doneUteis,

Ute=Cp(T 3 −T 4 )ηtkJ/kg ( 2. 21 )

The revised compressor work doneUceis,

Uce=Cp(T 2 −T 1 )

1

ηc

kJ/kg ( 2. 22 )

The revised heat input from the fuelUfeis,

Ufe=Cp(T 3 −T 2 e)kJ/kg ( 2. 23 )

where,

T 2 e=T 1

(

rpβ− 1 +ηc

ηc

)

From (2.19),
T 4 =T 3 rpδ ( 2. 24 )

and
T 2 =T 1 rpβ ( 2. 25 )

Substituting forT 2 ,T 2 eandT 4 gives the resulting output work doneUouteto be,

Uoute=Ute−Uce=Cp(T 3 −T 3 rpδ)ηt−Cp

(

T 1 rpβ−T 1

ηc

)

=Cp

[

T 3 ( 1 −rδ)ηt−

T 1

ηc

(rpβ− 1 )

]

kJ/kg (2.26)

To find the maximum value ofUoutedifferentiateUoutewith respect toγpand equate the result
to zero. The optimum value ofγpto give the maximum value ofUouteis,

rpmax=

(

T 1

T 3 ηcηt

)d

( 2. 27 )