Page 78 - Advanced Gas Turbine Cycles

54 Advanced gas turbine cycles
There are two approaches to integrating this equation:
(a) the three terms can be integrated separately to give a p, T, $ relation; and
(b) two of the three terms can be brought together if an expression for d@dT is known; a
more familiar polytropic p, T type of relation can then be obtained.
In the first approach [T' - T,,,,]/T in Eq. (4.14) may be written approximately as
[ 1 - (Tcomp/T)], for a process which does not deviate too far from the original (uncooled)
isentropic expansion; and further (TcomdT') may be approximated to T2/T3 = x/8. Then
Eq. (4.14) may be integrated to give
T/p(y-'yy = C/[1 + $I6, (4.15)

where 6 = 1 - (x/f3). The cooling is carried out over the full turbine expansion, to an exit
state (PI, TE), so

i& = TE/Tl = (O/x)/[l + I/+$ < (Wx). (4.16)
In the second approach, a value for & is not assumed but a relationship for d$/dT is
determined from semi-empirical expressions for the amount of cooling air that is
required in an (elementary) turbine blade row. One such relationship, derived in Ref. [5],
gives
&llc1/[1 + $1 = -AdT/T, (4.17)

where A = 2Cw+[1 - (x/@)]/[@(y - 1)M3 = 2Cw+6/[@(y - 1)M,3, in which Cand w+
are parameters obtained from the definition of the blade cooling effectiveness, Mu is the
blade Mach number and @ = cPAT/U2 is the stage loading coefficient, with AT the
(positive) temperature drop across the stage.
Eq. (4.14) can then be integrated to give
TIPu = constant, (4.18)
where I+ = (y - 1)/$1 - A) and it follows that

i& = TE/TI = 8/r'. (4.19)

4.2.1.4. The turbine exit condition (for reversible cooled cycles)
There is a link between the thermal efficiency and the turbine exit temperature TE. It
results from expressing the thermal efficiency of the cycle in the form
7 = [ 1 - QA/&] = 1 - (l/X), (4.20)
and it has been argued that this equation is valid for all the reversible cycles considered
above (except for the second step cooling by throttled compressor delivery air,
[CWIRCZT).
The heat supplied is QB = cp[T3 - T2], and for each of these reversible cycles the heat
rejected will be QA = cp( 1 + I/+)(TE - TI), Thus, the efficiency is given by