Compressor & Turbine

Effect of an Inefficient Compressor and Turbine
Frictional losses in the compressor raise the output temperature. Similarly the losses in the turbine
raise the exhaust temperature. These losses are quantified by modifying the temperatures T2 and T4
to account for their increases.
The compression ratio (P2/P1) of the compressor is usually given by the manufacturer and
therefore the temperature of the air leaving the compressor is easily found from (2.13). If the efficiency
of compression ηc is known e.g. 90% and that of the turbine ηt is known e.g. 85% then a better
estimate of the output energy can be calculated. In this situation T2 becomes T2e and T4 becomes
T4e, as follows:-
T2e = T2
ηc
+
1 − 1
ηc
T1 and T4e = T4ηt + (1 − ηt)T3 (2.18)
These would be the temperatures measurable in practice. In (2.14) and (2.15) the pressure
ratios are theoretically equal, and in practice nearly equal, hence:
T2
T1
= T3
T4
= rp
β (2.19)
Where rp is the pressure ratio
P2
P1
or
P3
P4
In practice the temperatures T1 and T3 are known from the manufacturer or from measuring
instruments installed on the machine. The pressure ratio rp is also known. The ratio of specific heats
is also known or can be taken as 1.4 for air. If the compressor and turbine efficiencies are taken into
account then the practical cycle efficiency ηp of the gas turbine can be expressed as:
ηp = T3(1 − rp
δ)ηcηt − T1(rp
β − 1)
T3ηc − T1(rp − 1 + ηc)
(2.20)
which has a similar form to (2.17) for comparison.
2.2.1.1 Worked example
A light industrial gas turbine operates at an ambient temperature T1 of 25◦C and the combustion
temperature T3 is 950◦C. The pressure ratio rp is 10.
If the overall efficiency is 32% find the efficiency of the compressor assuming the turbine
efficiency to be 86%.
From (2.20),
T1 = 273 + 25 = 298◦K
T3 = 273 + 950 = 1223◦K
rp
δ = 10−0.2857 = 0.51796 and rp
β = 10+0.2857 = 1.93063
Therefore,
ηp = 0.32 = 1223(1.0 − 0.51796)ηc(0.86) − 298(1.93063 − 1.0)
1223ηc − 298(1.93063 − 1.0 + ηc)
Transposing for ηc results in ηc = 0.894. Hence the compressor efficiency would be 89.4%.
2.2.2 Maximum Work Done on the Generator
If the temperatures T2e and T4e 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 rp.
The revised turbine work done Ute is,
Ute = Cp(T3 − T4)ηt kJ/kg (2.21)
The revised compressor work done Uce is,
Uce = Cp(T2 − T1)
1
ηc
kJ/kg (2.22)
The revised heat input from the fuel Uf e is,
Uf e = Cp(T3 − T2e) kJ/kg (2.23)
where,
T2e = T1
rp
β − 1 + ηc
ηc

From (2.19),
T4 = T3rp
δ (2.24)
and
T2 = T1rp
β (2.25)
Substituting for T2, T2e and T4 gives the resulting output work done Uoute to be,
Uoute = Ute − Uce = Cp(T3 − T3rp
δ)ηt − Cp
T1rp
β − T1
ηc

= Cp T3(1 − rδ)ηt − T1
ηc
(rp
β − 1) kJ/kg (2.26)
To find the maximum value of Uoute differentiate Uoute with respect to γp and equate the result
to zero. The optimum value of γp to give the maximum value of Uoute is,
rpmax =
T1
T3ηcηt
d
(2.27Where
d = 1
2δ
which when substituted in (2.26) gives the maximum work done Uoutemax.
2.2.2.1 Worked example
Find rpmax for the worked example in sub-section 2.2.1.1.
Given that,
T1 = 298 K, T3 = 1223◦C,
r = 1.4, ηt = 0.86 and ηc = 0.894
d = γ
2(1 − γ )
= 1.4
2(1.0 − 1.4)
= −1.75
rpmax =  298
1223(0.894)(0.86)

−1.75
= 0.3169−1.75 = 7.4
2.2.3 Variation of Specific Heat
As mentioned in sub-section 2.2 the specific heat Cp changes with temperature. From Reference 4,
Figure 4.4, an approximate cubic equation can be used to describe Cp in the range of temperature
300 K to 1300 K when the fuel-to-air ratio by mass is 0.01, and for the air alone for compression, as
shown in Table 2.1. The specific heat for the compressor can be denoted as Cpc and for the turbine
Cpt . The appropriate values of Cpc and Cpt can be found iteratively from the cubic expression and
equations (2.24) and (2.25). At each iteration the average of T1 and T2 can be used to recalculate Cpc,
and T3 and T4 to recalculate Cpt . The initial value of γ can be taken as 1.4 in both cases, and Cv
can be assumed constant at 0.24/1.4 = 0.171 kcal/kg K. The pressure ratio is constant. Having found
suitable values of Cpc and Cpt it is now possible to revise the equations for thermal efficiency ηpa
and output energy Uoutea, where the suffix ‘a’ is added to note the inclusion of variations in specific
heat Cp.
Table 2.1. Specific heat Cp as a cubic function of absolute temperature
K in the range 373 K to 1273 K Cp = a + bT + cT 2 + dT 3
Fuel-air Cubic equation constants
ratio
a × 100 b × 10−4 c × 10−7 d × 10−10
0.0 0.99653 −1.6117 +5.4984 −2.4164
0.01 1.0011 −1.4117 +5.4973 −2.4691
0.02 1.0057 −1.2117 +5.4962 −2.5218
The energy equations for the compressor and turbine become,
Ucea = Cpc(T2 − T1)
1
ηc
 kJ/kg (2.28)
and
Utea = Cpt (T3 − T4)
1
ηt
 kJ/kg (2.29)
Also assume that the specific heat Cpf of the fuel–air mixture is the value corresponding to
the average value of T2 and T3, see Reference 4, sub-section 4.7.1, (2.23).
Hence the fuel energy equation becomes, from (2.23),
Uf ea = Cpf (T3 − T2ea) kJ/kg (2.30)
Where
T2ea = T1(rp
βc − 1 + ηc)
ηc
(2.31)
Where rc and rt apply to the compressor and turbine and are found from Cpc, Cpt and Cv.
The work done on the generator is now,
Uoutea = Cpt T3(1 − rp
δt )ηt − CpcT1
ηc
(rp
βt − 1) (2.32)
and
T4ea = T3(ηt rp
δc + 1 − ηt )
From Uf ea and Uoutea the thermal efficiency ηpa can be found as,
ηpa = Uoutea
Uf ea
(2.33)