Gas Turbine Power Cycles [2]

3. The Effect of Friction on The Joule Cycle

3.1 TURBINE
The isentropic efficiency for a gas turbine is given by:
ηi = (Actual change in enthalpy)/(Ideal change in enthalpy)
ηi = (Actual change in temperature)/(Ideal change in temperature)

3.2 COMPRESSOR
For a compressor the isentropic efficiency is inverted and becomes as follows.
ηi = (Ideal change in enthalpy)/(Actual change in enthalpy)
hi = (Ideal change in temperature)/(Actual change in temperature)

Remember that friction always produces a smaller change in temperature than for the ideal
case. This is shown on the T-s diagrams (fig.4a and 4b).

The power output from the turbine is hence
P(out) = m cp (T3 – T4’) ηi

The power input to the compressor is hence

P(in) = m cp (T2’ – T1)/ηi

3.3 THE CYCLE WITH FRICTION
It can be seen that the effect of friction on the gas turbine cycle is reduced power output
and increased power input with an overall reduction in nett power and thermal efficiency.
Figs. 5a and 5b show the effect of friction on T-s and p-h diagrams for the Joule cycle.

Note the energy balance which exists is
P(in) + Φ(in) = P(out) + Φ(out) P(nett) = P(out) - P(in) = Φ(nett) = Φ(in) - Φ(out)

Worked Example No.3
A Joule Cycle uses a pressure ratio of 8. Calculate the air standard efficiency. The
isentropic efficiency of the turbine and compressor are both 90%. The low pressure in
the cycle is 120 kPa. The coldest and hottest temperatures in the cycle are 20oC and
1200oC respectively. Calculate the cycle efficiency with friction and deduce the
change. Calculate the nett power output. γ = 1.4 and cp = 1.005 kJ/kg K. Take the mass
flow as 3 kg/s.

Solution

No friction ηth = 1 - rp1/γ -1 = 0.448 or 48.8 %
With friction T2' = 293 x 8 0.286 = 531 K
ηi = 0.9 = (531-293)/(T2-293) T2 = 531 K
T4' = 1473/8 0.286 = 812.7 K
ηi = 0.9 = (1473-T4)/(1473-812.7) T4= 878.7
ηth = 1 - Φ(out)/Φ(in) = 1 - (T4-T1)/(T3-T2)
ηth= 0.36 or 36%
The change in efficiency is a reduction of 8.8%
Φ(in) = m cp (T3-T2) = 3x1.005 x (1473-557) = 2760 kW
Nett Power Output = P(nett) = ηth x Φ(in) = 0.36 x 2760 = 994 kW

Self Assessment Exercise ELF No. 3
A gas turbine uses a standard Joule cycle but there is friction in the compressor and
turbine. The air is drawn into the compressor at 1 bar 15oC and is compressed with an
isentropic efficiency of 94% to a pressure of 9 bar. After heating, the gas temperature
is 1000oC. The isentropic efficiency of the turbine is also 94%. The mass flow rate is
2.1 kg/s. Determine the following.
1. The net power output.
2. The thermal efficiency of the plant.
γ = 1.4 and cp = 1.005 kJ/kg K.
(Answers 612 kW and 40.4%)

4. Variants of The Basic Cycle
In this section we will examine how practical gas turbine engine sets vary from the basic
Joule cycle.

4.1 GAS CONSTANTS
The first point is that in reality, although air is used in the compressor, the gas going
through the turbine contains products of combustion so the adiabatic index and specific
heat capacity is different in the turbine and compressor.

4.2 FREE TURBINES
Most designs used for gas turbine sets use two turbines, one to drive the compressor and a
free turbine. The free turbine drives the load and it is not connected directly to the
compressor. It may also run at a different speed to the compressor.
Fig.6a. shows such a layout with turbines in parallel configuration.

Fig.6b shows the layout with series configuration.

4.3 INTERCOOLING
This is not part of the syllabus for the power cycles but we will come across it later when
we study compressors in detail. Basically, if the air is compressed in stages and cooled
between each stage, then the work of compression is reduced and the efficiency increased.
The layout is shown on fig. 7a.

4.4 REHEATING
The reverse theory of intercooling applies. If several stages of expansion are used and the
gas reheated between stages, the power output and efficiency is increased. The layout is
shown on fig. 7b.

Worked Example No.4
A gas turbine draws in air from atmosphere at 1 bar and 10oC and compresses it to 5
bar with an isentropic efficiency of 80%. The air is heated to 1200 K at constant
pressure and then expanded through two stages in series back to 1 bar. The high
pressure turbine is connected to the compressor and produces just enough power to
drive it. The low pressure stage is connected to an external load and produces 80 kW of
power. The isentropic efficiency is 85% for both stages.
Calculate the mass flow of air, the inter-stage pressure of the turbines and the thermal
efficiency of the cycle.
For the compressor γ = 1.4 and for the turbines γ = 1.333.
The gas constant R is 0.287 kJ/kg K for both.
Neglect the increase in mass due to the addition of fuel for burning.

Solution

Hence cp = 1.005 kJ/kg K for the compressor and 1.149 kJ/kg K for the turbines.

COMPRESSOR

Power input to compressor = m cp (T2-T1)
Power output of h.p. turbine = m cp (T3-T4)
Since these are equal it follows that
1.005(489.8-283)=1.149(1200-T4)

T4 =1019.1 K

HIGH PRESSURE TURBINE

LOW PRESSURE TURBINE

NETT POWER
The nett power is 80 kW hence
80 = m cp(T4-T5) = m x 1.149(1019.1 - 854.5) m = 0.423 kg/s

HEAT INPUT
Φ(in) = m cp (T3-T2) = 0.423 x 1.149 (1200 - 489.8) = 345.2 kW
THERMAL EFFICIENCY
ηth = P(nett)/Φ(in) = 80/345.2 = 0.232 or 23.2%

Self Assessment Exercise No. 4
A gas turbine draws in air from atmosphere at 1 bar and 15oC and compresses it to 4.5
bar with an isentropic efficiency of 82%. The air is heated to 1100 K at constant
pressure and then expanded through two stages in series back to 1 bar. The high
pressure turbine is connected to the compressor and produces just enough power to
drive it. The low pressure stage is connected to an external load and produces 100 kW
of power. The isentropic efficiency is 85% for both stages.
For the compressor γ = 1.4 and for the turbines γ = 1.3. The gas constant R is 0.287
kJ/kg K for both.
Neglect the increase in mass due to the addition of fuel for burning.
Calculate the mass flow of air, the inter-stage pressure of the turbines and the thermal
efficiency of the cycle.
(Answers 0.642 kg/s and 20.1 %)