Evaluating Performance of a Brayton Cycle with Irreversibilities
Reconsider Example 9.4, but include in the analysis that the turbine and compressor each have an isentropic efficiency of 80%. Determine for the modified cycle (a) the thermal efficiency of the cycle, (b) the back work ratio, (c) the net power developed, in kW.
Known An air-standard Brayton cycle operates with given compressor inlet conditions, given turbine inlet temperature, and known compressor pressure ratio. The compressor and turbine each have an isentropic efficiency of 80%.
Find Determine the thermal efficiency, the back work ratio, and the net power developed, in kW.
Schematic and Given Data:
Engineering Model
1. Each component is analyzed as a control volume at steady state.
2. The compressor and turbine are adiabatic.
3. There are no pressure drops for flow through the heat exchangers.
4. Kinetic and potential energy effects are negligible.
5. The working fluid is air modeled as an ideal gas.
Analysis
a. The thermal efficiency is given by
\eta=\frac{\left(\dot{W}_{ t } / \dot{m}\right)-\left(\dot{W}_{ c } / \dot{m}\right)}{\dot{Q}_{\text {in }} / \dot{m}}
The work terms in the numerator of this expression are evaluated using the given values of the compressor and turbine isentropic efficiencies as follows:
The turbine work per unit of mass is
\frac{\dot{W}_{ t }}{\dot{m}}=\eta_{ t }\left(\frac{\dot{W}_{ t }}{\dot{m}}\right)_{ s }
where \eta_{ t } is the turbine efficiency. The value of \left(\dot{W}_{ t } / \dot{m}\right)_{ s } is determined in the solution to Example 9.4 as 706.9 kJ/kg. Thus,
1 \frac{\dot{W}_{ t }}{\dot{m}}=0.8(706.9)=565.5 kJ / kg
For the compressor, the work per unit of mass is
\frac{\dot{W}_{ c }}{\dot{m}}=\frac{\left(\dot{W}_{ c } / \dot{m}\right)_{ s }}{\eta_{ c }}
where \eta_{ c } is the compressor efficiency. The value of \left(\dot{W}_{ c } / \dot{m}\right)_{ s } is determined in the solution to Example 9.4 as 279.7 kJ/kg, so
\frac{\dot{W}_{ c }}{\dot{m}}=\frac{279.7}{0.8}=349.6 kJ / kg
The specific enthalpy at the compressor exit, h_{2}, is required to evaluate the denominator of the thermal efficiency expression. This enthalpy can be determined by solving
\frac{\dot{W}_{ c }}{\dot{m}}=h_{2}-h_{1}
to obtain
h_{2}=h_{1}+\dot{W}_{ c } / \dot{m}
Inserting known values
h_{2}=300.19+349.6=649.8 kJ / kg
The heat transfer to the working fluid per unit of mass flow is then
\frac{\dot{Q}_{\text {in }}}{\dot{m}}=h_{3}-h_{2}=1515.4-649.8=865.6 kJ / kg
where h_{3} is from the solution to Example 9.4.
Finally, the thermal efficiency is
\eta=\frac{565.5-349.6}{865.6}=0.249(24.9 \%)
b. The back work ratio is
bwr =\frac{\dot{W}_{ c } / \dot{m}}{\dot{ W }_{ t } / \dot{m}}=\frac{349.6}{565.5}=0.618(61.8 \%)
c. The mass flow rate is the same as in Example 9.4. The net power developed by the cycle is then
2 \dot{W}_{\text {cycle }}=\left(5.807 \frac{ kg }{ s }\right)(565.5-349.6) \frac{ kJ }{ kg }\left|\frac{1 kW }{1 kJ / s }\right|=1254 kW
1 The solution to this example on a cold air-standard basis is left as an exercise.
2 Irreversibilities within the turbine and compressor have a significant impact on the performance of gas turbines. This is brought out by comparing the results of the present example with those of Example 9.4. Irreversibilities result in an increase in the work of compression and a reduction in work output of the turbine. The back work ratio is greatly increased and the thermal efficiency significantly decreased. Still, we should recognize that the most significant irreversibility of gas turbines by far is combustion irreversibility.
Skills Developed
Ability to…
• sketch the schematic of the basic air-standard gas turbine and the T–s diagram for the corresponding Brayton cycle with compressor and turbine irreversibilities.
• evaluate temperatures and pressures at each principal state and retrieve necessary property data.
• calculate the thermal efficiency and back work ratio.
Quick Quiz
What would be the thermal efficiency and back work ratio if the isentropic turbine efficiency were 70% keeping isentropic compressor efficiency and other given data the same? Ans. η = 16.8%, bwr = 70.65%.
