Analyzing the Ideal Brayton Cycle
Air enters the compressor of an ideal air-standard Brayton cycle at 100 kPa, 300 K, with a volumetric flow rate of 5 m ^{3} / s. The compressor pressure ratio is 10. The turbine inlet temperature is 1400 K. Determine (a) the thermal efficiency of the cycle, (b) the back work ratio, (c) the net power developed, in kW.
Known An ideal air-standard Brayton cycle operates with given compressor inlet conditions, given turbine inlet temperature, and a known compressor pressure ratio.
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. The control volumes are shown on the accompanying sketch by dashed lines.
2. The turbine and compressor processes are isentropic.
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.
1 Analysis Table A-22The analysis begins by determining the specific enthalpy at each numbered state of the cycle. At state 1, the temperature is 300 K. From , h_{1}=300.19 kJ/kg and p_{ r 1}=1.386.
Since the compressor process is isentropic, the following relationship can be used to determine h_{2}:
p_{ r 2}=\frac{p_{2}}{p_{1}} p_{ r 1}=(10)(1.386)=13.86
Then, interpolating in Table A-22, we obtain h_{2}=579.9 kJ/kg.
The temperature at state 3 is given as T_{3}=1400 K. With this temperature, the specific enthalpy at state 3 from Table A-22 is h_{3}=1515.4 kJ / kg \text { . Also, } p_{ r 3}=450.5.
The specific enthalpy at state 4 is found by using the isentropic relationship
p_{ r 4}=p_{ r 3} \frac{p_{4}}{p_{3}}=(450.5)(1 / 10)=45.05
Interpolating in Table A-22, we get h_{4}=808.5 kJ/kg.
a. The thermal efficiency is
\eta=\frac{\left(\dot{W}_{ t } / \dot{m}\right)-\left(\dot{W}_{ c } / \dot{m}\right)}{\dot{Q}_{ in } / \dot{m}}
=\frac{\left(h_{3}-h_{4}\right)-\left(h_{2}-h_{1}\right)}{h_{3}-h_{2}}=\frac{(1515.4-808.5)-(579.9-300.19)}{1515.4-579.9}
=\frac{706.9-279.7}{935.5}=0.457(45.7 \%)
b. The back work ratio is
2 bwr =\frac{\dot{W}_{ c } / \dot{m}}{\dot{W}_{ t } / \dot{m}}=\frac{h_{2}-h_{1}}{h_{3}-h_{4}}=\frac{279.7}{706.9}=0.396(39.6 \%)
c. The net power developed is
\dot{W}_{\text {cycle }}=\dot{m}\left[\left(h_{3}-h_{4}\right)-\left(h_{2}-h_{1}\right)\right]
To evaluate the net power requires the mass flow rate \dot{m} , which can be determined from the volumetric flow rate and specific volume at the compressor inlet as follows:
\dot{m}=\frac{( AV )_{1}}{v_{1}}
Since v_{1}=(\bar{R} / M) T_{1} / p_{1}, this becomes
\dot{m}=\frac{( AV )_{1} p_{1}}{(\bar{R} / M) T_{1}}=\frac{\left(5 m ^{3} / s \right)\left(100 \times 10^{3} N / m ^{2}\right)}{\left(\frac{8314}{28.97} \frac{ N \cdot m }{ kg \cdot k }\right)(300 K )}
= 5.807 kg/s
Finally,
\dot{W}_{\text {cycle }}=(5.807 kg / s )(706.9-279.7)\left(\frac{ kJ }{ kg }\right)\left|\frac{1 kW }{1 kJ / s }\right|=2481 kW
1 The use of the ideal gas table for air is featured in this solution. A solution also can be developed on a cold air-standard basis in which constant specific heats are assumed. The details are left as an exercise, but for comparison the results are presented for the case k = 1.4 in the following table:
| Parameter | Air-Standard Analysis | Cold Air-Standard Analysis, k = 1.4 |
| T_{2} | 574.1 K | 579.2 K |
| T_{4} | 787.7 K | 725.1 K |
| \eta | 0.457 | 0.482 |
| bwr | 0.396 | 0.414 |
| \dot{W}_{\text {cycle }} | 2481 kW | 2308 kW |
2 The value of the back work ratio in the present gas turbine case is significantly greater than the back work ratio of the simple vapor power cycle of Example 8.1.
Skills Developed
Ability to…
• sketch the schematic of the basic air-standard gas turbine and the T–s diagram for the corresponding ideal Brayton cycle.
• evaluate temperatures and pressures at each principal state and retrieve necessary property data.
• calculate the thermal efficiency and back work ratio.
Quick Quiz
Determine the rate of heat transfer to the air passing through the combustor, in kW. Ans. 5432 kW.
