EGR 334 Thermodynamics Chapter 9: Sections 7-8 Lecture 36: Reheat and Intercooling of Gas Turbine Systems Quiz Today?

Today’s main concepts: Be able to explain the concept and purpose of using reheat in a gas turbine. Be able to explain the concept and purpose of using intercooling in a gas turbine system. Be able draw and explain Brayton cycles with reheat and intercooling. Reading Assignment: Read Chapter 10 Homework Assignment: Problem 9:80

Process 1 – 2 : Isentropic compression of air (compressor). Sec 9.5 : Modeling Gas Turbine Power Plants Last time: Introduced to the Gas Turbine Power Plant Working fluid is air Heat transfer from an external source (assumes there is no reaction) Process 1 – 2 : Isentropic compression of air (compressor). http://www.utpb.edu/research-grants/ht3r/description/components/brayton-cycle-lab/ Process 2 – 3 : Constant pressure heat transfer to the air from an external source (combustion) Process 3 – 4 : Isentropic expansion (through turbine) Process 4 – 1 : Completes cycle by a constant volume pressure in which heat is rejected from the air

How a Jet Engine Works: Jet Engine http://www.youtube.com/watch?NR=1&feature=endscreen&v=ON0sVe1yeOk Power Station Gas Turbine Images from: http://library.thinkquest.org/C006011/english/sites/gasturbine.php3?v=2

Possible Enhancements to the Brayton Cycle: 1) Use of Regenerator to preheat combustion air. 2) Reheating air between successive turbine stages. 3) Intercooling of air between successive compressor stages.

Brayton Cycle with Regeneration: The exhaust air out of the turbine contains significant waste heat that would normally be discarded to the atmosphere. Use of regenerator can make use of this heat that would be discarded to preheat the air on its way to the combustor, allowing the burned fuel to be used more efficiently.

Sec 9.8 : Regenerative Gas Turbines with Reheat and Intercooling To limit the temperature of the gas entering the turbine, air is provided in excess to the primary combustion chamber. The excess air limits the temperature of the combustion process by providing a heat sink for the heat produced during combustion. This has the added advantage of having the second turbine run at lower pressure. Processes 2-3, a-b, and 4-1 occur isobarically.

Sec 9.8 : Regenerative Gas Turbines with Reheat and Intercooling It is not practice to achieve such cooling within a compressor. Consequently, the compressor is spit with a heat-exchanger in between. Improved efficiency may also be achieved by decreasing the work required for the compressors.

Sec 9.8 : Regenerative Gas Turbines with Reheat and Intercooling Putting all three enhancements together, the diagram show a process combining regeneration, reheat, and intercooling.

The volumetric flow rate entering the cycle. Example (9.74): Air enters the compressor of a gas turbine at 100 kPa, 300 K. The air is compressed in two stages to 900 kPa, with intercooling to 300 K between the stages at a pressure of 300 kPa. The turbine inlet temperature is 1480 K and the expansion occurs in two stages, with reheat to 1420 K between the stages at a pressure of 300 kPa. The compressor and turbine stage efficiencies are 84 and 82% respectively, The net power developed is 1.8 W. Determine State 1 a b 2 3 c d 4 T (K) 300 1480 1420 P (kPa) 100 900 Pr h (kJ/kg) The volumetric flow rate entering the cycle. The thermal efficiency of the cycle. The back work ratio.

The thermal efficiency of the cycle. The back work ratio. a b Example (9.74): The compressor and turbine stage efficiencies are 84 and 82% respectively, The net power developed is 1.8 W. Determine The thermal efficiency of the cycle. The back work ratio. State 1 a b 2 3 c d 4 T (K) 300 1480 1420 p (kPa) 100 900 pr 1.3860 568.8 478.0 h (kJ/kg) 300.19 1611.79 1539.44 Find state a, Process 1 – a is Isentropic compression From the table haS = 411.26 kJ/kg Using the isentropic compressor efficiency:

The thermal efficiency of the cycle. The back work ratio. a b Example (9.74): The compressor and turbine stage efficiencies are 84 and 82% respectively, The net power developed is 1.8 W. Determine The thermal efficiency of the cycle. The back work ratio. State 1 a b 2 3 c d 4 T (K) 300 1480 1420 P (kPa) 100 900 pr 1.3860 568.8 478.0 h (kJ/kg) 300.19 432.42 1611.79 1539.44 Find state a, Process b – 2 is Isentropic compression From the table h2S = 411.26 kJ/kg Using the isentropic compressor efficiency:

The thermal efficiency of the cycle. The back work ratio. a b a b Example (9.74): The compressor and turbine stage efficiencies are 84 and 82% respectively, The net power developed is 1.8 W. Determine The thermal efficiency of the cycle. The back work ratio. State 1 a b 2 3 c d 4 T (K) 300 1480 1420 p (kPa) 100 900 pr 1.3860 568.8 478.0 h (kJ/kg) 300.19 432.42 1611.79 1275.4 1539.44 State 1 a b 2 3 c d 4 T (K) 300 1480 1420 p (kPa) 100 900 pr 1.3860 568.8 478.0 h (kJ/kg) 300.19 432.42 1611.79 1539.44 Find state a, Process 3 – c is Isentropic expansion From the table hcS = 1201.5 kJ/kg Using the isentropic turbine efficiency:

Find state a, Process d – 4 is Isentropic expansion Example (9.74): The compressor and turbine stage efficiencies are 84 and 82% respectively, The net power developed is 1.8 W. Determine State 1 a b 2 3 c d 4 T (K) 300 1480 1420 p (kPa) 100 900 pr 1.3860 568.8 478.0 h (kJ/kg) 300.19 432.42 1611.79 1275.4 1539.44 1216.77 State 1 a b 2 3 c d 4 T (K) 300 1480 1420 p (kPa) 100 900 pr 1.3860 568.8 478.0 h (kJ/kg) 300.19 432.42 1611.79 1275.4 1539.44 Find state a, Process d – 4 is Isentropic expansion From the table h4S = 1145.94 kJ/kg Using the isentropic turbine efficiency:

The thermal efficiency of the cycle. The back work ratio. Example (9.74): The compressor and turbine stage efficiencies are 84 and 82% respectively, The net power developed is 1.8 W. Determine State 1 a b 2 3 c d 4 T (K) 300 1480 1420 p (kPa) 100 900 pr 1.3860 568.8 478.0 h (kJ/kg) 300.19 432.42 1611.79 1275.4 1539.44 1216.77 The thermal efficiency of the cycle. The back work ratio. Determine the mass flow rate

The volumetric flow rate. The thermal efficiency of the cycle. Example (9.74): Air enters the compressor of a gas turbine at 100 kPa, 300 K. The air is compressed in two stages to 900 kPa, with intercooling to 300 K between the stages at a pressure of 300 kPa. The turbine inlet temperature is 1480 K and the expansion occurs in two stages, with reheat to 1420 K between the stages at a pressure of 300 kPa. The compressor and turbine stage efficiencies are 84 and 82% respectively, The net power developed is 1.8 W. Determine State 1 a b 2 3 c d 4 T (K) 300 1480 1420 p (kPa) 100 900 pr 1.3860 568.8 478.0 h (kJ/kg) 300.19 432.42 1611.79 1275.4 1539.44 1216.77 The volumetric flow rate. The thermal efficiency of the cycle. The back work ratio.

The volumetric flow rate. The thermal efficiency of the cycle. Example (9.74): Air enters the compressor of a gas turbine at 100 kPa, 300 K. The air is compressed in two stages to 900 kPa, with intercooling to 300 K between the stages at a pressure of 300 kPa. The turbine inlet temperature is 1480 K and the expansion occurs in two stages, with reheat to 1420 K between the stages at a pressure of 300 kPa. The compressor and turbine stage efficiencies are 84 and 82% respectively, The net power developed is 1.8 W. Determine State 1 a b 2 3 c d 4 T (K) 300 1480 1420 p (kPa) 100 900 pr 1.3860 568.8 478.0 h (kJ/kg) 300.19 432.42 1611.79 1275.4 1539.44 1216.77 The volumetric flow rate. The thermal efficiency of the cycle. The back work ratio.

The volumetric flow rate. The thermal efficiency of the cycle. Example (9.74): Air enters the compressor of a gas turbine at 100 kPa, 300 K. The air is compressed in two stages to 900 kPa, with intercooling to 300 K between the stages at a pressure of 300 kPa. The turbine inlet temperature is 1480 K and the expansion occurs in two stages, with reheat to 1420 K between the stages at a pressure of 300 kPa. The compressor and turbine stage efficiencies are 84 and 82% respectively, The net power developed is 1.8 W. Determine State 1 a b 2 3 c d 4 T (K) 300 1480 1420 p (kPa) 100 900 pr 1.3860 568.8 478.0 h (kJ/kg) 300.19 432.42 1611.79 1275.4 1539.44 1216.77 The volumetric flow rate. The thermal efficiency of the cycle. The back work ratio.

end of slides for Lecture 36