Gas Power Cycles
Multi-Cylinder Spark Ignition Engine
Power Cycles Ideal Cycles, Internal Combustion Otto cycle, spark ignition Diesel cycle, compression ignition Brayton cycles Combined cycle
Ideal Cycles Assumptions Air is the working fluid, circulated in a closed loop, is an ideal gas exhaust and air intake are substituted with heat transfer from the system to the surroundings combustion is replaced by heat transfer from an external source to the system all processes are internally reversible. gas specific heat is constant Cycle does not involve any friction Pipes connecting components have no heat loss Neglecting changes in kinetic and potential energy
Carnot Cycle
Carnot Cycle
Otto Cycle Nicolaus August Otto the inventor of the four-stroke cycle was born on 14th June 1831 in Germany. In 1862 he began first experiments with four-strokes engines. The first four-stroke engines is shown. He died on 26th January 1891 in Cologne
Engine Terms Bottom-dead center (BDC) – piston position where volume is maximum Top-dead center (TDC) – piston position where volume is minimum Clearance volume – minimum cylinder volume (VTDC = V2) Compression ratio (r) Displacement volume ME 152 8
Engine Terms Mean effective pressure (MEP)
Processes of Otto Cycle Four internally reversible processes 1-2 Isentropic compression 2-3 Constant-volume heat addition 3-4 Isentropic expansion 4-1 Constant-volume heat rejection
Otto Cycle
Ideal Otto Cycle (cont.) 1st law for this cycle: energy conversion efficiency is:
Ideal Otto Cycle (cont.) for an isentropic process: in case of an ideal gas: Compression ratio
Ideal Otto Cycle (cont.)
Diesel Cycle Rudolf Diesel (1858 – 1913) was born in Paris in 1858. After graduation he was employed as a refrigerator engineer. However, his true love was in engine design. In 1893, he published a paper describing an engine with combustion within a cylinder, the internal combustion engine. In 1894, he filed for a patent for his new invention, the diesel engine. He operated his first successful engine in 1897.
Spark or Compression Ignition Spark (Otto), air-fuel mixture compressed (constant-volume heat addition) Compression (Diesel), air compressed, then fuel added (constant-pressure heat addition)
Early CI Engine Cycle vs Diesel Cycle FUEL Fuel injected at TC A I R Fuel/Air Mixture Combustion Products Actual Cycle Intake Stroke Compression Stroke Power Stroke Exhaust Stroke Qin Qout Air Diesel Cycle TC BC Compression Process Const pressure heat addition Process Expansion Process Const volume heat rejection Process
Diesel Cycle 1-2......isentropic (Adiabatic) compression. 2-3......Addition of heat at constant pressure. 3-4...... isentropic (Adiabatic) expansion. 4-1......Rejection of heat at constant volume.
Diesel Cycle Analysis Thermal efficiency Heat addition (process 2-3, P = const) Heat rejection (process 4-1, v = const)
Cold-Air Standard Thermal Efficiency Otto Cycle Diesel Cycle
Diesel Cycle
Brayton Cycle (Joule Cycle) Usually used in gas turbines Basis of jet engines
Operation of an open cycle gas turbine engine 2.8 Brayton cycle : The Ideal Cycle for Gas Turbine Engines Operation of an open cycle gas turbine engine
Brayton Cycle Four internally reversible processes 1-2 Isentropic Compression (compressor) 2-3 Constant-pressure heat addition 3-4 Isentropic expansion (turbine) 4-1 Constant-pressure heat rejection
Brayton Cycle Gas turbine cycle Open vs closed system model
Brayton Cycle Analyze as steady-flow process So With cold-air-standard assumptions
Brayton Cycle Since processes 1-2 and 3-4 are isentropic, P2 = P3 and P4 = P1 where
Brayton Cycle
Actual Gas-Turbine Cycles For actual gas turbines, compressor and turbine are not isentropic
Regeneration
Regeneration Regenerator Effectiveness Use heat exchanger called regenerator Counter flow Regenerator Effectiveness
The Brayton Cycle: Reheat A high pressure and a low pressure turbine are used in a Reheat Brayton Cycle . A reheater is a heat exchanger that increases the power output without increasing the maximum operating temperature but it does not increase the efficiency of the cycle The capital cost to build a reheater alone cannot be justified because the thermal efficiency does not increase.
Brayton with Intercooling, Reheat, & Regeneration
T1 = T3 T6 = T8 P2 = P4 P1 = P3
Decrease the total compression work Improve the back work ratio If the number of compression and expansion stages is increased, the ideal turbine cycle with intercooling, reheating and regeneration approaches the Ericsson cycle
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.