The sum of the electric power (WE1 + WE2) generated by generator 1 and 2 will...

   

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The sum of the electric power (WE1 + WE2) generated by generator 1 and 2 will feed the demand from the 150,000 homes. Notice how their efficiencies are different: generator 1 has an efficiency of 0.94 while generator 2 has an efficiency of 0.92. The gas turbine generates power (W₁) using air. Air enters the compressor in state 1 and exists in state 2. This compressor is not adiabatic and loses some heat (Qc). After that, air passes through a combustion chamber where is receives heat (₁) and leaves in state 3. Next, air enters the gas turbine, which is adiabatic, produces power and exists at state 4. The power generated by the gas turbine goes to generator 1 to become electric power. The air leaving the gas turbine at state 4 is hot. Thus, it has a significant amount of energy. Instead of wasting all this energy, we will extract some of it by means of the steam turbine. First, the energy of the air at state 4 is passed to water in the boiler. This energy is Q2, and it is given by: where ma is the mass flow rate of air (kg/s), h4 is the enthalpy of air at state 4 and h, is the enthalpy of air at state 1. Heat 2 is transferred to water, which goes from state 5 to state 6. After that, water goes through the steam turbine, which is adiabatic, exiting at state 7. Next, water goes through the condenser, where it loses heat in the amount of OCD. Now it is at state 8. Finally, water enters the pump to increase its pressure. The pump is adiabatic and brings water from state 8 to state 5. Known parameters/variables for our power generation system. The pressure and temperature of air entering the compressor (state 1) are 100 kPa and 300 K, respectively. ● ● 2=ma (124-11) ● The heat loss at the compressor. Qc. is 2 kW. The temperature of air after the compressor (state 2) is 580 K. The maximum air temperature allowed to enter the gas turbine is 1300 K. Indeed, this will be the temperature at state 3. The temperature of air leaving the gas turbine at state 4 is 675 K. The pressure of water entering and leaving the boiler is 7 MPa. The maximum water temperature allowed to enter the steam turbine is 1000 C. Indeed. we will reach this temperature at state 6. Water leaves the turbine as saturated vapor at a pressure of 50 kPa (state 7). Water leaves the condenser at 50 kPa and 50 C (state 8). Assumptions You may treat air as an ideal gas. You may treat water passing through the pump (only here) as an incompressible substance. You may neglect a change in internal energy for water before and after the pump. The entire system is running in steady state conditions. Part 1 - Energy consumption in residential homes. Your first task is to find your average power consumption (kW) for the months of March, July. and November. You can obtain this data from your electric bill. Keep in mind your electric bill will present numbers of energy consumption (not power) in kWh. You will have to calculate the average power consumption (kW) in your house for those 3 months. Next, imagine 150,000 other homes in Tampa consume the same power as you. Calculate the total power for all homes for the aforementioned months. Air Compressor Wc Qc ma 2 Air Water Wp Pump Combustion chamber 8 Q₁ m 5 mw Water Boiler Air ma 3 Water Gas turbine (T1) Steam turbine (T2) Water mw Generator 1, 7/₁ = 0.94 WEI Condenser S QCD Figure 2: Combined cycle for our project. Generator 2, 72 ||| WE2 = 0.92 The sum of the electric power (WE1 + WE2) generated by generator 1 and 2 will feed the demand from the 150,000 homes. Notice how their efficiencies are different: generator 1 has an efficiency of 0.94 while generator 2 has an efficiency of 0.92. The gas turbine generates power (W₁) using air. Air enters the compressor in state 1 and exists in state 2. This compressor is not adiabatic and loses some heat (Qc). After that, air passes through a combustion chamber where is receives heat (₁) and leaves in state 3. Next, air enters the gas turbine, which is adiabatic, produces power and exists at state 4. The power generated by the gas turbine goes to generator 1 to become electric power. The air leaving the gas turbine at state 4 is hot. Thus, it has a significant amount of energy. Instead of wasting all this energy, we will extract some of it by means of the steam turbine. First, the energy of the air at state 4 is passed to water in the boiler. This energy is Q2, and it is given by: where ma is the mass flow rate of air (kg/s), h4 is the enthalpy of air at state 4 and h, is the enthalpy of air at state 1. Heat 2 is transferred to water, which goes from state 5 to state 6. After that, water goes through the steam turbine, which is adiabatic, exiting at state 7. Next, water goes through the condenser, where it loses heat in the amount of OCD. Now it is at state 8. Finally, water enters the pump to increase its pressure. The pump is adiabatic and brings water from state 8 to state 5. Known parameters/variables for our power generation system. The pressure and temperature of air entering the compressor (state 1) are 100 kPa and 300 K, respectively. ● ● 2=ma (124-11) ● The heat loss at the compressor. Qc. is 2 kW. The temperature of air after the compressor (state 2) is 580 K. The maximum air temperature allowed to enter the gas turbine is 1300 K. Indeed, this will be the temperature at state 3. The temperature of air leaving the gas turbine at state 4 is 675 K. The pressure of water entering and leaving the boiler is 7 MPa. The maximum water temperature allowed to enter the steam turbine is 1000 C. Indeed. we will reach this temperature at state 6. Water leaves the turbine as saturated vapor at a pressure of 50 kPa (state 7). Water leaves the condenser at 50 kPa and 50 C (state 8). Assumptions You may treat air as an ideal gas. You may treat water passing through the pump (only here) as an incompressible substance. You may neglect a change in internal energy for water before and after the pump. The entire system is running in steady state conditions. Part 1 - Energy consumption in residential homes. Your first task is to find your average power consumption (kW) for the months of March, July. and November. You can obtain this data from your electric bill. Keep in mind your electric bill will present numbers of energy consumption (not power) in kWh. You will have to calculate the average power consumption (kW) in your house for those 3 months. Next, imagine 150,000 other homes in Tampa consume the same power as you. Calculate the total power for all homes for the aforementioned months. Air Compressor Wc Qc ma 2 Air Water Wp Pump Combustion chamber 8 Q₁ m 5 mw Water Boiler Air ma 3 Water Gas turbine (T1) Steam turbine (T2) Water mw Generator 1, 7/₁ = 0.94 WEI Condenser S QCD Figure 2: Combined cycle for our project. Generator 2, 72 ||| WE2 = 0.92

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Here are the stepbystep workings Part 1 My average power consumption March 1 Look at my March electr... View the full answer