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Chapter 2: Sections 6 and 7 . Lecture 04: Energy Analysis of Cycles. Quiz Today?. Today’s Objectives:. Be able to explain what a Power Cycle is Be able to explain what a Refrigeration Cycle is Be able to explain what a Heat Pump Cycle is
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Chapter 2: Sections 6 and 7 Lecture 04: Energy Analysis of Cycles Quiz Today?
Today’s Objectives: • Be able to explain what a Power Cycle is • Be able to explain what a Refrigeration Cycle is • Be able to explain what a Heat Pump Cycle is • Be able to calculate thermal efficiencies for Power and Heat Pumps Be able to calculate COP for Refrigeration Cycles. • Be able to intelligently discuss method of energy storage. Reading Assignment: • No Reading Assignment Homework Assignment: From Chap 2: Problems 73, 75, 84, 90
Sec 2.4: Energy Transfer by Heat Recall past concepts: Work, W: W > 0 : Work done BY the system W < 0 : Work done ON the system Q > 0 : Heat transferred TO the system Q < 0 : Heat transferred FROM the system Heat, Q: 1st Law of Thermodynamics: system Qin Wout ΔE
Sec 2.2.1: Sign Convention Work done by Gas System W > 0 : Expansion of Gas W < 0 : Compression of Gas Run Animation
Sec 2.6: Energy Analysis of Cycles Energy Analysis of Cycles A cycle is any process that returns to its original state. • Ecycle= Qcycle - Wcycle • therefore: Ecycle= 0 • and Qcycle = Wcycle
Sec 2.6: Energy Analysis of Cycles Schematic models of Cycles: Refrigeration (Heat pump) Cycle Power cycle
Sec 2.6.2: Power Cycles Power Cycle • Wcycle= Qcycle = Qin - Qout • Qin : Heat in from hot body • (chemical reaction, geothermal) • Qout : Heat out to cold body • (surroundings) • Qin > Qout W > 0 • Thermal efficiency • when η = 1, efficiency = 100% so want to minimize Qout
Sec 2.6.3: Refrigeration and Heat Pump Cycles Refrigeration Cycle • Wcycle= Qcycle = Qout - Qin • Qout : Heat out to warm body • Qin : Heat in from cold body • W > 0 Qout> Qin • Refrigeration Coefficient of Performance
Sec 2.6.3: Refrigeration and Heat Pump Cycles Heat Pump Cycle • Wcycle= Qcycle = Qout - Qin • Qout : Heat out to warm body • Qin : Heat in from cold body • W > 0 > Qout> Qin • Heat Pump Coefficient of Performance
A quick comparison Refrigeration Cycle Heat Pump Cycle Objective is to do Work to Add heat to a system. Objective is to do Work to Remove heat from system.
Example 1 (Problem 2.76): A gas within a piston-cylinder assembly undergoes a thermodynamic cycle consisting of three processes: Process 1-2 : Constant volume, V = 0.028 m3, U2-U1=26.4 kJ Process 2-3: Expansion with PV = constant, U3 = U2 Process 3-1: Constant pressure, P = 1.4 bar, W31 = -10.5 kJ There are no significant changes in kinetic or potential energy. • Sketch to cycle on a PV diagram • Calculate the net work for the cycle, in kJ • Calculate the heat transfer for process 2-3, in kJ • Calculate the heat transfer for process 3-1, in kJ Is this a power cycle or a refrigeration cycle?
Example 2 (Problem2.83): A power cycle has a thermal efficiency of 40% and generates electricity at a rate of 100 MW. The electricity is valued at $0.08 per kWh. Based on the cost of fuel, the cost to supply Qin is $4.50 per GJ. For 8000 hours of operation annually, determine, in $ • the value of electricity generated per year and • the annual fuel cost • Is operation profitable?
Example 3 (Problem2.91): A heat pump maintains a dwelling at 68oF. When operating steadily, the power input to the heat pump is 5 hp, and the heat pump receives energy by heat transfer from 55oF well water at a rate of 500 BTU/min. a) Determine the COP. b) Evaluating electricity at $0.10 per kW-hr, determine the cost of electricity in a month when the heat pump operate for 300 hr.
End of Lecture 04 • Slides which follow show solutions to example problems
Example (2.76): A gas within a piston-cylinder assembly undergoes a thermodynamic cycle consisting of three processes: Process 1-2 : Constant volume, V = 0.028 m3, U2-U1=26.4 kJ Process 2-3: Expansion with PV = constant, U3 = U2 Process 3-1: Constant pressure, P = 1.4 bar, W31 = -10.5 kJ There are no significant changes in kinetic or potential energy. • Sketch to cycle on a PV diagram • Calculate the net work for the cycle, in kJ • Calculate the heat transfer for process 2-3, in kJ • Calculate the heat transfer for process 3-1, in kJ Is this a power cycle or a refrigeration cycle? p 3 3 V
Example (2.76): Process 1-2 : Constant volume, V = 0.028 m3, U2-U1=26.4 kJ Process 2-3: Expansion with PV = constant, U3 = U2 Process 3-1: Constant pressure, P = 1.4 bar, W31 = -10.5 kJ • Sketch to cycle on a PV diagram 2 1.4 P(bar) 1 3 0.0028 V(m3)
Example (2.76): Process 1-2 : Constant volume, V = 0.028 m3, U2-U1=26.4 kJ Process 2-3: Expansion with pV= constant, U3 = U2 Process 3-1: Constant pressure, p= 1.4 bar, W31 = -10.5 kJ • Calculate the net work for the cycle, in kJ for Process 1-2: for Process 3-1: for Process 2-3:
Example (2.76): Process 1-2 : Constant volume, V = 0.028 m3, U2-U1=26.4 kJ Process 2-3: Expansion with pV= constant, U3 = U2 Process 3-1: Constant pressure, p= 1.4 bar, W31 = -10.5 kJ • Calculate the net work for the cycle, in kJ • We now have, • Then, • Since Wcycle >0, the cycle is a power cycle
Example (2.76): A gas within a piston-cylinder assembly undergoes a thermodynamic cycle consisting of three processes: Process 1-2 : Constant volume, V = 0.028 m3, U2-U1=26.4 kJ Process 2-3: Expansion with pV= constant, U3 = U2 Process 3-1: Constant pressure, p= 1.4 bar, W31 = -10.5 kJ There are no significant changes in kinetic or potential energy. • Calculate the heat transfer for process 2-3, in kJ • Find Q23, using an energy balance for Process 2-3. • 0 0 0
Example (2.76): A gas within a piston-cylinder assembly undergoes a thermodynamic cycle consisting of three processes: Process 1-2 : Constant volume, V = 0.028 m3, U2-U1=26.4 kJ Process 2-3: Expansion with pV = constant, U3 = U2 Process 3-1: Constant pressure, p = 1.4 bar, W31 = -10.5 kJ There are no significant changes in kinetic or potential energy. • Calculate the heat transfer for process 3-1, in kJ • 0 0 • so • o • to find U31: • then:
Example (2.76): A gas within a piston-cylinder assembly undergoes a thermodynamic cycle consisting of three processes: Process 1-2 : Constant volume, V = 0.028 m3, U2-U1=26.4 kJ Process 2-3: Expansion with pV = constant, U3 = U2 Process 3-1: Constant pressure, p = 1.4 bar, W31 = -10.5 kJ There are no significant changes in kinetic or potential energy. • Double check math on heat calculations • We can also calculate the efficiency of this cycle
Example (2.83): A power cycle has a thermal efficiency of 40% and generates electricity at a rate of 100 MW. The electricity is valued at $0.08 per kWh. Based on the cost of fuel, the cost to supply Qin is $4.50 per GJ. For 8000 hours of operation annually, determine, in $ • the value of electricity generated per year • the waste heat returned to the environment • the annual fuel cost • Is operation profitable? Qin • Fuel • Wout=100 MW • Air Qout
Example (2.83): A power cycle has a thermal efficiency of 40% and generates electricity at a rate of 100 MW. The electricity is valued at $0.08 per kWh. Based on the cost of fuel, the cost to supply Qin is $4.50 per GJ. For 8000 hours of operation annually, determine, in $ • the value of electricity generated per year and • next find the heat generated and the heat returned to the environment
Example (2.83): A power cycle has a thermal efficiency of 40% and generates electricity at a rate of 100 MW. The electricity is valued at $0.08 per kWh. Based on the cost of fuel, the cost to supply Qin is $4.50 per GJ. For 8000 hours of operation annually, determine, in $ • Cost of fuel? • Is operation profitable? • Profit = Revenue – Costs • So, this could be profitable, but the calculation ignore other costs such as capital and labor.
Example 3 (Problem2.91): A heat pump maintains a dwelling at 68oF. When operating steadily, the power input to the heat pump is 5 hp, and the heat pump receives energy by heat transfer from 55oF well water at a rate of 500 BTU/min. a) Determine the COP. b) Evaluating electricity at $0.10 per kW-hr, determine the cost of electricity in a month when the heat pump operates for 300 hr. Principle: COP for heat pump (written in terms of power) where: and therefore:
Example 3 (Problem2.91): A heat pump maintains a dwelling at 68oF. When operating steadily, the power input to the heat pump is 5 hp, and the heat pump receives energy by heat transfer from 55oF well water at a rate of 500 BTU/min. a) Determine the COP. b) Evaluating electricity at $0.10 per kW-hr, determine the cost of electricity in a month when the heat pump operate for 300 hr. Principle: Cost = Cost of energy * Power * time where: therefore: