EGR 334 Thermodynamics Chapter 4: Section 10-12

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EGR 334 Thermodynamics Chapter 4: Section 10-12
Lecture 18: Integrated Systems and System Analysis Quiz Today?

Today’s main concepts:
Be able to explain what an integrated system is Be able to describe the components of some common integrated systems Apply mass balance, energy balance, and continuity to streams of flow through integrated systems. Reading Assignment: No New Reading Assignment for Wed. Read Chap 5 for Friday. Homework Assignment: Problems from Chap 4: 95, 98, 102

Terms for Chap 5 Class Discussion :
Spontaneous Heat Transfer Clausius Statement Kelvin-Planck Statement Entropy Statement Irreversible vs. Reversible Internally Reversible Process Carnot Corollaries Carnot Efficiency Max. heat pump efficiency Max. refrigeration cycle COP Carnot Cycle Clausius Inequality You may want to create a summary sheet to help you discuss each of the concepts.

System Integration Engineers creatively combine components to achieve some overall objective, subject to constraints such as minimum total cost. This engineering activity is called system integration. The simple vapor power plant of Fig 4.16 provides an illustration.

Integrated Thermodynamics Systems:
Sec 4.11: System Integration Integrated Thermodynamics Systems: System Integration : Combine components to make a useful cycle Some common systems: - Power Plant - Refrigerator - Heat Pump Components: Pipes Nozzles/Diffusers Turbines Compressors/Pumps Heat Exchanger Throttling

Power Plant Cycle 1. Boil Water Burn something Nuclear Reaction
Sec 4.11: System Integration Power Plant Cycle 1. Boil Water Burn something Nuclear Reaction Geothermal 2. Use the steam as it rises to turn a turbine.

Power Plant Cycle 3.Condense Steam (to recycle water)
Sec 4.11: System Integration Power Plant Cycle 3.Condense Steam (to recycle water) using a heat exchanger 4. Pump water back to boiler.

Power Plant Cycle 2. Turbine 3. Condenser 4. Pump 1. Boiler
Sec 4.11: System Integration Power Plant Cycle 2. Turbine 3. Condenser 4. Pump 1. Boiler

Power Plant Cycle QCV WCV 1. Boiler 2. Turbine WCV 4. Pump QCV
Sec 4.11: System Integration Power Plant Cycle QCV WCV 1. Boiler 2. Turbine WCV . 4. Pump QCV 3. Condenser

Refrigeration cycle 1.Condenser
Sec 4.11: System Integration Refrigeration cycle 1.Condenser We know that condensing something will remove heat from the fluid. 3. Evaporator To have something to condense, we must have evaporated something. (Both are heat exchangers)

Refrigeration cycle 1. Condenser 4. Compressor 3. Evaporator
Sec 4.11: System Integration Refrigeration cycle 1. Condenser 4. Compressor 3. Evaporator 2. Throttling valve/ Expander

Refrigeration Cycle 1. Condenser QCV WCV 4. Compressor 3. Evaporator
Sec 4.11: System Integration Refrigeration Cycle 1. Condenser WCV QCV 2. Throttling valve/ Expander 4. Compressor 3. Evaporator QCV

Power required for the compressor, in hp,
Sec 4.11: System Integration Example: (4.103) A simple gas turbine power cycle operating at steady state with air as the working substance is shown in the figure. The cycle components include an air compressor mounted on the same shaft as the turbine. The air is heated in the high-pressure heat exchanger before entering the turbine. The air exiting the turbine is cooled in the low- pressure heat exchanger before returning to the compressor. KE and PE effects are negligible. The compressor and turbine are adiabatic. Using the ideal gas model for air, determine the Power required for the compressor, in hp, Power output of the turbine, in hp, Thermal efficiency of the cycle

Power required for the compressor, in hp,
Sec 4.11: System Integration Example: (4.103) Power required for the compressor, in hp, Power output of the turbine, in hp, Thermal efficiency of the cycle state 1 2 3 4 AV (ft3/min) 30,000 p (atm) p2>p1 p3=p2 p4=p1 T (°R) 520 650 2000 980 h (BTU/lb) state 1 2 3 4 AV (ft3/min) 30,000 p (atm) p2>p1 p3=p2 T (°R) 520 650 2000 980 h (BTU/lb) 124.27 155.51 504.71 236.02 From Table A-22E : Ideal Gas Properties of Air Assumptions Steady State KE=  PE = 0 Turbine and compressor are Adiabatic (QCV= 0) No work in Heat Ex. (WCV = 0) Air is modeled as an ideal gas state Q W Compressor ? Heat Ex 1 Turbine Heat Ex 2

Power required for the compressor, in hp,
Sec 4.11: System Integration Example: (4.103) Power required for the compressor, in hp, Power output of the turbine, in hp, Thermal efficiency of the cycle state 1 2 3 4 AV (ft3/min) 30,000 p (atm) p2>p1 p3=p2 T (°R) 520 650 2000 980 h (BTU/lb) 124.27 155.51 504.71 236.02 state Q W Compressor ? Heat Ex 1 Turbine Heat Ex 2 Using Continuity Eq Ideal Gas Eq. so and

Power required for the compressor, in hp,
Sec 4.11: System Integration Example: (4.103) Power required for the compressor, in hp, Power output of the turbine, in hp, Thermal efficiency of the cycle state 1 2 3 4 AV (ft3/min) 30,000 p (atm) p2>p1 p3=p2 T (°R) 520 650 2000 980 h (BTU/lb) 124.27 155.51 504.71 236.02 state Q W Compressor -1687 Heat Ex 1 Turbine ? Heat Ex 2 state Q W Compressor ? Heat Ex 1 Turbine Heat Ex 2 Energy Balance (Compressor):

Power required for the compressor, in hp,
Sec 4.11: System Integration Example: (4.103) Power required for the compressor, in hp, Power output of the turbine, in hp, Thermal efficiency of the cycle state 1 2 3 4 AV (ft3/min) 30,000 p (atm) p2>p1 p3=p2 T (°R) 520 650 2000 980 h (BTU/lb) 124.27 155.51 504.71 236.02 state Q W Compressor -1687 Heat Ex 1 Turbine 14,505 Heat Ex 2 Energy Balance (Same as Compressor):

Power required for the compressor, in hp,
Sec 4.11: System Integration Example: (4.103) Power required for the compressor, in hp, Power output of the turbine, in hp, Thermal efficiency of the cycle state 1 2 3 4 AV (ft3/min) 30,000 p (atm) p2>p1 p3=p2 T (°R) 520 650 2000 980 h (BTU/lb) 124.27 155.51 504.71 236.02 state Q W Compressor -1687 Heat Ex 1 18,851 Turbine 14,505 Heat Ex 2 state Q W Compressor -1687 Heat Ex 1 Turbine 14,505 Heat Ex 2 Find Q23 Energy Balance around heat exchanger:

Sec 4.9: Heat Exchangers (Revisited)
Example: (4.78) As sketched in the figure, a condenser using river water to condense steam with a mass flow rate of 2x105 kg/h from saturated vapor to saturated liquid at a pressure of 0.1 bar is proposed for an industrial plant. Measurements indicate that several hundred meters upstream of the plant, the river has a volumetric flow rate of 2x105 m3/h and a temperature of 15°C. For operation at steady state and ignoring changes in KE and PE, determine the river-water temperature rise, in °C, downstream of the plant traceable to use of such a condenser, and comment.

Example: (4.78) Sec 4.9: Heat Exchangers (Revisited) state Pl,i Pl,e
R,i R,e AV (m3/h) 2x105 m (kg/h) x 1 p (bar) 0.1 T (°C) 45.81 15 ? h (kJ/kg) 2584.7 191.83 62.99 state Pl,i Pl,e R,i R,e Av (m3/h) 2x105 m (kg/h) x 1 P (bar) 0.1 T (°C) 45.81 15 ? h (kJ/kg) state Pl,i Pl,e R,i R,e Av (m3/h) 2x105 m (kg/h) x 1 P (bar) 0.1 T (°C) 15 ? h (kJ/kg) Look up intensive properties for water from Tables based on known property values From Table A-2 : Properties of Saturated Water

Example: (4.78) Sec 4.9: Heat Exchangers (Revisited) state Pl,i Pl,e
R,i R,e Av (m3/h) 2x105 m (kg/h) 2x108 x 1 P (bar) 0.1 T (°C) 45.81 15 ? h (kJ/kg) 2584.7 191.83 62.99 Using the energy balance simplified for a heat exchanger From Table A-2 : Properties of Saturated Water (p817 & 819)

Using this value, the exit temperature may be found from Table A-2.
Sec 4.9: Heat Exchangers (Revisited) Example: (4.78) state Pl,i Pl,e R,i R,e Av (m3/h) 2x105 m (kg/h) x 1 P (bar) 0.1 T (°C) 45.81 15 ? h (kJ/kg) 2584.7 191.83 62.99 Plugging known values: From Table A-2 : Properties of Saturated Water (p817 & 819) Using this value, the exit temperature may be found from Table A-2. TR,e = 15.6°C  giving a temperature rise of 0.6 °C

end of Lecture 18 Slides