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

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Presentation on theme: "EGR 334 Thermodynamics Chapter 4: Section 10-12 Lecture 18: Integrated Systems and System Analysis Quiz Today?"— Presentation transcript:

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

2 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: Homework Assignment: No New Reading Assignment for Wed. Read Chap 5 for Friday. Problems from Chap 4: 95, 98, 102

3 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 3 You may want to create a summary sheet to help you discuss each of the concepts.

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

5 5 Sec 4.11: System Integration 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 Integrated Thermodynamics Systems:

6 6 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.

7 7 Sec 4.11: System Integration Power Plant Cycle 3.Condense Steam (to recycle water) using a heat exchanger 4. Pump water back to boiler.

8 8 Sec 4.11: System Integration Power Plant Cycle 1. Boiler 2. Turbine 3. Condenser 4. Pump

9 W CV. 9 Sec 4.11: System Integration Power Plant Cycle 1. Boiler 2. Turbine 3. Condenser 4. Pump Q CV W CV

10 10 Sec 4.11: System Integration 1.Condenser We know that condensing something will remove heat from the fluid. 3. Evaporator Refrigeration cycle To have something to condense, we must have evaporated something. (Both are heat exchangers)

11 11 Sec 4.11: System Integration Refrigeration cycle 1. Condenser 2. Throttling valve/ Expander 3. Evaporator 4. Compressor

12 12 Sec 4.11: System Integration Refrigeration Cycle 1. Condenser 2. Throttling valve/ Expander 3. Evaporator 4. Compressor W CV Q CV

13 13 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 Sec 4.11: System Integration (a)Power required for the compressor, in hp, (b)Power output of the turbine, in hp, (c)Thermal efficiency of the cycle

14 state1234 AV (ft 3 /min)30,000 p (atm)1p 2 >p 1 p 3 =p 2 1 T (°R) h (BTU/lb) Example: (4.103) Sec 4.11: System Integration (a)Power required for the compressor, in hp, (b)Power output of the turbine, in hp, (c)Thermal efficiency of the cycle state1234 AV (ft 3 /min)30,000 p (atm)1p 2 >p 1 p 3 =p 2 p 4 =p 1 T (°R) h (BTU/lb) stateQW Compressor0? Heat Ex 10 Turbine0? Heat Ex 20 From Table A-22E : Ideal Gas Properties of Air Assumptions Steady State  KE=  PE = 0 Turbine and compressor are Adiabatic (Q CV = 0) No work in Heat Ex. (W CV = 0) Air is modeled as an ideal gas

15 15 Example: (4.103) Sec 4.11: System Integration (a)Power required for the compressor, in hp, (b)Power output of the turbine, in hp, (c)Thermal efficiency of the cycle stateQW Compressor0? Heat Ex 10 Turbine0? Heat Ex 20 state1234 AV (ft 3 /min)30,000 p (atm)1p 2 >p 1 p 3 =p 2 1 T (°R) h (BTU/lb) and so Using Continuity Eq. Ideal Gas Eq.

16 16 Example: (4.103) Sec 4.11: System Integration (a)Power required for the compressor, in hp, (b)Power output of the turbine, in hp, (c)Thermal efficiency of the cycle stateQW Compressor0? Heat Ex 10 Turbine0? Heat Ex 20 state1234 AV (ft 3 /min)30,000 p (atm)1p 2 >p 1 p 3 =p 2 1 T (°R) h (BTU/lb) Energy Balance (Compressor): stateQW Compressor Heat Ex 10 Turbine0? Heat Ex 20

17 17 Example: (4.103) Sec 4.11: System Integration (a)Power required for the compressor, in hp, (b)Power output of the turbine, in hp, (c)Thermal efficiency of the cycle stateQW Compressor Heat Ex 10 Turbine014,505 Heat Ex 20 state1234 AV (ft 3 /min)30,000 p (atm)1p 2 >p 1 p 3 =p 2 1 T (°R) h (BTU/lb) Energy Balance (Same as Compressor):

18 18 Example: (4.103) Sec 4.11: System Integration (a)Power required for the compressor, in hp, (b)Power output of the turbine, in hp, (c)Thermal efficiency of the cycle stateQW Compressor Heat Ex 10 Turbine014,505 Heat Ex 20 state1234 AV (ft 3 /min)30,000 p (atm)1p 2 >p 1 p 3 =p 2 1 T (°R) h (BTU/lb) Find Q 23 Energy Balance around heat exchanger: stateQW Compressor Heat Ex 118,8510 Turbine014,505 Heat Ex 20

19 19 Example: (4.78) As sketched in the figure, a condenser using river water to condense steam with a mass flow rate of 2x10 5 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 m 3 /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. Sec 4.9: Heat Exchangers (Revisited)

20 20 Example: (4.78) Sec 4.9: Heat Exchangers (Revisited) statePl,iPl,eR,iR,e Av (m 3 /h)2x10 5 m (kg/h)2x10 5 x1000 P (bar)0.1 T (°C)15? h (kJ/kg) From Table A-2 : Properties of Saturated Water statePl,iPl,eR,iR,e Av (m 3 /h)2x10 5 m (kg/h)2x10 5 x1000 P (bar)0.1 T (°C) ? h (kJ/kg) statePl,iPl,eR,iR,e AV (m 3 /h)2x10 5 m (kg/h)2x10 5 x1000 p (bar)0.1 T (°C) ? h (kJ/kg) Look up intensive properties for water from Tables based on known property values

21 21 Example: (4.78) Sec 4.9: Heat Exchangers (Revisited) From Table A-2 : Properties of Saturated Water (p817 & 819) statePl,iPl,eR,iR,e Av (m 3 /h)2x10 5 m (kg/h)2x10 5 2x10 8 x1000 P (bar)0.1 T (°C) ? h (kJ/kg) Using the energy balance simplified for a heat exchanger

22 22 Example: (4.78) Sec 4.9: Heat Exchangers (Revisited) From Table A-2 : Properties of Saturated Water (p817 & 819) statePl,iPl,eR,iR,e Av (m 3 /h)2x10 5 m (kg/h)2x10 5 x1000 P (bar)0.1 T (°C) ? h (kJ/kg) Using this value, the exit temperature may be found from Table A-2. T R,e = 15.6°C  giving a temperature rise of 0.6 °C Plugging known values:

23 23 end of Lecture 18 Slides


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