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Lecture Objectives: Introduce HW3 Learn about sorption chillers
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Sequence of operation (PRC research facility) Control logic: Mixture in zone 1: IF (( TM<TSP) & (DPTM<DPTSP) ) heating and humidifying Heater control: IF (TSP>TSA) increase heating or IF (TSP<TSA) decrease heating Humidifier: IF (DPTSP>DPTSA) increase humidifying or IF (DPTSP<DPTSA) decrease humid. Mixture in zone 2: IF ((TM>TSP) & (DPTM<DPTSP) ) cooling and humidifying Cool. coil cont.: IF (TSP TSA) decrease cooling Humidifier: IF (DPTSP>DPTSA) increase humidifying or IF (DPTSP<DPTSA) decrease hum. Mixture in zone 3: IF ((DPTM>DPTSP) ) cooling/dehumidifying and reheatin Cool. coil cont.: IF (DPTSP>DPTSA) increase cooling or IF (DPTSP<DPTSA) decrease cooling Heater control: IF (TSP>TSA) increase heating or IF (TSP<TSA) decrease heating Set Point (SP) Mixture 2 Mixture 3 Mixture 1 DBT SP DPT SP
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HW3 Writhe a sequence of operation instruction list for the air handling unit from HW2 (problems: 3&4)
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HW3 You will need to define: A)what environmental variable/condition change - Temperature and RH of ambient air, Q cooling, Q heating, ….. B)which variables can you control -recirculation rate, recirculation position, Total flow rate, T CC,… C) which variable will you control - ….. D) how are you going to “move” between different operation schemes E) writhe “if - else” set of instruction for different controlled devices while considering different operation scheme (see the example I gave you in class)
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Absorption Cycle Same as vapor compression but NO COMPRESSOR Replace compressor
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Absorption cooling cycle Relatively simple thermodynamics with addition of mixtures (water – ammonia) Heat Rich solution of H2OH2O H2OH2O H 2 O + NH 3
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Mixtures (T-x diagram) For P= 4 bar Dew point curve Bubble point curve Saturated vapor Saturated liquid Mixture of liquid and vapor
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Impact of Pressure
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h-x diagram Isotherms are showmen only in liquid region h fg for H 2 O h fg for NH 3
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Composition of h-x diagram Saturated vapor line at p 1 Equilibrium construction line at p 1 x1x1 A Adding energy B Isotherm at P 1 and T 1 Used to determine isotherm line in mixing region! Start from x1; move up to equilibrium construction line; move right to saturated vapor line; determine 1’; connect 1 and 1’. X 1’ mass fraction of ammonia in saturated vapor 1e1e
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h-x diagram at the end of your textbook you will find these diagrams for 1) NH 3 -H 2 O 2) H 2 O-LiBr LiBr is one of the major liquid descants in air-conditioning systems
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Adiabatic mixing in h-x diagram (Water – Ammonia) From the textbook (Thermal Environmental Eng.; Kuehen et al)
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Absorption cooling cycle Heat Rich solution of H2OH2O H2OH2O H 2 O + NH 3
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Mixing of two streams with heat rejection (Absorber) Q Heat rejection m1m1 m2m2 m3m3 x x3x3 1 2 3’ 3 Mass and energy balance: From mixture equation: (1) (2) (3) Substitute into (2) Substitute into (3) From adiabatic mixing (from previous slide) m1m1 m2m2 m3m3 =pure NH 3 (x 2 =1) Mixture of 1 and 2 cooling mixture of H 2 O and NH 3
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Change of pressure (pump) 1 2 p1p1 ≠p 2 p2p2 p1p1 Saturated liquid at m1m1 =m 2 x1x1 =x 2 x 1 =x 2 1 2 Saturated liquid at p 1 Sub cooled liquid at p 2
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Heat transfer with separation into liquid and vapor (Generator) Apply mass and energy balance In the separator : Defines points 3 and 4 in graph Heat Sub cooled liquid Saturated liquid Saturated vapor We can “break” this generator into 2 units Q 12 m1m1 =m 2 m4m4 m3m3 Separator Apply mass and energy balance In the heat exchanger defines point 2 in graph Q 12 /m 1 x 1 sub cooled liquid mixture =2 V 2L=2L= heating
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Heat rejection with separation into liquid and vapor (Condenser) m1m1 =m 2 Saturated vapor m1m1 m2m2 1 Saturated vapor at p 1 Saturated liquid at p 1 x 1 =x 2 2 x1x1 =x 2 p1p1 =p 2 Q 1-2 /m 1 heat rejection
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Throttling process (Expansion valve) m1m1 =m 2 x1x1 =x 2 p1p1 ≠p 2 1 2 x1x1 =x 2 p1p1 ≠p 2 p1p1 p2p2 h1h1 =h 2 1 2 Saturated liquid at T2T2 T1T1 2V2V 2L2L Saturated vapor Saturated liquid
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Simple absorption system 3L3L 3V3V 3 LLP
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Saturated vapor at p 2 =p 3 =p 4 Saturated liquid at p 2 =p 3 =p 4 Saturated liquid at p 1 =p 5 =p 6 =p 3_LLP Simple absorption system 1 2 3 3V3V 3L3L 4 5 1’ 5L5L 6 5V5V 3 LLP mixing Useful cooling energy Needed thermal energy
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Heat transfer with separation into liquid and vapor (Generator) Q 12 =m 2 m3m3 x 1 mixture =2 V 2L=2L= heating Q 12 m1m1 =m 2 m4m4 m3m3 Separator Q 12 /m 1 x 1 sub cooled liquid mixture =2 V 2L=2L= How to move point 4 to right ?
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Heat rejection with separation into liquid and vapor (Enrichment NH 3 in the vapor mixture) Q 12 m1m1 =m 2 m3m3 Separator Q 12 /m 1 x 1 sub cooled liquid mixture 4=2 V 1 cooling 2 2L2L 5 Q 45 /m 4 6=5 V cooling 7 8 m8m8 x8x8 x8x8 This is our point isotherm
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