DEPARTMENT OF MECHANICAL ENGINEERING REFRIGERATION & AIR CONDITIONING LAB TRIPLE FLUID VAPOR ABSORPTION SYSTEM Aim: To find COP of Triple Fluid Vapor Absorption System (TF_VAS) for 1 ohm and 2 ohm heat input. Theory: The triple fluid vapor absorption system uses refrigerant (NH 3 ), absorbent (H 2 O) and carrier gas (H 2 ). The system does not have any mechanical liquid pump. Circulation of working fluids is enabled using a bubble pump, density differences and gravity flow. The difference in partial pressures in condenser and evaporator for refrigerant (NH 3 ), results in cooling effect. In this experiment, the unit is operated at two different heat input levels and the performance is observed. The heat balance of the system is also considered. Measure Parameters: Solution Heat Exchanger Outlet (strong solution side): Condenser Inlet (vapor at rectifier outlet) D: Solution Heat Exchanger Outlet (weak solution side): Solution Heat Exchanger Inlet (strong solution side): Condenser Outlet (liquid at evaporator inlet): Evaporator Outlet (absorber inlet): Evaporator Inlet (from condenser outlet) F: Bubble Pump Outlet (generator outlet) C: t1 t2 t3 t4 t5 t6 t7 t8 Instrumentation: Temperature : Thermocouples are mounted at important points to measure the temperature. A digital multi channel thermometer indicates temperatures at these points.
Calculations: Locating various State Points on Enthalpy Concentration Diagram for NH 3 /H 2 O mixture and then using them to calculate the performance of the cycle. Liquid at Outlet of the Condenser: 5 t 5 WFS 5 0.999 Condenser Outlet temperature This concentration is assumed, it should be verified based on rectifier outlet condition p 5 Enthalpy of liquid h l5 Vapor at Outlet of the Rectifier: 2 t 2 p 2 Condenser inlet temperature is same as that in the condenser
WFV 2 0.999 Concentration at the rectifier outlet. This is higher than that assumed in the condenser and using this new concentration, condense pressure may be recalculated if needed Enthalpy of vapor h v2 Vapor at Inlet of the Evaporator: 7 t 7 Evaporator inlet temperature WFV 7 Concentration is 1, since the liquid refrigerant being throttled is 0.9995. p 7 Enthalpy of saturated vapour h v7 Vapor at Outlet of the Evaporator: 6 p 6 =p 7 t 6 Evaporator outlet temperature WFV 6 =WFV 2 = Concentration is 0.9995. Since, the carrier gas is being cooled vaporizes 0.9995 the entire refrigerant stream, so its composition has to same as that of condenser liquid outlet. Enthalpy of saturated vapor h v6 Heat Picked - up by Refrigerant in the Evaporator: Refrigerant enters the evaporator through the throttling device at state point 7. Enthalpy at 7 is same as enthalpy at 5, at the condenser outlet. Refrigerant may leave the evaporator in the form of superheated vapor/saturated vapor/two phase mixture state point 6, depending on its temperature and pressure. Concentration of saturated vapor at t 6 and p 6 is > 0.998. Since, this is less than 0.9995 at condenser outlet, state point 5, all the liquid refrigerant would be vaporized at t 6 and p 6. This cooling effect delivered at temperature substantially above the evaporator temperature, up to the condenser temperature can be gainfully used to cool saturated condensate, 5. Saturated liquid refrigerant exiting the condenser is subcooled to a lower temperature. This reduces the enthalpy at the entrance of the evaporator and results in increasing cooling effect delivered per unit mass flow rate through the evaporator and thus leads to 5 to 10% increase in cooling capacity of the absorption refrigeration system. This also results in increased cooling Coefficient of Performance, COP. Since, heat exchanged in the rhe, between 5-7 and evaporator outlet-6, is internal to the cycle, the evaporator heat duty can easily be evaluated by taking the
heat balance across 5-7. This will be valid because there is no external heat or work interaction between 5 and 6 other than the evaporator duty, Qe. Solution at outlet of the absorber: 4 p 6 =p 4 t 4 Absorber outlet temperature WFS 4 Concentration at outlet of absorber. Enthalpy of liquid h l4 Solution at Outlet of the Generator: C/8 p 8 =p 5 T 8 Generator outlet temperature WFS 8 Concentration at generator outlet. WFS 1 = WFS 4 Concentration split is the change in concentration of the absorbing or Split = WFS 1 WFS 8 generating solution Enthalpy of solution h l8 Mass and Concentration Balance Across the Absorber: Discussion Solution weak in refrigerant, weak ammonia solution, exiting the generator at C / 8 is subcooled to 3 in the solution heat exchanger. This subcooled weak solution enters the absorber. Low pressure ammonia vapor from the evaporator enters the absorber as 6 along with the carrier gas. Two phase refrigerant from the evaporator vapourizes completely as it cools the carrier gas entering the evaporator from the absorber and also as it subcools the liquid refrigerant at condenser outlet 5 on its way to the evaporator inlet F / 7. This low pressure vapour 6, is absorbed in the absorber in weak solution 3. After absorption of ammonia vapour the resulting strong solution exits the absorber as 4. Mass balance across the absorber: Concentration balance across the absorber: m6 + m3 = m4 m6.wfv6 + m3.wfs3 = m4.wfs4 Heat of absorption, Qa, that is equivalent to heat of condensation plus the heat of mixing, is released in the absorber. This heat is rejected to the ambient air. WFS3= WFS8 m6= 0.000037 kg/s, Mass flow rate at 6 can be selected to match the generator heat duty with the electrical heat input, eg. V2/R = (12 V)2/1 _ = 144 W.
m6= m2=m5=m7 m1= m4 m3= m4- m6 m8= m3 Heat Balance Across the Absorber: Discussion Heat of absorption, Qa, that is equivalent to heat of condensation plus the heat of mixing, is released in the absorber. This heat is rejected in the ambent air through the absorber, condenser and rectifier. Mass balance across the absorber: m6 + m3 = m4 Energy balance on ammonia side in absorber: m6.hv6 + m3.hl3 = m4.hl4 + Qa Solution at Inlet of the Generator: 1 p 1, t 1 WFS 1 t sat, Temperature at generator inlet Concentration at generator inlet Saturtaion temeprature h l1 Enthalpy of solution Solution at Inlet of the absorber: 3 p 3, t 3, Temperature at absorber inlet h l3 Enthalpy of solution Energy Balance Across the Solution Heat Exchanger Qshe.c= hl.1x m1 - hl.4x m4 Qshe.h= hl.8xm8 - hl.3x m3 Find the mismatch.
Mass and Concentration Balance across the Rectifier: Discussion Since, vapor pressure of water at the generator temperature is not insignificant, water vapour is also present in the ammonia vapour generated from the generator. Refrigerant vapour exiting the generator at 8 is cooled in the rectifier to 2. This helps in reducing the amount of water vapour present in the refrigerant vapour being sent to the condenser. On cooling, the vapour stream exiting the generator, water vapour preferentially condenses in the form of ammonia/water solution of concentration far weaker than that of the vapour. This solution, drains back to the generator due to slope of the rectifier. Reducing the water content in the vapour, before it enters the condenser, helps in reducing the quantity of unevaporated refrigerant stream at the outlet of the evaporator. If the quantity of unevaporated refrigerant stream is high, state at 6 will be wet. There will be unevaporated refrigerant at the outlet of the refrigerant heat exchanger at 6. Cooling effect delivered in the evaporator per unit refrigerant circulated through the condenser and the evaporator will reduce. This will also result in reduced COP and require higher heat input to the cycle. Mass and Concentration Balance Across the Rectifier: 8v - 2-8l p 8, t 8 and Temperature across rectifier. WFV 8v, WFS 8 Concentration Enthalpy of vapor h v8 Mass and Concentration Balance Across the Rectifier: 8v - 2-8l Mass balance across the rectifier: m8v = m2 + m8l Concentration balance across the absorber: m8v.wfv8v = m2.wfv2 + m8l.wfs81 Heat of rectification, Qr, that is equivalent to heat of condensation plus the heat of mixing, is released in the rectifier. This heat is rejected in the room air. m 8v = m 2 + m 8l Energy Balance Across the Rectifier: 8v - 2-8l Note: This heat may be multiplied with 1.1 to account for non-equilibrium conditions between the v.13 and l.14. In this example equilibrium is assumed and hence 1.1 multiplier is not used. Qr =(hv.8 x m8v - hl.8 x m8l - hv.2 x m2)
Energy Balance Across the Condenser: 2-5 Qc= hv.2 x m2 - hl.5 x m5 Energy Balance Across the Generator: 2-1 - 8 Qg = Qc + Qr + m 5.h l.5 - m 1.h l.1 + m 8.h l.8 dm g = m 8v + m 8 - m 8l - m 1 Qg = m 8v.h v.8 + m 8.h l.8 - m 8l.h l.8 - m 1.h l.1 Energy Balance Across the Absorber: 3-6 - 4 Qa = h l.3. m 3 - h l.4. m 4 + h v.6. m 6 Energy Balance Across the Evaporator: 5-6 Qe = m 5 ( h v.6 - h l.5 ) Overall Heat Balance Heatin := Qe + Qg Heatout := Qa + Qc + Qr