ATD NOTES

1AIR COMPRESSOR Compressors are used for producing high compressed air. The working principle of an reciprocating air compressor is similar to an I.C. Engine. The air is sucked into a cylinder during suction stroke & compressed to a high pressure & delivered at the end of the Compression stroke. CLASSIFICATION: Compressors are classified as 1) Single acting & Double acting compressor. 2) Single stage & Multi stage compressor. In a single acting reciprocating compressor, the suction, compression and delivery of the air takes place on one side of the piston. In a double acting reciprocating compressor the suction, compression and delivery of the air takes place on both side of the piston. In a single stage compressor, the compression is carried out is done only one cylinder. In a multistage compressor, the compression is carried out is more than one cylinder. Uses of Compressed air:The compressed air has wide application in industry as well as in commercial equipments. 1) It is used in operating drills & hammers in road building. 2) Excavating 3) Tunneling & mining 4) Starting diesel engines & 5) Operating brakes in buses, trucks & trains. 6) A large quantity of air at moderate pressure is used in smelting of various metals such as melting iron, in blowing converters & cupola work. 7) Large quantities of air are used in the air conditioning & drying. 8) Used in air lift pumps for pumping water from deep bore wells. 9) Used in foundary for sand blasting Work done in a single stage air compressor:i).Without clearance 2 The P-V diagram for a single stage air compressor without clearance is as shown in fig:- Process 4-1 :- The suction of air is drawn at the pressure P1, Inlet value opens, volume V1 Process 1-2 :- The compression of air is polytropically. Inlet and outlet closed. PVn=C1, Pr P1 to P2 , volume decreases from V1 to V2. Temperature increases from T1 to T2. Process 2-3 :- The discharge of air at a pressure P2. Outlet value opens. Compressed air Volume is V2 & temp T2. b) with clearance volume :All reciprocating compressors will havw a clearance volume. The clearance Volume is that volume which remain in the cylinder after the piston has reached the end of its inward stroke. At the end of the delivery strike, the high pressure air is left in the clearance volume as shown in P-V diagram. The next cycle starts only when the air pressure falls to atmosphere pressure. This is given by Expansion curve 3-4. As summing the compression & expansion of the air follow thw same law. : X = 3265 + 6217 – 5417 = P2V2 + P2 – P1V1 - P1V1 n-1 = (P2 – P1V1) + P2V2 – P1V1 n-1 1/ n – 1 = (P2 – P1V1) ( 1 + ) = n (P2 – P1V1) n-1 =Work done n P1V1 P2V2 - 1 ( n – 1) P1V1 P1V1n = P2V2n for polytropic process V2 V1 = P1 P2 :W = n n-1 1/n = P2 P1 P1V1 -1/n P2 P2 P1 P1 -1/n - 1 ----------- (1) 3 = n P1V1 n-1 P2 P1 W= n mRT1 n-1 n-1/n P2 P1 -1 n-1/n - 1 -1 Where m is the mass of air delivered per cycle. If the air delivered at temp T2 is required then. T2/T1 = (P2/P1) n-1/n or T2=T1 (P2/P1) n-1/n b) With Clearance Volume :All reciprocating compressors will have a clearance Volume. The clearance volume is that volume which remain in the cylinder after the piston has reached the end of its inward stroke. At the end of the delivery stroke, the light pressure air is left in the clearance volume as shown in PV diagram. The next cycle starts only when the air pressure fath to atmosphere pressure. This is given by expansion curve 3-4. Assuming the compression & expansion of the air follow the same law. Work done / cycle = Area 1 2 3 4 1 4 W=Wc – We Area 12561 – Area 34653 W+ n P1V1 [P2] n-1/n – 1] - n P4V4 [P3] n-1 [P1] n-1 n-1/n [P4] As P3 P2 and P4 = P1 W= n P1 (V1 - V4) n–1 P2 n-1/n -1 P1 W= n P1Va P2 n-1 P1 n-1/n -1 Where m1 = the actual mass of air delivered per cycle Work done / Kg of air delivered W cycle = n n-1 RT1 P2 P1 n-1 ---------------------- (2) -1 The gas is first compressed is a low pressure cylinder at a pressure P1 where it is compressed to pressure P2 and is then discharged into the intercooler. The air is called is intercooler at a constant pressure P2before passing it to the second stage. If the temperature of air leaving the intercooler is less than the original temperature of air, the cooling is said to be incomplete is the point ’d’ doesn’t lie on isothermal line of the inter cooling process is complete, then P1V1 = P2V2 5 for in complete inter cooling:- WD by L.P Cylinder = W LP = n P1V1 n–1 P2 n-1/n P1 & W.D by H.P.Cylinder = WHP = n P2V2 n–1 P3 - 1 n-1/n P4 - 1 Work done per cycle W= WLP + WHP W= n P1V1 P2 n–1 P1 W= n n–1 P1V1 P2 P1 n-1/n n-1/n - 1 + n n–1 P2V2 P3 P2 - 1 + P2V2 P3 P2 n-1/n - 1 n-1/n - 1 If intercooling is perfect then P1V1 = P2V2 WD by L.P Cylinder = P1V1 n P2 n – 1 P1 n-1/n + P3 n-1/n -2 P2 Condition for Maximum efficiency or Minimum work required for a two stages air compressor with Inter/ cooler The work required to drive the compressor will be minimum when the point d lies on the isothermal line. WORK Done /Cycle = n P1V1 n–1 P2 P1 n-1/n + P3 P2 n-1/n -2 of initial pressure P1 and the final pressure P3 are fixed, the inter mediate pressure P2 can be determined by differentiating the above eqn with respect to P2 & equating it to zero Let n-1/n = z constant 6 W = Z Cost P2 P1 Z + P3 P2 Z -2 Diff wrt P2 For minimum work done dw =0 dP2 Z -Z W = Z Constant ( P2 P1 + (-Z) P2-Z-1 P3Z ) = 0 dw = Constant (Z P2 Z-1 P1-Z P2 (-Z-1) P3Z) = 0 dP2 0 = (Z P2 Z-1 P1-Z - Z P2 (-Z-1) P3Z) (Z P2 Z-1 P3-Z - Z P2 (-Z-1) P1Z) P3-Z P2 (-Z-1) _____________ P2 (Z-1) . P2 (Z+1) P1 Z P2 (-Z-1) P3-Z ___ P2 2Z P1 Z P2 2Z = P3Z P1Z P22Z = (P3P1) Z P22 = P1P3 P2= √P1P3 Hence for Maximum efficiency, the intermediate pressure is the geometric of the initial & final pressure Volumetric efficiency of an air compressor The volumetric efficiency of an air compressor is the ratio of free air delivered to the displacement of the compressor or It is the ratio of volume of free air inhaled (effective swept volume) at NTP during suction stroke to the swept volume of piston. 7 Therefore Volumetric Efficiency = Effective swept volume at STP Swept Volume Because of clearance volume, Volumetric efficiency is always less than unity, as percentage varies from 60% to 85% STP - std temp Pr NTP - std temp Pr 150 & 1.03 kgf/cm2 (1.03 x 104 x 10 kg/m2) (1.03 N/m2) 00 & 1.03 kgf/cm2 Clearance ratio = Clearance Volume = Vc Swept Volume Clearance ratio C = V3 V1 – V3 Volumetric Efficiency = V1 – V4 V1 – V3 Expression For Volumetric n Vs = Vc Vs 8 Let P1, VA & T1 be properties at state point (1). Pa, Va & Ta be the properties at STP(i.e., ambient conditions) [ In practice the air that is sucked in during the suction stroke gets heated up while passing through the hot values & coming in contact with hot cylinder walls. There is a wire drawing effect through the values resulting in drop in pressure. Thus the ambient conditions are different from conditions obtaining at state 1. Clearance ratio C = Vc Vs PaVa = P1VA Ta T1 Va = P1 Ta . VA Pa T1 From figure, VA = Vc + Vs – V4 For Process 3-4:P3V3n = P4V4n ∴ V3 = V4 P4 1/n P3 i.e., V4 = V3 P3 1/n P4 V4 = Vc P3 P4 Va = Vc +Vs – Vc P2 1/n η vol = Va = ______ VS = P1 . Ta Va Pa T1 ____________ VS P1.Ta VC+VS –VC (P2/P1 ) 1/n ___ ___ _________________ Pa. T1 Vs η vol = P1 .Ta C+1-C (P2/P1)1-n Pa . T1 if P1 = Pa & T1 =Ta we get 1/n ∴ (V3=Vc) 9 η vol = 1 + C –C (P2 /P1) 1/n for the pressure (for expansion) P3 V3 n =P4V4n P3 /p4 = (v4/v3)n V4/v3 = (p2/p1)1/n -------------- (1) η vol = V1 – V4 = V1-V3 (P2/P1) 1/n V1 – V3 V1-V3 = V1 – V3 (P2/P1)1/n V3/C η vol = CV1 C(P2/P1)1/n = C V3(1/C+1) - C(P2/P1)1/n V3 Volumetric Efficiency = 1 + c – c(p2/p1)i/n η vol 1 + c – c(v1/v2) The Volumetric efficiency decreases due to following conditions:(1) Very high speed (2) Leakage (past) through the piston (3) Too large a clearance Volume (4) Obstruction at inlet values (5) Overheating of air by contact with hot cylinder walls (6) Inertia effect of air in suction pipe. Methods adopted for increasing Isothermal Efficiency The following methods are used to achieve nearly isothermal compression for high speed compressors. The final temperature T2 is reduced during compression, so that actual work approaches more closely that of Isothermal Compression. (1) Spray injection :- 10 In this method water is sprayed into the cylinder, at the end of compression stroke reducing the temperature of air. The adiabatic equation PV1.4 = Constant reduces to PV1.2 = Constant. Disadvantages :(1) It requires special gear for injection. (2) The injected water interferes with the cylinder lubrication & attacks cylinder walls & values. (3) The water mixed with air should be separated before using the air. (2)Water Jacketing :The water is circulated around the cylinder through the water jacket which helps to cool the air during Compression. (3)Inter cooling :Water Jacketing is not much effective when the speed of the compressor is high and pressure ratio required is also high with single stage Compression. Inter cooling is used in addition to the water jacketing by dividing the Compression process into two or more stages. Before taking for second stage, air is cooled in an inter cooler. (4)External fins :Effective cooling can be achieved for small capacity air compressor with the use of fins on the external surface of the compressor. Multi stage Compression :If high pressure is to be delivered by a single machine, then it will require heavy working parts in order to accommodate the high pressure ratio through the machine. This will increase the balancing problem & the high torque fluctuation will require a heavier flywheel installation. Such disadvantages can be overcome by Multi stage Compression. 11 Advantages :(1) The air can be cooled at pressures intermediate between intake and delivery pressure. (2) The power required to drive a multi stage machine is less than would be required by a single stage machine delivering the same quantity of air at the same delivery pressure. (3) Multi stage machines have better Mechanical balance. (4) The pressure range(& also temp) may be kept within the desirable limits. This results in (i) reduced losses due to air leakage. (ii) improved lubrication, due to lower temperature. (iii) improved volumetric efficiency. (5) The cylinder , in asingle stage machine must be robust enough to withstand the delivery pressure. The down pressure cylinders of a multi stage machine is lighter in construction. Disadvantages :More expensive in initial cost. REFRIGERATION The process of producing the effect of cooling & maintaining low temperature as long as required is called Refrigeration. Net Refrigerating Effect (N) :The amount of heat extracted from a body in a given amount of time is called Net Refrigerating Effect (N). Co-efficient of performance is the measure of performance of refrigerator & is the ratio of net refrigerating effect to that of work done. COP = N > 1 W N & W are measured theoretically, the COP measured is called Theoretical COP. Theoretical COP = Th. N Th. W In practice, N & W are measured when refrigerator is working & COP thus obtained is called Actual COP. 12 Actual COP = Act.N Act.W The ratio of actual COP to theoretical COP is called Relative COP. Relative COP = Act. COP Th. Cop which indicates the performance of machine & is a measure of deviation of actual performance from theoretical performance. Capacity of Refrigerator :-(Unit of Refrigeration) It is measured in terms of tonnes of refrigeration (T.R) 1 tonnes refrigeration = amount of heat extracted to produce one tones of ice from 00c water to 00 ice in a period of 24 hrs. This equivalent amount of heat is found to be 210kJ/min. i.e., 1 T.R = 210kJ/min 2 T.R = 420 kJ/min Air Refrigeration System :- 13 COP = N/W N = net refrigerating effect (process 4-1) = CP(T1-T4) kJ/kg air W = compressor work – expn work = n n–1 = n n-1 P2V2 – P1V1 - n P3V3 – P4V4 n-1 R [(T2 – T1) – (T3 – T4)] if PV=RT 14 Theoretical COP = N/W Process 1-2 :- Air is compressed adiabatically/ polytropically inside r.p air compressor. Process 2-3 :- High pressure and temperature air is cooled in air cooler under constant pressure by circulating cooling water externally in the air cooler. Process 3-4 :- High pressure & relatively low temperature air is expanded behind the piston of air expansion cylinder, because of which temperature will be lowered to about -800C. Process 4-1 :- Low temperature air enters into the cold chamber. In cold chamber, the articles to be cooled are kept (vegetables, fruits, medicine, etc..) which are at atmospheric temperature. Low temperature air extracts heat from the articles, produces refrigeration effect & comes out of cool chamber at high temperature. In open-air system, low temperature air comes in direct contact with the articles kept in cold chamber. Pressure in the cold chamber is of 1 atmosphere. Normally moisture content associated with the articles will be carried away by air & moisture becomes ice when temperature becomes very low. In dense air refrigeration system, as in fig, low temperature air flows inside coils. There is no direct contact between air & articles to be cooled. Hence there is no chance of ice formation. Pressure of air can be more than 1 atmosphere as it is closed circuit. Hence, compact refrigerator is possible. Shafts of compressor &expansion cylinder are connected to common shaft. Hence, network supplied to compressor is Compressor Work – Expansion Work. N = net refrigerating effect = Cp( T1-T4) W = compressor work – expansion work = n P2V2 – P1V1 n–1 = n - n n-1 R [(T2 – T1) – (T3 – T4)] n-1 COP = N/W = Cp (T1 - T4 ) n n-1 R [(T2 – T1) – (T3 – T4)] P3V3 – P4V4 15 Pb 1:-It is required to produce 10 tonnes of ice from water at 250C to ice at -50C in 24hrs. Assume relative COP of 90%, find HP required to run the compressor. Air is compressed from 1atm to 4atm in an open air system. In a dense air refrigeration system pr ratio will be 4 to 16 Sol:- T1 = 10+273 = 283K T3 = 30+273 = 303K P1/P2 = 4 Applying Gas law to 1-2:T2 = P2 T1 P1 n-1/n T2 = 283(4)1.35-1/1.35 = 405.4K Similarly T3 = P2 T4 T4 n-1/n P1 = 303 = 220K (-52.950C) (4)1.3-1/1.3 N = Cp (T1 - T4) kg/kg of air = 1.01(283 – 220) = 63.63 kJ/kg of air W = n R(T2 – T1) – n n–1 = R(T3 – T4) n-1 1.35 (0.294)[(405.4 – 283)] – 1.3 (0.0294)[303-220] 1.35-1 = 33.10 kJ/kg 1.3-1 16 Th COP = Th N/ Th W = 63.63/33.10 = 1.924 Actual COP = 0.9(Th. COP)= 0.9(1.924) = 1.731 cool 0 25 C 00C water cool 00C ice cool -50C Heat to be extracted from 1kg of water at 250C to produce 1kg of ice at -50C 1 x 4.2 x(25.0) +336 + 1 x 2.1 x (0 + 5)= 451.5 kJ/kg of ice Actual N = (10 x 1000 x 451.5)/ 24 kJ/kg = 18.8125 x 104 kJ/hr Act COP = Ac. N/ Ac.W 1.73 = 18.8125 x 104/ Act. W=x Act W = x / 3600 kW = 30.21 kW Act N = 18.25 x 104/60 x 212 = 14 TR (1 TR =212 kJ/min) VAPOUR COMPRESSION REFRIGERATION SYSTEM :In VCR system, vapours like ammonia, SO2, CO2 & frcon are used & they have better thermodynamic properties when compared to air. 17 Fairly dry NH3 vapour is compressed adiabatically so that heat of compression dries up, the fairly dry ammonia vapour (Process 1-2) Ammonia vapour is condensed in the coils of condenser to liquid NH3 by circulating external cooling water (process 2-3) Liquid NH3 is throttled to a very low pressure through restricted passage of expansion value corresponding temperature will be around -15oC.(process 3-4) Low temperature NH3 wet vapour is circulated in coils of evaporator. In evaporator articles to be cooled are kept . Low temperature NH3 extracts heat from articles & produces the refrigeration effect (process 4-1) Cycle is completed and repeated again. 18 W = (h2-h1) kJ/kg N = (h1-h4) kJ/kg 3-4 is throttling h3 = h4 N = h1 - h3 Th COP = N/W = h1 - h3 / h2-h1 To improve the performance of Simple Refrigeration Cycle :(1)SUPERHEATING OF VAPOUR AFTER COMPRESSION :- It can be seen from the T-S & P-h diagram, because of superheating of the refrigerant after compression, the increase in N is (h1-h5) instead of (h11-h5). Increase in W is (h2h1)instead of (h3-h11). It is found in practice, the rate at which W increases is more than the rate at which N increases. Hence COP=N/W, decreases. It is always easy to compress the vapour alone 19 when compared to that of liquid refrigerant. It is desirable to have vapour compression instead of liquid compression. (2) SUB COOLING OR UNDER COOLING:- An under cooler is added between a condenser and expansion value. In the U.C a part of low temperature refrigerant is circulated. Liquid Refrigerant is passed through U.C so that refrigerant is sub-cooled to below to saturation temperature from 3-4. From 4-5 it is throttle as usual in expansion value. It is seen from P-h & T-S diagrams that N increases to (h1-h5)instead of (h1-h51) whereas W remains same as (h2-h1). Hence effect of under cooling is to improve the COP=N/W. (3) COMBINED SUPER HEATING & UNDER COOLING :- It can be seen from T-S & P-h diagram that N increases both in super heating & under heating. N= (h1-h6) instead of (h11-h61) 20 W= (h2-h1) only once. W= (h2-h1) instead of (h3-h11) It is found that the rate at which W increases is less than the rate at which N increases. Hence, COP= N/w, increases. Wet Compression During any part of compression, if the refrigerant is wet, then it is called as Wet compression as in fig 1. Performance of compressor is poor during wet compression. If entire compression less in super heated region, it is called as dry compression. Dry compression of refrigerant is always advisible, which improves performance of compressor. Pb1: A refrigerating plant works b/w temperature limits of -5 & 25oC. Working fluid ammonia has a dry fraction of 0.62 entry to compressor. If m/c has relative COP of 55%. Calculate the amount of ice formed during a period of 24hrs. Ice is to be formed at 00C from water at 150C & 6.4 kg of NH3 is circulated /min. Sp. heat of H2O = 4.187 kJ/kg Latent Heat of ice = 335 kJ/kg Show process on T-S & P-h diagram. Obtain properties from ammonia tables or use P-h chart. Sol: From NH3 tables:- 21 At sat temperature -50C hf hg sf sg 1-2 adiabatic -50C 158.1 1437.95 0.62985 5.4037 S1=S2 25oC 298.8 1466.99 1.12345 5.0355 Sf1+x1 Sfg1 = Sf2 + x2 Sfg2 Sf1+x1(Sg1-Sf1) = Sf2 + x2(Sg2-Sf1) 0.62985+0.62(5.4037-0.62985)= 1.12345+x2(5.0355-1.12345) x2 = 0.6304 h1= hf1+x1(hg1-hf1) = 158.1+0.62(1437.95-158.1) kJ/kg of refrigerant = 951.61 kJ/kg h2 = hf2+x2(hg2-h2f) = 298.8+ 0.63(1466.99-298.8) = 1034.76 kJ/kg h3 = 298.8 kJ/kg (enthalpy of liquid at high temp.) N = (h1-h4) = (h1-h3) = 652.81 kJ/kg W = (h2-h1) = 83.23 kJ/kg Th COP = N/W = 652.81/83.23 = 7.84 Rel COP = Act COP/Th COP Act COP = 0.55(7.84) = 4.312 Act COP = Act N/Act W Heat to be extracted from water to produce 1kg ice from 150C to 00C. Cool 0 15 C water Cool 0 0C water mst + latent heat = 1(4.2) (15-0) + 335 = 398 kJ/kg of ice Actual N = (0.55 X 652.81 X 6.4) kJ/min 00C ice 22 = 2297.89 kJ/min ∴ ice produced/min, = 2297.89/398 = 5.77 kg/min = 346.4 kg/hr = 8313.97 kg/day = 8.31 tonnes/day Pb2:- A vapour refrigerating system using NH3 as refrigerant operates b/w evaporater temp of 60C & condenser temp of 230C. Vapour leaving the evaporator is dry & saturated. There is no under cooling of liquid in condenser. Determine (1) COP (2) Power per ton of refrigeration in kW (3) mass flow rate of NH3 for 10TR. Use foll. Prop of NH3. Enthalpy Entropy Temp hf hg Sf Sg 0 23 C 528.36 1707.01 4.5801 8.5613 0 -6 C 392.28 1679.37 4.0979 8.9170 To find x1, 1-2 : adiabatic S1 = S2 Sf1+x1(Sg1-Sf1)=Sg2 4.5801+x1(8.5613-4.5801) = 8.9170 x1 = 0.925 h1 = hf1+x1(hg1 – hf1) = 4.2[93.4+0.925(399.85-93.4)] = 1582.84 h2 = 4.2 X 406.43 = 1707.01 h3 = h4 = hf2 = 125.8 X 4.2 = 528.36 N = h1 – h3 = 1054.48 kJ/kg W = h2 – h1 = 124.17 kJ/kg 23 Th COP = N/W = 8.5 1 TR = 212 kJ/min COP = 212/W = 8.4 W = 212/8.4 X 1/60 = 0.42 kW TR reqd – 10 TR = 2120 kJ/min (heat to be removed) ∴ mass flow rate of NH3 = 2120/1053.3 = 2.01kg/min PART B:In above problem, if there is an under cooling of 30C, before expansion. Determine change in theoretical COP. Repeat to h1, h2, h3 h3 – h4 = Cp liquid( T3-T4) 528.36 – h4 = (0.64 x 4.2) (23 -20) h4 = 520.3 kJ/kg N = (h1-h5) = (h1-h4) = 1062.54 kJ/kg W = h2 – h1 = 124.17 kJ/kg Th COP = N/W = 8.6 (8.557) Pb 3:- Temperature range in a frcon -12 plant is -60C to 270C. compression is isotropic & there is no cooling of liquid. Find COP assuming that the refrigerant, (1) after compression is dry & saturated.. (2) leaving the compressor is dry & saturated properties of frcon12 are as follows:- 24 Temp -60C 270C hf 413 445 hg 571 585 sf 4.17 4.28 sg 4.76 4.75 Cp 0.641 0.71 Vapour Absorption Type Refrigeration System : It is found in practice that ammonia vapour readily dissolves in water to form strong solution of ammonia. It is also observed that when this solution is heated, NH3 vapours are readily evolved out of solution. This principle is made use of in vapour absorption system. It is also called as “Aqua- Ammonia System”. Improved or modified vapour absorption refrigeration system A simple V.A.R system consists of (i)an absorber (ii)a pump (ii)a generator (iv)a condenser (v)expansion value (vi)evaporator 25 To improve the performance of system, as in fig, following additional devices are added :(i)heat exchanger (ii)analyzer (iii)rectifier (iv)moisture absorber NH3 wet vapour enters into absorber and mixes with water to form strong NH3 solution. During mixing process heat is generated(exothermic process) & capacity of absorption is reduced. To keep the temperature of water cool in absorber, cooling water is circulated externally as in fig. Strong NH3 solution is prepared by using an ordinary centrifugal pump. Therefore strong & pressurized NH3 solution enters the generator through a heat exchanger. In heat exchanger, NH3 solution is preheated by absorbing heat from warm, weak solution coming from generator & going to absorber through a pressure reducing value as in fig. In the generator, pre heated strong NH3 solution is heated by using electrical means or steam heating or solar heating. Therefore of heating NH3 vapour is separated at same pressure. In the analyzer, water particles associated with NH3 vapour are separated. In rectifier, cooling water is circulated to condense any water vapour & this water comes back to generator. Strong pressurized NH3 vapour is condensed to liquid NH3 as it is passed through coils of condenser. Heat of condensation is taken away by cooling water circulated externally. Any traces of water particles are completely removed in moisture absorption otherwise, moisture at low temperature becomes ice & may choke the passage. NH3 liquid is now throttled to low pressure as it passes through a expansion value & corresponding temperature will be around -150C. This low temperature NH3 vapour enters coils of evaporator & extracts heat from articles to be cooled & kept in evaporator. ( Refrigeration effect is performed here) NH3 vapour leaving the evaporator enters the absorber. Cycle is completed and repeated. ∴ COP = Heat extracted in evaporator Pump work + Heat supplied in generator 26 In practice, in most cases. Pump work is very small & is neglected. Vap. Compression System (i)Presence of compressor always makes Sound & consumes more power. Vap. Absorption System (i) Pump is always noiseless & consumes only Fraction of compressor works. (ii)Reciprocating motion of compression (ii)Rotary motion requires less maintainance and involves regular maintainance like lubrication. almost no lubrication. (iii)Part load performance of compressor is poor & hence COP varies with the load coming over the evaporator. (iii)COP almost remains constant with load. (iv)Size of compressor increases with quantity of NH3 to be compressed. (iv)Size of pump remains almost constant as pump is a non-positive type of displacement device. SOLVED QUESTION PAPERS ON STEAM GENERATION AND STEAM NOZZLES :(1)Draw a neat sketch of throttling calorimeter and explain how dryness fraction of steam is determined. Clearly explain its limitations. Figure shows a throttle calorimeter which is connected to the steam main pipe. 27 Steam is throttled down to lower pressure and it comes out in a super heated condition. Pressure and temperature of steam after throttling is measured using a water manometer and a thermometer respectively. Let P1 – Gauge pressure of steam at entrance.(bar) x1 – Dryness fraction of steam in the steam main. Pa – Atmospheric pressure(bar) Ts1 – Saturation temperature of steam at (P1+Pa) (K) hfg1 – Latent heat of steam at (P1+Pa) (kJ/kg) hf1 – Sensible heat of steam at (P1+Pa) (kJ/kg) hW – manometer reading above atmospheric pressure(m H2O) P2 – Absolute pressure of steam at exit. = Pa + hW . 1.03 13.6 (bar) 76 Ts2 – saturation temperature of steam at P2. (K) Tsup2 – Temperature of steam after throttling (K) Cps – Specific heat of super heated steam (kJ/kgK) hg2 – Enthalpy of saturated steam at P2. ( kJ/kg) From energy balance, Enthalpy before throttling = Enthalpy after throttling hf1 + x1 . hfg1 = hg2 + Cps (Tsup2 – Ts2) x1 = hg2 + Cps(Tsup2 – Ts2)-hf1 hfg1 If Tsup2 = Ts2, then x1= hg2 – hf1 hfg1 28 Limitations: (i)If the steam whose dryness fraction is to be determined is very wet, then throttling to atmospheric pressure may not be sufficient to ensure super heated steam at exit. (ii)This calorimeter cannot be used if the dryness fraction of the steam is above 0.95. (2)Steam pipeline supplies steam to a turbine at 20 bar in superheated condition. When coming out of the generator enthalpy of steam was 3200kJ/kg. due to losses for some reason steam supplied to the turbine contained only 2700kJ/kg of heat. Determine (i)The temperature of steam at the steam generator outlet. (ii) The quantity of steam supplied to the turbine. At P=20 bar, from the steam tables hf= 908.6 kJ/kg, hfg=1888.6 kJ/kg, hg = 2797.2 kJ/kg Ts=212.40C = 485.4K Assume the specific heat of superheat of steam, Cps = 2.25 kJ/kgK Therefore, the enthalpy of the steam from the steam generator; h1 = hg + Cps (Tsup1-Ts) 3200 = 2797.2 + 2.25(Tsup1 – 485.4) Tsup1 = 3200-2797.2/2.25 + 485.4 (i)The temperature of the steam at the steam generator outlet: [Tsup1 = T1= 664.42K=391.420C] At the turbine entry; h2 < hg. Therefore the steam is wet.. The enthalpy of steam at the turbine entry; h2 = hf + x2. hfg 2700 = 908.6 + x2 x 1888.6 x2 = (2700-908.6)/1888.6 = 0.9485 29 The quantity of the steam at the turbine entry is wet with a dryness fraction, x2 = 0.9485 (3)Describe the process of formation of steam and give its graphical representation also. Formation of steam: Consider a cylinder filled with 1kg of water at 00C with volume Vf m3 under a piston which can freely move upwards and downwards in it. Further, the piston is loaded with a weight as in fig.As heat is imparted to water, a rise in temperature will be noticed and this rise will continue till boiling point is reached. 30 The temperature at which water starts boiling depends upon the pressure and as such for each pressure, there is a different boiling point. This boiling temperature is known as the temperature of formation of steam or Saturation temperature Ts. The increase in volume of water upto boiling point is negligible. When heating of water is continued after saturation point is reached, it will be noted that there will be an increase in volume which indicates that steam formation is taking place. The heat being supplied does not show any rise of temperature but changes water into vapour state(steam) and is known as Latent heat or Hidden heat. So long as the steam is in contact with water, it is called wet steam(fig b). If heating of steam is further progressed such that all the water particles associated with the steam are evaporated, the steam so obtained is called dry and saturated steam. (Figc)During the latent heat addition to steam, the volume increases from Vf to Vg producing work. Further if the supply of heat to the dry and saturated steam is continued at constant pressure there will be an increase in temperature and volume of steam. The steam so obtained is called Super heated steam and it behaves as a perfect gas. The temperature of the superheated steam will be Tsup and it will be above Ts. The volume of the steam decreases fron Vg to Vsup.(fig d). The temperature heat addition diagram shopwn by fig(e) gives the graphical representation of formation of steam. ( 4 ) In a separating – throttling calorimeter the total quantity of steam passed was 40 k.g. and 2.2 kg of water was collected from separator . Steam pressure before throttling are 1200C and 107 .88 kpa . Determine the dryness fraction of steam before entering to calorimeter. Specific heat of super heated steam may be considered as 2.09 kJ /k.g K. P 1 = P2 = 1.47 MPa = 14.7 bar ms = 40 kg 31 mW = 2.2 kg P3 = 107.88 kPa = 107.88/1000 MPa = 107.88/1000 x 10 bar P3 = 1.078 bar T3 = 1200C = 393K CPs = 2.09 kJ/kgK From the steam table; with suitable data interpolations At P1=P2 = 14.7 bar Ts = 1970C hf = 840.5 kJ/kg , hfg = 1948.5 kJ/kg , hg = 2789 kJ/kg At P3 = 1.078 bar Ts = 99.70C hf = 418.5 kJ/kg , hfg = 2257.9 kJ/kg , hg = 2676.4 kJ/kg Consider the throttling calorimeter, since Ts < 1200C the steam is super heated after throttling. Tsup3 = 1200C = 393K Ts3 = 99.70C =372.7K Also; Enthalpy of steam before throttling = Enthalpy of steam after throttling h2 = h3 hf2 + x2 . hfg2 = hg3 + Cps ( Tsup3 –Ts3 ) 840.5 + x2 . 1948.5 = 2676.4 + 2.09 ( 393 – 372.7 ) x2 = (2676.4 + 43.054 – 840.5)/1948.5 therefore fraction of steam after throttling, x2 = 0.9643 Consider the separating calorimeter, dryness fraction of steam in the steam main, x1 is equal to dryness of steam in the separating calorimeter, x2. Also; x1 = x2 . ms/ ms + mw 32 x1 = 0.9643 x 40 / 40 + 2.2 = 0.9140 Therefore the dryness fraction of steam before entering to calorimeter x1 = 0.914 (5)A rigid vessel contains 1kg of wet steam at 5bar. Heat is added to wet steam to increase its pressure to 10 bar and temperature to 5000C. Determine : (i) The dryness fraction (ii) Initial specific volume (iii) Final internal energy of steam. Assume Cps = 2.25 kJ/kgK Mass of steam = m= 1kg Initial pressure of steam P1 = 5 bar Final pressure of steam P2 = 10 bar Final steam temperature T2 = 5000C = 773K Specific heat of steam Cps = 2.25 kJ/kgK From steam tables; At P1 = 5 bar; Ts1 = 151.80C = T1 (initial temperature) hf1 = 640.1 kJ/kg hg1 = 2747.5kJ/kg hfg1 = 2107.4 kJ/kg vf1 = 0.001093 m3 / kg , vg1 = 0.375 kJ/kg At P2 = 10 bar ; Ts2 = 179.90C hf2 = 762.6 kJ/kg , hg2 = 2776.2 kJ/kg hfg2 = 2013.6 kJ/kg vf2 = 0.001127 m3/kg , vg2 = 0.194 m3/kg Since Ts2 < T2 , the steam is superheated in its final state. 33 Tsup2 = T2 = 773K Ts2 = 179.90C = 452.9K Ts1 = 151.8 = 424.8K For a constant volume heat addition process with superheated steam in the final state, we have the relation, x1 . vg1 = vsup2 = vg2 . Tsup2/Ts2 x1 = vg2/vg1 . Tsup2/Ts2 = 0.194/0.375 x 773/452.9 (i)The dryness fraction of steam : x1 = 0.8829 (ii) The initial specific volume of steam : v1 = vf1 + x1 ( vg1 – vf1 ) v1 = 0.001093 + 0.8829 ( 0.375 – 0.001093 ) v1 = 0.3312 m3/kg Also; the specific volume of superheated steam; vsup = vg2 . Tsup2/Ts2 vsup = 0.194 x 773/452.9 vsup = 0.3311 m3/kg (iii) The final internal energy of steam ; u2 = hsup – P2. vsup2 u2 = hg2 + Cps (Tsup2 – Ts2) – P2 vsup2 = 2776.2 + 2.25 ( 773 – 452.9 ) – 10 x 0.3311 u2 = 3493.114 kJ/kg (6)Define : (i)Superheated steam (ii)Internal energy (iii)Wet steam (iv)Specific volume 34 (v)Dryness fraction (i)Superheated steam : When dry steam is heated after it has become dry & saturated to a temperature above the saturation temperature under a given constant pressure, it is called Superheated steam. (ii)Internal energy : It is defined as the actual energy stored in the steam. The internal energy (u) of steam can be found by subtracting work of evaporation from the total heat or enthalpy of steam (h). Therefore, Internal energy of 1kg of steam at pressure P, u = h-P vg/J kJ/kg J=1 in S.I units vg – specific volume of dry saturated steam. (m3/kg) (iii)Wet steam : It is a two – phase mixture at saturation temperature consisting of dry saturated steam and water particles in thermal equilibrium with each other under a constant pressure. (iv)Specific volume : Specific volume of steam is the volume occupied by 1kg of steam under a constant pressure and temperature. (v)Dryness fraction : It is defined as the ratio of the mass of actual dry steam to the mass of steam containing it. If, ms – mass of dry steam is contained in the steam considered (kg) mW - weight of water particles in suspension in the steam considered (kg) Then, dryness fraction, x = ms / (ms + mW) It indicates the fraction of dry steam present in a given wet steam. (7)Determine the density of 1kg of steam initially at a pressure if 10 bar having a dryness fraction 0.78. If 500kJ/kg of heat is added at constant pressure, determine the condition & internal energy of the final state of steam. Take Cps = 2.1 kJ/kg K. Given: Mass of steam, m = 1 kg Pressure, P = 10 bar Dryness fraction initially, x1 = 0.78 35 Constant pressure heat addition, h = 500kJ/kg Specific heat of dry saturated steam, Cps = 2.1 kJ/kg K From steam tables : At P= 10 bar Ts = 179.90C = 452.9 K vf = 0.0011274 m3/kg , vg = 0.19430 m3/kg hf = 762.6 kJ/kg , hfg = 2013.6 kJ/kg , hg = 2776.2 kJ/kg Specific volume of steam in its initial state ; v1 = vf1 + x1 ( vg1 – vf1 ) = 0.0011274 + 0.78 ( 0.19430 – 0.0011274 ) v1 = 0.1518 m3/kg Therefore density of steam in its initial state, ρ = m = 1 v1 v1 = 1 0.1518 ρ = 6.5876 kg/m3 The enthalpy of steam in its initial state ; h1 = hf1 + x1 . hfg1 = 762.6 + 0.78 x 2013.6 h1 = 2333.208 kJ/kg When heat is added at constant pressure, the final enthalpy of steam ; h2 = h1 + h = 2333.208 + 500 h2 = 2833.208 kJ/kg Since h2 > hg , the steam is superheated in its final state hence x2 = 1 The enthalpy of super heated steam ; hsup = hg +Cps (Tsup2 – Ts) = h2 36 2833.208 = 2776.2 + 2.1 ( Tsup2 – 452.9 ) Tsup2 = 2833.208 – 2776.2 + 452.9 2.1 Therefore, temperature of steam in its final state ; Tsup2 = 480.046 K = 207.040C The internal energy of final state of steam ; u2 = h2 – P vg = 2833.208 – 10 x 0.1943 u2 = 2831.265 kJ/kg (8)Determine the mass of 0.15 m3 of wet steam at a pressure of 4 bar and dryness fraction 0.8. Also calculate the heat of 1 m3 of steam. Given : Volume of wet steam, V1 = 0.15 m3 Steam pressure, P = 4 bar Dryness fraction, x = 0.8 From steam tables, at P = 4 bar ; Ts = 143.630C , vf = 0.0010839 m3/kg hf = 604.7 kJ/kg , vg = 0.46220 m3/kg hfg = 2132.9 kJ/kg , hg = 2737.6 kJ/kg Let m1 be the mass of 0.15 m3 of wet steam. Then, V1 = m1 [ vf + x . vfg] m1 = V = 0.0010839 + 0.8 (vg – vf) vf + x . vfg = 0.15 0.15 0.0010839 + 0.8 ( 0.4622 – 0.0010839 ) Therefore mass of 0.15 m3 of wet steam at 4 bar, m1 = 0.4054 kg 37 Let m2 be the mass of 1m3 of wet steam at 4 bar and dryness fraction 0.8. Then ; V2 = m2[ vf + x . vfg ] m2 = v2 vf + x ( vg – vf ) = 1 0.0010839 + 0.80.4622 – 0.0010839 ) m2 = 2.7028 kg The heat of 1m3 of steam; h = m2 [hf + x . hfg ] = 2.7028 [ 604.7 + 0.8 x 2132.9 ] h = 6246.224 kJ (9)Wet steam of dryness fraction 0.8 at a pressure of 8 bar is heated under constant volume until the steam becomes fully dry. Find the final pressure and hence the heats transfer during the process. Given: Initial pressure of steam, p1 = 8bar Initial dryness fraction, x1 = 0.8 Let the final pressure of steam be p2. At the end of constant volume heat addition process the steam is fully dry therefore x2 = 1 From steam tables, at p1 = 8 bar Ts1 = 170.410C, vf1 = 0.0011150 m3/kg, vg1 = 0.24026 m3/kg hf1 = 720.9 kJ/kg, hg1 = 2767.5 kJ/kg, hfg1 = 2046.5 kJ/kg Initial enthalpy of the steam; h1 = hf1 + x1 . hfg1 = 720.9 + 0.8 x 2046.5 = 2358.1 kJ/kg The initial specific volume of steam; 38 v1 = vf1 + x1 . vfg1 = vf1 + x1 (vg1 – vf1) = 0.001115 + 0.8 (0.24026 – 0.001115 ) = 0.19243 m3/kg Since heat addition is under constant volume; v1 = v2 v1 = vf2 + x ( vg2 – vf2 ) = vf2 + 1 ( vg2 – vf2 ) v1 = vg2 vg2 = 0.19243 m3/kg From the steam tables, comparing the above values of vg with an appropriate value of pressure, with suitable approximation, we get the final pressure of steam for vg = 0.19243 m3/kg ; P2 = 10.2 bar Likely; at P2 = 10.2 bar hf2 = 764.6 kJ/kg, hfg2 = 2012.2 kJ/kg, hg2 = 2776.8 kJ/kg Final enthalpy of steam; h2 = hg2 = 2776.8 kJ/kg Therefore, heat transfer during the constant volume process, dh = h2 – h1 = 2776.8 – 2358.1 dh = 418.7 kJ/kg (10)Steam at 20 bar and 4000C expands to a turbine to 0.1 bar isotropically. Steam flow rate is 5 kg/s. Calculate (i)Work done (ii)Power developed. Given: Initial pressure of steam, P1 = 20 bar Initial temperature of steam, T1 = 4000C = 673K 39 Final pressure of steam, P2 = 0.1 bar Steam flow rate, m = 5 kg/s From the steam tables; At P1 = 20 bar; Ts1 = 212.370C = 485.37 K vf1 = 0.0011766 m3/kg vg1 = 0.099549 m3/kg hf1 = 908.6 kJ/kg, hg1 = 2797.2 kJ/kg hfg1 = 1888.7 kJ/kg, Sg1 = 6.3377 kJ/kg Since T1 is greater than Ts1 the steam is superheated in its initial state. T1 = Tsup1 = 673K Assume; Specific heat of saturated steam, Cps = 2.25 kJ/kg K Enthalpy of super heated steam, hsup1 = hg1 +Cps ( Tsup1 – Ts1 ) = 2797.2+ 2.25 ( 673 – 485.37 ) = 3219.36 kJ/kg Specific volume of superheated steam; vsup1 = vg1 . Tsup1 = 0.099549 x 673 Ts1 485.37 = 0.138 kJ/kg Therefore, internal energy of superheated steam; u1 = hsup1 – P1 . vsup1 = 3219.36 – 20 x 0.138 = 3216.6 kJ/kg From the steam tables; At P2 = 0.1 bar; Ts2 = 45.830C = 318.83K vf2 = 0.0010102 m3/kg vg2 14.6737 m3/kg hf2 = 191.8 kJ/kg, hg2 = 2584.8 kJ/kg hfg2 = 2392.9 kJ/kg, sf2 = 0.6493 kJ/kg k 40 Since the expansion of steam is isentropic; Entropy of superheated steam, s1 = Entropy of steam after expansion, s2 Cps ln( 673/ 485.37 ) = sf2 + x2 . hfg2/ Ts2 2.25 ln(673/485.37)+ 6.3377 = 0.6493 + x2 . 2392.9/318.83 Therefore dryness fraction of steam in its final state, x2 = 0.8559 The steam is wet after expansion . Specific volume of wet steam v2 = vf2 + x2 (vg2 – vf2) = 0.0010102 + 0.3559(14.6737 – 0.0010102) = 12.56 m3/kg Therefore internal energy of wet steam; u2 = [ hf2 + x2 . hfg2] p2.x2. vg2 = [191.8+0.8559x2392.9] 0.1x0.8551x14.6737 u2 = 2238.627 kJ/kg 1) The work done by isentropic expansion of steam through the turbine W = u1 – u2. W = 3216.6 – 2238.627 W = 977.973 kJ/kg 2) Power developed : P = m x W = 5 x 977.973 = 4889.865 kJ/s P = 4889.865 kW P = 4.889 MW (11)Describe with a neat sketch a separating – throttling calorimeter for measuring the dryness fraction of the steam. A Separating – Throttling Calorimeter Arrangement 41 Figure shows an arrangement for separating – throttling calorimeter. Wet steam from the main pipe passes through the sample pipe at a pressure P1 and dryness fraction x1 into a separating calorimeter. A certain mass ( mW ) of water particles is separated from the sample by the separator, and it is measured by a weighing machine. The remaining steam has a higher dryness fraction x2 and is throttled down to lower pressure P3 at the throttling calorimeter. Thus the steam after throttling becomes super heated with a temperature T3. The steam is then passed into a bucket calorimeter to measure the mass of dry steam (ms). At the throttling calorimeter. Enthalpy of steam before throttling, h2 = enthalpy of steam after throttling, h3 hf2 + x2 . hfg2 = hg3 + Cps ( T3 – Ts3) x2 = hg3 + Cps ( T3 – Ts3) – hf2 hfg2 At the separating calorimeter, the mass of dry vapour in the steam sample drawn from the main pipe = x2 . ms = Mass of dry steam leaving the separating calorimeter. Therefore, dryness fraction of the steam in the main; x1 = x2 . m3 m3 + mW The values of enthalpies at their respective steam pressures are read from the steam tables. (12)Define dryness fraction of steam. Sketch and describe a method to determine dryness fraction. Dryness fraction (x) is defined as the ratio of the mass of actual dry steam (ms) to the mass of steam containing it. i.e., x = ms ms + mw Determination of dryness fraction using separating calorimeter: 42 Figure shows a separating calorimeter. Steam is drawn from the main pipe through the sample tube and then passed into the separating calorimeter. The inlet pressure of steam is P1 and its dryness fraction is x1. In the separating calorimeter, the steam is passed on perforated trays. Due to the inertia of the water droplets, the moisture from the steam is separated mechanically. The mass of the water droplets (mW) collected is measured using a weighing machine. A peizometer is used to measure the pressure P2 of the steam after passing into the separating calorimeter. However, the drop in pressure is negligible and hence P1 approx. equal to P2. The steam then goes into the bucket calorimeter, from which the mass of dry steam (ms) can be found. Therefore, the dryness fraction of steam in the steam main; x1 = ms ms + mW (13)Show that dA = 1 . dP 1 - 1 for relationship b/w area, velocity and pressure in nozzle flow. A r P M2 Consider a convergent nozzle as shown below. A saturated steam flows through the nozzle. Let : P – Steam pressure (bar) V – Specific volume ( m3/kg) T – Steam temperature (K) h – Enthalpy of steam (kJ/kg) V – Steam velocity (m/s) A – cross sectional area of the nozzle (m2) 43 Subscripts 1 and 2 represent the inlet and outlet conditions respectively. We have the mass flow rate, m = ρ A V Also ρ = density = 1/v Assume mass flow rate to be constant. Then, m = AV/v = C Taking Natural logarithm; ln A + ln V – ln v = ln C Differentiating; 1/A . dA + 1/V . dV – 1/v .dv = 0 dA/A = dv/v – dV/V -----------(1) Consider the expansion of steam through the nozzle to be adiabatic. Then; PVr = c1 constant r – Index for expansion process. Taking natural logarithm; ln p + r . ln v = ln 0 Differentiating; 1 . dp + r . 1 P v dv = - 1 v . dv = 0 . dp r (2) p From the steady flow energy equation m10 h1 + v12 + gz1 + dϴ = m20 2 dt h2 + v22 + gz2 2 Since the nozzle is insulated; the rate of heat transfer, dQ = 0 Dt Since there is no work transfer, dW = 0 dt Assume the inlet and outlet to be at the same elevation i.e., z1 = z2 Therefore adopting all the above considerations we get; + dW dt 44 h1 + v12 2 = h2 + v22 2 (14)Derive an expression for the optimum pressure ratio for maximum discharge through the steam nozzle. Derive an expression for the mass flow rate of steam through a nozzle in terms of pressure ratio and determine the condition for maximum discharge. Show that the maximum discharge of steam through the nozzle takes place when the ratio of steam pressure at the throat to the inlet pressure is given by : ( p2/ p1 ) = (2 / n + 1) . n / n-1 , where n is the index of expansion. Consider a convergent nozzle as shown below: (14)Derive an expression for the optimum pressure ratio for maximum discharge through the steam nozzle. 45 Derive an expression for the mass flow rate of steam through a nozzle in terms of pressure ratio and determine the condition for maximum discharge. Show that the maximum discharge of steam through the nozzle takes place when the ratio of steam pressure at the throat to the inlet pressure is given by : ( p2/ p1 ) = (2 / n + 1) . n / n-1 , where n is the index of expansion. Consider a convergent nozzle as shown below: Let: P – Steam pressure (K) V – Specific volume (m3/kg) T1 – Steam temperature (K) h – Enthalpy of steam (kJ/ kg) v – Velocity of steam flow (m/s) A – Cross sectional area of the nozzle (m2) Subscripts 1 & 2 represent the inlet & outlet conditions respectively. Consider the expansion of steam through the nozzle to be polytropic. Then, the work done in expansion of steam, ( P1V1 – P2V2 ) = Rankine Cycle area W= n n–1 Where n – index of expansion. Also, Work done in expansion = Gain in kinetic energy of steam = n ( P1V1 – P2V2 ) = V22 – V12 n–1 2 Since V1 << V2, V1 can be neglected. ∴ V22 = n 2 n–1 ( P1V1 – P2V2 ) 46 Or V22 = 2n P1V1 1 - P2 n–1 . V2 P1 (1) V1 For a polytropic process: PV n = constant P1 V1n = P2 V2 n P2 = V1 P1 n V2 V1 = P2 V2 P1 1/n We have the pressure ratio, rp = P2 P1 V1 = rp 1/n (2) V2 Also V2 = V1 . 1 (3) rp 1/n (2) in (1): V22 = 2n . P1 V1 [ 1 – rp . 1 rp 1/n n-1 v2 = √ 2n . P1 V1 [ 1 – rp n-1/n ] n–1 We have the mass flow rate, m0 = ρ A V Where, ρ = density = 1/v ( for unit mass) Also; From continuity equation; m0 = A1 V1 = A2 V2 V1 ] V2 (4) 47 m0 = A2 V2 Consider, V2 From (3) & (4); √ 2n . P1V1[ 1 – rp n-1/n ] m0 = A2 V1 . rp -1/n or m0 = n-1 √ 2n . A2 P1V1. rp2/n [ 1 – rp n-1/n ] V12 n-1 m0 = A2 √ 2n . n-1 m0 = A2 √ 2n . n-1 P1 [ rp 2/n - rp 2/n + n-1/n ] V1 P1 [ rp 2/n - rp n + 1/n ] (5) V1 (1) gives the expression for mass flow rate of steam through the nozzle in terms of pressure ratio rp . Condition for maximum discharge: Discharge through the nozzle is maximum when dm0 = 0 drp Therefore, differentiating (5) w.r.t rp, we write the following expression excluding the constants; 2 . rp 2/n-1 – ( n+1) rp n+1/n – 1 = 0 n n 2 . rp 2-n/n = ( n+1) rp n+1-n/n – 1 = 0 n n 1/n – (2-n)/n 2 = rp n+1 rp1-2+n/n = 2 n+1 rpn-1/n = 2 n+1 48 rpc = ( 2 )n/n-1 (6) n+1 The pressure ratio rpc is given by (6) is optimum pressure. Therefore, the discharge through the nozzle will be maximum when rp = rpc (15)Explain the concept of super saturation flow of steam. Explain the super saturated or metastable flow of steam through a nozzle and significance of Wilson’s line. Explain Wilson’s line. Concept of super saturation flow of steam / meta stable flow of steam through a nozzle : When steam flows through a nozzle, it is normally expected that the discharge of steam through the nozzle would be slightly less than the theoretical value. But, during experiments on the flow of wet steam, it has been observed that the discharge is slightly greater than that calculated by the formula. This phenomenon is explained as follows: The converging part of the nozzle is so short and the steam velocity so high that the molecules of steam have insufficient time to collect and form droplets so that normal condensation does not take place. Such rapid expansion is said to be metastable and produces a super-saturated state. In this state of super saturation the steam is under cooled to a temperature less than that corresponding to its pressure; consequently the density of steam increases and hence the weight of discharge. Wilson line: Prof. Wilson through experiments showed that dry saturated steam, when suddenly expanded in the absence of dust, does not condense until its density is about 8 times that of the saturated vapour of the same pressure. This effect is discussed below: 49 In the figure above, the point 1 represents initial state of the steam. The steam expands isentropically without any condensation to point 2, which is on the superheat constant pressure curve AB produced. At point 2, the limit of super saturation is reached and steam reverts to its normal condition at 3 at the same enthalpy value as 2, and at the same pressure. The steam continues expanding is entropic ally to lower pressure to point 4 instead of 41 which would have been reached if thermal equilibrium had been maintained. Consequently, enthalpy drop is reduced and the condition of the final steam is improved. The limiting condition of under cooling at which condensation commences and is assumed to restore conditions of normal thermal equilibrium is called the ‘Wilson line’. The problems on super saturated flow cannot be solved by Mollier chart unless Wilson line is drawn on it. (16)Dry steam at 10 bar at 100m/s enters the nozzle & leaves it with velocity of 300m/s at 5 bar. For 16 kg/s of steam mass flow rate, determine heat drop in a nozzle & the final state of steam leaving nozzle assuming heat losses to surroundings as 10 kJ/s. Given: Inlet pressure, P1 = 10 bar Inlet velocity, V1 = 100m/s Exit pressure, P2 = 5 bar Exit velocity, V2 = 300m/s Mass flow rate of steam, m0 = 16kg/s Rate of heat lost to the surroundings, dQ / dt = - 10 kJ/s Therefore total heat lost to the surroundings, Q = - dQ/dt = - 10 = -0.625 kJ/kg mo 16 From the steady flow energy equation; m10 [ h1 + V12 / 2 + gz1 ] + Q = m20 [ h2 + V22/2 + gz2 ] + W The mass flow rate is assumed to be constant. Therefore m10 = m20 = m0 The nozzle is assumed to be horizontal. Therefore z1 = z2 There is no net work transfer for the nozzle therefore. W = 0 (1) 50 In (1); h1 + V12 + Q = h2 + V22 2 2 h1 – h2 = V22 – V12 - Q 2 h1 – h2 = 3002 – 1002 - ( - 0.625 x 103) = 40.625 kJ/kg 2 Therefore heat drop in the nozzle, h1 – h2 = 40.625 kJ/kg From the steam tables: Inlet 1 Inlet 2 P(bar) 10 5 Ts(0C) 179.88 151.85 hf(kJ/kg) 762.6 640.1 hfg(kJ/kg) 2013.6 2107.4 hg(kJ/kg) 2776.2 2747.5 At the nozzle inlet, the steam is dry saturated & x1 = 1 Also; h1 = hg1 = 2776.2 kJ/kg h1 – h2 = 40.625 Enthalpy of steam at exit, h2 = h1 – 40.625 = 2776.2 – 40.625 = 2735.575 kJ/kg Since h2 < hg2, the condition of steam at the exit is wet. Also; h2 = hf2 + x2 . hfg2 Dryness fraction; x2 = h2 – hf2/hfg2 x2 = [2735.575 – 640.1 ] / 2107.4 = 0.9943 (17)In a nozzle steam expands from 12bar in 3000C to 6 bar flow rate of 5 kg. Determine the throat & exit area. If exit velocity is 500 m/s and velocity at inlet to the nozzle is negligible, also find the coefficient of velocity at exit. 51 Given: Inlet pressure, P1 = 12 bar Inlet temperature, T1 = 3000C Exit pressure, P3 = 6 bar Exit velocity, V21 = 500m/s Velocity at inlet is negligible. Throat area, A2 =? Mass flow rate of steam, m0 = 5kg/s Exit area, A3 =? Co efficient of velocity at exit, Cv =? From the steam tables: At P1 = 12 bar; Ts1 = 187.960C = 460.96 hf1=798.4kJ/kg hfg1=1984.2kJ/kg hg1=2782.7kJ/kg Vg1=0.16321 m3/kg At P3 = 6 bar Ts3 = 158.840C hf3=670.4kj/kg 52 hfg3=2085kJ/kg hg3=2755.5kJ/kg Vg3=0.31546kJ/kg Since T1>Ts1, the steam is superheated at entry. Assume Cps = 2.25 kJ/kg Then, h1=hg1 + Cps ( T1 – Ts1) = 2782.7+2.25(573-460.96) h1= 3034.79 kJ/kg The optimum pressure ratio; rpc = P2/P1 = (2/n+1)n/n-1 The index of expansion for superheat steam, n=1.3 Therefore, P2=P1(2/n-1)n/n-1 = 12(2/1.3+1) 1.3/1.3-1 P2 = 6.55 bar (pressure at throat) At P2 = 6.5 bar Ts2 = 161.90C Vg2=0.29407 m3/kg hf2=683.75 kJ/kg hfg2=2074.95 kJ/kg hg2 = 2758.75 kJ/kg Assume the steam to be dry saturated at the throat after expansion. x2 = 1 & h2 = hg2 = 2758.75kJ/kg Applying steady flow energy equation b/w the inlet & the throat; m10 [ h1 + V12 / 2 + gz1 ] + Q = m20 [ h2 + V22/2 + gz2 ] + W Assume the mass flow rate to be constant throughout, m10 = m20 53 Consider the nozzle to be horizontal, z1 = z2 There is no heat transfer Q=0 & no work transfer W=0 Therefore modified SFEE: h1 + V12/2 = h2 + V22/2 The inlet velocity is negligible h1 = h2 + V22/2 V2 = √2 h1 – h2 x 103 V2 = √293034.79 – 2758.75) x 103 V2 = 743.02 m/s We have the mass flow rate, m0 = ρ A V Where, ρ = density = 1/v ( for unit mass) Also; From continuity equation; m0 = A1 V1 = A2 V2 V1 Consider, V2 m0 = A2 V2 V2 A2 = m0 x vg2 ;v2 = vg2 V2 = 5 x 0.29407 = 1.9788x10-3 m2 743.02 Therefore the throat area A2 = 1978.88 mm2 Assume the steam to be dry saturated even at the exit. x3 = 1 & h3 = hg3 = 2755.5 kJ/kg Applying the modified SFEE b/w the throat & the exit h2 = V22/2 = h3 + V32/2 54 V32/2 = (h2 – h3) x 103 + V22/2 V3 = √2{(2758.75-2755.5) x 103 + 743.02 2/2}\V3 = 747.38 m/s (ideal velocity) From (1), m0 = A3V31/V3 But V3 = Vg3 = 0.31546 m3/kg A3 = m0 x V3/V31 = 5x 0.31546/500 Exit area,A3 =3.1546 x 10-3 m2 A3 = 3154.6 mm2 The coefficient of velocity at exit Cv=V31/V3 =500/747.38 Cv=0.669 (18)Dry air at a temperature of 270C & pressure of 20 bar enters a nozzle and leaves it at a pressure of 4 bar. Find the mass of air discharged, if the area of the nozzle is 200 mm2. Consider a convergent nozzle as shown below: Inlet pressure, P1 = 20 bar Inlet temperature, T1 = 270C= 300K Exit pressure, P2 = 4 bar Exit area, A2 = 200mm2 = 200 x 10-6 m2 From the steady flow energy equation; m10 [ h1 + V12 / 2 + gz1 ] + Q = m20 [ h2 + V22/2 + gz2 ] + W The mass flow rate of air is assumed to be same at inlet and exit. Therefore m10 = m20 = m0 Consider the nozzle to be horizontal, z1 = z2 55 There is no heat transfer Q=0 & no work transfer W=0 h1 = h2 + V22/2 ; V22 = 2 (h1 - h2 ) V2 = √2 h1 – h2 x 103 (1) Assume the change in enthalpy, dh = Cp.dT (h1 - h2 ) = Cp (T1 – T2 ) (2) Cp for air = 1.005 kJ/kgK Assume the expansion of steam through the nozzle to be isentropic. For an isentropic process, P1 T1 –r/r-1 = P2 T2 –r/r-1 ϒ= 1.4 for air T2 = T1 (P2/P1) r-1/r T2 = 300(4/20)1.4-1/1.4 T2 = 189.42 K In (2); (h1 - h2 ) = 1.005 (300 – 189.42 ) = 111.1329 kJ/kg V2 = √ 2 x 111.1329 x 1000 V2 = 471.45 m/s We have the gas law, PV = RT R = 287.14 J/kg K for air P2 V2 = R T2 Or V2 = R T2 / P2 = 287.14 x 189.42/4 x 10 5 The specific volume of air at exit, V2 = 0.1359 m3/kg The mass flow rate of air, m0 = ρ2 A2 V2 = A2 V2/V2 = 200 x 10 -6 x 471.45/ 0.1359 = 0.6938 kg/s (19)The inlet steam to a convergent divergent nozzle is at 22 bar and 2600C. The exit pressure is 4 bar. Assuming frictionless flow up to the throat and a nozzle efficiency of 85%, calculate (i)flow rate for a throat area = 32.2 cm and (ii)Exit area. 56 Given: Inlet pressure, P1 = 22 bar Inlet temperature, T1 = 2600C= 533K Exit pressure, P3 = 4 bar Steam flow is frictionless up to the throat. Nozzle efficiency, η =85% = 0.85 Throat area, A2 = 32.2 cm2 = 32.2 x 10 -4 m2 At P1 = 22 bar; Ts1 = 217.20C= 490.2K; hg1 = 2799.1 kJ/kg At P3 = 4 bar; vg3 = 0.462 m3/kg hg3 = 2737.6 kJ/kg Since the saturation temperature Ts1 < T1 the steam is superheated at inlet. Assume Cps = 2.25 kJ/kg K h1 = hg1 + Cps (T1 - Ts1 ) = 2799.1 + 2.25 ( 533 – 490.2 ) = 2895.4 kJ/kg We have, the optimum pressure ratio; rpc = P2 /P1 = ( 2/ n+1) n/n-1 For super heated steam; n = 1.3 = 22 ( 2/1.3 + 1) 1.3/1.3 -1 57 Throat pressure, P2 = 12 bar. Vg2 = 0.163 m3/kg hg2 = 2782.7 kJ/kg Assume the steam to be dry saturated at throat and exit. Then; h2=hg2=2782.7 kJ/kg V2=vg2=0.163 m3/kg h3=hg3=2737.6kJ/kg V3=vg3=0.462m3/kg Applying steady flow energy equation b/w the inlet and the throat; m10 [ h1 + V12 / 2 + gz1 ] + Q = m20 [ h2 + V22/2 + gz2 ] + W Assume the mass flow rate to be constant throughout, m10 = m20 Consider the nozzle to be horizontal, z1 = z2 There is no heat transfer Q=0 & no work transfer W=0 Therefore modified SFEE: h1 + V12/2 = h2 + V22/2 The inlet velocity is negligible h1 = h2 + V22/2 V2 = √2 h1 – h2 x 103 V2 = √2(2895.4 – 2782.7) x 103 V2 = 474.76 m/s The mass flow rate of air, m0 = ρ2 A2 V2 = A2 V2/V2 = 32.2 x 10 -6 x 474.76/ 0.163 (i)Steam mass flow rate: m0 = 9.378 kg/s Applying the modified SFEE b/w the throat & the exit h2 + V22/2 = h3 + V32/2 58 V32/2 = (h2 – h3) x 103 + V22/2 V3 = √2{(2782.7-2737.6) x 103 + 474.76 /2}\V3 = 561.78 m/s We have the nozzle efficiency, η =(V31)2 – V22 V32 – V22 V31 = √0.859V32 – V22) + V22 V31 = √0.85(561.78 2 – 474.76 2 ) + 474.76 2 V31 = 550 m/s Also; m0 = A3 V31 / V3 A3 = m0 x V3 / V3 A3 = (9.378 x 0.462) / 550 = 7.8775 x 10 -3 m2 (ii)Exit area; A3 = 7877.52 mm2 (20) The inlet conditions of steam to a convergent divergent nozzle is 2.2 MN/m2 & 2600C. The exit pressure is 0.4 MN/m2. Assuming a nozzle efficiency of 85% b/w throat & exit, determine: (i) The flow rate of steam for a throat area of 33cm2 and (ii) exit area Given: Inlet pressure, P1 = 2.2 MN/m2 = 22 bar 59 Inlet temperature, T1 = 2600C= 533K Exit pressure, P3 =0.4 MN/m2 = 4 bar Throat area, A2 = 33 cm2 = 33 x 10 -4 m2 From the steam tables: At P1 = 22 bar; Ts1 = 217.20C= 490.2K; hg1 = 2799.1 kJ/kg Since the saturation temperature Ts1 < T1 the steam is superheated at inlet. Assume Cps = 2.25 kJ/kg K h1 = hg1 + Cps (T1 - Ts1 ) = 2799.1 + 2.25 ( 533 – 490.2 ) = 2895.4 kJ/kg We have, the optimum pressure ratio; rpc = P2 /P1 = ( 2/ n+1) n/n-1 P2 = P1 ( 2/ n+1) n/n-1 N = 1.3 for superheated steam. P2 = 22 ( 2/ 1.3 + 1) 1.3/1.3-1 = 12 bar At P2 = 12 bar; Assume the steam to be dry saturated at the throat. h2=hg2=2782.7 kJ/kg V2=vg2=0.163 m3/kg From the steady flow energy equation; m10 [ h1 + V12 / 2 + gz1 ] + Q = m20 [ h2 + V22/2 + gz2 ] + W Assume the mass flow rate of air is assumed to be same throughout. Therefore m10 = m20 Neglect the inlet velocity V1. Consider the nozzle to be insulated. Therefore Q = 0 Consider the nozzle to be horizontal, z1 = z2 60 There is no heat transfer Q=0 & no work transfer W=0 We get, h1 = h2 + V22/2 ; V22 = 2 (h1 - h2 ) V2 = √2 h1 – h2 x 103 = √ 2 ( 2895.4 – 2782.7 ) x 103 V2 = 474.76 m/s From the continuity equation, the flow rate, m0 = ρAV = AV/v = 33 x 10 -4 x 474.76 / 0.163 (i)The flow rate; , m0 =9.612 kg/s Applying the modified SFEE b/w the throat & the exit h2 + V22/2 = h3 + V32/2 V32/2 = (h2 – h3) x 103 + V22/2 V3 = √2000{(2782.7-2737.6) x 103 + 474.76 /2}\V3 = 561.78 m/s Assume the steam to be dry saturated at the exit. Then at P3 = 0.4 MN/m2 = 4 bar h3=hg3=2737.6kJ/kg V3=vg3=0.462m3/kg Also, the nozzle efficiency b/w the throat and the exit. η = 85% = 0.85 η =(V31)2 – V22 V32 – V22 V31 = √0.859V32 – V22) + V22 V31 = √0.85(561.78 2 – 474.76 2 ) + 474.76 2 V31 = 550 m/s Also; m0 = A3 V31 / V3 A3 = m0 x V3 / V3 61 A3 = (9.612 x 0.462) / 550 = 8.074 x 10 -3 m2 (ii)Exit area; A3 = 8074 mm2 STEAM TURBINES: Discuss the differences b/w an impulse and a reaction turbine. IMPULSE TURBINE (1)The steam flows through the nozzles & impinges on the moving blades. (2)The steam impinges on the on the buckets with kinetic energy. (3)The steam may or may not be admitted over the whole circumference. (4)The steam pressure remains constant during its flow through the moving blades. (5)The relative velocity of steam while gliding over the blades remains constant (assuming no friction). (6)The blades are symmetrical (7)The number of stages required are less for the same power developed. (8)The turbine rotates with high speed. Hence compounding is required. (9)The turbine is compact & requires less space for installation. (10)Used for small capacity power plants. REACTION TURBINE (1)The steam flows first through guide mechanism & then through the moving blades. (2)The steam glides over the moving vanes with pressure and kinetic energy. (3)The steam must be admitted over the whole circumference. (4)The steam pressure is reduced during its flow through the moving blades. (5)The relative velocity of steam while gliding over the moving blades increases (assuming no friction). (6)The blades are not symmetrical. (7)The number of stage required are more of the same power developed. (8)The turbine rotates with low speed. Hence compounding is not required. (9)The turbine is of larger size & requires more space for installation. (10)Used for medium & large capacity power plants. (2)Explain the pressure compounded and velocity compounded impulse steam turbine showing pressure and velocity variations along the axis of the turbine. Pressure compounded impulse turbines: In pressure compounding of impulse turbined, rings of fixed nozzles are incorporated between the rings of moving blades. The steam at boiler pressure enters the first set of nozzle and expands partially. The kinetic energy of steam thus obtained is absorbed by the moving blades(stage1). The steam then expands partially in the second set of nozzles where its pressure again falls and the velocity increases. The kinetic energy so obtained is absorbed by the second ring of moving blades(stage2). This is repeated in stage 3 and steam finally leaves the turbine at low velocity and pressure. The number of stages (or pressure reductions) depends on the number of rows of nozzles through which the steam must pass. This method of compounding is used in Rateau and 62 Zoelly turbine. This is the most efficient turbine since the speed ratio remains constant, but is expensive during to a large number of stages. Velocity compound impulse turbines: In velocity compounding of an impulse turbine, the expansion of steam takes place in a nozzle or a set of nozzles from the boiler pressure to the condenser pressure. The impulse carries two or three rows of moving blades, which are separated by rings of fixed or guide blades in the reverse manner. The steam after expanding through nozzles, enters the first ring of moving blades at a high velocity. A portion of this high velocity is absorbed by this blade ring and the remaining is passed on to the next ring of fixed blades. The fixed blades change the direction of steam and direct it to the second ring of moving blades, without altering the velocity. After passing through the second ring of moving blades, a further portion of velocity is absorbed. The steam is now directed by the second ring of moving blades and then enters into the condenser. In velocity compounding, the number of stages used are less, yet its efficiency is less. 63 (3)What do you mean by compounding of turbines? Explain the velocity compounding. In impulse turbines, if the steam is expanded from the boiler pressure to condenser pressure in one stage, the speed of rotor becomes tremendously high which results in practical incompatibilities. There are several methods of reducing this speed to a lower value. All these methods utilize a multiple system of rotor in series, keyed on a common shaft and the steam pressure or jet velocity is absorbed in stages as the steam flows over the blades. This is known as Compounding. (4)What are the methods of governing a steam turbine? Describe any one method of governing steam turbine. Methods of governing a steam turbine: 1) Throttle governing. 2) Nozzle governing. 3) By-pass governing. 4) Combination of throttle and nozzle governing and throttle and By-pass governing methods. Throttle governing of steam turbine: In throttle governing of a steam turbine, the turbine outlet is controlled by varying the quantity of steam entering into the turbine. This method is also known as Servomotor method. 64 As shown in the figure, the centrifugal governor is driven from the main shaft of turbine. The control value controls the direction of flow of oil(which is pumped by gear pump) either in the pipe AA or BB. The servomotor or relay value has a piston(whose motion towards left or right depends upon the pressure of the oil flowing through the pipes AA & BB), & is connected to a spear which moves inside the nozzle. When the turbine is running at its normal speed, the positions of piston in the relay, cylinder, control value, flyballs of centrifugal governor will be in their normal positions as shown in the figure. he oil pumped by the gear pump into the control value will come back into the oil sump as both pipes AA & BB are closed by the two wings of control value. When the load on the turbine increases its speed decreases. As a result, the speed of the governor decreases, and hence the flyballs rotate with lesser radius & lower the sleeve. The lever which is connected to the sleeve moves up about the pivot, and lifts the control value rod. This upward movement opens the mouth of pipe AA(still keeping mouth of pipe BB closed). The oil under pressure will rush from the control value to the right side of the piston in the servomotor through the pipe AA, and will push the piston and hence the spear which will open more area of the nozzle. This will increase the rate of steam flows into the turbinr. As a result, the speed of the turbine will increase upto the normal value. When the load on the turbine decreases its speed increases. The speed of the governor also increases causing the walls to rotate with larger radius, thus lifting the sleeve. This causes the lever to move down about the pivot, and hence pushing down the control value rod. This downward movement opens the mouth of pipe AA. The oil under pressure rushes into left side of the piston in the servo motor from the control value, pushing the piston towards right. Thus, the spear moves into the nozzle causing decrease in area of flow. This decrease of the rate of steam flow into the turbine & as a result the speed of the turbine will decrease down to the normal value. 65 (5)Describe the different classifications of turbines and their working. Classification of steam turbines: 1)According to the mode of steam action : (i)Impulse turbine: It runs by the impulse of steam jet. In this turbine, steam is first expanded through a nozzle and then the steam jet impinges on the turbine blades that are mounted on the circumference of the wheel. (ii)Reaction turbine: In these turbines, the runner is rotated by the reactive forces of steam jets. In a reaction turbine, the steam enters the wheel under pressure and flows over the blades. The steam, while gliding, propels the blades and make them to move. 2)According to the direction of steam flow: (i)Axial turbines: In those, steam flows in a direction parallel to the axis of the turbine. (ii)Radial turbines: In these, steam flows in a direction perpendicular to the axis of the turbine. 3)According to the method of governing: (i) Turbines with throttle governing: In these, fresh steam enters through one or more simultaneously operated throttle values. (ii) Turbines with nozzle governing: In these, fresh steam enters through two or more consecutively opening regulators. (iii) Turbines with by pass governing: In these, the turbines besides being fed to the first stage, is also fed to one, teo or even three intermediate stages of the turbine. 4)According to their usage in industries: (i) Stationary turbines with constant speed of rotation: They are primarily used for driving alternators. (ii) Stationary steam turbines with variable speed: They are used for driving turbo-blowers, air – circular pumps etc. (iii) Non-stationary turbines with variable speed: They are used in steamers, ships and railway locomotives. (6)Discuss the compounding of steam turbines. Compounding of steam turbines: 66 1) Velocity compounding 2) Pressure compounding 3) Pressure- velocity compounding: This method is the combination of velocity and pressure compounding. The total drop in steam pressure is divided into stages and the velocity obtained in each stage is also compounded. The rings of nozzle are fixed at the beginning of each stage and pressure remains constant during each stage. This method of compounding is used in Curtis and Moore turbine. (7)Explain the methods of governing of steam turbines. Methods of governing steam turbines: 1) Throttle governing: This is the most widely used method particularly on small turbines, because its initial cost is less and mechanism is simple. The principle of throttle governing is to throttle the steam whenever there is reduction of load compared to design load for maintaining speed and vice versa. 2) Nozzle governing: It is the more efficient form of governing which is carried out by means of nozzle control. In this method of governing, the nozzles are grouped into 3, 5 or more groups and supply of steam to each group is controlled by regulating values. Under full load conditions the load on the turbine becomes more or less than the design value, the supply of steam to a group of nozzles may be varied accordingly so as to restore the original speed. Nozzle control can only be 67 applied to the first stage of a turbine. It is suitable for simple impulse turbine and larger units which have an impulse stage followed by an impulse-reaction turbine. 3) By-pass governing: It is desirable to have full admission of steam in the high pressure stages for the steam turbines to work at the design load. However, at the maximum load (which is greater than the design load), the additional steam required cannot be passed through the first stage since additional nozzles are not available. By-pass regulation allows for this in a turbine which is throttle governed, by means of a second by-pass value in the first stage nozzle. This value opens when the throttle value is opened by a definite amount. Steam is by-passed through the second value to a lower stage in the turbine. When by-pass value operates, it is under the control of the turbine governor. The secondary and tertiary supplies of steam in the lower stages of steam in the work output in these stages out with a loss in efficiency. (8)Draw a neat sketch of regenerative cycle and with the help of T-S diagram analyse the cycle. Regenerative Cycle: In an ideal regenerative cycle, the dry saturated steam from the boiler enters the turbine at a higher temperature and then expands isentropically to a lower temperature. Now the condensate from the condenser is pumped back and circulated around the turbine casing in the direction 68 opposite to the steam flow in the turbine. The condensate steam is thus heated before entering into the boiler. Such a system of heating is known as Regenerative Heating. However, due to loss of heat, the expansion in the steam turbine is no more isentropic. From the T-S diagram, it may observed that the steam expansion in the turbine follows the path BC, which is exactly parallel to EA, which shows the regenerative process. Further, the heat transferred to the liquid is equal to the heat transferred from the steam. Heat transfer to liquid is represented by area EAGF, and that from the steam by area BPQC. The heat is supplied to the working fluid at a constant temperature in the process AB. This is represented by area ABPG. The heat is rejected from the working fluid at constant temperature shown by curves CE. This is represented by the area CQFE which is equal to the area RPGO. (9)Draw a neat sketch of re-heat cycle with the help of a T-S diagram analyse the cycle. Reheat cycle: 69 The T-S diagram shown above represents an ideal reheating process. The steam at state point1(pressure P1 and temperature T1) enters the turbine to a certain pressure P2 and temperature T2. From this state point 2 the whole of steam is drawn out of the turbine and is reheated in a reheater to a temperature T3. This reheated steam is then readmitted to the turbine where it is expanded isentropically to condenser pressure P3. (10)Derive the expression for maximum blade efficiency in a single stage impulse turbine. Expression for maximum blade efficiency in a single stage impulse turbine: Let: U Linear velocity of moving blade (m/s) V1 Absolute velocity of steam entering the moving blade (m/s) V2 Absolute velocity of steam leaving the moving blade (m/s) VW1 Tangential component of V1 Vf1 Axial component of V1 Vr1 Relative velocity of steam entering the moving blade. VW2 Tangential component of V2 Vf2 Axial component of V2 Vr2 Relative velocity of steam leaving the moving blade. α Nozzle (or jet ) angle. θ Inlet angle of moving blade. 70 ϕ Outlet angle of moving blade. β Fixed blade angle. From the velocity diagram; VW = PQ = MP + MQ = Vr1 Cos θ + Vr2 Cos ϕ = Vr1 Cos θ 1 + Vr2 . Cos ϕ Vr1 Cos θ We have, the blade velocity coefficient, K = Vr2 Vr1 Take Cos ϕ = Z, a constant Cos θ ∴ Vw = Vr1 Cos θ [ 1 + K . Z ] But Vr1 Cos θ = MP = ∠P - ∠M = V1 Cos α – U ∴ Vw = ( V1 Cos α - U ) (1 + K.Z ) We know that, Blade efficiency, η b = 2 U VW V12 η b = 2 U ( V1 Cos α - U ) (1 + K.Z ) V12 = 2 U V1 (Cos α – U ) (1 + K.Z ) V1 V12 = 2 (U/V1) ( Cos α - U/V1 )( 1 + K.Z ) Take U/V1 = ρ, speed ratio η b = 2 P ( Cos α – ρ)( 1 + K.Z ) For a particular impulse turbine, α, K & Z are assumed to be constants. Differentiating (1) w.r.t P, we get; (1) 71 d η b = 2 ( Cos α – ρ)( 1 + K.Z ) – 2 ρ ( 1 + K.Z ) dρ Equating the above expression to zero, we get; 2 ( 1 + K.Z ) [ Cos α – ρ] = 0 Cos α – 2ρ = 0 ρ = Cos α 2 This is the condition for the maximum value of η b. Substituting this value of ρ in (1), we get, η b = 2 Cos α (Cos α - Cos α ) ( 1 + K.Z ) 2 2 = Cos α . Cos α ( 1 + K.Z ) 2 ( η b )max = Cos2 α ( 1 + K.Z ) 2 Assuming the blades to be symmetric i.e, θ = ϕ, and that there is no friction in the fluid passage; We get Z = 1 and K = 1 ( η b )max = ( η b )max = Cos2 α ( 1 + 1 ) 2 Cos2 α The above expression gives the maximum blade efficiency in a single stage impulse turbine. (11)Write short notes on the following: a)Improvement of steam turbine efficiency. b)Binary vapour cycle c)Velocity diagram for impulse and reaction turbines. d)Degree of reaction. a) Improvement of steam turbine efficiency: (i)Regenerative cycle: In this cycle, the dry steam from boiler enters the turbine at a higher temperature, and then expands isentropically to a lower temperature. Now, the condensate from the condenser is pumped back and circulated around the turbine casing in the direction opposite 72 to the steam flow in the turbine. The condensate steam is thus heated before entering into the boiler. Such a system of heating is known as Regenerative Heating. REGENERATIVE CYCLE (ii)Reheat cycle: In a reheat cycle , the steam enters the turbine in a super heated state. The steam then expands isentropically through the first stage of turbine after which it becomes wet. The wet steam is reheated at a constant pressure upto the same temperature at entry until it becomes super heated. The steam again expands isentropically which flowing through the next stage of the turbine. 73 (b)Binary vapour cycle: Fig (a) shows a binary vapour plant which uses vapours of mercury and water for its operative. Fig (b) shows the T-S diagram for the binary vapour cycle. The line AB represents the evaporation of liquid mercury. The mercury vapour at B has a much higher temperature than the steam at same pressure. The mercury vapours are now expanded isentropically in a mercury turbine as represented by the line BC. The condensation of mercury is shown by the line CD. 74 During condensation, the latent heat is utilized for evaporating a corresponding amount of steam. The line DA represents the heating of mercury. Thus mercury has completed a cycle ABCDA. The steam cycle is represented by 1-2-3-4-5-1. The line 1-2 represents the evaporation of water by the condensing mercury. The line 2-3 represents the superheating of steam by the flue gases. The steam is now expanded isentropically through the steam turbine as shown by the line 3-4. The condensation of the exhaust steam is represented by the line 4-5. The heating of feed water is represented by 5-1. This completes the steam cycle. ©Velocity diagram for impulse and reaction turbines: 75 Let: U Linear velocity of moving blade (m/s) V1 Absolute velocity of steam entering the moving blade (m/s) V2 Absolute velocity of steam leaving the moving blade (m/s) VW1 Tangential component of V1 Vf1 Axial component of V1 Vr1 Relative velocity of steam entering the moving blade. VW2 Tangential component of V2 Vf2 Axial component of V2 Vr2 Relative velocity of steam leaving the moving blade. α Nozzle (or jet ) angle. θ Inlet angle of moving blade. ϕ Outlet angle of moving blade. β Fixed blade angle. For a simple turbine, θ = ϕ also C Vr1 Velocity diagram for reaction turbine: 76 For a reaction turbine blade, Vr2 > Vr1. Also, for a Breson’s reaction turbine, θ = β & ϕ = α Vr1 = V2 & Vr2 = V1 (d)Degree of reaction: The degree of reaction of a reaction turbine stage is defined as the ratio of heat drop over moving blades to the total heat drop in the stage. Thus the degree of reaction is given by; Rd = Heat drop in moving blades Heat drop in the stage Rd = ∆ hm ∆ hf + ∆ hm (12)A single row steam turbine develops 115 kW at a blade speed of 180 m/s. The steam flow rate is 2 kg/s. The steam leaves the nozzle at 400 m/s. The velocity co-efficient of blade is 0.9. Steam leaves the blade axially. Determine the nozzle angle & the blade angle, assuming no shock. Given: 77 Power developed, P=115 kW Mean blade speed, U=180 m/s Steam flow rate, m°=2kg/s Inlet velocity of steam, V1=400m/s Blade velocity co-efficient,K=0.9 Steam leaves the blade axially. ∴ V2 = Vf2 & β = 900 & VW2 = 0 θ=? ϕ=? α=? Taking a scale of 1cm = 50m/s, we draw a combined velocity diagram as shown below: Also; P = m° VW U = m° (VW1 +VW2 ) U = 115 x 103 = 2 ( VW1 + 0 ) x 180 VW1 = 319.44 m/s From the diagram, Vr1 = 5.6 x 50 = 280 m/s K = Vr2 / Vr1 Vr2 = K x Vr1 = 0.9 x 280 = 252 m/s From the diagram, hence we find the following angles; Nozzle angle, α = 390 Inlet blade angle, θ = 620 Outlet blade angle, ϕ = 440 78 (13)The simple impulse steam turbine has a mean blade speed of 200m/s. The nozzle are inclined at 200 to the plane of rotation of the blades. The steam velocity from the nozzle is 600m/s. The turbine uses 3500 kg/hr of steam. The absolute velocity at the exit is along the axis of the turbine. Determine (i)the inlet and exit angles of the blade.(ii)The power output of the turbine. (iii)The axial thrust(per kg of steam per second). Given: Power developed, P=115 kW Mean blade speed, U=180 m/s Steam flow rate, m°=2kg/s Inlet velocity of steam, V1=400m/s Blade velocity co-efficient,K=0.9 Steam leaves the blade axially. ∴ V2 = Vf2 & β = 900 & VW2 = 0 θ=? ϕ=? α=? Taking a scale of 1cm = 50m/s, we draw a combined velocity diagram as shown below: Also; P = m° VW U = m° (VW1 +VW2 ) U = 115 x 103 = 2 ( VW1 + 0 ) x 180 VW1 = 319.44 m/s From the diagram, Vr1 = 5.6 x 50 = 280 m/s K = Vr2 / Vr1 Vr2 = K x Vr1 = 0.9 x 280 = 252 m/s 79 From the diagram, hence we find the following angles; Nozzle angle, α = 390 Inlet blade angle, θ = 620 Outlet blade angle, ϕ = 440 (13)The simple impulse steam turbine has a mean blade speed of 200m/s. The nozzle are inclined at 200 to the plane of rotation of the blades. The steam velocity from the nozzle is 600m/s. The turbine uses 3500 kg/hr of steam. The absolute velocity at the exit is along the axis of the turbine. Determine (i)the inlet and exit angles of the blade.(ii)The power output of the turbine. (iii)The axial thrust(per kg of steam per second). Given: Simple impulse turbine ∴ θ = ϕ (blades are symmetrical) Mean blade speed, U = 200m/s Nozzle angle, α = 200 Inlet velocity of steam, Vi = 600m/s Steam flow rate m° = 3500 kg/hr = 0.9722 kg Taking a scale of 1 cm = 100 m/s We draw a combined velocity diagram as shown below: The absolute velocity at the exit is along the axis of the turbine. ∴ V2 = V2f, β = 900 , VW2 = 0 From the diagram, Vr1 = 4.2 x 100 = 420 m/s θ = 300 = ϕ Vf1 = 2.1 x 100 = 210m/s 80 VW1 = 3.6 x 100 = 360m/s Vr2 = 2.3 x 100 = 230m/s V2 = Vf2 = 1.7 x 100 = 170m/s ∴ (𝑖) Inlet and exit angles of the blades: θ = ϕ = 300 (ii) Power output of the turbine, P = [m° ( Vw1 + VW2 ) U] / 1000 = [0.9722(360+0)200]/1000 P = 70 kW (iii) The axial thrust; Fa = ( Vf1 – Vf2 ) = ( 210 – 170 ) Fa = 40 N/kg/s (14) In a De-laval turbine, steam issues from the nozzle with a velocity of 850m/s. The nozzle angle is 200, the mean blade velocity is 350m/s and the inlet and outlet angles of blade are equal ( blades are symmetrical ). The mass of steam flowing through the turbine is 100kg/min. Calculate (i)Blade angle.(ii)Relative velocity of steam entering the blades.(iii)Tangential force on the blades. (iv)Power developed. (v)Diagram efficiency (vi)Axial thrust. Take blade velocity co-efficient as 0.8 Given: De-laval turbine Inlet steam velocity, V1 = 850m/s Nozzle angle, α = 200 Mean blade velocity, U =350m/s The blades are symmetrical i.e, θ = ϕ Steam flow rate, m0 = 100 kg/min = 1.667 kg/s Taking a scale of 1cm = 100m/s, we draw a combined velocity diagram as shown below: From the diagram; Vw1 = 8x100=800m/s θ = ϕ = 330 81 Vr1 = 5.3x100=530m/s Vf1 = 2.9x100=290m/s The blade velocity co – efficient, K = Vr2/Vr1 =0.8 Vr2 = 0.8 Vr1 = 0.8 x 530 = 424 m/s Vf2 ≅ V2 = 2.3 x 100 = 230 m/s Vw2 = 0.1 x 100 = 10 m/s (i)Blade angles; θ = ϕ = 330 (ii)Relative velocity of steam entering the moving blades; Vr1 = 530m/s (iii)Tangential force on the blades; Ft = m0 [ Vw1 + Vw2 ] = 1.667 [ 800+100] Ft = 1500-3 N (iv)Power developed; P = m0 [ Vw1 + Vw2 ] U kW 1000 = 1.667 [ 800 + 10 ] 350 1000 P = 472.59 kW (v)Diagram efficiency; ηd = m0 [ Vw1 + Vw2 ] U m0 V12 = 2 x 1.667 [ 800 + 10 ] 350 1.667 x 8502 ηd = 0.7847 0r 78.47 % (vi) Axial thrust; Fa = m0 ( Vf1 – Vf2 ) 82 = 1.667 [290-230] Fa = 100.02 N (15) In a simple impulse turbine the nozzles are inclined at 200 to the direction of motion of moving blade. The steam leaves the nozzle at 375m/s. The blade speed is 165m/s. Find the inlet and outlet angles of the blades. The relative velocity of steam as it flows over the blades is reduced by friction by 15%. Calculate the power developed for a flow rate of 10 kg/s. Given: Simple impulse turbine- the blades are symmetrical hence θ = ϕ = ? Nozzle angle, α = 200 Inlet velocity of steam, V1 = 375m/s Mean blade speed, U = 165m/s Vr1 – 15/100 . Vr1 = Vr2 Vr2/ Vr1 = 0.85 K Vr2 = 0.85 Vr1 Steam flow rate, m0 = 10 kg/s Taking a scale of 1cm = 50 m/s, we draw a combined velocity diagram as shown below; In the diagram; Vr1 = 4.6 x 50 = 230m/s Vr2 = 230 x 0.85 = 195m/s From the diagram; θ = ϕ = 290 Vw1 = 7.1 x 50 = 355m/s Vw2 = 0.15 x 50 = 7.5m/s 83 Therefore, the inlet and outlet angles of the blade, θ = ϕ = 290 The power developed, P = m0 [ Vw1 + Vw2 ] U kW 1000 = 10 [ 355 + 7.5 ] 165 1000 P = 598.125 kW (16)Steam enters an impulse turbine having nozzle angle of 200 at a velocity of 450 m/s. The exit angle of moving blade is 200 and the relative velocity of steam may be assumed to remain same over the moving blades. If the blade speed is 180m/s, determine : (i)Blade angle at inlet (ii)Power produced for a mass flow rate of 2 kg/s Given: Nozzle angle, α = 200 Inlet velocity of steam, V1 = 450 m/s Mean blade speed, U = 180m/s Exit angle of moving blade, ϕ = 200 Vr1 = Vr2 Steam flow rate, m0 = 2 kg/s Taking a scale of 1cm = 50m/s we obtain a combined velocity diagram as shown below: 84 From the diagram; Vw1 = 8.3 x 50 = 415 m/s Vw2 = 1.7 x 50 = 85 m/s θ = 320 therefore, (i) Blade angle at inlet; θ = 320 (ii)Power produced; P = m0 [ Vw1 + Vw2 ] U kW 1000 = 2 [ 415 + 85 ] 180 1000 P = 180 kW (17)A simple impulse turbine has one ring of moving at 150m/s. the absolute velocity of steam from the stage is 85m/s, at an angle of 800 to the tangential direction. The blade velocity co efficient is 0.82, and the flow of steam through the stage is 2.5 kg/s. If the blades are equiangular, Determine: (i)Blade angles. (ii)Nozzle angle (iii)Absolute velocity of steam issuing from the nozzle. (iv)Axial thrust. Given: Mean blade speed, U = 150m/s Exit velocity of steam, V2 = 85m/s Angle made by the steam jet to the tangential direction at exit, β = 800 Blade velocity coefficient, K =0.82 Steam flow rate, m0 = 2.5 kg/s Blades are equiangular, θ = ϕ Taking a scale of 1cm = 50 m/s, we construct a combined velocity diagram as shown below: 85 In the diagram, Vr2 = 3.7 x 50 = 185 m/s Therefore, K = Vr2 / Vr1 Vr1 = Vr2 / K = 185/0.82 = 225.61 m/s From the diagram, θ = ϕ = 260 α = 160 Vw1=7.1x50 = 355m/s V1=7.3x50 = 365 m/s Vw2=0.21x50 = 10.5 m/s Vf1=2x50 = 100m/s Vf2=1.6x50 = 80m/s (i)Blade angles; θ = ϕ = 260 (ii)Nozzle angle; α = 160 (iii)Absolute velocity of steam issuing from the nozzle; V1 = 365 m/s (iv)Axial thrust; Fa = m0 ( Vf1 – Vf2 ) = 2.5 [100 – 80 ] = 50N The following data refers to a particular stage of Parson’s reaction turbine: Speed = 1500 rpm Mean diameter of rotor = 1m Stage efficiency = 80% Speed ratio = 0.7 86 Blade outlet angle = 200 Determine the isentropic enthalpy drop in the stage. Given: Speed, N = 1500 rpm Mean diameter of rotor = 1m Stage efficiency = 80% Speed ratio = 0.7 Blade outlet angle = ϕ = 200 For a Parson’s reaction turbine, Degree of reaction = 0.5 Inlet blade angle, θ = Outlet angle of steam jet to the blade, β. Outlet blade angle, ϕ = Inlet angle of steam jet, α Relative velocity of steam at entry, Vr1 = Absolute velocity of steam at exit, V2 Relative velocity of steam at exit, Vr2 = Absolute velocity of steam at entry, V1 Therefore, ϕ = 200 = α The mean blade speed, U = π DN/60 = [3.142x1x1500]/60 = 78.55 m/s Also; U/V1 = 0.7 V1 = U/0.7 = 78.55/0.7 = 112.214 m/s = Vr2 Taking a scale of 1cm = 25 m/s, we draw a combined velocity diagram as shown: 87 From the diagram; Vw1 = 4.25 x 25 = 106.25 m/s Vw2 = 1.2 x 25 = 30 m/s Therefore, assuming unit mass flow rate, the power developed by the stage; P = m0 [ Vw1 + Vw2 ] U kW 1000 = 1[ 105.25 + 30 ]x 78.55 1000 P = 10.702 kW Therefore, the efficiency, ηstage = [ Vw1 + Vw2 ] U = 0.8 1000 x ∆h ∆h = 10.702 / 0.8 = 13.3775 kJ/kg Therefore, the isentropic heat drop, ∆h = 13.3775 kJ/kg (19)Ina 50% reaction turbine stage running at 50 revolutions per second the exit angles are 300 and inlet angles are 500. The mean rotor diameter is 1m. The steam flow rate is 10,000 kg/min and the stage efficiency is 85%. Determine: (i)Power output of the stage. (ii)Specific enthalpy drop in the stage velocity of the steam when it flows over the moving blades. Given: 50% reaction turbine stage speed, N = 50 revolutions per second. Exit angles, ϕ = 30 degrees = α also; V1 = Vr2 & V2 = Vr1 Inlet angles, θ = 500 β. Mean rotor diameter, D =1m Steam flow rate,m0 = 10,000 kg/min = 166.67 kg/s Stage efficiency, ηstage = 85% = 0.85 88 Therefore mean blade speed, U = π DN = 3.142 x 1 x 50 = 157.1 m/s Taking a scale of 1cm = 50 m/s, we draw a combined velocity diagram as shown below: From the diagram, Vw1 = 6.25 x 50 = 312.5 m/s Vw2 = 2.9 x 50 = 145 m/s Vr1 = 4.9 x 50 = 245m/s Vr2 = 7.1 x 50 = 355m/s (i)Power output of the stage: P = m0 [ Vw1 + Vw2 ] U kW 1000 = 166.57[ 312.5 + 145 ]x 157.1 1000 P = 11979.1 kW = 11.979 MW (ii)Specific enthalpy drop in the stage∆h ηstage = [ Vw1 + Vw2 ] U = 0.85 x 1000 x ∆h ∆h = [ 312.5 + 145 ]x 157.1 1000 x 0.85 ∆h = 84.556 kJ/kg (iii) Percentage increase in relative velocity of steam, % Vr = [Vr2 – Vr1] x 100 89 Vr1 = [355 – 245] x 100 245 % Vr = 44.897 (20)At a stage of reaction turbine, rotor diameter is 1.4m & speed ratio is 0.7. If the blade outlet angle is 200 & the rotor speed is 3000 rpm, find the blade inlet angle & diagram efficiency. Also find the % increase in diagram efficiency & rotor speed if the turbine is designed to run at best theoretical speed. Given: Rotor diameter, D = 1.4 m Speed ratio, U / V1 = 0.7 Blade outlet angle, ϕ = 20 degrees = α Rotor speed, N = 3000 rpm Inlet blade angle, θ = ? Diagram efficiency, ηd =? Mean blade speed V = 𝜋 DN = 3.142 x 1.4 x 3000 60 60 = 219.94 m/s U = 0.7 V1 = U V1 0.7 = 219.94 = 314.2 m/s 0.7 For a reaction turbine, θ = β α=ϕ Vr2 = V1 Vr1 = V2 Taking a scale of 1cm = 50 m/s, we draw a combined velocity diagram as shown below. 90 From the diagram, Θ = β = 550 V1 = 6.3 x 50 = 315 m/s Vr1 = 2.6 x 50 = 130 m/s = v2 Vr2 = 6.3 x 50 = 315 m/s Vw1= 5.9 x 50 = 295 m/s Vw2 = 1.5 x 50 = 75 m/s ∴ Blade inlet angle , θ = 550 ∴ Diagram efficiency,𝜂d = U1 [Vw1 + Vw2 ] V12 - Vr12 2 𝜂d = 219.9 + [ 295 + 75 ] 315 2 - 130 2 2 𝜂d = 0.8963 Or 89.605% The condition for the turbine to run at its best theoretical speed: The speed ratio, ρ = Cos α ρ = Cos 200 = 0.9396 ρ = U1 U1 = ρ V1 = 0.9396 x 314.2 = 295.22 m/s V1 Also: U1 = 𝜋 DN Nmax = 60 U1 = 60 x 295 𝜋D 60 The maximum rotor speed, Nmax = 3.142 x 1.4 4026.825 rpm 91 The diagram efficiency, 𝜂d2 = U1 [Vw1 + Vw2 ] V12 - Vr12 2 = 295.22 [ 295 + 75 ] 3152 - 1302 2 = 0.934 or 93.4 % Therefore, the percentage crease in diagram efficiency, % 𝜂d = 𝜂d2 – ηd1 = 93.4 – 89.63 = 3.765 % In a two row velocity compounded impulse turbine, the steam from the nozzle issues at a velocity of 600 m/s. The nozzle angle is 200 to the plane of rotation of the wheel. The mean diameter of the rotor of moving blades have equiangular blades. N = 3000 rpm, D = 1m. The intermediate row of fixed guide blades makes the steam flow again at 200 to the second moving blade ring. The frictional losses in each row are 3% . Find : 1) Inlet and outlet angles of moving blades of each row. 2) Inlet blade angles of the guide vanes. 3) Power output from the first and second moving blade rings for unit mass flow rate. Given : Two row velocity compounded impulse turbine. Inlet velocity of steam, V1 = 600 m/s 92 Nozzle angle = 200 The moving blades are equiangular ∴θ=ϕ The mean diameter of rotor, D = 1 m. Speed , N = 3000 rpm. ∴ Mean blade speed, U = = 𝜋 DN 60 = 3.142 x 1x 3000 60 = 157.1 m/s The steam jet angle to the second row of moving blades, α2 = 200 Frictional losses in each row = 3% Blade velocity coefficient, K1 = K2 = K3 = 1 – 0.03 = 0.97 Taking a scale of 1 cm = 50 m/s, to draw a combined velocity diagram for each stage as shown: In the diagram; Vr1 = 9.1 x 50 = 455 m/s K1 = Vr2 Vr2 = K1. Vr1 = 0.97 x 455 = 441.35 m/s Vr1 V2 = 6.2 x 50 = 310 m/s K2 = V11 / V2 V11 = K2 . V2 = 0.97 x 310 = 300.7 m/s K3 = Vr21 m/s Vr11 Vr21 = K3 . Vr11 = 0.97 x 3.3 x 50 = 160.05 93 From the diagram; Vw1 = 11.2 x 50 = 560 m/s , Vw21 = 0 Vw2 = 4.6 x 50 = 230 m/s , Vw11= 5.6 x 50 = 280 m/s θ1= ϕ1 = 270 θ2= ϕ2 = 390 β1 = 400 (i)Inlet and outlet angles of moving blade of each row; θ1= ϕ1 = 270 θ2= ϕ2 = 390 (ii)Inlet blade angles of the guide vanes; β1 = 400 (iii)Power output; Steam flow rate = 1kg/s From 1st row of moving blades; P1 = m0 [ Vw1 + Vw2 ] U kW 1000 = 1 [ 560 + 230 ] x 157.1 1000 P1 = 124.109 kW From 2nd row of moving blades; P2 = m0 [ Vw11 + Vw21 ] U kW 1000 = 1 [ 280 + 0 ] x 157.1 1000 P2 = 43.988 kW 94 With the helps of a simple sketch explain the working of a thermo – electric refrigeration. (Jan 2010) (Jul 2008) Thermo – electric refrigeration: Figure (a) shows an action thermo electric refrigerator. Figure(b) shows the working principle of such a generator. When two dissimilar metals are joined together and their joints are kept at different temperatures, an electromotive force is produced. However, when the direction of flow of electrons is reversed in the thermoelectric circuit by externally applying a potential difference in the reverse direction, a refrigeration effect can be credited. This is called peltier effect and forms the basis for thermoelectric refrigeration. Here electrons act as the working fluid. Heat is absorbed from the refrigerated space in the amount of QL and rejected to the warmer environment (atmosphere) in the amount of QH. the difference between these two heats is the net electrical work that needs to be supplied. 95 i-e. We = QH - QL A practical thermoelectric refrigeration circuit uses semiconductor materials instead of mental wires. Breifly discuss the most commonly used refrigerants. (Jan 2010, Dec 2007/Jan 2008) Commonly used refrigerants: 1 AIR : Properties: (i) Easily available and no cost is involved. (ii) Completely non – tonic and hence safe. (iii) C.O.P of air cycle operating between temperatures of 80 o c and – 15o c is 1.67 USES: (i) Air is one of the oldest refrigerants and was widely used. (ii) Because of low C.O.P, it is used for only those purpose cohere the operating efficiency is secondary as in aircraft refrigeration. 2 AMMONIA (NH3) : Properties : (i) Highly tonic and flammable. (ii) Has excellent thermal properties. (iii) Has the highest refrigeration effect per kg of refrigerant. (iv) How volumetric displacement. (v) How cost and high efficiency. (vi) How weight of liquid circulated per tone of refrigeration. USES : (i) Widely used in large industrial and commercial reciprocating compression systems where toxicity is secondary. (ii) Extensively used in ice plants, packing plants, large cold storages, etc… (iii) Widely used as refrigerant in vapour absorption refrigeration systems 3 Carbon Dioxide (CO2) : Properties: (i) colourless and obourless gas, and heavier then air (ii) Non- toxic, non – flammable, non- explosive and non – corrosive. (iii) Has extremely high operating pressures. (iv) Gives very low refrigeration effect. USES ; (i) uses are limited because of high power consumption per tone of refrigeration and high operating pressures. (ii) formerly used for marine refrigeration, theater air-conditioning systems, and for hotel and institutional refrigeration. 96 4 Dichloro - dilfuoro methane (R-12/Freon-12) Properties : (i) Non – toxic, non – flammable and non explosive. Hence it is most suitable refrigerant. (ii) Fully oil miscible and hence simplifies the problem of oil return . (iii) Does not break even under extreme operating conditions. (iv) Condenses under atmospheric conditions and at moderate pressures. USES : (i) suitable for high, medium and low temperature applications. (ii) Used for domestic applications. (iii) Since it is an electric insulator, it is universally used in sealed type compressors. 5 Monochloro – difluoro methane (R-22/From-22) Properties : (i) Miscible with oil at condenser temperature but tries to separate at evaporator temperature when the system is called for very low temperature applications(90oc). In such cases oil seperators must be incorporated to return the oil from the evaporator. (ii) Pressures in the evaporator and condenses at standard tone of refrigeration are 2.9 bar (ab5.) and 11.9 bar (abs). (iii) Discharge temperature is high and hence requires water cooling of cylinder and the compressor head. USES : (i) Universally used in commercial and industrial low temperature systems. Breifly discuss the propertices of refrigerants. (jan 2010,Dec 2008/ Jan 2009,Jun/Jul2008, Dec2007/Jan 2008) Properties of refrigerants: An ideal refrigerant should posses the following properties: Thermodynamic properties: (i) Low boiling point. (ii) Low freezing point. (iii)Moderate positive pressures in condenser and evaporator. (iv) High saturation temperature. (v) High latent heat of vapourisation. Chemical properties: (i) Non - toxic. (ii) Non – flammable and non – explosive. (iii)Non – corrosive. (iv) Chemically inert. (v) Non – irritating and odorless. Physical properties (i) Low specific volume of vapour. (ii) Low specific heat. (iii)High thermal conductivity. 97 (iv) Low viscosity (v) High electrical insulation. Other Properties (i) Ease of leakage location. (ii) Availability and low cost. (iii) Ease of handling (iv) High C.O.P (Coefficient of Performance) (v) How power consumption per tone of refrigeration. (vi) How pressure ratio and pressure difference. Explain briefly the properties of refrigerants (Dec 2008/Jan 2009) 1 2 3 4 5 6 Important Properties of Refrigerants Freezing Point : As the refrigerant must operate in a cycle above its freezing point, it is evident that the freezing point of the refrigerant must be lower than system temperatures. Condenser and evaporator pressures : The evaporating pressure should be as near atmospheric as possible. If it is too low, it would result in a large volume of the suction vapour. If it is too high, overall high pressures including condenses pressure would necessitate the requirement of a stronger equipment, which increases the cost. A positive pressure is required to prevent air and moisture from entering into the system. Critical temperature and pressure : Generally, for high C.O.P. the critical temperature must be very high. The critical pressure should be low so as to give low condensing pressure. Latent heat of vapourisation : It should be as large as possible to reduce the weight of the refrigerant to be circulated in the system. This reduces the initial cost of the refrigerant and the size of the system will also be small. Thermal conductivity and viscosity : For a high heat transfer co- efficient a high thermal conductivity of the refrigerant is desirable. Also, for a high heat transfer coefficient, low viscosity is desirable. Action with oil and with materials of construction : no chemical reaction between refrigerant and the lubricating oil of compressor should take place. Also, the refrigerant should not react with the materials used for the refrigeration system. With the helps of a neat sketch explain the working of a pulse tube refrigeration (Dec 2008/Jan 2009) Pulse tube refrigeration : 98 In pulse tube refrigeration, sudden expansion and release of gas are employed to get the refrigeration effect. A simple circuit of pulse tube refrigeration is shows in Fig (a). It consists of 99 a high pressure gas source at the temperature close to the ambient value. The compressed gas is supplied to the pulse tube through a suitable value mechanism. During the pressure building process, the high pressure gas enters the pulse tube and acts as a fictitious piston. Thus, the gas present inside the pulse tube gets compressed resulting in increase in the temperature verying from minimum at the left end to the maximum at the right end (Fig.(a)). Thereafter the heat transfer to the cooling medium reduces the temperature to Tn, the temperature of the cooling medium. The supply of the high pressure gas stops after the inlet value is closed. Thereafter the exhaust value opens and the exhaust phase begins resulting in continuous decrease in temperature as shows by the dotted lines (Fig. (b)), giving a lowest temperature T1. It is evident that the air leaving the cold end is at a temperature much lower than that of the ambient value. Hence a regenerator is provided (Fig. (a)). During the admission of compressed air the cold air from the pulse tube adsorbs heat and thus the temperature of the compressed gas is lowered up to T11 (Fig.(c)). By this means the temperature is further lowered of the order of -83oc, and was achieved by means of a single stage pulse tube when the heat sink or cooling is accomplished at about 6oc. Define C.O.P Explain working of vapour compression refrigeration system, with flow diagram and T-S diagram. (Jun/Jul 2008, Jan/Feb 2005) The co- efficient of performance (C.O.P) is defined as the ratio of heat absorbed by the refrigerant which passing through the evaporator to the work input required to compress the refrigerant in the compressor. In short, it is the ration of heat extracted to the work input to a refrigeration system. Vapour Compression Refrigeration System Fig (a) 100 Figure (a) shows the flow diagram of the vapour compression refrigeration cycle, and Figure (b) shows the T-s diagram for such a cycle. Vapour Absorption Refrigeration System: Ammonia – water absorption refrigeration system: Fig – (a) 101 The vapour at low temperature and pressure (state 2) enters the Compressor where it is compressed isentropically and subsequently its temperature and pressure increase considerable (state 3). This vapour after leaving the compressor enters the condens where it is condensed into high pressure liquid (state 4) and is collected in a receiver tank. From the receiver tank, it passes through the expansion value, when it is throttled has a low temperature (state 1). finally it passes an to evoporator where it extracts heat from the fluid being refrigerated and vapourises to low pressure vapour (state 2). Explain working of an Ammonia – water absorption refrigeration system with flow diagram and T-s diagram (Dec 2007/ Jan 2008) Sketch and explain the operation of a vapour obsorption refrigeration system. (Jan/Feb 2004) Figure (a) shows the flow diagram for a Ammonia – water obsorption refrigeration system. Figure (B) shows the T-S diagram for the corresponding cycle. The solubility of ammonia in water at low temperatures and pressures is higher than it is at higher temperatures and pressures. The ammonia vapour leaving the evaporator at point 2 is evadily obsorbed in the low temperature hot solution in the obsorber. This process is 102 accompanied by the evejection of heat. The ammonia in water solution (strong solution) is pumped to the higher pressure and is heated in the generator. Due to reduced solubility of ammonia in water at the higher pressure and temperature, the vapour is evemoved from the solution. The vapour than passes to the condenser and the weakened ammonia in water solution is returned to the absorber. The ammonia vapour loses its temperature in the condenser and then is collected in the receiver. From the receiver it is expanded through the expansion value to a lower pressure. Thus this low, pressure, low temperature ammonia vapour again enters into the eveaprator to repeat the cycle. A heat exchanger is located between the generator and the absorber. Here, the strong solution pumped from the absorber to the generator is heated after receiving heat from the weak solution returning to the absorber from the generator, and the latter gets cooled. With the help of a simple sketch, explain the working of a steam jet refrigeration system. (Dec 2007 / jan 2008) Steam jet refrigeration system : Figure shows a steam jet evefrigeration system. In this system, the steam from the boiler also called motive steam expands through the nozzle of an ejector. This high velocity vapour imparts the momentum to the vapour of the flash chamber and thereby the flash vapour moves along with the notice steam through the ejector. This process is called entrainment. The vapour mixture is then compressed in the ejector to the condenser pressure where circulating water couses its condensation. The condensate is pumped back to the boiler while extra water is purged into atmosphere. 103 The cool water of the flash chamber is pumped through the load which may be the space to be cooled or refrigerated. The warmed up water due to the heat load is sprayed into the flash chamber. Since the cooling is caused as a result of vapourization of water, the make up water is supplied through a float value, keeping a constant water level in the flash chamber. 9. Describe the thermodynamic cycle commonly used for refrigeration define co – efficienof performance (Jan / Feb 2005 ) The most commonly used thermodynamic cycle for refrigeration is the vapour compression system. Refer Q – 6. 10. Derive an expression for C.O.P for an air refrigeration system working an reversed brayton cycle. (Dec 2006) Explain air refrigeration system working an Bell Coleman air cycle derive C.O.P equation in terms of working temperature. (Jan / Feb 2004) Sketch and explain the working of air refrigeration system working on Bell – Coleman cycle and show the C.O.P is given by : 104 T1 C.O.P = ------------------ T2 - T1 Note : The reversed brayton cycle is save as the Bell – Coleman cycle. Conventionally Bell Coleman cycle refers to a closed cycle with expansion and compression taking place in reciprocating expander and compressor respectively, and heat rejection and heat absorption taking place in condenser and evaporator respectively. Expression for C.O.P for an air refrigeration system working on reversed Brayton cycle : In a reversed Brayton cycle, it is assumed that, 105 (i) Adsorption and rejection of heat are constant pressure process. (ii) Compression and expansion are isentropic process. Considering ‘m’ kg of air and that its specific heat is invariant with temperature and pressure, we derive the following. Heat absorbed in refrigerator, QL = CP (T3 – T2) ------------------------------------Heat rejected in cooler, QH = M CP (TS – T1) The work output of the expander or turbine in supplied to the compressor Thus, q2-3 = dh = h3 –h2. And, q4-1 = dh = h1 – h4 The net work 2 3 4 1 W = ∅dq = ∫1 𝛿 q+∫2 𝛿 q+∫3 𝛿 q+∫4 𝛿 q =q2-3 + q4-1 Q4-1 is the refrigeration effect. COP = q4 -1 = h1-h2 W1 W COP = 1 (h2-h3 ) h1-h4 - 1 _ _ _ _ (1) Since T2 = (P2/P1) = T3 T1 T4 We get T2 – T3 = T1 – T4 T1 T1 Or T2 – T3 = T2 = P2 T1 – T4 = T1 P1 r-1/r Using the above relations,1 is reduce to ; COP = T1 T2-T1 106 Bell Coleman air refrigeration cycle: This system comprises of a cooler (heat exchanger), cold chamber and reciprocating air compressor and expander. Air at state is compressor isentropically up to state 2. The constant pressure energy rejection accurs during the process 2-3 until temperature T3 ( very close to the ambient value is reached). Finally, the air at state 3 is allowed to expand down to pressure P1, state. The low temperature air passes through the cold chamber where the constant pressure energy transfer waises air to state 1, completing the cycle. The work output of the expander is supplied to the compressor. 11. A standard vapour compression refrigeration system produces 20 tonnes of refrigeration using From – 12as the refrigerant operating between the condenser temperature of 40o C and an evaporator temperature of – 25oC. determine: (i) Net refrigerating effect (kj/kg) (ii) Power supplied (iii) C.O.P (iv) Heat evejcted (in kw) Given : standard vapour compression refrigeration system. Capacity = 20 tonnes. 107 Refrigerant : From – 12 Condenser temperature, T4 = 40oC = T3 Evaporator temperature, T2 = -25oC = T1 For from – 12, From the thermodynamic data hand book (T.D.H.B), page -82, H2 = 178 KJ/KG H3 = 217 KJ/KG H4 = 74 KJ/KG = h1 1. The net refrigeration effect, R.E = h2 – h1 = 178 – 74 = 104 KJ/ Kg. R.E = 104 KJ / KG. Capacity = mo(R.E) ---------------- = 20 210 o The mass flow rate, m = 210 * 20 --------------- = 40.38 kg/min 104 2. Power required to run the compressor Pc = mo (h3 –h2) Pc = 40.38 (217 – 178) -------------------------- = Pc = 26.247KW 60 Assume mechanical efficiency, Mm = 0.85 Pc => ppm = pc = 26.247 ---------------------Ppm 0.85  Power to drive the prime mover, Ppm => 30.878 KW. 3. Co-efficient of performance (cop): COP = refrigeration effect = h2-h1 --------------------------------Work input h3-h2 108 = 104 ------ = COP = 2.667 217 - 178 4. Heat rejected, => QH QH = mo(h3 – h7) --------------60 = 40.38(217-74) -----------------60 = 96.239 KJ/kg 12. In a vapour compression refrigerator using Ammonia as the refrigerant, the condenser and evaporator temperature are 30oC and -15oC respectively. The liquid emerging from the condenser is subcooled by 4oC. the isentropic efficiency of compressor is 76% and mass flow rate of NH3 is 0.9 kg / min. assuming the refrigerant is duly and saturated vapour. Determine: (i) Capacity of refrigerator (IN TONS) (ii) Cofficient of performance (COP) (iii) Power required to drive the compressor. Given : vapour compression refrigerator refrigerant Ammonia. Condenser temperature, T3 = 30oC Evaporator temperature, T3 = -15oC Subcooling of refrigerant by 4o C. T14 = 30-4 = 26Oc Isentropic efficiency of compressor 𝜂ci= 0.76. mass flow rate, Mo = 0.9 kg/min From T.D.H.B, for NH3 H14 = h11 = 300kj/kg H2 = 1430 kg/kg H3 = 1540 kj/kg Refrigerating effect, R.E = h2 – h11 =1430 – 300 = 1130kj/kg (i) Capacity = m X RE 60 X 3.5 = 0.9 X 1130 = 4.84 tones 6 X 3.5 109 (ii) Coefficient of performance COP = h2 – h11 h13 – h2 𝜂ci = h3 – h2 h13 – h2 = 1130 144.736 COP = 7.81 h31 – h2 = 𝜂ci (h3 – h2) = 0.76 (1540-1430 = 144.73 KJ /KG Assume a mechanical efficiency 𝜂m =0.85 𝜂 m = Pc = Pm = P c Ppm 𝜂m 1 Ppm = m(h 3 – h2) = 0.9 X (144.73) 60 X 0.85 60 X 0.85 Ppm = 4.4 KW 13. A simple vapour compression plant produces 2 tons of refrigeration. The enthalpy values at inlet to the compressor, at the exit of compressor, and at exit of condenser are 300.41, 384.38, and 92.34 kj/kg respectively. Estimate : (i) The refrigerant flow rate. (ii) C.O.P (iii) The power required to drive the compressor (iv) The rate of heat rejection in condenser. (Jan 2010) Given : simple vapour compression plant. Capacity = 2 tons. H2 = 300.41kj/kg H3 = 385.38 kj/kg H4 = 92.34 kj/kg = h1 We have capacity 110 =mo X refrigerating effect (R.E) ---------------------------------------- (tones) 210 2 = mo (h2 – h1) -----------------210 Mo = 2 X 210 = 2 X 210 ------------------H2 – h1 300.41 – 92.34 (i) The refrigerant flow rate, o (m = 2.0185 kg/min ) per ton of refrigeration The C.O.P = h2 - h1 300.41-92.34 = -----------------------------------h3 - h2 384.38-300.41 C.O.P = 2.478 (iii) power required to drive the compressor : pc = mo (h3 –h2) Pc = 2.0185(385.38 – 300.41) ---------------------------------------= pc = 2.824 KW 60 (ii) Rate of heat rejection in the condenser: Qrej = mo (h3 –h4) ---------------(kj/s) 60 Qrej = 2.0185(385.38-92.34) -------------------------------60 = 9.8247 KJ/s 14. A simple vapour compression plant produces 5 tonnes of refrigeration. The enthalphy values at inlet to compressor, at the exit of compressor and at the exit of condenser are 183.19,209.41 and 74.59 kj/kg Respectively. Estimate: (i) The refrigerant flow rate (ii) C.O.P (iii) The power required to drive the compressor (iv) The rate of heat rejection in condenser. (Dec 2007/Jan2008) 111 Given: Simple vapour compression plant capacity = 5tonnes = 4.5 tones. ---------------------(1 ton = 0.9 tonne) H2 = 183.19 kj/kg H3 = 209.41 kj/kg H4 = 74.59 kj/kg = h1 We have capacity = mo refrigerating effect ------------------------------------210 4.5 = mo (h2 – h1) ----------------210 o M = 210 X 4.5 210 X 4.5 ----------------- = --------------h2 – h1 183.19 – 74.59 (i) The refrigerant flow rate ( mo = 8.7016 kg / min ) per ton of refrigeration (ii) The C.O.P h2 – h1 183.19 – 74.59 --------= ------------------C.O.P = 4.142 h3 – h2 209.41 – 183.19 (iii) Power require to drive the compresses Pc = mo ( h3 – h2) 8.7016(209.41 – 183.19 ---------------- = -------------------------------60 60 = pc = 3.8 kw Assuming a mechanical efficiency of 0.85, we get: Mm = Pc ----Ppm 112 Ppm = Pc = 3.8 ------------ = 4.47 kw Mm o.85 Power of the prime mover to drive the compressor Ppm = 4.47kw (iv) Rate of heat rejection in condenser: Qrej = Mo (h3 – h4) 8.7016(209.41-74.59 --------------- = -----------------------60 60 Qrej = 19.55KJ/s 15. A vapour compression refrigeration cycle works between a condenser pressure of to bar and evaporator pressure of 1 bar. The refrigerant From – 12 leaves the evaporator at -20oC and the condenser at 30oC. Determine: (i) The COP of refrigeration (ii) Power required per ten of refrigeration (iii) The bore and stroke of the compressor cylinder if it runs at 250 rpm Assume volumetric efficiency of 90% the stroke is 1.2 times the bore. Given : vapour compression refrigeration cycle. Condenser pressure, p2 = 10 bar. Evaporator pressure P1 = 1 bar Refrigerant From – 12 Evaporator temperature, T2 = -20oC Condenser temperature, T3 = 30oC Compressor speed, N = 250 rpm Volumetric efficiency, nv = 90% = 0.9 Stroke, l = 1.2 Bore, D From the thermodynamic hand book, page 82, we observe the following for the above data given. 113 We get (i) h11 = 55kj/kg = h14 H12 = 182kj/kg H3 = 220kj/kg The COP of refrigeration COP = h12 - h1 --------------= 1 H3 - h 2 = COP = 3.0789 182 – 65 ----------220 - 182 We have capacity = mo(h12 - h11) --------------------60 X 3.5 For capacity = 1 ton => 1 = mo (182 – 65) ----------------------60 X 3.5 = mo = 1.8 kg/ min. 16. A 2- ton refrigeration unit uses NH3 as refrigerant. The working pressure limits are 2 bar and 10 bar respectively. The refrigerant is dry and saturated before it enters the compressor. After compression, the energy rejected by the refrigerant in the condenser is 1550 KJ/kg. the liquid emerging from the condenser is sub cooled by 10oC find: (i) C.O.P (ii) Mass flow rate of refrigerant (iii) Power required to drive the compressor if the mechanical efficiency is 0.8 (iv) Isentropic efficiency of the compressor. Given : capacity = 2ton Refrigerant: NH3 114 Operating pressure limits, P1 = 2 bar . p2 = 10 bar Energy rejected in the condenser, Qrej = 1550 kj/kg. (ii) Power required per ton of refrigeration: Pc = mo(h3 – h12) 1.8(220 – 182) ---------= -------------60 60 = Pc = 1.14kw (iii) The bore and stroke of the compressor We have 𝜂v = mo.V2 𝜋 D2 X L X N 4 60 V2 = 1.8 X 0.12 3.14 X D2 X 1.2D X 250 4 60 = 0.9 D3 = 1.0184 10-3 m3 D = 0.1m = 1000.6 mm L = 1.2D = 1.2 100.6 L = 120.72 mm Liquid emerging from condenser is subcooled by 10oC T14 = T4 – 10 From the page 80 of the thermodynamic data hand book, for NH3, we abserve the following from the chart. 115 Also Qrej = h3 – h14 = 1550 = h13 = h14 + 1550 = 260+1550 = 1810 kj/kg From the chart h11 = h14 = 260 kj/kg h2 = 1429 kj/kg h3 = 1520 kj/kg The C.O.P = h2 – h11 = 1420 – 260 -------------------------= C.O.P = 2.974 1 H 3 – h2 1810 – 1420 (ii) Mass flow rate of the refrigeration capacity = mo (h2 – h11) ------------------- (tones) 210 o 2 = m (1420 – 260) ----------------------210 o M = 0.362 kg / min (iii) The power consumed by the compressor , Pc = mo(h13 – h2) ---------60 = 0.362(1810 – 1420) -------------------60 = 2.353 kw = Pc ------ = 0.8 (given) Ppm The power required to drive the compressor , Ppm = Pc = 2.353 -----------0.8 0.8 = Ppm = 2.941 kw (i) (iv) The isentropic efficiency of the compresses : Mi = h3 - h2 -----------h13 – h2 1520 – 1420 = ------------1810 – 1420 mi = Wideal ------------Wactual 116 Mi = 0.2564 0r 25. 64% 20 tones of ice is to be produced from water at 20oC to at -6oC in 24 hours, when the temperature range in the compressor is -15oC to 25oC. the condition of vapour is dry at the end of compression. Assume the relative C.O.P of ).8 and calculate the power of the compressor. Take Cp of ice as 2.1 kj/kgk, and latent heat of ice is 335 kj/kg. used the following properties of the refrigerant. Temp 25oC -15oC Enthalpy(h) 100.04 - 54.55 Liquid Enthalpy(S) 0.347 -2.1338 Enthalpy(h) 1319.2 1304.99 Given: capacity = 20 tones T2 = -15oC T3 = 25oC (COP)rel = 0.8 Cpice = 2.1kj/kgk. Hfgice = 335 kj/kg Observing the p-h diagram; H3 = 1319.2 kj/kg H4 = 100/04 kj/kg = h1 For isentropic process 2 – 3; S2 = s3 Sf2 = x2 . sfg2 = 4.4852 -2.1338 + x2 (5.058 –(-2.1338)) = 4.4852. X2 = 0.09 H2 = hf2 +x2 (hg2 – hf2) = -54.55+0.92(130.99-(-54.55) H2 = 1196.22kj/kg Capacity = mo (h2 – h1) ---------------210 20 210 -------------------------- = mo (1196.22 – 100.04) Refrigerant flow rate Mo = 3.83kg/min (based on theoretical refrigeration effect) (COP)rel COP actual -----------------COP ideal (COP)actual = (COP) real COPideal = 0.8 h2 –h1 X ----------- Vapour Enthalpy(S) 4.4852 5.0585 117 h3 – h2 (1196.22 – 100.04) X ------------------------------ = 7.1296 (1318.2 – 1196.2) Actual refrigerating effect (RE)actual = 7.1296 (h3 – h2) 7.1296(1319.2 – 1196.22) = 876.798kj/kg Capacity = (RE) actual X mo 20 X 1000 ----------------(Cpw(20-0+hfice+Cpi(0+6)) 24 X 60 X 60 = 0.8 = 876.798 m0 mo 876.798 = 20 X 1000 -------------------( 4.18 X 20+335+2.1 6) 24 X 60 X 60 o m = 0.1138 kg/s The power required to run the compresser: Pc = mo (h3 – h2) 0.1138(1319.2 – 1196.22) Pc = 13.99kw kw. 118