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REFRIGERATION PROBLEMS Anu.pptx

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Numerical problems for Refrigeration
TABLE B-1 Thermodynamic Properties of Saturated Ammonia
An open air refrigeration system operating between pressures of 16 bar and 1 bar, required to produce 33.5 kW of refrigeration. The temperature of air leaving the refrigerated room is -50C and that leaving the air cooler is 300C. Assuming no losses and clearances, calculate for the theoretical cycle (a) mass flow rate of air circulated per minute (b) Piston displacement of the compressor and expender (c) Net work (d) COP Solution: Given: 𝑃2 𝑃1 = π‘Ÿπ‘ = 16 Total refrigeration effect produced=33.5 kW T1=273-5=268K T3=30+273=303K
(a) mass flow rate of air circulated per minute Total refrigeration effect=mCp(T1-T4) 33.5 = π‘š Γ— 1.005 Γ— 268 βˆ’ 𝑇4 ----------(1) Processes 1-2 and 3-4 are isentropic : Using p-v-t relationships for isentropic process; 𝑇2 𝑇1 = 𝑃2 𝑃1 π›Ύβˆ’1 𝛾 ∴ π‘»πŸ = πŸπŸ”πŸ– Γ— πŸπŸ” 𝟎.πŸ’ 𝟏.πŸ’ = πŸ“πŸ—πŸ. πŸ–π‘² Similarly, 𝑇3 𝑇4 = 𝑃3 𝑃4 π›Ύβˆ’1 𝛾 = 𝑃2 𝑃1 π›Ύβˆ’1 𝛾 ∴ π‘»πŸ’ = πŸ‘πŸŽπŸ‘ Γ· πŸπŸ” 𝟎.πŸ’ 𝟏.πŸ’ = πŸ“πŸ—πŸ. πŸ–π‘² = πŸπŸ‘πŸ•. πŸπŸπ‘²
Substituting T4 in equation (1): 33.5 = π‘š Γ— 1.005 Γ— 268 βˆ’ 137.22 π‘š = 33.5 131.436 Γ— 60 = 15.3π‘˜π‘”/π‘šπ‘–π‘› (b) Piston displacement of the compressor and expender Compression process is 1-2: ∴ π‘ƒπ‘–π‘ π‘‘π‘œπ‘› π‘‘π‘–π‘ π‘π‘™π‘Žπ‘π‘’π‘šπ‘’π‘›π‘‘ π‘œπ‘“ π‘π‘œπ‘šπ‘π‘Ÿπ‘’π‘ π‘ π‘œπ‘Ÿ, 𝑉1 = π‘šπ‘…π‘‡1 𝑃1 𝑉1 = 15.3 Γ— 0.287 Γ— 268 1 Γ— 100 = 11.768π‘š3/π‘šπ‘–π‘› π‘˜π‘”/π‘šπ‘–π‘› Γ— π‘˜π½/π‘˜π‘”πΎ Γ— 𝐾 π‘˜π‘/π‘š2 = π‘˜π‘” Γ— π‘˜π‘ βˆ’ π‘š Γ— 𝐾 Γ— π‘š2 π‘˜π‘ Γ— π‘šπ‘–π‘› Γ— π‘˜π‘”πΎ
Expansion process is 3-4: ∴ π‘ƒπ‘–π‘ π‘‘π‘œπ‘› π‘‘π‘–π‘ π‘π‘™π‘Žπ‘π‘’π‘šπ‘’π‘›π‘‘ π‘œπ‘“ 𝑒π‘₯π‘π‘Žπ‘›π‘‘π‘’π‘Ÿ, 𝑉4 = π‘šπ‘…π‘‡4 𝑃4 = π‘šπ‘…π‘‡4 𝑃1 𝑉4 = 15.3 Γ— 0.287 Γ— 137.22 1 Γ— 100 = 6.025π‘š3/π‘šπ‘–π‘› π‘˜π‘”/π‘šπ‘–π‘› Γ— π‘˜π½/π‘˜π‘”πΎ Γ— 𝐾 π‘˜π‘/π‘š2 = π‘˜π‘” Γ— π‘˜π‘ βˆ’ π‘š Γ— 𝐾 Γ— π‘š2 π‘˜π‘ Γ— π‘šπ‘–π‘› Γ— π‘˜π‘”πΎ (c) Net work 𝑁𝑒𝑑 π‘€π‘œπ‘Ÿπ‘˜ = π‘š π‘€π‘π‘œπ‘šπ‘. βˆ’ 𝑀𝑒π‘₯π‘π‘Žπ‘›. 𝑁𝑒𝑑 π‘€π‘œπ‘Ÿπ‘˜ = π‘šπΆπ‘ƒ 𝑇2 βˆ’ 𝑇1 βˆ’ 𝑇3 βˆ’ 𝑇4 𝑁𝑒𝑑 π‘€π‘œπ‘Ÿπ‘˜ = 15.3 60 Γ— 1.005 591.8 βˆ’ 268 βˆ’ 303 βˆ’ 137.22 ∴ 𝑡𝒆𝒕 π’˜π’π’“π’Œ = πŸ’πŸŽ. πŸ“π’Œπ‘Ύ
(d) COP π‘ͺ𝑢𝑷 = 𝑻𝒐𝒕𝒂𝒍 π’“π’†π’‡π’“π’Šπ’ˆπ’†π’“π’‚π’•π’Šπ’π’ 𝒆𝒇𝒇𝒆𝒄𝒕 𝑡𝒆𝒕 π’˜π’π’“π’Œ = πŸ‘πŸ‘. πŸ“ π’Œπ‘Ύ πŸ’πŸŽ. πŸ“ π’Œπ‘Ύ = 𝟎. πŸ–πŸπŸ•
A dense air machine operates between 17 bar and 3.4 bar. The temperature of air after the cooler is 150C and after the refrigerating coil is 60. Determine the (a)Air circulation per minute (b) work of the compressor and the expander/TR (c) Theoretical COP and (d) HP/TR Solution: Given: 𝑃2 𝑃1 = π‘Ÿπ‘ = 17 3.4 = 5 T1=273+6=279K T3=273+15=288K
(a) air circulated per minute Total refrigeration effect=mCp(T1-T4) 𝑇𝑅𝐸 = π‘š Γ— 1.005 Γ— 279 βˆ’ 𝑇4 ----------(1) Processes 1-2 and 3-4 are isentropic : Using p-v-t relationships for isentropic process; 𝑇2 𝑇1 = 𝑃2 𝑃1 π›Ύβˆ’1 𝛾 ∴ π‘»πŸ = πŸπŸ•πŸ— Γ— πŸ“ 𝟎.πŸ’ 𝟏.πŸ’ = πŸ’πŸ’πŸ. πŸ–πŸ–π‘² Similarly, 𝑇3 𝑇4 = 𝑃3 𝑃4 π›Ύβˆ’1 𝛾 = 𝑃2 𝑃1 π›Ύβˆ’1 𝛾 ∴ π‘»πŸ’ = πŸπŸ–πŸ– Γ· πŸ“ 𝟎.πŸ’ 𝟏.πŸ’ = πŸ“πŸ—πŸ. πŸ–π‘² = πŸπŸ–πŸ. πŸ–πŸ‘π‘²
π΄π‘ π‘ π‘’π‘šπ‘–π‘›π‘” 𝑇𝑅𝐸 π‘“π‘œπ‘Ÿ 1 π‘‘π‘œπ‘› π‘œπ‘“ π‘Ÿπ‘’π‘“π‘Ÿπ‘–π‘”π‘’π‘Ÿπ‘Žπ‘‘π‘–π‘œπ‘› = 3.51 π‘˜π‘Š π‘š = 𝑇𝑅𝐸 𝐢𝑃 𝑇1 βˆ’ 𝑇4 = 3.51 Γ— 60 1.005 279 βˆ’ 181.83 ∴ π’Ž = 𝟐. πŸπŸ“πŸ”π’Œπ’ˆ/π’Žπ’Šπ’ (b) work of compressor and the expander/TR π‘ŠπΆπ‘œπ‘šπ‘. = π‘š Γ— 𝐢𝑃 Γ— 𝑇2 βˆ’ 𝑇1 = 2.156 60 Γ— 1.005 441.88 βˆ’ 279 ∴ 𝑾π‘ͺπ’π’Žπ’‘.= πŸ“. πŸ–πŸ– π’Œπ‘Ύ π‘Š 𝑒π‘₯π‘π‘Žπ‘›. = π‘š Γ— 𝐢𝑃 Γ— 𝑇3 βˆ’ 𝑇4 = 2.156 60 Γ— 1.005 288 βˆ’ 181.83 ∴ 𝑾𝒆𝒙𝒑𝒂𝒏.= πŸ‘. πŸ–πŸ‘πŸ’ π’Œπ‘Ύ (c) Theoretical COP 𝐢𝑂𝑃 = 𝑇𝑅𝐸 𝑁𝑒𝑑 π‘€π‘œπ‘Ÿπ‘˜ = 3.51 π‘˜π‘Š 5.88 βˆ’ 3.834 π‘˜π‘Š ∴ π‘ͺ𝑢𝑷 = 𝟏. πŸ•πŸ (d) HP/TR π‘·π’π’˜π’†π’“/𝒕𝒐𝒏 = π‘Š (π‘π‘œπ‘šπ‘. ) βˆ’ π‘Š (𝑒π‘₯π‘π‘Žπ‘›. ) = 5.88 βˆ’ 3.884 = 𝟐. πŸŽπŸ’πŸ” 𝐀𝐖
An air refrigerating machine working on Bell-Coleman cycle takes in air into the compressor at 1 bar and -50C. It is compressed in the compressor to 5 bar and is cooled to 250C at constant pressure. In the expander it is expanded to 1bar. The isentropic efficiency of the compressor and the expander are 85% and 90% respectively. Calculate : (a) Refrigerating capacity of the system if the mass flow rate is 30kg/m (b) Power required to run the compressor (c) COP Solution: Given: P1=P4=P4’= 1 bar, P2’=P2=P3=5 bar T1=273-5=268K T3=273+25=298K Ξ·c=85% and Ξ·e=90% m= 30 kg/min
(a) Refrigerating capacity of the system if the mass flow rate is 30kg/min 𝑇𝑅𝐸 = π‘š Γ— 𝐢𝑃 Γ— 𝑇1 βˆ’ 𝑇4 𝑇𝑅𝐸 = 30 60 Γ— 1.005 Γ— 268 βˆ’ 𝑇4 ---------(1) Processes 1-2’ and 3-4’ are isentropic : Using p-v-t relationships for isentropic process; 𝑇2β€² 𝑇1 = 𝑃2β€² 𝑃1 π›Ύβˆ’1 𝛾 ∴ π‘»πŸβ€² = πŸπŸ”πŸ– Γ— πŸ“ 𝟎.πŸ’ 𝟏.πŸ’ = πŸ’πŸπŸ’. πŸ“π‘² Similarly, 𝑇3 𝑇4β€² = 𝑃3 𝑃4β€² π›Ύβˆ’1 𝛾 = 𝑃2β€² 𝑃1 π›Ύβˆ’1 𝛾 ∴ π‘»πŸ’β€² = πŸπŸ—πŸ– Γ· πŸ“ 𝟎.πŸ’ 𝟏.πŸ’ = πŸ“πŸ—πŸ. πŸ–π‘² = πŸπŸ–πŸ–. πŸπŸ”π‘²
Using isentropic efficiency formula for compressor and expander: πœ‚πΆ = 𝑇2β€² βˆ’ 𝑇1 𝑇2 βˆ’ 𝑇1 0.85 = 424.5 βˆ’ 268 𝑇2 βˆ’ 268 ∴ π‘»πŸ = πŸ’πŸ“πŸ. πŸπ‘² πœ‚πΈ = 𝑇3 βˆ’ 𝑇4 𝑇3 βˆ’ 𝑇4β€² 0.90 = 298 βˆ’ 𝑇4 298 βˆ’ 188.16 ∴ π‘»πŸ’ = πŸπŸ—πŸ—. πŸπŸ“π‘² substituting π‘»πŸ’ in equation (1): 𝑇𝑅𝐸 = 30 60 Γ— 1.005 Γ— 268 βˆ’ 199.15 = πŸ‘πŸ’. πŸ” 𝐀𝐖
Refrigeration capacity in terms of ton is obtained as below: 1 ton= 3.51 kW ∴ 34.6 Γ· 3.51 = πŸ—. πŸ–πŸ” 𝒕𝒐𝒏 (b) Power required to run the compressor π‘ƒπ‘œπ‘€π‘’π‘Ÿ = π‘€π‘œπ‘Ÿπ‘˜ π‘‘π‘œπ‘›π‘’ 𝑖𝑛 π‘˜π‘Š = π‘Š π‘π‘œπ‘šπ‘π‘Ÿπ‘’π‘ π‘ π‘œπ‘Ÿ βˆ’ π‘Šπ‘’π‘₯π‘π‘Žπ‘›π‘‘π‘’π‘Ÿ π‘ƒπ‘œπ‘€π‘’π‘Ÿ = π‘šπΆπ‘ƒ 𝑇2 βˆ’ 𝑇1 βˆ’ 𝑇3 βˆ’ 𝑇4 π‘ƒπ‘œπ‘€π‘’π‘Ÿ = 30 60 Γ— 1.005 452.2 βˆ’ 268 βˆ’ 298 βˆ’ 199.15 ∴ π‘·π’π’˜π’†π’“ = πŸ’πŸ. πŸ–πŸ— π’Œπ‘Ύ (c) COP 𝐢𝑂𝑃 = 𝑇𝑅𝐸 𝑁𝑒𝑑 π‘€π‘œπ‘Ÿπ‘˜ = 34.6 π‘˜π‘Š 42.89 π‘˜π‘Š ∴ π‘ͺ𝑢𝑷 = 𝟎. πŸ–πŸ
An ideal vapor compression refrigeration system operates on Freon- 12 between the temperature limits of 300C and -100C without undercooling. Vapor leaving the compressor is dry saturated. Determine (a) the power input to the compressor if the refrigerating capacity of the unit is 1.5tons (b) COP (c) Condenser duty Solution: (a) the power input to the compressor if the refrigerating capacity of the unit is 1.5tons π‘ƒπ‘œπ‘€π‘’π‘Ÿ = π‘šπ‘Ÿ β„Ž2 βˆ’ β„Ž1 Mass of refrigerant is calculate from TRE equation: 𝑇𝑅𝐸 = π‘šπ‘Ÿ Γ— 𝐢𝑃 Γ— 𝑇1 βˆ’ 𝑇4 𝑇𝑅𝐸 = π‘šπ‘Ÿ Γ— β„Ž1 βˆ’ β„Ž4 1.5 Γ— 3.51 = π‘šπ‘Ÿ Γ— β„Ž1 βˆ’ β„Ž4 -------(1)
To calculate β€˜β„Ž1’: Point 1 on p-h diagram is within the vapor dome: This means at 1, the refrigerant is wet vapor of quality β€˜π‘₯1’ ∴ β„Ž1 = β„Žπ‘“ + π‘₯1β„Žπ‘“π‘” at T=-10Β°C In thermodynamics data handbook, refer Table-B3 for Thermodynamic properties of saturated Freon-12: At Temp=-100C , 𝒉𝒇= 26.851 kJ/kg and π’‰π’‡π’ˆ= 156.21 kJ/kg ∴ β„Ž1 = 26.851 + π‘₯1156.21 at T=-10Β°C---------(2) To find β€˜π‘₯1’ : Process 1-2 is isentropic compression: ∴ 𝑠1 = 𝑠2 and 𝑠2 = 𝑠𝑔 at T=30Β°C. π΄π‘™π‘ π‘œ, 𝑠2 = 𝑠1 = 𝑠𝑓 + π‘₯1𝑠𝑓𝑔 at T=-10Β°C---------(3)
In thermodynamics data handbook, refer Table-B3 for Thermodynamic properties of saturated Freon-12: At Temp=300C , π’”π’ˆ= 0.6848 kJ/kg K= π’”πŸ. Substituting 𝑠2 in equation (3); 𝑠2 = 𝑠1 = 0.6848 = 𝑠𝑓 + π‘₯1𝑠𝑓𝑔 at T=-10Β°C-------(4) In thermodynamics data handbook, refer Table-B3 for Thermodynamic properties of saturated Freon-12: At Temp=-100C , 𝒔𝒇= 0.1079 kJ/kg K and π’”π’‡π’ˆ= 0.5936 kJ/kg K. Substituting these in equation (4); 0.6848 = 𝑠𝑓 + π‘₯1𝑠𝑓𝑔 at T=-10Β°C 0.6848 = 0.1079 + π‘₯10.5936 ∴ π’™πŸ = 𝟎. πŸ—πŸ•πŸπŸ–. Substitute π’™πŸ in equation (2) β„Ž1 = 26.851 + π‘₯1156.21 β„Ž1 = 26.851 + 0.9718 Γ— 156.21 π’‰πŸ = πŸπŸ•πŸ–. πŸ”πŸ” π’Œπ‘±/π’Œπ’ˆ
π’‰πŸ‘= π’‰πŸ’. Because, process 3-4 is isenthalpic process And π’‰πŸ‘= 𝒉𝒇 at T=30Β°C. In thermodynamics data handbook, refer Table-B3 for Thermodynamic properties of saturated Freon-12: At Temp=300C , 𝒉𝒇= 64.539 kJ/kg ∴ π’‰πŸ‘ = πŸ”πŸ’. πŸ“πŸ‘πŸ— π’Œπ‘± π’Œπ’ˆ = π’‰πŸ’ Substituting the values in equation (1) 1.5 Γ— 3.51 = π‘šπ‘Ÿ Γ— 178.66 βˆ’ 64.539 π’Žπ’“ = 𝟎. πŸŽπŸ’πŸ” π’Œπ’ˆ/𝒔 π‘ƒπ‘œπ‘€π‘’π‘Ÿ = π‘šπ‘Ÿ β„Ž2 βˆ’ β„Ž1 π‘ƒπ‘œπ‘€π‘’π‘Ÿ = 0.046 β„Ž2 βˆ’ 178.66 ------ (5); π’‰πŸ= π’‰π’ˆ at T=30Β°C. In thermodynamics data handbook, refer Table-B3 for Thermodynamic properties of saturated Freon-12: At Temp=300C , π’‰π’ˆ= 199.475 kJ/kg. substituting in eq. (5) π‘ƒπ‘œπ‘€π‘’π‘Ÿ = 0.046 199.475 βˆ’ 178.66 π‘·π’π’˜π’†π’“ = 𝟎. πŸ—πŸ” π’Œπ‘Ύ
(b) COP 𝐢𝑂𝑃 = 𝑇𝑅𝐸 𝑁𝑒𝑑 π‘€π‘œπ‘Ÿπ‘˜ = β„Ž1 βˆ’ β„Ž4 β„Ž2 βˆ’ β„Ž1 𝐢𝑂𝑃 = 178.66 βˆ’ 64.539 199.475 βˆ’ 178.66 ; ∴ π‘ͺ𝑢𝑷 = πŸ“. πŸ’πŸ– (c) Condenser duty πΆπ‘œπ‘›π‘‘π‘’π‘›π‘ π‘’π‘Ÿ 𝑑𝑒𝑑𝑦 = π‘šπ‘Ÿ Γ— β„Ž2 βˆ’ β„Ž3 πΆπ‘œπ‘›π‘‘π‘’π‘›π‘ π‘’π‘Ÿ 𝑑𝑒𝑑𝑦 = 0.046 Γ— 199.475 βˆ’ 64.539 ∴ π‘ͺ𝒐𝒏𝒅𝒆𝒏𝒔𝒆𝒓 π’…π’–π’•π’š = πŸ”. 𝟐 π’Œπ‘Ύ
An ammonia plant having a capacity of 15 tons is working at an evaporator temperature of -60C and a condenser temperature of 300C. The refrigerant is superheated to 50C before entering the compressor. A two cylinder single acting compressor is used which runs at 900rpm. Determine (i) COP (ii) Mass of ammonia (iii) Bore and stroke of the cylinder assuming that L= 1.5xD and neglect the clearance. Solution: (a) COP 𝐢𝑂𝑃 = 𝑇𝑅𝐸 𝑁𝑒𝑑 π‘€π‘œπ‘Ÿπ‘˜ = β„Ž1βˆ’β„Ž4 β„Ž2βˆ’β„Ž1 ---------(1) β„Ž1 = β„Ž1β€² + 𝐢𝑃(π‘ π‘’π‘π‘’π‘Ÿβ„Žπ‘’π‘Žπ‘‘π‘’π‘‘ π‘£π‘Žπ‘π‘œπ‘Ÿ π‘Ÿπ‘’π‘“.) 𝑇1 βˆ’ 𝑇1β€² β„Ž1β€² = β„Žπ‘” at T= -60C In thermodynamics data handbook, refer Table-B1 for Thermodynamic properties of saturated Ammonia: At Temp=-6Β°C , β„Ž1β€² = β„Žπ‘”= 1436.8 kJ/kg;
In thermodynamics data handbook, refer Table-B9 for specific heat of superheated Ammonia vapor: At Pressure 3.4 bar, CP=2.3818 kJ/kg K (From Table-B1 at Temp=-6Β°C, Abs. Pressure=3.4125 bar) ∴ 𝐢𝑃(π‘ π‘’π‘π‘’π‘Ÿβ„Žπ‘’π‘Žπ‘‘π‘’π‘‘ π‘£π‘Žπ‘π‘œπ‘Ÿ π‘Ÿπ‘’π‘“.)= 2.3818 kJ/kg K Given that the refrigerant is superheated to 50C before entering the compressor; ∴ 𝑇1 = 5°𝐢 π‘Žπ‘›π‘‘ 𝑇1β€² = βˆ’ 5°𝐢, π‘€β„Žπ‘–π‘β„Ž 𝑖𝑠 π‘‘β„Žπ‘’ π‘’π‘£π‘Žπ‘π‘œπ‘Ÿπ‘Žπ‘‘π‘œπ‘Ÿ π‘‘π‘’π‘šπ‘. 𝑆𝑒𝑏𝑠𝑑𝑖𝑑𝑒𝑑𝑖𝑛𝑔 𝑖𝑛: β„Ž1 = β„Ž1β€² + 𝐢𝑃(π‘ π‘’π‘π‘’π‘Ÿβ„Žπ‘’π‘Žπ‘‘π‘’π‘‘ π‘£π‘Žπ‘π‘œπ‘Ÿ π‘Ÿπ‘’π‘“.) 𝑇1 βˆ’ 𝑇1β€² β„Ž1 = 1436.8 + 2.3818 5 βˆ’ βˆ’5 π’‰πŸ = πŸπŸ’πŸ”πŸŽ. πŸ”πŸπŸ– π’Œπ‘±/π’Œπ’ˆ π’‰πŸ‘= π’‰πŸ’. Because, process 3-4 is isenthalpic process And π’‰πŸ‘= 𝒉𝒇at T=30Β°C. In thermodynamics data handbook, refer Table-B1 for Thermodynamic properties of saturated Ammonia: At Temp=30Β°C , 𝒉𝒇 = 322.9 kJ/kg ∴ π’‰πŸ‘= π’‰πŸ’=322.9 kJ/kg
To find β„Ž2: Point β€˜2’ is in the superheated vapor region and We cannot find its value directly from the saturated tables. ∴ β„Ž2 = β„Ž2β€² + 𝐢𝑃 π‘ π‘’π‘π‘’π‘Ÿβ„Žπ‘’π‘Žπ‘‘π‘’π‘‘ π‘£π‘Žπ‘π‘œπ‘Ÿ π‘Ÿπ‘’π‘“. 𝑇2 βˆ’ 𝑇2β€² βˆ’ βˆ’(2) To find β€˜π‘‡2’: We know that, process 1-2 is isentropic compression ∴ 𝑠1 = 𝑠2; But, 𝑠1 is also in the superheated vapour region. ∴ 𝑠1 = 𝑠1β€² + 𝐢𝑃(π‘ π‘’π‘π‘’π‘Ÿβ„Žπ‘’π‘Žπ‘‘π‘’π‘‘ π‘£π‘Žπ‘π‘œπ‘Ÿ π‘Ÿπ‘’π‘“.)𝑙𝑛 𝑇1 𝑇1β€² ----------(3) i.e., using In thermodynamics data handbook, refer Table-B1 for Thermodynamic properties of saturated Ammonia: At Temp=-6Β°C , 𝑠1β€² = 𝑠𝑔= 5.4173 kJ/kg K; substituting in eq. (3) ∴ 𝑠1 = 5.4173 + 2.3818𝑙𝑛 273+5 273βˆ’5 ; ∴ 𝑠1 = 5.504 π‘˜π½ π‘˜π‘”K = 𝑠2
Similarly, 𝑠1 = 𝑠2 = 𝑠2β€² + 𝐢𝑃(π‘ π‘’π‘π‘’π‘Ÿβ„Žπ‘’π‘Žπ‘‘π‘’π‘‘ π‘£π‘Žπ‘π‘œπ‘Ÿ π‘Ÿπ‘’π‘“.)𝑙𝑛 𝑇2 𝑇2β€² 𝑖. 𝑒. , 5.504 = 𝑠2β€² + 𝐢𝑃(π‘ π‘’π‘π‘’π‘Ÿβ„Žπ‘’π‘Žπ‘‘π‘’π‘‘ π‘£π‘Žπ‘π‘œπ‘Ÿ π‘Ÿπ‘’π‘“.)𝑙𝑛 𝑇2 𝑇2β€² In thermodynamics data handbook, refer Table-B1 for Thermodynamic properties of saturated Ammonia: At Temp=30Β°C , 𝑠2β€² = 𝑠𝑔= 4.9805 kJ/kg K; ∴ 5.504 = 4.9805 + 2.3818𝑙𝑛 𝑇2 303 ; ∴ 𝑇2 = 377.48𝐾 Substituting in eq.(2); β„Ž2 = β„Ž2β€² + 𝐢𝑃 π‘ π‘’π‘π‘’π‘Ÿβ„Žπ‘’π‘Žπ‘‘π‘’π‘‘ π‘£π‘Žπ‘π‘œπ‘Ÿ π‘Ÿπ‘’π‘“. 𝑇2 βˆ’ 𝑇2β€² In thermodynamics data handbook, refer Table-B1 for Thermodynamic properties of saturated Ammonia: At Temp=30Β°C , β„Ž2β€² = β„Žπ‘”= 1467.9 kJ/kg; 𝑇2β€²= 273+30=303 K
β„Ž2 = 1467.9 + 2.3818 377.48 βˆ’ 303 ∴ π’‰πŸ= πŸπŸ”πŸ’πŸ“. πŸ‘ 𝐀𝐉/𝐀𝐠 Substituting in eq.(1): 𝐢𝑂𝑃 = β„Ž1βˆ’β„Ž4 β„Ž2βˆ’β„Ž1 = πŸπŸ’πŸ”πŸŽ.πŸ”πŸπŸ–βˆ’πŸ‘πŸπŸ.πŸ— πŸπŸ”πŸ’πŸ“.πŸ‘βˆ’πŸπŸ’πŸ”πŸŽ.πŸ”πŸπŸ– ∴ π‘ͺ𝑢𝑷 = πŸ”. πŸπŸ” (b) Mass of ammonia Using TRE equation: 𝑇𝑅𝐸 = π‘šπ‘Ÿ Γ— 𝐢𝑃 Γ— 𝑇1 βˆ’ 𝑇4 𝑇𝑅𝐸 = π‘šπ‘Ÿ Γ— β„Ž1 βˆ’ β„Ž4 15 Γ— 3.51 = π‘šπ‘Ÿ Γ— 1460.618 βˆ’ 322.9 ∴ π’Žπ’“= 𝟎. πŸŽπŸ’πŸ”πŸ π’Œπ’ˆ/𝒔
(c) Bore and stroke of the cylinder assuming that L= 1.5xD In order to find cylinder dimensions, we need to evaluate the stroke volume of the compressor. 𝑉 𝑠 = π‘šπ‘Ÿ Γ— 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐 π‘£π‘œπ‘™π‘’π‘šπ‘’ π‘Žπ‘‘ π‘‘β„Žπ‘’ π‘π‘œπ‘šπ‘π‘Ÿπ‘’π‘ π‘ π‘œπ‘Ÿ 𝑖𝑛𝑙𝑒𝑑 = πœ‹ 4 Γ— 𝐷2 Γ— 𝐿 Γ— 𝑁 60 π‘š3 /𝑠 𝑉 𝑠 = π‘šπ‘Ÿ Γ— 𝑣1 = πœ‹ 4 Γ— 𝐷2 Γ— 𝐿 Γ— 𝑁 60 Γ— 2------- (4) Points 1 and 1’ are on the same constant pressure line; ∴ 𝑃1𝑣1 𝑇1 = 𝑃1′𝑣1β€² 𝑇1β€² ; 𝑃1=𝑃1β€²; ∴ 𝑣1 = 𝑇1 𝑇1β€² Γ— 𝑣1β€²; 𝑣1β€²= 𝑣𝑔 at -6Β°C = 0.3599 m3/kg. ∴ 𝑣1 = 273 + 5 273 βˆ’ 6 Γ— 0.3599; ∴ π’—πŸ = 𝟎. πŸ‘πŸ•πŸ’π’ŽπŸ‘/π’Œπ’ˆ Substituting in eq. (4); 𝑉 𝑠 = 0.0462 Γ— 0.374 = πœ‹ 4 Γ— 𝐷2 Γ— 1.5𝐷 Γ— 900 60 Γ— 2 ∴ 𝑫 = 𝟎. πŸŽπŸ•πŸ– π’Ž 𝒂𝒏𝒅 𝑳 = 𝟎. πŸπŸπŸ– π’Ž (Since it is two-cylinder compressor)
A food storage required a refrigeration system of 12 tons of capacity at an evaporator temperature of -100C and condenser temperature of 250C. The refrigerant ammonia is sub-cooled by 50C before passing through the throttle valve. The vapor leaving the coil is 0.97 dry. Find the power required to drive the compressor if the actual COP is 60% of theoretical COP. Solution: Actual power required to drive the compressor is found by actual COP formula: πΆπ‘‚π‘ƒπ‘Žπ‘π‘‘π‘’π‘Žπ‘™ = π΄π‘π‘‘π‘’π‘Žπ‘™ π‘…π‘’π‘“π‘Ÿπ‘–π‘”π‘’π‘Ÿπ‘Žπ‘‘π‘–π‘œπ‘› 𝑒𝑓𝑓𝑒𝑐𝑑 π΄π‘π‘‘π‘’π‘Žπ‘™ π‘π‘œπ‘€π‘’π‘Ÿ π‘Ÿπ‘’π‘žπ‘’π‘–π‘Ÿπ‘’π‘‘ -----(1) Given actual COP=60% of theoretical COP; πΆπ‘‚π‘ƒπ‘Žπ‘π‘‘π‘’π‘Žπ‘™ = 0.60 Γ— πΆπ‘‚π‘ƒπ‘‡β„Žπ‘’π‘œπ‘Ÿπ‘’π‘‘π‘–π‘π‘Žπ‘™-------- (2) πΆπ‘‚π‘ƒπ‘‡β„Žπ‘’π‘œπ‘Ÿπ‘’π‘‘π‘–π‘π‘Žπ‘™ = β„Ž1βˆ’β„Ž4 β„Ž2βˆ’β„Ž1 --------(3)
To find β€˜β„Ž1’: Point no. β€˜1’ is in the wet vapor region on p-h plot. ∴ β„Ž1 = β„Žπ‘“ + π‘₯1β„Žπ‘“π‘”----- (4) In thermodynamics data handbook, refer Table-B1 for Thermodynamic properties of saturated Ammonia: At Temp=-10Β°C: β„Žπ‘“= 135.2 kJ/kg; β„Žπ‘“π‘”= 1296.8 kJ/kg Given The vapor leaving the coil is 0.97 dry. ∴ π‘₯1 = 0.97; substituting in eq. (4); β„Ž1 = 135.2 + 0.97 Γ— 1296.8 ∴ β„Ž1= 1393.1 π‘˜π½/π‘˜π‘”
To find β€˜β„Ž2’: Point β€˜2’ is in the superheated vapor region on p-h Plot. Hence, we do not get β„Ž2 directly from the table. ∴ β„Ž2 = β„Ž2β€² + 𝐢𝑃 π‘ π‘’π‘π‘’π‘Ÿβ„Žπ‘’π‘Žπ‘‘π‘’π‘‘ π‘£π‘Žπ‘π‘œπ‘Ÿ π‘Ÿπ‘’π‘“. 𝑇2 βˆ’ 𝑇2β€² ---(5) To find 𝑇2: Process 1-2 is isentropic; ∴ 𝑠1 = 𝑠2 Since we know β€˜π‘₯1’; 𝑠1= 𝑠𝑓 + π‘₯1𝑠𝑓𝑔 In thermodynamics data handbook, refer Table-B1 for Thermodynamic properties of saturated Ammonia: At Temp=-10Β°C: 𝑠𝑓= 0.5440 kJ/kg K; 𝑠𝑓𝑔= 4.9290 kJ/kg K; ∴ 𝑠1= 0.5440 + 0.97 Γ— 4.9290; ∴ π’”πŸ= πŸ“. πŸ‘πŸπŸ“ π’Œπ‘± π’Œπ’ˆπ‘² = π’”πŸ 𝑠2 = 𝑠2β€² + 𝐢𝑃(π‘ π‘’π‘π‘’π‘Ÿβ„Žπ‘’π‘Žπ‘‘π‘’π‘‘ π‘£π‘Žπ‘π‘œπ‘Ÿ π‘Ÿπ‘’π‘“.)𝑙𝑛 𝑇2 𝑇2β€²
In thermodynamics data handbook, refer Table-B1 for Thermodynamic properties of saturated Ammonia: At Temp=26Β°C:𝑠2β€² = 𝑠𝑔= 5.0244 kJ/kg K; In thermodynamics data handbook, refer Table-B9 for specific heat of superheated Ammonia vapor: At Pressure approx. 10.3 bar, CP=2.7167 kJ/kg K (From Table-B1 at Temp=26Β°C, Abs. Pressure=10.3397 bar) ∴ 𝐢𝑃(π‘ π‘’π‘π‘’π‘Ÿβ„Žπ‘’π‘Žπ‘‘π‘’π‘‘ π‘£π‘Žπ‘π‘œπ‘Ÿ π‘Ÿπ‘’π‘“.)= 2.7167 kJ/kg K ∴ 𝑠2= 5.325 = 5.0244 + 2.7167𝑙𝑛 𝑇2 273 + 26 ∴ π‘»πŸ = πŸ‘πŸ‘πŸ’ 𝑲 π’‰πŸβ€² = π’‰π’ˆ 𝒂𝒕 𝑻 = πŸπŸ”Β°π‘ͺ = πŸπŸ’πŸ”πŸ“. πŸ” π’Œπ‘±/π’Œπ’ˆ Substituting in eq. (5); β„Ž2 = β„Ž2β€² + 𝐢𝑃 π‘ π‘’π‘π‘’π‘Ÿβ„Žπ‘’π‘Žπ‘‘π‘’π‘‘ π‘£π‘Žπ‘π‘œπ‘Ÿ π‘Ÿπ‘’π‘“. 𝑇2 βˆ’ 𝑇2β€² β„Ž2 = 1465.6 + 2.7167 334 βˆ’ 299 ∴ π’‰πŸ= πŸπŸ“πŸ”πŸŽ. πŸ”πŸ– π’Œπ‘±/π’Œπ’ˆ
To find β€˜β„Ž4’: Point β€˜4’ is in the wet vapor region and the quality β€˜x4’ is not known. Hence, we cannot use β„Ž4 = β„Žπ‘“ + π‘₯4β„Žπ‘“π‘” WKT, π’‰πŸ‘ = π’‰πŸ’; Point β€˜3’ is in the sub-cooled liquid region and we cannot Obtain the value for β€˜3’ directly from the tables. Hence, Assume 3’ on the saturated liquid line as shown in the plot. π’‰πŸ‘ = π’‰πŸ‘β€² βˆ’ π‘ͺ𝑷 π‘³π’Šπ’’π’–π’Šπ’… 𝒓𝒆𝒇. π‘»πŸ‘β€² βˆ’ π‘»πŸ‘ βˆ’βˆ’βˆ’ βˆ’ πŸ” π’‰πŸ‘β€² = 𝒉𝒇 𝒂𝒕 𝑻 = πŸπŸ”Β°π‘ͺ = πŸ‘πŸŽπŸ‘. πŸ” π’Œπ‘±/π’Œπ’ˆ In thermodynamics data handbook, refer Table-B12 for specific heat of liquid Ammonia : CP(liquid)=4.69 kJ/kg K Given The refrigerant ammonia is sub-cooled by 50C ∴ π‘»πŸ‘β€² βˆ’ π‘»πŸ‘ = πŸ“Β°π‘ͺ Substituting in eq.(6): π’‰πŸ‘ = πŸ‘πŸŽπŸ‘. πŸ”πŸ” βˆ’ πŸ’. πŸ”πŸ— πŸ“ = πŸπŸ–πŸŽ. 𝟐𝟏 π’Œπ‘± π’Œπ’ˆ = π’‰πŸ’
Substituting in eq.(3): πΆπ‘‚π‘ƒπ‘‡β„Žπ‘’π‘œπ‘Ÿπ‘’π‘‘π‘–π‘π‘Žπ‘™ = 1393.1 βˆ’ 280.21 1560.68 βˆ’ 1393.1 ∴ π‘ͺπ‘Άπ‘·π‘»π’‰π’†π’π’“π’†π’•π’Šπ’„π’‚π’= πŸ”. πŸ”πŸ’ Substituting in eq.(2): π‘ͺ𝑢𝑷𝒂𝒄𝒕𝒖𝒂𝒍 = 𝟎. πŸ”πŸŽ Γ— πŸ”. πŸ”πŸ’ = πŸ‘. πŸ—πŸ–πŸ’ Substituting in eq.(1): 3.984 = 12 Γ— 3.51 π΄π‘π‘‘π‘’π‘Žπ‘™ π‘π‘œπ‘€π‘’π‘Ÿ π‘Ÿπ‘’π‘žπ‘’π‘–π‘Ÿπ‘’π‘‘ ∴ 𝑨𝒄𝒕𝒖𝒂𝒍 π’‘π’π’˜π’†π’“ π’“π’†π’’π’–π’Šπ’“π’†π’… = 𝟏𝟎. πŸ“πŸ• π’Œπ‘Ύ
An ideal vapor compression system with Freon -12 as the working fluid operates with an evaporator temperature of -300C and a condenser pressure of 9.61 bar. The fluid leaves the condenser sub-cooled by 130C. The compression is adiabatic with an efficiency of 75%. Calculate the COP of the machine. Solution: (a) COP 𝐢𝑂𝑃 = β„Ž1βˆ’β„Ž4 β„Ž2βˆ’β„Ž1 -------- (1) To find β€˜β„Ž1’: Assuming that point β€˜1’ is on saturated-vapor line: β„Ž1=β„Žπ‘” at -30Β°C; In thermodynamics data handbook, refer Table-B3 for Thermodynamic properties of saturated Freon-12: At Temp=-30Β°C:π’‰π’ˆ= 174.076 kJ/kg=π’‰πŸ
To find β€˜β„Ž2’: Point β€˜2’ is in the superheated vapor region and hence Cannot be found directly from the saturated tables. Given that the compressor is having an efficiency of 75%; We need to use the efficiency formula for compressor;; 𝜼π‘ͺπ’π’Žπ’‘π’“π’†π’”π’”π’π’“ = π’‰πŸβ€²βˆ’π’‰πŸ π’‰πŸβˆ’π’‰πŸ ------- (2) where 1-2’ represent ideal isentropic compression and 1-2 represent actual compression. To find β€˜π’‰πŸβ€²β€™, assume point 2’’ situated on the saturated vapour line such that π’‰πŸβ€²β€²=π’‰π’ˆ at 9.61 bar condenser pressure. In thermodynamics data handbook, refer Table-B3 for Thermodynamic properties of saturated Freon-12: At Pressure=9.61 bar :π’‰π’ˆ= 203.051 kJ/kg=π’‰πŸβ€²β€² and Temp. T=40Β°C= π‘»πŸβ€²β€². In thermodynamics data handbook, refer Table-B10 for specific heat of superheated Freon-12 vapor: At Pressure approx. 9.61 bar, CP=0.746 kJ/kg K
∴ π’‰πŸβ€² = π’‰πŸβ€²β€² + π‘ͺ𝑷(𝒔𝒖𝒑𝒆𝒓𝒉𝒆𝒂𝒕𝒆𝒅 𝒗𝒂𝒑𝒐𝒓 𝒓𝒆𝒇.) π‘»πŸβ€² βˆ’ π‘»πŸβ€²β€² π’‰πŸβ€² = πŸπŸŽπŸ‘. πŸŽπŸ“πŸ + 𝟎. πŸ•πŸ’πŸ” π‘»πŸβ€² βˆ’ πŸ‘πŸπŸ‘ ------(3) To find β€˜π‘»πŸβ€²β€™: Process 1-2’ is isentropic compression: hence, 𝑠1=𝑠2β€² =𝑠𝑔 at -30Β°C In thermodynamics data handbook, refer Table-B3 for Thermodynamic properties of saturated Freon-12: At Temp.=-30Β°C:π’”π’ˆ= 0.7165 kJ/kgK=π’”πŸ=π’”πŸβ€² At Pressure 9.61 bar: π’”π’ˆ= 0.6820 kJ/kgK=π’”πŸβ€²β€² And, 𝑠2β€² = 𝑠2β€²β€² + 𝐢𝑃(π‘ π‘’π‘π‘’π‘Ÿβ„Žπ‘’π‘Žπ‘‘π‘’π‘‘ π‘£π‘Žπ‘π‘œπ‘Ÿ π‘Ÿπ‘’π‘“.)𝑙𝑛 𝑇2β€² 𝑇2β€²β€² ∴ 0.7165 = 0.6820 + 0.746𝑙𝑛 𝑇2β€² 273+40 ; ∴ π‘»πŸβ€² = πŸ‘πŸπŸ•. πŸ–πŸ 𝐊 ; substitute in eq.(3); π’‰πŸβ€² = πŸπŸŽπŸ‘. πŸŽπŸ“πŸ + 𝟎. πŸ•πŸ’πŸ” πŸ‘πŸπŸ•. πŸ–πŸ βˆ’ πŸ‘πŸπŸ‘ ∴ π’‰πŸβ€²= πŸπŸπŸ’. 𝟏 π’Œπ‘±/π’Œπ’ˆ
Substituting in eq(2): 𝜼π‘ͺπ’π’Žπ’‘π’“π’†π’”π’”π’π’“ = π’‰πŸβ€²βˆ’π’‰πŸ π’‰πŸβˆ’π’‰πŸ ; 0.75 = 214.1βˆ’174.076 β„Ž2βˆ’174.076 ; ∴ π’‰πŸ= πŸπŸπŸ•. πŸ’πŸ’ π’Œπ‘±/π’Œπ’ˆ To find β€˜π’‰πŸ’β€™: WKT, π’‰πŸ‘ = π’‰πŸ’; Point β€˜3’ is in the sub-cooled liquid region and we cannot Obtain the value for β€˜3’ directly from the tables. Hence, Assume 3’ on the saturated liquid line as shown in the plot. π’‰πŸ‘ = π’‰πŸ‘β€² βˆ’ π‘ͺ𝑷 π‘³π’Šπ’’π’–π’Šπ’… 𝒓𝒆𝒇. π‘»πŸ‘β€² βˆ’ π‘»πŸ‘ βˆ’βˆ’βˆ’ βˆ’ πŸ’ π’‰πŸ‘β€² = 𝒉𝒇 𝒂𝒕 𝑷 = πŸ—. πŸ”πŸπ’ƒπ’‚π’“ = πŸ•πŸ’. πŸ“πŸπŸ• π’Œπ‘± π’Œπ’ˆ (π‘“π‘Ÿπ‘œπ‘š π‘‡π‘Žπ‘π‘™π‘’ 𝐡3) In thermodynamics data handbook, refer Table-B12, specific heat of liquid Freon-12 : CP(liquid)=0.94 kJ/kg K Given The refrigerant Freon-12 is sub-cooled by 130C ∴ π‘»πŸ‘β€² βˆ’ π‘»πŸ‘ = πŸπŸ‘Β°π‘ͺ Substituting in eq.(4): π’‰πŸ‘ = πŸ•πŸ’. πŸ“πŸπŸ• βˆ’ 𝟎. πŸ—πŸ’ πŸπŸ‘ = πŸ”πŸ. πŸ‘πŸ π’Œπ‘± π’Œπ’ˆ = π’‰πŸ’
Substituting in eq.(1); 𝐢𝑂𝑃 = β„Ž1 βˆ’ β„Ž4 β„Ž2 βˆ’ β„Ž1 𝐢𝑂𝑃 = 174.076 βˆ’ 62.31 227.44 βˆ’ 174.076 ∴ π‘ͺ𝑢𝑷 = 𝟐. πŸŽπŸ—
Refrigeration is desired at -100C. Cooling water having a mean temperature of 300C is available. Assume that a 50C difference is required for heat flow both in condenser and the evaporator. The refrigerant is Fr-12. The compressor has an isentropic efficiency of 75%, volumetric efficiency of 80% and the piston displacement is 6m3/min. Neglecting the other losses determine the (i) Tonnage of refrigeration (ii) Shaft power (iii) COP. Solution: (a) Tonnage of refrigeration WKT, 𝑇𝑅𝐸 = π‘šπ‘Ÿ Γ— β„Ž1 βˆ’ β„Ž4 in kW To express TRE in tons; we have, 1 ton=3.51 kW; ∴ 𝑇𝑅𝐸 𝑖𝑛 π‘‘π‘œπ‘›π‘  = π‘šπ‘Ÿ β„Ž1βˆ’β„Ž4 60Γ—3.51 -----(1) where π‘šπ‘Ÿ is mass of refrigerant in kg/s
To find β€˜β„Ž1’: Assuming that point β€˜1’ is on saturated-vapor line: β„Ž1=β„Žπ‘” at -30Β°C; In thermodynamics data handbook, refer Table-B3 for Thermodynamic properties of saturated Freon-12: At Temp=-15Β°C:π’‰π’ˆ= 180.946 kJ/kg=π’‰πŸ To find β€˜β„Ž2’: Point β€˜2’ is in the superheated vapor region and hence Cannot be found directly from the saturated tables. Given that the compressor is having an efficiency of 75%; We need to use the efficiency formula for compressor; 𝜼π‘ͺπ’π’Žπ’‘π’“π’†π’”π’”π’π’“ = π’‰πŸβ€²βˆ’π’‰πŸ π’‰πŸβˆ’π’‰πŸ ------- (2) where 1-2’ represent ideal isentropic compression and 1-2 represent actual compression.
To find β€˜π’‰πŸβ€²β€™, assume point 2’’ situated on the saturated vapour line such that π’‰πŸβ€²β€²=π’‰π’ˆ at 35Β°C condenser temperature. In thermodynamics data handbook, refer Table-B3 for Thermodynamic properties of saturated Freon-12: At Temp.=35Β°C :π’‰π’ˆ= 201.299 kJ/kg=π’‰πŸβ€²β€² and Pressure =8.477 bar. In thermodynamics data handbook, refer Table-B10 for specific heat of superheated Freon-12 vapor: At Pressure approx. 8.477 bar, CP=0.733 kJ/kg K ∴ π’‰πŸβ€² = π’‰πŸβ€²β€² + π‘ͺ𝑷(𝒔𝒖𝒑𝒆𝒓𝒉𝒆𝒂𝒕𝒆𝒅 𝒗𝒂𝒑𝒐𝒓 𝒓𝒆𝒇.) π‘»πŸβ€² βˆ’ π‘»πŸβ€²β€² π’‰πŸβ€² = 𝟐𝟎𝟏. πŸπŸ—πŸ— + 𝟎. πŸ•πŸ‘πŸ‘ π‘»πŸβ€² βˆ’ πŸπŸ•πŸ‘ + πŸ‘πŸ“ ------(3) To find β€˜π‘»πŸβ€²β€™: Process 1-2’ is isentropic compression: hence, 𝑠1=𝑠2β€² =𝑠𝑔 at -15Β°C
In thermodynamics data handbook, refer Table-B3 for Thermodynamic properties of saturated Freon-12: At Temp.= -15Β°C:π’”π’ˆ= 0.7046 kJ/kgK=π’”πŸ=π’”πŸβ€². Also, At Temp.= 35Β°C:π’”π’ˆ= 0.6834 kJ/kgK=π’”πŸβ€²β€² And, 𝑠2β€² = 𝑠2β€²β€² + 𝐢𝑃(π‘ π‘’π‘π‘’π‘Ÿβ„Žπ‘’π‘Žπ‘‘π‘’π‘‘ π‘£π‘Žπ‘π‘œπ‘Ÿ π‘Ÿπ‘’π‘“.)𝑙𝑛 𝑇2β€² 𝑇2β€²β€² ∴ 0.7046 = 0.6834 + 0.733𝑙𝑛 𝑇2β€² 273+35 ; ∴ π‘»πŸβ€² = πŸ‘πŸπŸ• 𝐊 ; substitute in eq.(3); π’‰πŸβ€² = 𝟐𝟎𝟏. πŸπŸ—πŸ— + 𝟎. πŸ•πŸ‘πŸ‘ πŸ‘πŸπŸ• βˆ’ πŸπŸ•πŸ‘ + πŸ‘πŸ“ ∴ π’‰πŸβ€²= πŸπŸŽπŸ•. πŸ— π’Œπ‘±/π’Œπ’ˆ Substituting in eq(2): 𝜼π‘ͺπ’π’Žπ’‘π’“π’†π’”π’”π’π’“ = π’‰πŸβ€²βˆ’π’‰πŸ π’‰πŸβˆ’π’‰πŸ ; 0.75 = 207.9βˆ’180.946 β„Ž2βˆ’180.946 ; ∴ π’‰πŸ= πŸπŸπŸ”. πŸ–πŸ– π’Œπ‘±/π’Œπ’ˆ
To find β€˜β„Ž4’: Process 3-4 is isenthalpic process. Hence, β„Ž3=β„Ž4 . π’‰πŸ‘= 𝒉𝒇 at T=35Β°C In thermodynamics data handbook, refer Table-B3 for Thermodynamic properties of saturated Freon-12: At Temp.=35Β°C :𝒉𝒇= 69.494 kJ/kg=π’‰πŸ‘=π’‰πŸ’ To find β€˜π’Žπ’“β€™: Mass of refrigerant circulated is found by the actual swept-volume of the compressor β€œπ‘‰ π‘Žπ‘ β€. Given volumetric efficiency=80% and 𝑉 𝑠= 6π‘š3 /π‘šπ‘–π‘› πœΌπ’—π’π’. = 𝑽𝒂𝒔 𝑽𝒔 ; i.e., 𝟎. πŸ– = 𝑽𝒂𝒔 πŸ” ; ∴ 𝑽𝒂𝒔 = 𝟎. πŸ– Γ— πŸ” = πŸ’. πŸ–π’ŽπŸ‘/π’Žπ’Šπ’ But, 𝑽𝒂𝒔 = π’Žπ’“ Γ— π’—πŸ where π’—πŸ is the specific volume at compressor inlet π’—πŸ=π’—π’ˆ at T= -15Β°C. ∴ π’—πŸ = π’—π’ˆ = 𝟎. πŸŽπŸ—πŸπŸŽπŸπŸ– π’ŽπŸ‘ /π’Œπ’ˆ (From Table B3)
∴ 𝑽𝒂𝒔= π’Žπ’“ Γ— π’—πŸ πŸ’. πŸ– π’ŽπŸ‘/π’Žπ’Šπ’ = π’Žπ’“ Γ— 𝟎. πŸŽπŸ—πŸπŸŽπŸπŸ–π’ŽπŸ‘/π’Œπ’ˆ ∴ π’Žπ’“= πŸ“πŸ. πŸ• π’Œπ’ˆ/π’Žπ’Šπ’ Substituting in eq. (1) 𝑇𝑅𝐸 𝑖𝑛 π‘‘π‘œπ‘›π‘  = π‘šπ‘Ÿ β„Ž1 βˆ’ β„Ž4 60 Γ— 3.51 𝑇𝑅𝐸 𝑖𝑛 π‘‘π‘œπ‘›π‘  = 52.7 180.946 βˆ’ 69.494 60 Γ— 3.51 𝑻𝑹𝑬 π’Šπ’ 𝒕𝒐𝒏𝒔 = πŸπŸ•. πŸ— 𝒕𝒐𝒏
(b) Shaft power From compressor work formula, Shaft power= power required to drive the compressor ∴ 𝑺𝒉𝒂𝒇𝒕 π’‘π’π’˜π’†π’“ = π’Žπ’“ πŸ”πŸŽ π’‰πŸ βˆ’ π’‰πŸ π’Œπ‘Ύ 𝑺𝒉𝒂𝒇𝒕 π’‘π’π’˜π’†π’“ = πŸ“πŸ. πŸ• πŸ”πŸŽ πŸπŸπŸ”. πŸ–πŸ– βˆ’ πŸπŸ–πŸŽ. πŸ—πŸ’πŸ” π’Œπ‘Ύ ∴ 𝑺𝒉𝒂𝒇𝒕 π’‘π’π’˜π’†π’“ = πŸ‘πŸ. πŸ“πŸ” π’Œπ‘Ύ (C) COP 𝐢𝑂𝑃 = β„Ž1 βˆ’ β„Ž4 β„Ž2 βˆ’ β„Ž1 𝐢𝑂𝑃 = 180.946 βˆ’ 69.494 216.88 βˆ’ 180.946 ∴ π‘ͺ𝑢𝑷 = πŸ‘. 𝟏𝟎

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REFRIGERATION PROBLEMS Anu.pptx .

  • 2.
  • 3.
  • 4. TABLE B-1 Thermodynamic Properties of Saturated Ammonia
  • 5. An open air refrigeration system operating between pressures of 16 bar and 1 bar, required to produce 33.5 kW of refrigeration. The temperature of air leaving the refrigerated room is -50C and that leaving the air cooler is 300C. Assuming no losses and clearances, calculate for the theoretical cycle (a) mass flow rate of air circulated per minute (b) Piston displacement of the compressor and expender (c) Net work (d) COP Solution: Given: 𝑃2 𝑃1 = π‘Ÿπ‘ = 16 Total refrigeration effect produced=33.5 kW T1=273-5=268K T3=30+273=303K
  • 6. (a) mass flow rate of air circulated per minute Total refrigeration effect=mCp(T1-T4) 33.5 = π‘š Γ— 1.005 Γ— 268 βˆ’ 𝑇4 ----------(1) Processes 1-2 and 3-4 are isentropic : Using p-v-t relationships for isentropic process; 𝑇2 𝑇1 = 𝑃2 𝑃1 π›Ύβˆ’1 𝛾 ∴ π‘»πŸ = πŸπŸ”πŸ– Γ— πŸπŸ” 𝟎.πŸ’ 𝟏.πŸ’ = πŸ“πŸ—πŸ. πŸ–π‘² Similarly, 𝑇3 𝑇4 = 𝑃3 𝑃4 π›Ύβˆ’1 𝛾 = 𝑃2 𝑃1 π›Ύβˆ’1 𝛾 ∴ π‘»πŸ’ = πŸ‘πŸŽπŸ‘ Γ· πŸπŸ” 𝟎.πŸ’ 𝟏.πŸ’ = πŸ“πŸ—πŸ. πŸ–π‘² = πŸπŸ‘πŸ•. πŸπŸπ‘²
  • 7. Substituting T4 in equation (1): 33.5 = π‘š Γ— 1.005 Γ— 268 βˆ’ 137.22 π‘š = 33.5 131.436 Γ— 60 = 15.3π‘˜π‘”/π‘šπ‘–π‘› (b) Piston displacement of the compressor and expender Compression process is 1-2: ∴ π‘ƒπ‘–π‘ π‘‘π‘œπ‘› π‘‘π‘–π‘ π‘π‘™π‘Žπ‘π‘’π‘šπ‘’π‘›π‘‘ π‘œπ‘“ π‘π‘œπ‘šπ‘π‘Ÿπ‘’π‘ π‘ π‘œπ‘Ÿ, 𝑉1 = π‘šπ‘…π‘‡1 𝑃1 𝑉1 = 15.3 Γ— 0.287 Γ— 268 1 Γ— 100 = 11.768π‘š3/π‘šπ‘–π‘› π‘˜π‘”/π‘šπ‘–π‘› Γ— π‘˜π½/π‘˜π‘”πΎ Γ— 𝐾 π‘˜π‘/π‘š2 = π‘˜π‘” Γ— π‘˜π‘ βˆ’ π‘š Γ— 𝐾 Γ— π‘š2 π‘˜π‘ Γ— π‘šπ‘–π‘› Γ— π‘˜π‘”πΎ
  • 8. Expansion process is 3-4: ∴ π‘ƒπ‘–π‘ π‘‘π‘œπ‘› π‘‘π‘–π‘ π‘π‘™π‘Žπ‘π‘’π‘šπ‘’π‘›π‘‘ π‘œπ‘“ 𝑒π‘₯π‘π‘Žπ‘›π‘‘π‘’π‘Ÿ, 𝑉4 = π‘šπ‘…π‘‡4 𝑃4 = π‘šπ‘…π‘‡4 𝑃1 𝑉4 = 15.3 Γ— 0.287 Γ— 137.22 1 Γ— 100 = 6.025π‘š3/π‘šπ‘–π‘› π‘˜π‘”/π‘šπ‘–π‘› Γ— π‘˜π½/π‘˜π‘”πΎ Γ— 𝐾 π‘˜π‘/π‘š2 = π‘˜π‘” Γ— π‘˜π‘ βˆ’ π‘š Γ— 𝐾 Γ— π‘š2 π‘˜π‘ Γ— π‘šπ‘–π‘› Γ— π‘˜π‘”πΎ (c) Net work 𝑁𝑒𝑑 π‘€π‘œπ‘Ÿπ‘˜ = π‘š π‘€π‘π‘œπ‘šπ‘. βˆ’ 𝑀𝑒π‘₯π‘π‘Žπ‘›. 𝑁𝑒𝑑 π‘€π‘œπ‘Ÿπ‘˜ = π‘šπΆπ‘ƒ 𝑇2 βˆ’ 𝑇1 βˆ’ 𝑇3 βˆ’ 𝑇4 𝑁𝑒𝑑 π‘€π‘œπ‘Ÿπ‘˜ = 15.3 60 Γ— 1.005 591.8 βˆ’ 268 βˆ’ 303 βˆ’ 137.22 ∴ 𝑡𝒆𝒕 π’˜π’π’“π’Œ = πŸ’πŸŽ. πŸ“π’Œπ‘Ύ
  • 9. (d) COP π‘ͺ𝑢𝑷 = 𝑻𝒐𝒕𝒂𝒍 π’“π’†π’‡π’“π’Šπ’ˆπ’†π’“π’‚π’•π’Šπ’π’ 𝒆𝒇𝒇𝒆𝒄𝒕 𝑡𝒆𝒕 π’˜π’π’“π’Œ = πŸ‘πŸ‘. πŸ“ π’Œπ‘Ύ πŸ’πŸŽ. πŸ“ π’Œπ‘Ύ = 𝟎. πŸ–πŸπŸ•
  • 10. A dense air machine operates between 17 bar and 3.4 bar. The temperature of air after the cooler is 150C and after the refrigerating coil is 60. Determine the (a)Air circulation per minute (b) work of the compressor and the expander/TR (c) Theoretical COP and (d) HP/TR Solution: Given: 𝑃2 𝑃1 = π‘Ÿπ‘ = 17 3.4 = 5 T1=273+6=279K T3=273+15=288K
  • 11. (a) air circulated per minute Total refrigeration effect=mCp(T1-T4) 𝑇𝑅𝐸 = π‘š Γ— 1.005 Γ— 279 βˆ’ 𝑇4 ----------(1) Processes 1-2 and 3-4 are isentropic : Using p-v-t relationships for isentropic process; 𝑇2 𝑇1 = 𝑃2 𝑃1 π›Ύβˆ’1 𝛾 ∴ π‘»πŸ = πŸπŸ•πŸ— Γ— πŸ“ 𝟎.πŸ’ 𝟏.πŸ’ = πŸ’πŸ’πŸ. πŸ–πŸ–π‘² Similarly, 𝑇3 𝑇4 = 𝑃3 𝑃4 π›Ύβˆ’1 𝛾 = 𝑃2 𝑃1 π›Ύβˆ’1 𝛾 ∴ π‘»πŸ’ = πŸπŸ–πŸ– Γ· πŸ“ 𝟎.πŸ’ 𝟏.πŸ’ = πŸ“πŸ—πŸ. πŸ–π‘² = πŸπŸ–πŸ. πŸ–πŸ‘π‘²
  • 12. π΄π‘ π‘ π‘’π‘šπ‘–π‘›π‘” 𝑇𝑅𝐸 π‘“π‘œπ‘Ÿ 1 π‘‘π‘œπ‘› π‘œπ‘“ π‘Ÿπ‘’π‘“π‘Ÿπ‘–π‘”π‘’π‘Ÿπ‘Žπ‘‘π‘–π‘œπ‘› = 3.51 π‘˜π‘Š π‘š = 𝑇𝑅𝐸 𝐢𝑃 𝑇1 βˆ’ 𝑇4 = 3.51 Γ— 60 1.005 279 βˆ’ 181.83 ∴ π’Ž = 𝟐. πŸπŸ“πŸ”π’Œπ’ˆ/π’Žπ’Šπ’ (b) work of compressor and the expander/TR π‘ŠπΆπ‘œπ‘šπ‘. = π‘š Γ— 𝐢𝑃 Γ— 𝑇2 βˆ’ 𝑇1 = 2.156 60 Γ— 1.005 441.88 βˆ’ 279 ∴ 𝑾π‘ͺπ’π’Žπ’‘.= πŸ“. πŸ–πŸ– π’Œπ‘Ύ π‘Š 𝑒π‘₯π‘π‘Žπ‘›. = π‘š Γ— 𝐢𝑃 Γ— 𝑇3 βˆ’ 𝑇4 = 2.156 60 Γ— 1.005 288 βˆ’ 181.83 ∴ 𝑾𝒆𝒙𝒑𝒂𝒏.= πŸ‘. πŸ–πŸ‘πŸ’ π’Œπ‘Ύ (c) Theoretical COP 𝐢𝑂𝑃 = 𝑇𝑅𝐸 𝑁𝑒𝑑 π‘€π‘œπ‘Ÿπ‘˜ = 3.51 π‘˜π‘Š 5.88 βˆ’ 3.834 π‘˜π‘Š ∴ π‘ͺ𝑢𝑷 = 𝟏. πŸ•πŸ (d) HP/TR π‘·π’π’˜π’†π’“/𝒕𝒐𝒏 = π‘Š (π‘π‘œπ‘šπ‘. ) βˆ’ π‘Š (𝑒π‘₯π‘π‘Žπ‘›. ) = 5.88 βˆ’ 3.884 = 𝟐. πŸŽπŸ’πŸ” 𝐀𝐖
  • 13. An air refrigerating machine working on Bell-Coleman cycle takes in air into the compressor at 1 bar and -50C. It is compressed in the compressor to 5 bar and is cooled to 250C at constant pressure. In the expander it is expanded to 1bar. The isentropic efficiency of the compressor and the expander are 85% and 90% respectively. Calculate : (a) Refrigerating capacity of the system if the mass flow rate is 30kg/m (b) Power required to run the compressor (c) COP Solution: Given: P1=P4=P4’= 1 bar, P2’=P2=P3=5 bar T1=273-5=268K T3=273+25=298K Ξ·c=85% and Ξ·e=90% m= 30 kg/min
  • 14. (a) Refrigerating capacity of the system if the mass flow rate is 30kg/min 𝑇𝑅𝐸 = π‘š Γ— 𝐢𝑃 Γ— 𝑇1 βˆ’ 𝑇4 𝑇𝑅𝐸 = 30 60 Γ— 1.005 Γ— 268 βˆ’ 𝑇4 ---------(1) Processes 1-2’ and 3-4’ are isentropic : Using p-v-t relationships for isentropic process; 𝑇2β€² 𝑇1 = 𝑃2β€² 𝑃1 π›Ύβˆ’1 𝛾 ∴ π‘»πŸβ€² = πŸπŸ”πŸ– Γ— πŸ“ 𝟎.πŸ’ 𝟏.πŸ’ = πŸ’πŸπŸ’. πŸ“π‘² Similarly, 𝑇3 𝑇4β€² = 𝑃3 𝑃4β€² π›Ύβˆ’1 𝛾 = 𝑃2β€² 𝑃1 π›Ύβˆ’1 𝛾 ∴ π‘»πŸ’β€² = πŸπŸ—πŸ– Γ· πŸ“ 𝟎.πŸ’ 𝟏.πŸ’ = πŸ“πŸ—πŸ. πŸ–π‘² = πŸπŸ–πŸ–. πŸπŸ”π‘²
  • 15. Using isentropic efficiency formula for compressor and expander: πœ‚πΆ = 𝑇2β€² βˆ’ 𝑇1 𝑇2 βˆ’ 𝑇1 0.85 = 424.5 βˆ’ 268 𝑇2 βˆ’ 268 ∴ π‘»πŸ = πŸ’πŸ“πŸ. πŸπ‘² πœ‚πΈ = 𝑇3 βˆ’ 𝑇4 𝑇3 βˆ’ 𝑇4β€² 0.90 = 298 βˆ’ 𝑇4 298 βˆ’ 188.16 ∴ π‘»πŸ’ = πŸπŸ—πŸ—. πŸπŸ“π‘² substituting π‘»πŸ’ in equation (1): 𝑇𝑅𝐸 = 30 60 Γ— 1.005 Γ— 268 βˆ’ 199.15 = πŸ‘πŸ’. πŸ” 𝐀𝐖
  • 16. Refrigeration capacity in terms of ton is obtained as below: 1 ton= 3.51 kW ∴ 34.6 Γ· 3.51 = πŸ—. πŸ–πŸ” 𝒕𝒐𝒏 (b) Power required to run the compressor π‘ƒπ‘œπ‘€π‘’π‘Ÿ = π‘€π‘œπ‘Ÿπ‘˜ π‘‘π‘œπ‘›π‘’ 𝑖𝑛 π‘˜π‘Š = π‘Š π‘π‘œπ‘šπ‘π‘Ÿπ‘’π‘ π‘ π‘œπ‘Ÿ βˆ’ π‘Šπ‘’π‘₯π‘π‘Žπ‘›π‘‘π‘’π‘Ÿ π‘ƒπ‘œπ‘€π‘’π‘Ÿ = π‘šπΆπ‘ƒ 𝑇2 βˆ’ 𝑇1 βˆ’ 𝑇3 βˆ’ 𝑇4 π‘ƒπ‘œπ‘€π‘’π‘Ÿ = 30 60 Γ— 1.005 452.2 βˆ’ 268 βˆ’ 298 βˆ’ 199.15 ∴ π‘·π’π’˜π’†π’“ = πŸ’πŸ. πŸ–πŸ— π’Œπ‘Ύ (c) COP 𝐢𝑂𝑃 = 𝑇𝑅𝐸 𝑁𝑒𝑑 π‘€π‘œπ‘Ÿπ‘˜ = 34.6 π‘˜π‘Š 42.89 π‘˜π‘Š ∴ π‘ͺ𝑢𝑷 = 𝟎. πŸ–πŸ
  • 17. An ideal vapor compression refrigeration system operates on Freon- 12 between the temperature limits of 300C and -100C without undercooling. Vapor leaving the compressor is dry saturated. Determine (a) the power input to the compressor if the refrigerating capacity of the unit is 1.5tons (b) COP (c) Condenser duty Solution: (a) the power input to the compressor if the refrigerating capacity of the unit is 1.5tons π‘ƒπ‘œπ‘€π‘’π‘Ÿ = π‘šπ‘Ÿ β„Ž2 βˆ’ β„Ž1 Mass of refrigerant is calculate from TRE equation: 𝑇𝑅𝐸 = π‘šπ‘Ÿ Γ— 𝐢𝑃 Γ— 𝑇1 βˆ’ 𝑇4 𝑇𝑅𝐸 = π‘šπ‘Ÿ Γ— β„Ž1 βˆ’ β„Ž4 1.5 Γ— 3.51 = π‘šπ‘Ÿ Γ— β„Ž1 βˆ’ β„Ž4 -------(1)
  • 18. To calculate β€˜β„Ž1’: Point 1 on p-h diagram is within the vapor dome: This means at 1, the refrigerant is wet vapor of quality β€˜π‘₯1’ ∴ β„Ž1 = β„Žπ‘“ + π‘₯1β„Žπ‘“π‘” at T=-10Β°C In thermodynamics data handbook, refer Table-B3 for Thermodynamic properties of saturated Freon-12: At Temp=-100C , 𝒉𝒇= 26.851 kJ/kg and π’‰π’‡π’ˆ= 156.21 kJ/kg ∴ β„Ž1 = 26.851 + π‘₯1156.21 at T=-10Β°C---------(2) To find β€˜π‘₯1’ : Process 1-2 is isentropic compression: ∴ 𝑠1 = 𝑠2 and 𝑠2 = 𝑠𝑔 at T=30Β°C. π΄π‘™π‘ π‘œ, 𝑠2 = 𝑠1 = 𝑠𝑓 + π‘₯1𝑠𝑓𝑔 at T=-10Β°C---------(3)
  • 19. In thermodynamics data handbook, refer Table-B3 for Thermodynamic properties of saturated Freon-12: At Temp=300C , π’”π’ˆ= 0.6848 kJ/kg K= π’”πŸ. Substituting 𝑠2 in equation (3); 𝑠2 = 𝑠1 = 0.6848 = 𝑠𝑓 + π‘₯1𝑠𝑓𝑔 at T=-10Β°C-------(4) In thermodynamics data handbook, refer Table-B3 for Thermodynamic properties of saturated Freon-12: At Temp=-100C , 𝒔𝒇= 0.1079 kJ/kg K and π’”π’‡π’ˆ= 0.5936 kJ/kg K. Substituting these in equation (4); 0.6848 = 𝑠𝑓 + π‘₯1𝑠𝑓𝑔 at T=-10Β°C 0.6848 = 0.1079 + π‘₯10.5936 ∴ π’™πŸ = 𝟎. πŸ—πŸ•πŸπŸ–. Substitute π’™πŸ in equation (2) β„Ž1 = 26.851 + π‘₯1156.21 β„Ž1 = 26.851 + 0.9718 Γ— 156.21 π’‰πŸ = πŸπŸ•πŸ–. πŸ”πŸ” π’Œπ‘±/π’Œπ’ˆ
  • 20. π’‰πŸ‘= π’‰πŸ’. Because, process 3-4 is isenthalpic process And π’‰πŸ‘= 𝒉𝒇 at T=30Β°C. In thermodynamics data handbook, refer Table-B3 for Thermodynamic properties of saturated Freon-12: At Temp=300C , 𝒉𝒇= 64.539 kJ/kg ∴ π’‰πŸ‘ = πŸ”πŸ’. πŸ“πŸ‘πŸ— π’Œπ‘± π’Œπ’ˆ = π’‰πŸ’ Substituting the values in equation (1) 1.5 Γ— 3.51 = π‘šπ‘Ÿ Γ— 178.66 βˆ’ 64.539 π’Žπ’“ = 𝟎. πŸŽπŸ’πŸ” π’Œπ’ˆ/𝒔 π‘ƒπ‘œπ‘€π‘’π‘Ÿ = π‘šπ‘Ÿ β„Ž2 βˆ’ β„Ž1 π‘ƒπ‘œπ‘€π‘’π‘Ÿ = 0.046 β„Ž2 βˆ’ 178.66 ------ (5); π’‰πŸ= π’‰π’ˆ at T=30Β°C. In thermodynamics data handbook, refer Table-B3 for Thermodynamic properties of saturated Freon-12: At Temp=300C , π’‰π’ˆ= 199.475 kJ/kg. substituting in eq. (5) π‘ƒπ‘œπ‘€π‘’π‘Ÿ = 0.046 199.475 βˆ’ 178.66 π‘·π’π’˜π’†π’“ = 𝟎. πŸ—πŸ” π’Œπ‘Ύ
  • 21. (b) COP 𝐢𝑂𝑃 = 𝑇𝑅𝐸 𝑁𝑒𝑑 π‘€π‘œπ‘Ÿπ‘˜ = β„Ž1 βˆ’ β„Ž4 β„Ž2 βˆ’ β„Ž1 𝐢𝑂𝑃 = 178.66 βˆ’ 64.539 199.475 βˆ’ 178.66 ; ∴ π‘ͺ𝑢𝑷 = πŸ“. πŸ’πŸ– (c) Condenser duty πΆπ‘œπ‘›π‘‘π‘’π‘›π‘ π‘’π‘Ÿ 𝑑𝑒𝑑𝑦 = π‘šπ‘Ÿ Γ— β„Ž2 βˆ’ β„Ž3 πΆπ‘œπ‘›π‘‘π‘’π‘›π‘ π‘’π‘Ÿ 𝑑𝑒𝑑𝑦 = 0.046 Γ— 199.475 βˆ’ 64.539 ∴ π‘ͺ𝒐𝒏𝒅𝒆𝒏𝒔𝒆𝒓 π’…π’–π’•π’š = πŸ”. 𝟐 π’Œπ‘Ύ
  • 22. An ammonia plant having a capacity of 15 tons is working at an evaporator temperature of -60C and a condenser temperature of 300C. The refrigerant is superheated to 50C before entering the compressor. A two cylinder single acting compressor is used which runs at 900rpm. Determine (i) COP (ii) Mass of ammonia (iii) Bore and stroke of the cylinder assuming that L= 1.5xD and neglect the clearance. Solution: (a) COP 𝐢𝑂𝑃 = 𝑇𝑅𝐸 𝑁𝑒𝑑 π‘€π‘œπ‘Ÿπ‘˜ = β„Ž1βˆ’β„Ž4 β„Ž2βˆ’β„Ž1 ---------(1) β„Ž1 = β„Ž1β€² + 𝐢𝑃(π‘ π‘’π‘π‘’π‘Ÿβ„Žπ‘’π‘Žπ‘‘π‘’π‘‘ π‘£π‘Žπ‘π‘œπ‘Ÿ π‘Ÿπ‘’π‘“.) 𝑇1 βˆ’ 𝑇1β€² β„Ž1β€² = β„Žπ‘” at T= -60C In thermodynamics data handbook, refer Table-B1 for Thermodynamic properties of saturated Ammonia: At Temp=-6Β°C , β„Ž1β€² = β„Žπ‘”= 1436.8 kJ/kg;
  • 23. In thermodynamics data handbook, refer Table-B9 for specific heat of superheated Ammonia vapor: At Pressure 3.4 bar, CP=2.3818 kJ/kg K (From Table-B1 at Temp=-6Β°C, Abs. Pressure=3.4125 bar) ∴ 𝐢𝑃(π‘ π‘’π‘π‘’π‘Ÿβ„Žπ‘’π‘Žπ‘‘π‘’π‘‘ π‘£π‘Žπ‘π‘œπ‘Ÿ π‘Ÿπ‘’π‘“.)= 2.3818 kJ/kg K Given that the refrigerant is superheated to 50C before entering the compressor; ∴ 𝑇1 = 5°𝐢 π‘Žπ‘›π‘‘ 𝑇1β€² = βˆ’ 5°𝐢, π‘€β„Žπ‘–π‘β„Ž 𝑖𝑠 π‘‘β„Žπ‘’ π‘’π‘£π‘Žπ‘π‘œπ‘Ÿπ‘Žπ‘‘π‘œπ‘Ÿ π‘‘π‘’π‘šπ‘. 𝑆𝑒𝑏𝑠𝑑𝑖𝑑𝑒𝑑𝑖𝑛𝑔 𝑖𝑛: β„Ž1 = β„Ž1β€² + 𝐢𝑃(π‘ π‘’π‘π‘’π‘Ÿβ„Žπ‘’π‘Žπ‘‘π‘’π‘‘ π‘£π‘Žπ‘π‘œπ‘Ÿ π‘Ÿπ‘’π‘“.) 𝑇1 βˆ’ 𝑇1β€² β„Ž1 = 1436.8 + 2.3818 5 βˆ’ βˆ’5 π’‰πŸ = πŸπŸ’πŸ”πŸŽ. πŸ”πŸπŸ– π’Œπ‘±/π’Œπ’ˆ π’‰πŸ‘= π’‰πŸ’. Because, process 3-4 is isenthalpic process And π’‰πŸ‘= 𝒉𝒇at T=30Β°C. In thermodynamics data handbook, refer Table-B1 for Thermodynamic properties of saturated Ammonia: At Temp=30Β°C , 𝒉𝒇 = 322.9 kJ/kg ∴ π’‰πŸ‘= π’‰πŸ’=322.9 kJ/kg
  • 24. To find β„Ž2: Point β€˜2’ is in the superheated vapor region and We cannot find its value directly from the saturated tables. ∴ β„Ž2 = β„Ž2β€² + 𝐢𝑃 π‘ π‘’π‘π‘’π‘Ÿβ„Žπ‘’π‘Žπ‘‘π‘’π‘‘ π‘£π‘Žπ‘π‘œπ‘Ÿ π‘Ÿπ‘’π‘“. 𝑇2 βˆ’ 𝑇2β€² βˆ’ βˆ’(2) To find β€˜π‘‡2’: We know that, process 1-2 is isentropic compression ∴ 𝑠1 = 𝑠2; But, 𝑠1 is also in the superheated vapour region. ∴ 𝑠1 = 𝑠1β€² + 𝐢𝑃(π‘ π‘’π‘π‘’π‘Ÿβ„Žπ‘’π‘Žπ‘‘π‘’π‘‘ π‘£π‘Žπ‘π‘œπ‘Ÿ π‘Ÿπ‘’π‘“.)𝑙𝑛 𝑇1 𝑇1β€² ----------(3) i.e., using In thermodynamics data handbook, refer Table-B1 for Thermodynamic properties of saturated Ammonia: At Temp=-6Β°C , 𝑠1β€² = 𝑠𝑔= 5.4173 kJ/kg K; substituting in eq. (3) ∴ 𝑠1 = 5.4173 + 2.3818𝑙𝑛 273+5 273βˆ’5 ; ∴ 𝑠1 = 5.504 π‘˜π½ π‘˜π‘”K = 𝑠2
  • 25. Similarly, 𝑠1 = 𝑠2 = 𝑠2β€² + 𝐢𝑃(π‘ π‘’π‘π‘’π‘Ÿβ„Žπ‘’π‘Žπ‘‘π‘’π‘‘ π‘£π‘Žπ‘π‘œπ‘Ÿ π‘Ÿπ‘’π‘“.)𝑙𝑛 𝑇2 𝑇2β€² 𝑖. 𝑒. , 5.504 = 𝑠2β€² + 𝐢𝑃(π‘ π‘’π‘π‘’π‘Ÿβ„Žπ‘’π‘Žπ‘‘π‘’π‘‘ π‘£π‘Žπ‘π‘œπ‘Ÿ π‘Ÿπ‘’π‘“.)𝑙𝑛 𝑇2 𝑇2β€² In thermodynamics data handbook, refer Table-B1 for Thermodynamic properties of saturated Ammonia: At Temp=30Β°C , 𝑠2β€² = 𝑠𝑔= 4.9805 kJ/kg K; ∴ 5.504 = 4.9805 + 2.3818𝑙𝑛 𝑇2 303 ; ∴ 𝑇2 = 377.48𝐾 Substituting in eq.(2); β„Ž2 = β„Ž2β€² + 𝐢𝑃 π‘ π‘’π‘π‘’π‘Ÿβ„Žπ‘’π‘Žπ‘‘π‘’π‘‘ π‘£π‘Žπ‘π‘œπ‘Ÿ π‘Ÿπ‘’π‘“. 𝑇2 βˆ’ 𝑇2β€² In thermodynamics data handbook, refer Table-B1 for Thermodynamic properties of saturated Ammonia: At Temp=30Β°C , β„Ž2β€² = β„Žπ‘”= 1467.9 kJ/kg; 𝑇2β€²= 273+30=303 K
  • 26. β„Ž2 = 1467.9 + 2.3818 377.48 βˆ’ 303 ∴ π’‰πŸ= πŸπŸ”πŸ’πŸ“. πŸ‘ 𝐀𝐉/𝐀𝐠 Substituting in eq.(1): 𝐢𝑂𝑃 = β„Ž1βˆ’β„Ž4 β„Ž2βˆ’β„Ž1 = πŸπŸ’πŸ”πŸŽ.πŸ”πŸπŸ–βˆ’πŸ‘πŸπŸ.πŸ— πŸπŸ”πŸ’πŸ“.πŸ‘βˆ’πŸπŸ’πŸ”πŸŽ.πŸ”πŸπŸ– ∴ π‘ͺ𝑢𝑷 = πŸ”. πŸπŸ” (b) Mass of ammonia Using TRE equation: 𝑇𝑅𝐸 = π‘šπ‘Ÿ Γ— 𝐢𝑃 Γ— 𝑇1 βˆ’ 𝑇4 𝑇𝑅𝐸 = π‘šπ‘Ÿ Γ— β„Ž1 βˆ’ β„Ž4 15 Γ— 3.51 = π‘šπ‘Ÿ Γ— 1460.618 βˆ’ 322.9 ∴ π’Žπ’“= 𝟎. πŸŽπŸ’πŸ”πŸ π’Œπ’ˆ/𝒔
  • 27. (c) Bore and stroke of the cylinder assuming that L= 1.5xD In order to find cylinder dimensions, we need to evaluate the stroke volume of the compressor. 𝑉 𝑠 = π‘šπ‘Ÿ Γ— 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐 π‘£π‘œπ‘™π‘’π‘šπ‘’ π‘Žπ‘‘ π‘‘β„Žπ‘’ π‘π‘œπ‘šπ‘π‘Ÿπ‘’π‘ π‘ π‘œπ‘Ÿ 𝑖𝑛𝑙𝑒𝑑 = πœ‹ 4 Γ— 𝐷2 Γ— 𝐿 Γ— 𝑁 60 π‘š3 /𝑠 𝑉 𝑠 = π‘šπ‘Ÿ Γ— 𝑣1 = πœ‹ 4 Γ— 𝐷2 Γ— 𝐿 Γ— 𝑁 60 Γ— 2------- (4) Points 1 and 1’ are on the same constant pressure line; ∴ 𝑃1𝑣1 𝑇1 = 𝑃1′𝑣1β€² 𝑇1β€² ; 𝑃1=𝑃1β€²; ∴ 𝑣1 = 𝑇1 𝑇1β€² Γ— 𝑣1β€²; 𝑣1β€²= 𝑣𝑔 at -6Β°C = 0.3599 m3/kg. ∴ 𝑣1 = 273 + 5 273 βˆ’ 6 Γ— 0.3599; ∴ π’—πŸ = 𝟎. πŸ‘πŸ•πŸ’π’ŽπŸ‘/π’Œπ’ˆ Substituting in eq. (4); 𝑉 𝑠 = 0.0462 Γ— 0.374 = πœ‹ 4 Γ— 𝐷2 Γ— 1.5𝐷 Γ— 900 60 Γ— 2 ∴ 𝑫 = 𝟎. πŸŽπŸ•πŸ– π’Ž 𝒂𝒏𝒅 𝑳 = 𝟎. πŸπŸπŸ– π’Ž (Since it is two-cylinder compressor)
  • 28. A food storage required a refrigeration system of 12 tons of capacity at an evaporator temperature of -100C and condenser temperature of 250C. The refrigerant ammonia is sub-cooled by 50C before passing through the throttle valve. The vapor leaving the coil is 0.97 dry. Find the power required to drive the compressor if the actual COP is 60% of theoretical COP. Solution: Actual power required to drive the compressor is found by actual COP formula: πΆπ‘‚π‘ƒπ‘Žπ‘π‘‘π‘’π‘Žπ‘™ = π΄π‘π‘‘π‘’π‘Žπ‘™ π‘…π‘’π‘“π‘Ÿπ‘–π‘”π‘’π‘Ÿπ‘Žπ‘‘π‘–π‘œπ‘› 𝑒𝑓𝑓𝑒𝑐𝑑 π΄π‘π‘‘π‘’π‘Žπ‘™ π‘π‘œπ‘€π‘’π‘Ÿ π‘Ÿπ‘’π‘žπ‘’π‘–π‘Ÿπ‘’π‘‘ -----(1) Given actual COP=60% of theoretical COP; πΆπ‘‚π‘ƒπ‘Žπ‘π‘‘π‘’π‘Žπ‘™ = 0.60 Γ— πΆπ‘‚π‘ƒπ‘‡β„Žπ‘’π‘œπ‘Ÿπ‘’π‘‘π‘–π‘π‘Žπ‘™-------- (2) πΆπ‘‚π‘ƒπ‘‡β„Žπ‘’π‘œπ‘Ÿπ‘’π‘‘π‘–π‘π‘Žπ‘™ = β„Ž1βˆ’β„Ž4 β„Ž2βˆ’β„Ž1 --------(3)
  • 29. To find β€˜β„Ž1’: Point no. β€˜1’ is in the wet vapor region on p-h plot. ∴ β„Ž1 = β„Žπ‘“ + π‘₯1β„Žπ‘“π‘”----- (4) In thermodynamics data handbook, refer Table-B1 for Thermodynamic properties of saturated Ammonia: At Temp=-10Β°C: β„Žπ‘“= 135.2 kJ/kg; β„Žπ‘“π‘”= 1296.8 kJ/kg Given The vapor leaving the coil is 0.97 dry. ∴ π‘₯1 = 0.97; substituting in eq. (4); β„Ž1 = 135.2 + 0.97 Γ— 1296.8 ∴ β„Ž1= 1393.1 π‘˜π½/π‘˜π‘”
  • 30. To find β€˜β„Ž2’: Point β€˜2’ is in the superheated vapor region on p-h Plot. Hence, we do not get β„Ž2 directly from the table. ∴ β„Ž2 = β„Ž2β€² + 𝐢𝑃 π‘ π‘’π‘π‘’π‘Ÿβ„Žπ‘’π‘Žπ‘‘π‘’π‘‘ π‘£π‘Žπ‘π‘œπ‘Ÿ π‘Ÿπ‘’π‘“. 𝑇2 βˆ’ 𝑇2β€² ---(5) To find 𝑇2: Process 1-2 is isentropic; ∴ 𝑠1 = 𝑠2 Since we know β€˜π‘₯1’; 𝑠1= 𝑠𝑓 + π‘₯1𝑠𝑓𝑔 In thermodynamics data handbook, refer Table-B1 for Thermodynamic properties of saturated Ammonia: At Temp=-10Β°C: 𝑠𝑓= 0.5440 kJ/kg K; 𝑠𝑓𝑔= 4.9290 kJ/kg K; ∴ 𝑠1= 0.5440 + 0.97 Γ— 4.9290; ∴ π’”πŸ= πŸ“. πŸ‘πŸπŸ“ π’Œπ‘± π’Œπ’ˆπ‘² = π’”πŸ 𝑠2 = 𝑠2β€² + 𝐢𝑃(π‘ π‘’π‘π‘’π‘Ÿβ„Žπ‘’π‘Žπ‘‘π‘’π‘‘ π‘£π‘Žπ‘π‘œπ‘Ÿ π‘Ÿπ‘’π‘“.)𝑙𝑛 𝑇2 𝑇2β€²
  • 31. In thermodynamics data handbook, refer Table-B1 for Thermodynamic properties of saturated Ammonia: At Temp=26Β°C:𝑠2β€² = 𝑠𝑔= 5.0244 kJ/kg K; In thermodynamics data handbook, refer Table-B9 for specific heat of superheated Ammonia vapor: At Pressure approx. 10.3 bar, CP=2.7167 kJ/kg K (From Table-B1 at Temp=26Β°C, Abs. Pressure=10.3397 bar) ∴ 𝐢𝑃(π‘ π‘’π‘π‘’π‘Ÿβ„Žπ‘’π‘Žπ‘‘π‘’π‘‘ π‘£π‘Žπ‘π‘œπ‘Ÿ π‘Ÿπ‘’π‘“.)= 2.7167 kJ/kg K ∴ 𝑠2= 5.325 = 5.0244 + 2.7167𝑙𝑛 𝑇2 273 + 26 ∴ π‘»πŸ = πŸ‘πŸ‘πŸ’ 𝑲 π’‰πŸβ€² = π’‰π’ˆ 𝒂𝒕 𝑻 = πŸπŸ”Β°π‘ͺ = πŸπŸ’πŸ”πŸ“. πŸ” π’Œπ‘±/π’Œπ’ˆ Substituting in eq. (5); β„Ž2 = β„Ž2β€² + 𝐢𝑃 π‘ π‘’π‘π‘’π‘Ÿβ„Žπ‘’π‘Žπ‘‘π‘’π‘‘ π‘£π‘Žπ‘π‘œπ‘Ÿ π‘Ÿπ‘’π‘“. 𝑇2 βˆ’ 𝑇2β€² β„Ž2 = 1465.6 + 2.7167 334 βˆ’ 299 ∴ π’‰πŸ= πŸπŸ“πŸ”πŸŽ. πŸ”πŸ– π’Œπ‘±/π’Œπ’ˆ
  • 32. To find β€˜β„Ž4’: Point β€˜4’ is in the wet vapor region and the quality β€˜x4’ is not known. Hence, we cannot use β„Ž4 = β„Žπ‘“ + π‘₯4β„Žπ‘“π‘” WKT, π’‰πŸ‘ = π’‰πŸ’; Point β€˜3’ is in the sub-cooled liquid region and we cannot Obtain the value for β€˜3’ directly from the tables. Hence, Assume 3’ on the saturated liquid line as shown in the plot. π’‰πŸ‘ = π’‰πŸ‘β€² βˆ’ π‘ͺ𝑷 π‘³π’Šπ’’π’–π’Šπ’… 𝒓𝒆𝒇. π‘»πŸ‘β€² βˆ’ π‘»πŸ‘ βˆ’βˆ’βˆ’ βˆ’ πŸ” π’‰πŸ‘β€² = 𝒉𝒇 𝒂𝒕 𝑻 = πŸπŸ”Β°π‘ͺ = πŸ‘πŸŽπŸ‘. πŸ” π’Œπ‘±/π’Œπ’ˆ In thermodynamics data handbook, refer Table-B12 for specific heat of liquid Ammonia : CP(liquid)=4.69 kJ/kg K Given The refrigerant ammonia is sub-cooled by 50C ∴ π‘»πŸ‘β€² βˆ’ π‘»πŸ‘ = πŸ“Β°π‘ͺ Substituting in eq.(6): π’‰πŸ‘ = πŸ‘πŸŽπŸ‘. πŸ”πŸ” βˆ’ πŸ’. πŸ”πŸ— πŸ“ = πŸπŸ–πŸŽ. 𝟐𝟏 π’Œπ‘± π’Œπ’ˆ = π’‰πŸ’
  • 33. Substituting in eq.(3): πΆπ‘‚π‘ƒπ‘‡β„Žπ‘’π‘œπ‘Ÿπ‘’π‘‘π‘–π‘π‘Žπ‘™ = 1393.1 βˆ’ 280.21 1560.68 βˆ’ 1393.1 ∴ π‘ͺπ‘Άπ‘·π‘»π’‰π’†π’π’“π’†π’•π’Šπ’„π’‚π’= πŸ”. πŸ”πŸ’ Substituting in eq.(2): π‘ͺ𝑢𝑷𝒂𝒄𝒕𝒖𝒂𝒍 = 𝟎. πŸ”πŸŽ Γ— πŸ”. πŸ”πŸ’ = πŸ‘. πŸ—πŸ–πŸ’ Substituting in eq.(1): 3.984 = 12 Γ— 3.51 π΄π‘π‘‘π‘’π‘Žπ‘™ π‘π‘œπ‘€π‘’π‘Ÿ π‘Ÿπ‘’π‘žπ‘’π‘–π‘Ÿπ‘’π‘‘ ∴ 𝑨𝒄𝒕𝒖𝒂𝒍 π’‘π’π’˜π’†π’“ π’“π’†π’’π’–π’Šπ’“π’†π’… = 𝟏𝟎. πŸ“πŸ• π’Œπ‘Ύ
  • 34. An ideal vapor compression system with Freon -12 as the working fluid operates with an evaporator temperature of -300C and a condenser pressure of 9.61 bar. The fluid leaves the condenser sub-cooled by 130C. The compression is adiabatic with an efficiency of 75%. Calculate the COP of the machine. Solution: (a) COP 𝐢𝑂𝑃 = β„Ž1βˆ’β„Ž4 β„Ž2βˆ’β„Ž1 -------- (1) To find β€˜β„Ž1’: Assuming that point β€˜1’ is on saturated-vapor line: β„Ž1=β„Žπ‘” at -30Β°C; In thermodynamics data handbook, refer Table-B3 for Thermodynamic properties of saturated Freon-12: At Temp=-30Β°C:π’‰π’ˆ= 174.076 kJ/kg=π’‰πŸ
  • 35. To find β€˜β„Ž2’: Point β€˜2’ is in the superheated vapor region and hence Cannot be found directly from the saturated tables. Given that the compressor is having an efficiency of 75%; We need to use the efficiency formula for compressor;; 𝜼π‘ͺπ’π’Žπ’‘π’“π’†π’”π’”π’π’“ = π’‰πŸβ€²βˆ’π’‰πŸ π’‰πŸβˆ’π’‰πŸ ------- (2) where 1-2’ represent ideal isentropic compression and 1-2 represent actual compression. To find β€˜π’‰πŸβ€²β€™, assume point 2’’ situated on the saturated vapour line such that π’‰πŸβ€²β€²=π’‰π’ˆ at 9.61 bar condenser pressure. In thermodynamics data handbook, refer Table-B3 for Thermodynamic properties of saturated Freon-12: At Pressure=9.61 bar :π’‰π’ˆ= 203.051 kJ/kg=π’‰πŸβ€²β€² and Temp. T=40Β°C= π‘»πŸβ€²β€². In thermodynamics data handbook, refer Table-B10 for specific heat of superheated Freon-12 vapor: At Pressure approx. 9.61 bar, CP=0.746 kJ/kg K
  • 36. ∴ π’‰πŸβ€² = π’‰πŸβ€²β€² + π‘ͺ𝑷(𝒔𝒖𝒑𝒆𝒓𝒉𝒆𝒂𝒕𝒆𝒅 𝒗𝒂𝒑𝒐𝒓 𝒓𝒆𝒇.) π‘»πŸβ€² βˆ’ π‘»πŸβ€²β€² π’‰πŸβ€² = πŸπŸŽπŸ‘. πŸŽπŸ“πŸ + 𝟎. πŸ•πŸ’πŸ” π‘»πŸβ€² βˆ’ πŸ‘πŸπŸ‘ ------(3) To find β€˜π‘»πŸβ€²β€™: Process 1-2’ is isentropic compression: hence, 𝑠1=𝑠2β€² =𝑠𝑔 at -30Β°C In thermodynamics data handbook, refer Table-B3 for Thermodynamic properties of saturated Freon-12: At Temp.=-30Β°C:π’”π’ˆ= 0.7165 kJ/kgK=π’”πŸ=π’”πŸβ€² At Pressure 9.61 bar: π’”π’ˆ= 0.6820 kJ/kgK=π’”πŸβ€²β€² And, 𝑠2β€² = 𝑠2β€²β€² + 𝐢𝑃(π‘ π‘’π‘π‘’π‘Ÿβ„Žπ‘’π‘Žπ‘‘π‘’π‘‘ π‘£π‘Žπ‘π‘œπ‘Ÿ π‘Ÿπ‘’π‘“.)𝑙𝑛 𝑇2β€² 𝑇2β€²β€² ∴ 0.7165 = 0.6820 + 0.746𝑙𝑛 𝑇2β€² 273+40 ; ∴ π‘»πŸβ€² = πŸ‘πŸπŸ•. πŸ–πŸ 𝐊 ; substitute in eq.(3); π’‰πŸβ€² = πŸπŸŽπŸ‘. πŸŽπŸ“πŸ + 𝟎. πŸ•πŸ’πŸ” πŸ‘πŸπŸ•. πŸ–πŸ βˆ’ πŸ‘πŸπŸ‘ ∴ π’‰πŸβ€²= πŸπŸπŸ’. 𝟏 π’Œπ‘±/π’Œπ’ˆ
  • 37. Substituting in eq(2): 𝜼π‘ͺπ’π’Žπ’‘π’“π’†π’”π’”π’π’“ = π’‰πŸβ€²βˆ’π’‰πŸ π’‰πŸβˆ’π’‰πŸ ; 0.75 = 214.1βˆ’174.076 β„Ž2βˆ’174.076 ; ∴ π’‰πŸ= πŸπŸπŸ•. πŸ’πŸ’ π’Œπ‘±/π’Œπ’ˆ To find β€˜π’‰πŸ’β€™: WKT, π’‰πŸ‘ = π’‰πŸ’; Point β€˜3’ is in the sub-cooled liquid region and we cannot Obtain the value for β€˜3’ directly from the tables. Hence, Assume 3’ on the saturated liquid line as shown in the plot. π’‰πŸ‘ = π’‰πŸ‘β€² βˆ’ π‘ͺ𝑷 π‘³π’Šπ’’π’–π’Šπ’… 𝒓𝒆𝒇. π‘»πŸ‘β€² βˆ’ π‘»πŸ‘ βˆ’βˆ’βˆ’ βˆ’ πŸ’ π’‰πŸ‘β€² = 𝒉𝒇 𝒂𝒕 𝑷 = πŸ—. πŸ”πŸπ’ƒπ’‚π’“ = πŸ•πŸ’. πŸ“πŸπŸ• π’Œπ‘± π’Œπ’ˆ (π‘“π‘Ÿπ‘œπ‘š π‘‡π‘Žπ‘π‘™π‘’ 𝐡3) In thermodynamics data handbook, refer Table-B12, specific heat of liquid Freon-12 : CP(liquid)=0.94 kJ/kg K Given The refrigerant Freon-12 is sub-cooled by 130C ∴ π‘»πŸ‘β€² βˆ’ π‘»πŸ‘ = πŸπŸ‘Β°π‘ͺ Substituting in eq.(4): π’‰πŸ‘ = πŸ•πŸ’. πŸ“πŸπŸ• βˆ’ 𝟎. πŸ—πŸ’ πŸπŸ‘ = πŸ”πŸ. πŸ‘πŸ π’Œπ‘± π’Œπ’ˆ = π’‰πŸ’
  • 38. Substituting in eq.(1); 𝐢𝑂𝑃 = β„Ž1 βˆ’ β„Ž4 β„Ž2 βˆ’ β„Ž1 𝐢𝑂𝑃 = 174.076 βˆ’ 62.31 227.44 βˆ’ 174.076 ∴ π‘ͺ𝑢𝑷 = 𝟐. πŸŽπŸ—
  • 39. Refrigeration is desired at -100C. Cooling water having a mean temperature of 300C is available. Assume that a 50C difference is required for heat flow both in condenser and the evaporator. The refrigerant is Fr-12. The compressor has an isentropic efficiency of 75%, volumetric efficiency of 80% and the piston displacement is 6m3/min. Neglecting the other losses determine the (i) Tonnage of refrigeration (ii) Shaft power (iii) COP. Solution: (a) Tonnage of refrigeration WKT, 𝑇𝑅𝐸 = π‘šπ‘Ÿ Γ— β„Ž1 βˆ’ β„Ž4 in kW To express TRE in tons; we have, 1 ton=3.51 kW; ∴ 𝑇𝑅𝐸 𝑖𝑛 π‘‘π‘œπ‘›π‘  = π‘šπ‘Ÿ β„Ž1βˆ’β„Ž4 60Γ—3.51 -----(1) where π‘šπ‘Ÿ is mass of refrigerant in kg/s
  • 40. To find β€˜β„Ž1’: Assuming that point β€˜1’ is on saturated-vapor line: β„Ž1=β„Žπ‘” at -30Β°C; In thermodynamics data handbook, refer Table-B3 for Thermodynamic properties of saturated Freon-12: At Temp=-15Β°C:π’‰π’ˆ= 180.946 kJ/kg=π’‰πŸ To find β€˜β„Ž2’: Point β€˜2’ is in the superheated vapor region and hence Cannot be found directly from the saturated tables. Given that the compressor is having an efficiency of 75%; We need to use the efficiency formula for compressor; 𝜼π‘ͺπ’π’Žπ’‘π’“π’†π’”π’”π’π’“ = π’‰πŸβ€²βˆ’π’‰πŸ π’‰πŸβˆ’π’‰πŸ ------- (2) where 1-2’ represent ideal isentropic compression and 1-2 represent actual compression.
  • 41. To find β€˜π’‰πŸβ€²β€™, assume point 2’’ situated on the saturated vapour line such that π’‰πŸβ€²β€²=π’‰π’ˆ at 35Β°C condenser temperature. In thermodynamics data handbook, refer Table-B3 for Thermodynamic properties of saturated Freon-12: At Temp.=35Β°C :π’‰π’ˆ= 201.299 kJ/kg=π’‰πŸβ€²β€² and Pressure =8.477 bar. In thermodynamics data handbook, refer Table-B10 for specific heat of superheated Freon-12 vapor: At Pressure approx. 8.477 bar, CP=0.733 kJ/kg K ∴ π’‰πŸβ€² = π’‰πŸβ€²β€² + π‘ͺ𝑷(𝒔𝒖𝒑𝒆𝒓𝒉𝒆𝒂𝒕𝒆𝒅 𝒗𝒂𝒑𝒐𝒓 𝒓𝒆𝒇.) π‘»πŸβ€² βˆ’ π‘»πŸβ€²β€² π’‰πŸβ€² = 𝟐𝟎𝟏. πŸπŸ—πŸ— + 𝟎. πŸ•πŸ‘πŸ‘ π‘»πŸβ€² βˆ’ πŸπŸ•πŸ‘ + πŸ‘πŸ“ ------(3) To find β€˜π‘»πŸβ€²β€™: Process 1-2’ is isentropic compression: hence, 𝑠1=𝑠2β€² =𝑠𝑔 at -15Β°C
  • 42. In thermodynamics data handbook, refer Table-B3 for Thermodynamic properties of saturated Freon-12: At Temp.= -15Β°C:π’”π’ˆ= 0.7046 kJ/kgK=π’”πŸ=π’”πŸβ€². Also, At Temp.= 35Β°C:π’”π’ˆ= 0.6834 kJ/kgK=π’”πŸβ€²β€² And, 𝑠2β€² = 𝑠2β€²β€² + 𝐢𝑃(π‘ π‘’π‘π‘’π‘Ÿβ„Žπ‘’π‘Žπ‘‘π‘’π‘‘ π‘£π‘Žπ‘π‘œπ‘Ÿ π‘Ÿπ‘’π‘“.)𝑙𝑛 𝑇2β€² 𝑇2β€²β€² ∴ 0.7046 = 0.6834 + 0.733𝑙𝑛 𝑇2β€² 273+35 ; ∴ π‘»πŸβ€² = πŸ‘πŸπŸ• 𝐊 ; substitute in eq.(3); π’‰πŸβ€² = 𝟐𝟎𝟏. πŸπŸ—πŸ— + 𝟎. πŸ•πŸ‘πŸ‘ πŸ‘πŸπŸ• βˆ’ πŸπŸ•πŸ‘ + πŸ‘πŸ“ ∴ π’‰πŸβ€²= πŸπŸŽπŸ•. πŸ— π’Œπ‘±/π’Œπ’ˆ Substituting in eq(2): 𝜼π‘ͺπ’π’Žπ’‘π’“π’†π’”π’”π’π’“ = π’‰πŸβ€²βˆ’π’‰πŸ π’‰πŸβˆ’π’‰πŸ ; 0.75 = 207.9βˆ’180.946 β„Ž2βˆ’180.946 ; ∴ π’‰πŸ= πŸπŸπŸ”. πŸ–πŸ– π’Œπ‘±/π’Œπ’ˆ
  • 43. To find β€˜β„Ž4’: Process 3-4 is isenthalpic process. Hence, β„Ž3=β„Ž4 . π’‰πŸ‘= 𝒉𝒇 at T=35Β°C In thermodynamics data handbook, refer Table-B3 for Thermodynamic properties of saturated Freon-12: At Temp.=35Β°C :𝒉𝒇= 69.494 kJ/kg=π’‰πŸ‘=π’‰πŸ’ To find β€˜π’Žπ’“β€™: Mass of refrigerant circulated is found by the actual swept-volume of the compressor β€œπ‘‰ π‘Žπ‘ β€. Given volumetric efficiency=80% and 𝑉 𝑠= 6π‘š3 /π‘šπ‘–π‘› πœΌπ’—π’π’. = 𝑽𝒂𝒔 𝑽𝒔 ; i.e., 𝟎. πŸ– = 𝑽𝒂𝒔 πŸ” ; ∴ 𝑽𝒂𝒔 = 𝟎. πŸ– Γ— πŸ” = πŸ’. πŸ–π’ŽπŸ‘/π’Žπ’Šπ’ But, 𝑽𝒂𝒔 = π’Žπ’“ Γ— π’—πŸ where π’—πŸ is the specific volume at compressor inlet π’—πŸ=π’—π’ˆ at T= -15Β°C. ∴ π’—πŸ = π’—π’ˆ = 𝟎. πŸŽπŸ—πŸπŸŽπŸπŸ– π’ŽπŸ‘ /π’Œπ’ˆ (From Table B3)
  • 44. ∴ 𝑽𝒂𝒔= π’Žπ’“ Γ— π’—πŸ πŸ’. πŸ– π’ŽπŸ‘/π’Žπ’Šπ’ = π’Žπ’“ Γ— 𝟎. πŸŽπŸ—πŸπŸŽπŸπŸ–π’ŽπŸ‘/π’Œπ’ˆ ∴ π’Žπ’“= πŸ“πŸ. πŸ• π’Œπ’ˆ/π’Žπ’Šπ’ Substituting in eq. (1) 𝑇𝑅𝐸 𝑖𝑛 π‘‘π‘œπ‘›π‘  = π‘šπ‘Ÿ β„Ž1 βˆ’ β„Ž4 60 Γ— 3.51 𝑇𝑅𝐸 𝑖𝑛 π‘‘π‘œπ‘›π‘  = 52.7 180.946 βˆ’ 69.494 60 Γ— 3.51 𝑻𝑹𝑬 π’Šπ’ 𝒕𝒐𝒏𝒔 = πŸπŸ•. πŸ— 𝒕𝒐𝒏
  • 45. (b) Shaft power From compressor work formula, Shaft power= power required to drive the compressor ∴ 𝑺𝒉𝒂𝒇𝒕 π’‘π’π’˜π’†π’“ = π’Žπ’“ πŸ”πŸŽ π’‰πŸ βˆ’ π’‰πŸ π’Œπ‘Ύ 𝑺𝒉𝒂𝒇𝒕 π’‘π’π’˜π’†π’“ = πŸ“πŸ. πŸ• πŸ”πŸŽ πŸπŸπŸ”. πŸ–πŸ– βˆ’ πŸπŸ–πŸŽ. πŸ—πŸ’πŸ” π’Œπ‘Ύ ∴ 𝑺𝒉𝒂𝒇𝒕 π’‘π’π’˜π’†π’“ = πŸ‘πŸ. πŸ“πŸ” π’Œπ‘Ύ (C) COP 𝐢𝑂𝑃 = β„Ž1 βˆ’ β„Ž4 β„Ž2 βˆ’ β„Ž1 𝐢𝑂𝑃 = 180.946 βˆ’ 69.494 216.88 βˆ’ 180.946 ∴ π‘ͺ𝑢𝑷 = πŸ‘. 𝟏𝟎