MAHALAKSHMI ENGINEERING COLLEGE

MAHALAKSHMI ENGINEERING COLLEGE TIRUCHIRAPALLI-621213. Department: Mechanical Subject Code: ME2202 U N IT - 1 Semester: III Subject Name: ENGG. THERMODYNAMICS 1. 1 kg of gas at 1.1 bar, 27 o C is compressed to 6.6 bar as per the law pv 1.3 const. Calculate w ork and heat transfer, if (1) When the gas is ethane (C 2 H 6 ) w ith molar mass of 30kg/k mol and c p of2.1 kj/kg K. (2) When the gas is argon (Ar) w ith molar mass of 40kg/k mol and c p of 0.52 kj/kg K. (AUC DEC 05) Given: Polytropic process Mass (m) 1kg Pressure (P 1 ) 1.1bar 1.1 100 110KN/m 2 Temperature (T 1 ) 27 C 27+273 300K Pressure (P 2 ) 6.6bar 6.6 100 660KN/m 2 Molar mass of ethane (m) 30Kg/K mol C p of ethane 2.1KJ/K mol Molar mass of argon 40Kg/K mol C p of argon 0.52KJ/Kg.K PV 1.3 C L.VIJAYAKUMAR /A.P-MECH Page 1

To Find: 1. Work Transfer (W) 2. Heat Transfer (Q) Solution: Step:1 Case:1 (Ethane) Work transfer (W) P 1 V 1 mrt 1 V 1 V 1 V 1 680.2363m 3 P 1 V 1 n P 2 V 2 V 2 V 2 802.2438m 3 Work transfer (W) W -1515516.892KJ Heat transfer (Q) W -1515516.892 Q -378879.223KJ L.VIJAYAKUMAR /A.P-MECH Page 2

Case:2 (Argon) Work transfer (W) P 1 V 1 mrt 1 V 1 V 1 V 1 906.9818m 3 P 1 V 1 n P 2 V 2 V 2 Work transfer (W) V 2 1166.0760m 3 Heat transfer (Q) W W -2232807.207KJ -2232807.207 Q 558201.8018KJ L.VIJAYAKUMAR /A.P-MECH Page 3

2. In an air compressor, air flow s steadily at the rate of 0.5 kg/sec. At entry to the compressor, air has a pressure of 105kPa and specific volume of 0.86 m 3 /kg and at exit of the compressor those corresponding values are 705 kpa and 0.16 m 3 /kg. Neglect Kinetic and Potential energy change. The internal energy of air leaking the compressor is 95 kj/kg greater than that of air entering. The cooling w ater in the compressor absorbs 60kJ/sec. of heat from the air. Find pow er required to derive the compressor. (AUC MAY 06) Given: Mass (m) 0.5 kg/s Entering velocity (C 1 ) 7 m/s Entering Pressure (P 1 ) 100 KPa 100 KN/m 2 Entering volume (V 1 ) 0.95 m 3 /Kg Leaving velocity (C 2 ) 5 m/s Leaving Pressure (P 2 ) 700 KPa 700 KN/m 2 Leaving volume (V 2 ) 0.19 m 3 /Kg Change in internal energy (U 2 - U 1 ) 90 KJ/Kg Heat absorbs (Q) 58 KW To Find: 1. Compute the rate of shaft w ork input to the air in KW (or) Work input 2. Ratio of inlet pipe to outlet pipe ( ) Solution: Step:1 m (P 1 V 1 + +Z 1.g) + Q m ((U 2 - U 1 )+P 2 V 2 + +Z 2.g) + W Assume Z 1 Z 2 0.5 ((100 0.95)+ ) + 58 0.5 (90+(700 0.19) + ) + W 47.51225+58 111.50625 + W L.VIJAYAKUMAR /A.P-MECH Page 4

W 105.51225-111.50625 W 5.994KW Note; - sign indicates that the w ork is done on the system Step:2 To find ratio of inlet to outlet dia ( ) From continuity equation 3.57 3.57 3.57 3.57 The ratio of inlet to outlet dia of the pipe( ) 1.8894 L.VIJAYAKUMAR /A.P-MECH Page 5

3. In an isentropic flow through nozzle, air flow s at the rate of 600 kg/hr. At inlet to the nozzle, Pressure is 2 MPa and temperature is 127 o C. The exit pressure is 0.5 MPa. Initial air velocity is 300 m/s determines (i) Exit velocity of air (ii) Inlet and exit area of nozzle. (AUC DEC 06) To Find: Given: Mass of flow rate (m 1 ) 600Kg/hr 600 3600 0.1666Kg/sec Inlet pressure (P 1 ) 2MPa 10^6 N/m 2 Inlet temp (T 1 ) 127 C + 273 400K Outlet pressure (P 2 ) 0.5MPa 10^6 N/ m 2 Inlet air velocity (C 1 ) 300m/s 1.Exit velocity of air (C 2 ) 2. inlet and exit area (d 1,d 2 ) Solution: Step:1 ( )^ 1.4859 T 2 1.4859 T 1 T 2 1.4859 400 L.VIJAYAKUMAR /A.P-MECH Page 6

T 2 594.3977 k Step:2 C 2 {2 m[c p (T 2 -T 1 ) + ] C 2 {2 0.1666[1.005(594.3977-400) + ] C 2 282.4937m/s Step:3 Inlet mass flow rate (m) P 1 V 1 mrt 1 V 1 V 1 V 1 0.0574 m 3 /Kg M 1 A 1 A 1 A 1 3.1876 10^-5 A 1 d 1^2 3.1876 10^-5 d 1^2 d 1 d 1 0.0063m 1000 L.VIJAYAKUMAR /A.P-MECH Page 7

d 1 6.3706mm Step:4 ( )^ ( )^ 2.692 V 2 2.692 V 1 V 2 2.692 0.0574 V 2 0.155m 3 /Kg A 2 A 2 A 2 3.1410 10^-5 d 2^2 3.1410 10^-5 d 2^2 D 2 0.01078m 1000 D 2 10.7883mm L.VIJAYAKUMAR /A.P-MECH Page 8

4. A centrifugal pump delivers 2750 kg of w ater per minute from initial pressure of 0.8 bar absolute to a final pressure of 2.8 bar absolute. The suction is 2 m below and the delivery is5 m above the centre of pump. If the suction and delivery pipes are of 15 cm and 10 cm diameter respectively, make calculation for pow er required to run the pump. (AUC DEC 06) Given: Mass (m) 2750 Kg/min 2750 60 45.8333Kg/s Initial pressure (P 1 ) 0.8bar 100 80KN/m 2 Final pressure (P 2 ) 2.8bar 100 280 N/m 2 Z 1-2m (below the centre of pump) Z 2 5m (above the centre of pump) Dia, d 1 15cm 100 0.15m d 2 10cm 100 0.1m To Find: Pow er (P) or Work (W) Solution: Step:1 The steady flow energy equation is, m [ + +P 1 V 1 ] m [ + + P 2 V 2 ] + W L.VIJAYAKUMAR /A.P-MECH Page 9

W m [ + +(P 1 V 1 - P 2 V 2 )] M d 1 2 C 1 ρ C 1 C 1 2.5936m/s C 2 C 2 5.8356m/s Step:2 W 45.83 [ + +(80 1000-280 1000)] W [-0.06867-0.01366-200 10^3] W 9166003.773KJ/Kg W W 91.6600KJ/Kg L.VIJAYAKUMAR /A.P-MECH Page 10

6. A blow er handles 1 kg/sec of air at 293 K and consumes a pow er of 15kw.Theinletandoutletvelocitiesofairarel00 m/sec and 150 m/sec respectively. Find the exit air temperature, assuming adiabatic conditions. Take C p of air as 1.005 kj/kg-k. (AUC DEC 07) Given: Mass(m) 1Kg/s Temp (T 1 ) 293K Pow er (P) or Work (W) 15 KW Inlet velocity (C 1 ) 100m/s Outlet velocity (C 2 ) 150m/s To Find: Exit air temp (T 2 ) Solution: Step:1 Note; C p 1.005KJ/Kg.k m (h 1 + +Z 1.g) + Q m (h 2 + +Z 2.g) + W Neglect datum head (Z 1 ) (Z 2 ) 0 Q 0 (Adiapatic process m (h 1 + ) m (h 2 + ) + W 1 (h 1 + ) 1 (h 2 + ) + 15 (h 1 h 2 ) [ ] + 15 (h 1 h 2 ) 21.25 C p (T 1 T 2 ) 21.25 L.VIJAYAKUMAR /A.P-MECH Page 11

1.005(293 T 2 ) 21.25 293 T 2 293 T 2 21.1442 293 21.1442 T 2 T 2 271.8557K 7. A room for four persons has tw o fans, each consuming 0.18 kw pow er and three 100W lamps. Ventilation ant at the rate of 0.0222 kg/sec enters w ith an enthalpy of 84 kj/kg and leaves w ith an enthalpy of 59 kj/kg. If each person puts out heat at the rate of0.175 kj/sec, determine the rate at w hich heat is to be removed by a room cooler, so that a steady state is maintained in the room. (AUC DEC 07) Given: No of person(n p ) 4 No of fan (n f ) 2 (W f ) 0.18KW (each) (W l ) 0.1KW (each) Mass of air (m) 80Kg/hr 80 3600 0.022Kg/s Enthalpy of air entering (h 1 ) 84KJ/Kg Enthalpy of air leaving (h 2 ) 59KJ/Kg Heat (Q p ) 630KJ/hr 630 3600 0.175KJ/s To Find: Rate of heat is to be removed Solution: Step:1 Rate of energy increase Rate of energy in flow - Rate of energy out flow L.VIJAYAKUMAR /A.P-MECH Page 12

Q 1 -Ƞ p Q p Q 1 -(4 0.175) Q 1-0.7KJ/s or KW Step:2 M(h 1 - h 2 ) 0.022(84-59) 0.55KJ/s Step:3 W electrical energy input W Ƞ f W f + Ƞ l W l W (2 0.18)+( 3 0.1) W 0.66KW Step:4 Rate of heat is to be removed (Q) Q Q 1 0.55 W Q - 0.7 0.55 0.66 Q - 1.916KW L.VIJAYAKUMAR /A.P-MECH Page 13

8. One liter of hydrogen at 273 K is adiabatically compressed to one half of its initial volume. in the change in temperature of the gas, if the ratio of tw o specific heats for hydrogen is 1.4. (AUC DEC 07) Given: adiabatic process Initial volume (V 1 ) 1lit 1 1000 0.001m 3 Temp (T 1 ) 273K Initial volume (V 2 ) one half of it s Initial volume (V 2 ) 0.001 2 0.0005 m 3 To Find: Change in temp of gas (T 2 T 1 ) Solution: Step:1 ( )^( ) ( )^( ) 1.3195 T 2 1.3195 T 1 T 2 1.3195 273 T 2 360.2256K Step:2 To find change in temp (T 2 T 1 ) L.VIJAYAKUMAR /A.P-MECH Page 14

T 2 T 1 360.2256 273 T 2 T 1 87.2256K 9. The velocity and enthalpy of fluid at the inlet of a certain nozzle are50 m/sec and2800 kj/kg respectively. The enthalpy at the exit of Nozzle is 2600 kj/kg. The nozzle is horizontal and insulated so that no heat transfer takes place from it' Find (1) Velocity of the fluid at exit of the nozzle (2) Mass flow rate, if the area at inlet of nozzle is 0.09 m 2 (3) Exit area of the nozzle, if the specific volume at the exit of the Nozzle is 0.495 m 3 /kg. (AUC DEC 07) Given: Inlet velocity of nozzle (C 1 ) 50m/s Inlet enthalpy of nozzle (h 1 ) 2800KJ/Kg h 1 2800 10^3J/Kg no of heat transfer take place (Q) 0 exit enthalpy (h 2 ) 2600KJ/Kg h 2 2600 10^3J/Kg To Find: 1. C 2 2. m 3. A 2 Solution: Step:1 To find velocity of the fluid at exit of the nozzle (C 2 ) C 2 [2 (h 1 - h 2 ) + C 1^2] C 2 [2 (2800-2600) 1000 + 50^2] L.VIJAYAKUMAR /A.P-MECH Page 15

C 2 634.4288m/s Step:2 Tofind mass flow rate (m) m V 1 Specific volume is not given so it is assumed to be V 1 1m 3 /Kg m m 4.5Kg/s Step:3 To find exit area of nozzle (A 2 ) A 2 A 2 0.0035m 2 L.VIJAYAKUMAR /A.P-MECH Page 16

10. A w ork done by substance in a reversible non-flow manner is in accordance w ith V (15/p) m 3, w here p is in bar. Evaluate the w ork done on or by the system as pressure increases from 10 to 100 bar. Indicate w hether it is a compression process or expansion process. If the change in internal energy is 500kJ, calculate the direction and magnitude of heat transfer. (AUC MAY 08) Given: reversible non-flow manner V (15/p) m 3 Pressure (P 1 ) 10bar Pressure (P 2 ) 100bar Change in internal energy (Δ u ) 500KJ To Find: The direction and magnitude of heat transfer (Q) Solution: Step:1 Heat transfer (Q) W+ Δ u L.VIJAYAKUMAR /A.P-MECH Page 17

Work done ʃ V dp ʃ 15/p dp 15 ʃ 1/p dp 15 [log p ] 15 [log 100- log 10] 15 [2-1] 15 [1] W 15KJ Direction: W 15 Positive sign indicates the expansion process Step:2 By first law of thermodynemics, Q W+ Δ u Q 15+ 500 Q 515KJ Q 515 KJ Positive sign indicates the expansion process L.VIJAYAKUMAR /A.P-MECH Page 18

11. In a Gas turbine installation, the gases enter the turbine at the rate of 5 kg/sec w ith a velocity of 500 m/sec and enthalpy of 900 kj/kg and leave the turbine w ith 150 m/sec, and enthalpy of 400 kj/kg. The loss of heat from the gases to the surroundings is 25kJ/kg. Assume R 0.285 kj/kg.k, Cp 1.004 kj/kg.k and inlet conditions to be at 100 Kpa and 27 o C.Determine the diameter of the inlet pipe. (AUC MAY 08) Given: Mass flow rate (m) 5Kg/s Inlet velocity (C 1 ) 50m/s Inlet enthalpy (h 1 ) 900KJ/Kg Outlet enthalpy (h 1 ) 400KJ/Kg Outlet velocity (C 1 ) 150m/s Heat loss (Q) -25KJ/Kg R 0.285KJ/Kg C p 1.004KJ/Kg Inlet pressure (P 1 ) 100KPa 1000 100000KN/m 2 Inlet temperature (T 1 ) 27 C + 273 300K L.VIJAYAKUMAR /A.P-MECH Page 19

To Find: Work done (W) Dia of the inlet pipe (d 1 ) Solution: Step: 1 m (h 1 + +Z 1.g) + Q m (h 2 + +Z 2.g) + W Z 2 Z 1 m (h 1 + ) + Q m (h 2 + ) + W 5 (900 + ) - 25 5 (400 + ) + W 4481.25 2056.25 + W 4481.25 2056.25 W W 2425KW Step: 2 To find dia of the inlet pipe (d 1 ) m V 1? P 1 V 1 mrt 1 V 1 V 1 V 1 4.2875 m 2 /s L.VIJAYAKUMAR /A.P-MECH Page 20

m A 1 A 1 A 1 0.4275 m 2 2 A 1 d 1 2 d 1 d 1 d 1 0.7377m d 1 d 1 0.7377m 1000 d 1 737.7736mm L.VIJAYAKUMAR /A.P-MECH Page 21

12. A frictionless piston-cylinder device contains 2 kg of nitrogen at 100 Kpa and 300 K. Nitrogen is now compressed slow ly according to the relation PV 1.4 C until it reaches a final temperature of 360 K. Calculate the w ork input during this process. (AUC DEC 09) Given: Polytropic process Molecular weight of nitrogen (m) 28 Ɣ 1.4 Inlet pressure (P 1 ) 100KN/m 2 Inlet temperature (T 1 ) 300K Outlet temperature (T 2 ) 360 K PV 1.4 C To Find: Work input (W) Solution: Step: 1 L.VIJAYAKUMAR /A.P-MECH Page 22

W Gas constant (R) 0.297KJ/ Kg.K W W W -89.1KJ Direction: W -89.1KJ Negative sign indicates that the work is done on the system. L.VIJAYAKUMAR /A.P-MECH Page 23