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2.Closed System does not permit……………………………. (A)Energy Transfer (B)Mass Transfer (C) Both A and B

1.A system undergoes a change of state during which 100 kJ of heat is transferred to it and it does 50 kJ of work. The system is brought to its original state through a process during which 120 kJ of heat is transferred to it. The work done by the system is (A)50 kJ (B)70kJ (C)170kJ

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2.Closed System does not permit……………………………. (A)Energy Transfer (B)Mass Transfer (C) Both A and B

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  1. 1.A system undergoes a change of state during which 100 kJ of heat is transferred to it and it does 50 kJ of work. The system is brought to its original state through a process during which 120 kJ of heat is transferred to it. The work done by the system is (A)50 kJ (B)70kJ (C)170kJ (D) None of the above Ans.(A) 170 kJ Solution: Net heat transfer =100+120=220kJ Work transfer=+50 kJ Hence work done by the system is=220-50 =170kJ.

  2. 2.Closed System does not permit……………………………. (A)Energy Transfer (B)Mass Transfer (C) Both A and B (D) None of above Ans. (A) Mass transfer

  3. 3.Which system is also called as non-flow system? (A) Open system (B) closed system (C) isolated system (D) all of the above Ans. (A) Closed system

  4. 4.Everything external to system is called……….. (A) Environment (B) Boundary (C) Both A and B (D) None of above Ans. (A)Environment

  5. 5.Air cooler can be consider as (A)Open system (B) Closed system (C) Isolated system (D) None of these Ans. (A)Open system

  6. 6.Properties Which do not depends on the mass is called (A)Intensive Property (B) Extensive Property (C )Massive Property (D) None of these Ans. (A)Intensive Property

  7. 7. Which of the following is not an extensive property? (A)Enthalpy (B)Specific enthalpy (C)Volume (D)Entropy Ans. (B)Specific enthalpy

  8. 8. Which of the following is an extensive property? (A)Enthalpy (B)Pressure (C)Temperature (D)Density Ans. (A)Enthalpy

  9. 9. Which of the following is an Intensive property? (A) Volume (B) Enthalpy (C) Specific Volume (D)Mass Ans. (C) Specific Volume

  10. 10.A system is said to be Thermal equilibrium when (A)There is no work transfer (B) There is no heat transfer (C)Internal energy of the system is zero (D)No change in macroscopic property Ans. (C)Internal energy of the system is zero

  11. 11.A process which is locus of all equilibrium point is called (A)reversible process (B)irreversible process (C)quasi-static process (D)poly-tropic process Ans.(D)poly-tropic process

  12. 12.Which of the following is not property (A)heat (B)volume (C)pressure (D)temperature Ans.(A)heat

  13. 13.Which of the following is the basis of temperature measurement? (A) first law of thermodynamics (B)kelvin-planck statement (C)clausis statement (D)zeroth law of thermodynamics Ans.(D)zeroth law of thermodynamics

  14. 14.A system is said to be Homogeneous if it consist of (A)One phase (B)Two phase (C)Three phase (D)More than three phase Ans.(A)One phase

  15. 15.A system is said to be Heterogeneous if it consist of (A)One phase (B)Two phase (C)Three phase (D)More than three phase Ans.(B)Two phase

  16. 16. The exact relation between Celsius and Fahrenheit is (A)(C/5)=((F-32)/9) (B) (C/5)=((F-42)/9) (C) (C/5)=((F-33)/9) (D) (C/5)=((F-52)/9) Ans.(C/5)=((F-33)/9)

  17. 17. On a piston of 20cm diameter a force of 5000N is uniformly applied. Pressure on the piston will be (A)1.591 Pa (B)1.591 kpa (C)1.591 bar (D)1.591 atm Ans. (C)1.591 bar Solution:Force/Area 5000/(3.14*0.2*0.2/4) =159100Pa=1.591 bar

  18. 18.Fahrenheit scale shows 108.90F,corresponding scale in 0C will be (A)43 (B)70 (C)23 (D)84 Ans. (A)43

  19. 19.Internal energy of a perfect gas depends on (A) temperature, specific heats and pressure (B) temperature, specific heats and enthalpy (C) temperature, specific heats and entropy (D) Temperature only. Ans.(D) Temperature only.

  20. 20.For cyclic process (A)Work transfer is zero (B)Heat transfer is zero (C)Change in internal energy is zero (D)All the three variable are zero. Ans.(C)Change in internal energy is zero

  21. 21.According to Kelvin-Planck’s statement of second law of thermodynamics • It is impossible to construct an engine working on a cyclic process, • whose sole purpose is to convert heat energy into work • (B) It is possible to construct an engine working on a cyclic process, • whose sole purpose is to convert the • heat energy into work • (C) It is impossible to construct a device which while working in a cyclic • process produces no effect other than the transfer of heat from a colder • body to a hotter body • (D) None of the above. Ans.(D) None of the above.

  22. 22.Which of the following statements is correct according to Clausius statement of • second law of Thermodynamics? • It is impossible to transfer heat from a body at a lower temperature to • a body at a higher temperature • (B) It is impossible to transfer heat from a body at a lower temperature to • a body at a higher temperature, without the aid of an external source. • (C) It is possible to transfer heat from a body at a lower temperature to • a body at a higher temperature by using refrigeration cycle • (D) None of the above. Ans.(B) It is impossible to transfer heat from a body at a lower temperature to a body at a higher temperature, without the aid of an external source.

  23. 23. The ideal gas equation is (A)PV=RT (B)PV=MRT (C)PVn=RT (D)PVn=MRT Ans.(B)PV=MRT

  24. 24. Mechanical efficiency is (A) (Break Power/Friction Power) (B) (Break Power/Indicated Power) (C) (Indicated Power /Break Power) (D) (Indicated Power/Friction Power) 24.(B) (Break Power/Indicated Power)

  25. 25. Which of the following is defined by second law of thermodynamics? (A)Internal energy (B)Enthalpy (C)Entropy (D)Heat Ans.(C)Entropy

  26. 26.The sequence of process that eventually returns the working substance to its original state is known as (A)Event (B)Thermodynamic cycle (C) Thermodynamic property (D)None of these Ans.(B)Thermodynamic cycle

  27. 27. 1TR= (A)232.6 KJ/min (B)252.6 KJ/min (C)262.6 KJ/min (D)275.6 KJ/min Ans.(A)232.6 KJ/min

  28. 28. Specific heat is the amount of heat required to raise the temperature (A) By unit degree of a substance (B) By unit degree of a unit mass (C) Of a unit mass by 10 (D) None of the above Ans.(B) By unit degree of a unit mass

  29. 29. According to Kelvin planks statement ,a perpetual motion machine of (A)First kind is possible (B)First kind is impossible (C)Second kind is impossible (D)Second kind is possible Ans.(C)Second kind is impossible

  30. 30. Total Degree of freedom for monoatomic gas is (A) 1 (B) 2 (C) 3 (D) 4 Ans.(C) 3

  31. 31. A system comprising of a single phase is known as (A) Open system (B)Closed system (C) Homogeneous system (D) Heterogeneous system. Ans.(C) Homogeneous system

  32. 32. Total Degree of freedom for Diatomic gas is (A) 5 (B) 2 (C) 3 (D) 4 Ans.(A) 5

  33. 33. Kelvin-Planck’s law deals with (A) conservation of energy (B) conservation of heat (C) conservation of mass (D) conversion of heat into work Ans.(D) conversion of heat into work

  34. 34. Enthalpy is (A)U+E (B)U+PT (C)U+PV (D)U+VT Ans.(C)U+PV

  35. 35.The correct relationship between specific heats is (A) Cp+Cv=R (B)Cp-Cv=R (C) Cp-Cv=P (D) Cp+Cv=P Ans.(B)Cp-Cv=R

  36. 36.………………..is the example of zeroth law of thermodynamics (A) Thermocouple (B) Thermometer (D) Thermister (D) RTD Ans.(B) Thermometer

  37. 37. 1cal= (A) 4.187 J (B)5.187 J (C)4.190 J (D)6.000 J Ans.(A) 4.187 J

  38. 38. 1 KW= (A)900 kcal/hr (B)1000 kcal/hr (C)860 kcal/hr (D)760 kcal/hr Ans.(C)860 kcal/hr

  39. 39. Efficiency of heat engine (A) (1-(Q2/Q1)) (B) (1-(Q1/Q2)) (C) 1 (D) None of these Ans.(A) (1-(Q2/Q1))

  40. 40. (C.O.P)pump= (A) 1 (B) (C.O.P)R+1 (C) (C.O.P)R-1 (D) 0 Ans.(B) (C.O.P)R+1

  41. 41.First law states the (A) Energy balance (B)Mass balance (C) Enthalpy Balance (D)Entropy balance Ans.(A) Energy balance

  42. 42.A cylinder contains 0.28 m3of oxygen at 3.5 bar. What will be its volume when • Expanded to 1.5 bar. At constant temperature. • 0.523m3 • 0.653m3 • 0.238m3 • 0.547m3 Ans : (B) 0.653 m3 Solution P1V1=P2V2 V2=P1V1/P2 =(3.5/1.5)*0.28 =0.653 m3

  43. 43.In an internal combustion engine, during the compression stroke the heat rejected to the cooling water is 50 kJ/kg and the work input is 100 kJ/kg. Calculate the increase in internal energy of the working fluid. (A)100 kJ (B)50KJ (C)150kJ (D)75 kJ Ans:(B) 50 kJ Solution: Heat rejected to the cooling water, Q = – 50 kJ/kg (–ve sign since heat is rejected) Work input, W = – 100 kJ/kg (–ve sign since work is supplied to the system) Using the relation, Q = (u2 – u1) + W – 50 = (u2 – u1) – 100 u2 – u1 = – 50 + 100 = 50 kJ/kg

  44. 44.10 kg of fluid per minute goes through a reversible steady flow process. The properties of fluid at the inlet are : p1 = 1.5 bar, ρ1 = 26 kg/m3, C1 = 110 m/s and u1 = 910 kJ/kg and at the exit are p2 = 5.5 bar, ρ2 = 5.5 kg/m3, C2 = 190 m/s and u2 = 710 kJ/kg. During the passage, the fluid rejects 55 kJ/s and rises through 55 meters. Determine :(i) The change in enthalpy (Δ h) ; (A)210kJ/kg (B) 135kJ/kg (C)105.77kJ/kg (D)204.65kJ/kg Ans:(C) 105.77 kJ/kg Solution: Δh = Δu + Δ(pv) Δ(pv) =p2v2- p1v1=p2/ρ2-p1/ρ1 =5.5*10^5/5.5-1.5*10^5/26 =105 × 0.9423J = 94.23 kJ Δu = u2 – u1 = (710 – 910) = – 200 kJ/kg Substituting the value in eqn. (i), we get Δh = – 200 + 94.23 = – 105.77 kJ/kg.

  45. 45.10 kg of fluid per minute goes through a reversible steady flow process. The properties of fluid at the inlet are : p1 = 1.5 bar, ρ1 = 26 kg/m3, C1 = 110 m/s and u1 = 910 kJ/kg and at the exit are p2 = 5.5 bar, ρ2 = 5.5 kg/m3, C2 = 190 m/s and u2 = 710 kJ/kg. During the passage, the fluid rejects 55 kJ/s and rises through 55 meters. Determine :Work done during the process (W) (A) – 39.46 kW. (B) 39.46 kW.(C)-45.36 kW. (D)45.36 kW. Ans:(A) – 39.46 kW The steady flow equation for unit mass flow can be written as Q = Δ KE + Δ PE + Δ h + W where Q is the heat transfer per kg of fluid Q = 55 kJ/s =55/(10/60)= 55 × 6 = 330 kJ/kg Now, ΔKE C22-C12/2=1902-1102/2=12000J =12kJ ΔPE = (Z2 – Z1) g = (55 – 0) × 9.81 = 539.5 J or ≈ 0.54 kJ/kg Substituting the value in steady flow equation, – 330 = 12 + 0.54 – 105.77 + W or W = – 236.77 kJ/kg. Work done per second = – 236.77*10/60 = – 39.46 kJ/s = – 39.46 kW.

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