heat, work and energy topic for thermodynamics

thermo

1. ME-THER THERMODYNAMICS

2. Learning Plan – Module 1 Week 1 – 6 ( 18 hrs Sept 1 – October 9) Week 1 – 2 (6 hrs Sept 2 and Sept 9 ) - Grading policies, course syllabus, expectations. 1 - Basic Principles, Concepts and Definitions Week 3 – 4 (6 hrs Sept 16 and 23 ) 2 - First Law of Thermodynamics – Conservation of Energy Week 5 – 6 (6 hrs Sept 30 and October 7 ) 3 - Ideal Gas Week 5 Oct 7 W – Preliminary Examination

3. Learning Plan – Module 1 Week 1 – 2 (6 hrs Sept 2 – Sept 9 ) Lesson 1 - Basic Principles, Concepts and Definitions Thermodynamics is a branch of physical sciences that treats various phenomena of energy and related properties of matter , especially the laws of transformation of heat into other forms of energy and vice versa. Systems of units Newton’s law states that acceleration of a body is directly proportional to force acting on it and inversely proportional to mass. a = k F/m F = ma / k k = ma / F where k is a proportional constant

4. Learning Plan – Module 1 cgs system: 1 dyne accelerates 1 gram mass at 1 cm / sec 2 mks system: 1 newton accelerates 1 kilogram mass at 1 m / sec 2 fps system: 1 pound-force accelerates 1 slug mass at 1 ft / sec 2 1g 1 dyne 1kg 1 newton 1 slug 1 lbf a a a 1 cm/sec 2 1 m/sec 2 1 ft /sec 2 k = 1 g-cm / dyne-sec 2 k = 1 kg-m/N-sec 2 k = 1 slug- ft /lbf-sec 2 Question: How many dynes in 1 newton , where 1N = 1 kg-m / sec 2 Answer : 1 x 10 5 dynes

5. Learning Plan – Module 1 Acceleration values 1 gram force accelerates a 1 gram mass at 980.66 cm/sec 2 1 kg force accelerates a 1 kg mass at 9.8066 m/sec 2 1 lb force accelerates a 1 lb mass at 32.174 ft /sec 2 1g m 1 g f 1kg m 1 kg f 1 lb m 1 lb f a a a 980.66 cm/sec 2 9.8066 m/sec 2 32.174 ft /sec 2 k= 980.66 g m -cm/ g f -s 2 k= 9.8066 kg m -m/ kg f -s 2 k= 32.174 lb m -ft /lb f -s 2 Question: If 1 N = 1 kg-m/sec 2 and 1 kgf = 9.8066 kg m -m/sec 2 , how many Newtons in 1 kg f ? Answer : 1 kg f = 9.8066 N

6. Learning Plan – Module 1 Solution: Relation between Kilogram force ( Kg f ) and Newton (N) k = 1 kg-m/N-sec 2 k = 9.8066 kg-m/ kg f - s 2 Equating k’s: 1 kg-m/N-sec 2 = 9.8066 kg-m/ Kg f - s 2 Therefore, 1 Kg f = 9.8066 N Or, 1 Kg f = 9.8066 kg-m/ s 2 x 1 N / 1 kg-m/ s 2 = 9.8066 N Question: How many Pound mass (lb m ) is 1 slug ? Ans : 32.174 lb m k = 32.174 lb m - ft / lb f - s 2 k = 1 slug- ft / lb f - s 2 Equating k’s: 1 slug - ft / lb f - s 2 = 32.174 lb m - ft / lb f - s 2

7. Learning Plan – Module 1 Acceleration A unit of force is one that produces 1 unit acceleration in a body of 1 unit of mass. F = mass x acceleration 1g 1 dyne 1kg 1 newton 1 slug 1 lbf a = 1 cm/sec 2 a = 1 m/sec 2 a = 1 ft /sec 2 1 dyne = 1 g x 1 cm/sec 2 1 lbm 1 poundal 1 Newton = 1 kg x 1 m/sec 2 1 lbf = 1 slug x 1 ft /sec 2 a = 1 ft /sec 2 1 poundal = 1 lbm x 1 ft /sec 2

8. Learning Plan – Module 1 Mass and Weight The mass of a body is the absolute quantity of matter in it. The weight of a body means the force of gravity on the body. F = ma / k or m / k = F / a = F g / g where g = acceleration due to gravity produced by force F g a = acceleration produced by another force F At or near the surface of the surface of the earth, k and g are numerically equal, so are mass and F g End of lesson 1 / week 1 – next Activity #1 / week 1 in Asynchronous period

9. Learning Plan – Module 1 Activity #1 – Lesson 1 / Week 1 Due: Sept 9, 2020 (week 2) Problems: What is the weight ( Kg f ) of a 66 kg-mass man at standard condition? m = 66 kg m g = 9.8066 m/s 2 What is the mass ( lb m ) of an object at standard condition which weighs 50 lb ? F g = 50 lb f g = 32.174 ft /s 2 Five masses where the acceleration due to gravity is 30.5 ft /s 2 are as follows: m1 = 500 gram mass, m2 = weighs 800 gram force, m3 = 15 poundals , m4 = 3 lbf and m5 = 0.01 slug mass. What is the total mass expressed in a) grams b) pounds c) slugs 30% presentation = 30 70% content / solution – accuracy = 70

10. Learning Plan – Module 1 Week 2 (3 hrs Sept 9 ) Lesson 1 - Basic Principles, Concepts and Definitions Specific Volume, Density and Specific Weight The density, ρ of a substance is its mass (not weight) per unit volume ρ = m / V (rho) The specific volume, ν is the volume per unit mass. ν = V / m = 1/ ρ ( nu) The specific weight , ɣ of a substance is the force due to gravity per unit volume. ɣ = Fg / V (gamma) Since the specific weight is to local acceleration of gravity , the density is to standard acceleration. ɣ / g = ρ / k

11. Learning Plan – Module 1 ɣ / g = ρ / k ρ = ɣ k /g ɣ = ρ g / k At or near the surface of the earth , k and g are numerically equal, so are ρ and ɣ Example: What is the specific weight of water at standard condition . g = 9.8066 m/s 2 ρ = 1000 kg / m 3 ɣ = ρ g / k = 1000 kg / m 3 x 9.8066 m / s 2 9.8066 kg - m / kg f - s 2 = 1000 kg f /m 3

12. Learning Plan – Module 1 Pressure The standard reference atmospheric pressure is: Po 760 mm Hg or 29.92 inches Hg at 32 o F or 14.696 psi absolute ( psi a ) or 14.7 psia 1 atmosphere ( atm ) Measuring Pressure By using manometers By using pressure gages

13. By using manometers

14. Learning Plan – Module 1 By using pressure gages

15. Learning Plan – Module 1

16. Learning Plan – Module 1 Gage pressure

17. Learning Plan – Module 1 Solution p g = ρ g h g / k P = F / A = N/ m 2 2 = 9.65 m / s 2 ( 1878 kg / m 3 ) (30 m ) 1 kg - m / N- s 2 = 543,680 N / m 2 = 543.68 KPa (gage) NOTE : 1 PASCAL = 1 N / m 2 1 KILO PASCAL ( kPa ) = 1000 N / m 2 Pabs = Po + Pg = Po + 543.68 kPa = Po + 543.68 kPa (abs)

18. Learning Plan – Module 1 Atmospheric Pressure

19. Learning Plan – Module 1 Problem 1: A vertical column of water will be supported to what height by standard atmospheric pressure? P = ɣ h

20. Learning Plan – Module 1 Atmospheric Pressure Problem Sp gr of Hg = 13.6 P = ɣ h ɣ Hg = sg ɣ w

21. Learning Plan – Module 1 Absolute Pressure

22. Learning Plan – Module 1 Absolute Pressure Ans : 54.5 psia = 375.78 kPa

23. Learning Plan – Module 1 Absolute Pressure 1 lbf = 1N / 0.2248 k = 1 1

24. Learning Plan – Module 1 Absolute Pressure 1 lbf = 1N / 0.2248 2 1

25. Learning Plan – Module 1 Temperature

26. Learning Plan – Module 1 Temperature

27. Learning Plan – Module 1 Temperature

28. Learning Plan – Module 1 Conservation of Mass

29. Learning Plan – Module 1 Conservation of Mass where ρ = 1/ ν m =

30. Learning Plan – Module 1 Conservation of Mass

31. Learning Plan – Module 1 Conservation of Mass m = V ρ m = A Ʋ ρ

32. Learning Plan – Module 1 Conservation of Mass ρ = m / V h = V / A

33. Learning Plan – Module 1 ACTIVITY 2 SOLVE THE FOLLOWING PROBLEMS 1 TO 11 AS YOUR ASYNCHRONOUS ACTIVITY FOR WEEK 3 (SEPT 16) AND WEEK 4 (SEPT 23) DUE ON SEPT 25, 2020 FRIDAY 5PM

34. Learning Plan – Module 1

35. Learning Plan – Module 1

36. Learning Plan – Module 1

37. Learning Plan – Module 1 DUE ON SEPT 25, 2020 FRI 5PM

38. Learning Plan – Module 1 Week 3 – 4 - Sept 16 and 23 Lesson 2 - First Law of Thermodynamics

39. 2 - Conservation of Energy Thermodynamics General Engineering OCM

40. OCM Conservation of Energy Thermodynamics General Engineering States that the energy of an isolated system remains constant . Energy cannot be created or destroyed in an isolated system. It can only be transformed from one form to another.

41. OCM Kinetic Energy (KE) Thermodynamics General Engineering Energy or stored capacity for performing work possessed by a moving body , by virtue of its momentum = F x d = N - m = Joule K E = ½ mV 2 kg – m 2 /s 2 = kg-m/s 2 - m Δ KE = K 2 – K 1 = ½ m ( V 2 2 – V 1 2 ) Where Δ KE = change in Kinetic energy V1 m m V2

42. OCM Potential Energy (PE) Thermodynamics General Engineering Energy due to its position or elevation P E = mgh = kg-m/s 2 – m = N – m = Joule Δ PE = P 2 – P 1 = mg ( h 2 – h 1 ) Where Δ PE = change in potential energy h Fg = mg 2 1

43. OCM Internal Energy (U) Thermodynamics General Engineering Energy stored within a body or substance by virtue of the activity and configuration of its molecules and of the vibration of the atoms within the molecules. Δ U = U 2 – U 1 (m mass) Joule Experiment,1843

44. OCM Heat (Q) Thermodynamics General Engineering Energy that is transferred between two systems by a virtue of a temperature difference Q is positive when heat is added to the body. Q is negative when heat is rejected by the body.

45. OCM Heat Transfer Thermodynamics General Engineering Three mechanisms: 1. Conduction – transfer of energy from the more energetic particles of a substance to the adjacent less energetic ones as a result of interaction between particles

46. OCM Heat Transfer Thermodynamics General Engineering Three mechanisms: 2. Convection – transfer of energy between a solid surface and the adjacent fluid that is in motion. It includes combined effects of conduction and fluid motion

47. OCM Heat Transfer Thermodynamics General Engineering Three mechanisms: 3. Radiation – transfer of energy due to the emission of electromagnetic waves or photons

48. OCM Heat Transfer Thermodynamics General Engineering Three mechanisms:

49. OCM Work (W) Thermodynamics General Engineering Energy transfer associated with a force acting through a distance . W = F x d = Joules W is positive when work is done by the system . W is negative when work is done on the system . W = ʃ 1 2 p dV = F x d ( N-m) Where p s pressure ( N/m 2 ) or Pascal and dV is differential Volume ( m 3 ) = W ( N-m ) or Joule

50. OCM Flow Work (W f ) Thermodynamics General Engineering Flow work or flow energy is work done in pushing a fluid across a boundary, usually into or out of a system. W f = FL = pAL W f = p V Δ W f = W f2 – W f1 = p 2 V 2 – p 1 V 1 Δ W f = change in flow work L Area of surface F V system Flow Work Boundary p

51. OCM Steady Flow Energy Equation Thermodynamics General Engineering Characteristics of steady flow system There is neither accumulation nor diminution of mass within the system There is neither accumulation nor diminution of energy within the system The state of the working substance at any point in the system remains constant

52. OCM Steady Flow Energy Equation Thermodynamics General Engineering Energy entering System = Energy Leaving System P 1 + K 1 + W f1 + U 1 + Q = P 2 + K 2 + W f2 + U 2 + W Q = (P 2 - P 1 )+(K 2 - K 1 )+ (W f2 - W f1 )+(U 2 - U 1 )+ W Q = Δ P+ Δ K + Δ W f + Δ U + W Where Q = Heat P = Potential energy = mgh K = Kinetic energy = ½ mV 2 W f = Flow work or Flow energy = pV = FL = pAL U = Internal energy W = Work = F d = pAL

53. OCM Steady Flow Energy Equation Thermodynamics General Engineering P 1 K 1 W f1 U 1 Q 2 2 System 1 h 1 h 2 W U 2 W f2 K 2 P 2 Datum level 1

54. OCM Enthalpy ( H ) Thermodynamics General Engineering It is a thermodynamic quantity equivalent to the total heat content of a system. H = U + pV Q = Δ U + W Q = Δ H = H 2 – H 1 The steady flow energy equation becomes P 1 + K 1 + ( W f1 +U 1 )+ Q = P 2 + K 2 + ( W f2 +U 2 )+ W P 1 + K 1 + H 1 + Q = P 2 + K 2 + H 2 + W Q = Δ P+ Δ K + Δ H + W

55. OCM Example Problem Thermodynamics General Engineering 1. During a steady flow process, the pressure of the working substance drops from 200 to 20 psia . The speed increases from 200 to 1000 ft /sec . The internal energy U of the open system decreases by 25 Btu/lb . the specific volume increases from 1 to 8 ft 3 /lb. No heat is transferred . Sketch an energy diagram. Determine the Work per lb . Is it done on ( - ) or by (+) the substance? Determine the Work in Hp for 10 lb per min. ( 1Hp = 42.4 Btu/min ) W p1 = 200 psia p2 = 20 psia v1 = 200 fps K1 K2 v2 = 1000 fps ν1 = 1 ft 3 / lb W f1 W f2 ν2 = 8 ft 3 / lb Δ U = -25 Btu/ lb U1 U2 Q = 0 Energy System P 1 + K 1 + W f1 + U 1 + Q = P 2 + K 2 + W f2 + U 2 + W + _

56. OCM Example Problem Thermodynamics General Engineering Assume 1 lbm of substance K1= ½ mV 2 /k = (200 ft / sec ) 2 / 2 (32.174 lbm - ft / lbf - sec 2 ) (778 lbf - ft / Btu ) = 0.80 Btu/ lbm K2 = ½ mV 2 /k = ( 1000) 2 / 2 (32.174) (778) = 19.97 Btu/ lbm W f1 = p V = (200 lbf / in 2 ) ( 144 i n 2 / ft 2 ) (1 ft 3 / lbm ) / 778 ft - lbf / Btu = 37.02 Btu/ lbm W f2 = p V = 20 (144) (8) / 778 = 29.61 Btu / lbm No heat is transferred , thus, Q = 0 , assume same elevation, no Δ P P 1 + K 1 + W f1 + U 1 + Q = P 2 + K 2 + W f2 + U 2 + W K 1 + W f1 + U 1 = K 2 + W f2 + U 2 + W 0.8 + 37.02 = 19.97 + 29.61 + (U 2 - U 1 ) + W 0.8 + 37.02 = 19.97 + 29.61 + (- 25 ) + W W = + 13.24 Btu / lbm ( positive so W is done by the substance) W = 13.24 Btu / lbm ( 10 lbm / min ) / 42.4 Btu / min / Hp = 3.12 Hp

57. OCM Example Problem Thermodynamics General Engineering 2. Steam is supplied to a fully loaded 100 hp turbine at 200 psia with U1 = 1163.3 Btu/ lb , V1 = 2.65 ft3/ lb and v1 = 400 ft /sec . Exhaust is at 1 psia with U2 = 925 Btu/ lb , V2 = 294 ft3/ lb and v2 = 1100 ft /sec . The heat loss from the steam in the turbine is 10 Btu/lb . Neglect Potential energy change, determine , Work per lb of steam Steam flow rate in lb /hour (1Hp = 2544 Btu/ hr ) = 42.4 x 60 P 1 + K 1 + W f1 + U 1 + Q = P 2 + K 2 + W f2 + U 2 + W K 1 + W f1 + U 1 + Q = K 2 + W f2 + U 2 + W K1= ½ mV 2 /k = 400 2 /2 (32.174)(778) = 3.2 Btu/ lbm K2 = ½ mV 2 /k = 1100 2 / 2 (32.174) (778) = 24.17 Btu/ lbm W f1 = p V = ( 200 ) ( 144) ( 2.65 ) / 778 = 98.10 Btu/ lbm W f2 = p V = ( 1 ) (144) ( 294 ) / 778 = 54.42 Btu/ lbm 3.2 + 98.10 + 1163.3 + (- 10 ) = 24.17 + 54.42 + 925 + W W = 251.01 Btu / lbm Steam Flow = 100 Hp (2544 Btu / hr / Hp ) / 251.01 Btu / lbm = 1014 lbm / hr

58. OCM Activity 3 Sept 23 and Sept 30 Thermodynamics General Engineering Steam enters a turbine with an enthalpy of 1292 Btu/ lb and leaves with an enthalpy of 1098 Btu/lb. The transferred heat is 13 Btu/lb. What is the Work in Btu/min and in Hp for a flow of 2 lb /sec. Ans. 512.3 Hp It could be (1292-1098)-13 *2 = 362 BTU/sec ->21720 BTU/min 2. A thermodynamic steady flow system receives 4.56 kg/min of fluid ( mass flow rate ) where p1 = 137.9 kPa , ν1 = 0.0388 m 3 /kg, v1 = 122 m/sec and U1 = 17.16 kJoules /kg. The fluid leaves the system at a boundary where p2 = 551.6 kPa , ν2 = 0.193 m 3 /kg, v2 = 183 m/sec and U2 = 52.8 kJoules /kg. During passage through the system, the fluid leaves 3,000 Joules/sec of Heat. Determine the Work. Ans. W = - 486 kJoules / min Note: 1 Joule = 1 N-m = 1 kg-m/s 2 - m

59. OCM Specific Heat Thermodynamics General Engineering It is the quantity of heat required to change the temperature of unit mass through one degree. Ration of specific heats k = c p /c v > 1

60. OCM Internal Energy of an Ideal Gas Thermodynamics General Engineering Based on Joule’s Law , Δ U = m c v (T 2 – T 1 ) whether the volume remains constant or not.

61. OCM Enthalpy of an Ideal Gas Thermodynamics General Engineering Based on Joule’s Law , Δ H = m c p (T 2 – T 1 ) whether the pressure remains constant or not.

62. OCM Constant Volume and Constant Pressure Specific Heat Thermodynamics General Engineering H = U + pV pV = RT c p dT = c v dT + R dT c p = c v + R c v = R/ k – 1 c p = kR / k - 1

63. Learning Plan – Module 1 Week 5 – 6 Sept 30 and Oct 7 Preliminary Examination – Oct 7 W Lesson 3 - Ideal Gas

64. 3 – The Ideal Gas Properties of Gases Thermodynamics General Engineering OCM

65. OCM Gas VS Vapor Thermodynamics General Engineering Vapor phase of a substance when it is above the critical temperature Has a single defined thermodynamic state at room temperature A gas that is not far from the state of condensation Mixture of two phases at room temperature

66. OCM Why is Ideal Gas called ideal? Thermodynamics General Engineering

67. OCM Why is Ideal Gas called ideal? Thermodynamics General Engineering Simply because it conforms to the simple perfect gas laws!

68. OCM Examples of Ideal Gases Thermodynamics General Engineering Air Nitrogen Oxygen Hydrogen Helium Argon Neon Krypton Heavier gases (carbon dioxide)

69. OCM Equation of State Thermodynamics General Engineering Any equation that relates the pressure, temperature and specific volume of a substance.

70. OCM Equation of State Thermodynamics General Engineering

71. OCM Gas Constant (R) Thermodynamics General Engineering It is different for each gas. R = R u /M where: R u = Universal gas constant M = Molar mass/mol. Weight

72. OCM Universal Gas Constant ( R u ) Thermodynamics General Engineering Same for all substances R u Values: 8.314 kJ/kMol.K 8.314 kPa.m3/kmol.K 0.08314 bar.m3/kmol.K 1.986 Btu/lbmol.R 10.73 psia.ft3/lbmol.R 1545 ft.lbf/lbmol.R

73. OCM Molar Mass (M) Thermodynamics General Engineering Mass of one mole, also called as gram-mole (gmol), of a substance in grams m = MN where: m = mass of system M = molar mass N = mole number

74. OCM Equation of State (Problem) Thermodynamics General Engineering 1. Determine the mass of the air in a room whose dimensions are 4m x 5m x 6m at 100 kPa and 25 degC. Gas constant of air is 0.287 kPa.m3/kg.K.

75. OCM Equation of State (Seatwork) Thermodynamics General Engineering 2. The volume of a 7 x 13 ft tank is 543.3 cu.ft . It contains air at 180 psig and 75 degF . How many 1 cu.ft . drums can be filled to 60 psig and 70 degF if its assumed that the air temperature in the tank remains at 75 degF . The drums have been sitting around in the atmosphere which is at 14.7 psia and 70 degF .

76. OCM Equation of State (Seatwork) Thermodynamics General Engineering 3. Two vessels X and Y of different sizes are connected by a pipe with a valve. Vessel X contains 164 L of air at 3246.87 kPa, 90 degC. Vessel Y, of unknown volume, contains air at 70.5 kPa, 5.6 degC. The valve is opened and when the properties have been determined, it is found that p m = 1500 kPa, t m = 40.5 degC. What is the volume of vessel B?

77. OCM Entropy (S, s) Thermodynamics General Engineering Degree of disorder or randomness in the system. It is the property of a substance which remains constant if no heat enters or leaves the substance, while it does work or alters its volume, but which increases or diminishes a small amount of heat enter or leave. Δ S = mc ln(T 2 /T 1 )

78. OCM Ideal Gas (Problem) Thermodynamics General Engineering 4. For a certain ideal gas R = 25.8 ft.lb/ lb.R and k = 1.09, What are the values of c p and c v ? What mass of this gas would occupy a volume of 15 cu.ft at 75 psia and 80 F? If 30 BTU are transferred to this gas at constant volume in (b), what are the resulting temperature and pressure?

79. OCM Ideal Gas (Problem) Thermodynamics General Engineering 5. The specific heat of superheated steam at approximately 150 kPa can be determined by the equation What is the enthalpy change between 300 ºC and 700 ºC for 3 kg of steam? Compare with the steam table (2565 kJ).

80. OCM Ideal Gas (Problem) Thermodynamics General Engineering 6. A 10 ft3 tank contains gas at a pressure of 500 psia, temperature of 85 degF and a weight of 25 pounds. A part of the gas was discharged and the temperature and pressure changed to 70 degF and 300 psia, respectively. Heat was applied and the temperature was back to 85 degF. Find the final weight, volume and pressure of the gas.

81. OCM Ideal Gas (Problem) Thermodynamics General Engineering 7. A motorist equips his automobile tires with a relief-type valve so that the pressure inside the tire will never willexceed 240 kPa (gage). He starts a trip with a pressure of 200 kPa (gage) and a temperature of 23 degC in the tires.During the long drive, the temperature of the air in the tires reaches 83 degC. Each tire contains 0.11 kg of air. Determine (a) the mass of air escaping each tire, (b) the pressure of the tire when the temperature returns to 23degree Celsius. Use R = 287.08 N.m/kg.K

82. OCM Ideal Gas (Problem) Thermodynamics General Engineering 8. An automobile tire contains 3730 cu in. of air at 32 psig and 80 degF. R = 53.342 lb.ft/lb.R (a) What mass of air is in the tire? (b) In operation, the air temperature increases to 145 degF. If the tire is inflexible, what is the resulting percentage increase in gage pressure? (c) What mass of the 145 degF air must be bled off to reduce the pressure back to its original value?

83. OCM Ideal Gas (Assignment) Thermodynamics General Engineering Guidelines: With Front Page Margin: 1x1x1x1 (Visible) Engineering Lettering Problem Statement and Solution Box your final answers and round off up to two (2) decimal places. CHAPTER 3: REVIEW PROBLEM No. 6 (A 6 m3 tank…) and No. 8 (A spherical balloon..)