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02 part3 work heat transfer first law

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- 1. Work Transfer S.Gunabalan Associate Professor Mechanical Engineering Department Bharathiyar College of Engineering & Technology Karaikal - 609 609. e-Mail : gunabalans@yahoo.com
- 2. Energy interaction in a closed system • A closed system and its surroundings can interact in two ways: – (a) by work transfer, – (b) by heat transfer. These may be called energy interactions and these bring about changes in the properties of the system.
- 3. work • work is the product of force and distance – The action of a force on a moving body is identified as work – N.m or Joule [1 Nm = 1 Joule], • Power is the rate at which work is done – J/s or watt
- 4. Work • Work - basic modes of energy transfer
- 5. Work • When work is done by a system, it is arbitrarily taken to be positive, and • when work is done on a system, taken to be negative
- 6. Displacement Work Quasi-static process : is an infinitesimal slow process, where each state of the system is pass through the equilibrium state. Work is a path function pdV-Work
- 7. Point-path functions Work is depends on the path of the process. work is a path function Thermodynamic properties (P,V,T) are all point functions. For a given state, there is a definite value for each property. The change in a thermodynamic properties are depends only on the initial and final states of the system.
- 8. Cyclic Process • the initial and final states of the system are the same. • The change in any property is zero • cyclic integral o f a property is always zero
- 9. Thermodynamic Processes • Constant pressure process (isobaric) • Pressure constant (p1= p2) • n = 0 • = ∆u + W (first law of thermodynamics) • ∆u = ( 2 − 1) • = ( 2 − 1) • W12 = ∫ = (v2-v1) Quasi-Static Processes =
- 10. Thermodynamic Processes • Constant Volume process (isochoric process) • Volume constant (V1= V2) • n = ∞ • ∆u = ( 2 − 1) • W12 = ∫ = (v1-v2) = 0 • = ∆u Quasi-Static Processes =
- 11. Thermodynamic Processes • Constant Temperature process (isothermal process) Process in which pV = C • (T1= T2) • n = 1 • ∆u = 2 − 1 = 0 • W12 = ∫ = P1V1 • = Quasi-Static Processes =
- 12. pdV-Work - Quasi-Static Processes • Process in which pV = C • 12 = ∫ – PV = P1V1 = C – P = P1V1 • 12 = ∫ P1V1 • 12 = P1V1 ∫ 1 • = P1V1 OR = P1V1 Quasi-Static Processes
- 13. Thermodynamic Processes Process in which pV n = C Polytrophic process 12 = – = 1 1 – P = 12 = 1 1 12 = 1 1 1 = 1 1
- 14. pdV-Work - Quasi-Static Processes Process in which pV n = C = 1 1 = 1 1 − + 1 2 1 = 1 1 2 − 1 − + 1 = 1 1 2 − 1 1 1 − + 1 Polytrophic process
- 15. pdV-Work - Quasi-Static Processes Process in which pV n = C = 1 1 2 − 1 1 1 − + 1 = 2 − 1 1 − + 1 1 1 = 2 2 = c = 2 2 2 − 1 1 − + 1 = 2 2 − 1 1 − + 1 = − − Polytrophic process
- 16. Thermodynamic Processes Process in which pV n = C = = = ∆ = − = ∆ + Also = (T2-T1) = * = = Polytrophic process
- 17. Thermodynamic Processes Process in which pV = C = No heat is transferred to or from the system It require perfect thermal insulation ∆ = = − − = -1= = Adiabatic process
- 18. Thermodynamic Processes Process in which pV = C = / = − + Hyperbolic process But not necessarily T = Constant
- 19. Free Expansion • Expansion of gas against vacuum is called free expansion • = 0 • = 0
- 20. Heat Transfer • Heat is defined as the form of energy that is transferred across a boundary by virtue of a temperature difference. • Energy transfer by virtue of temperature difference is called heat transfer (Q - Joules) • Conduction : The transfer of heat between two bodies in direct contact. • Radiation : Heat transfer between two bodies separated by empty space or gases through electromagnetic waves. • Convection: The transfer of heat between a wall and a fluid system in motion.
- 21. Heat Transfer • At constant pressure – = – = ( ) – = KJ • At constant Volume – = – heat capacity Heat Transfer is a boundary phenomenon, for isolated system Heat Transfer and work transfer is zero
- 22. Heat Transfer • Heat flow into a system – positive • heat flow out o f a system - negative
- 23. Short Answer • What is an indicator diagram ? – An indicator diagram is a trace made by a recording pressure gauge. • Define Quasi-static process ? – It is an infinitesimal slow process, where each state of the system is pass through the equilibrium state.
- 24. Questions -1 A 280mm diameter cylinder fitted with a frictionless leak proof piston contains 0.02 Kg of steam at a pressure of 0.6MPa and a temperature of 200 oC As the piston moves slowly outwards through a distance of 305 mm, the steam undergoes a fully resisted expansion during which the steam pressure p and the steam volume V are related by = , ℎ . The final pressure of the steam is 0.12MPa. Determine A) the value of n B) Work done by the steam • (Apr-May 2012 Pondicherry university)
- 25. Given data – Cylinder size : 280 diameter (D) – Mass of steam 0.02 Kg (m) – pressure of 0.6MPa (p1) – temperature of 200 oC (T1) – As the piston moves a distance of 305 mm (h) • = , ℎ . • The final pressure of the steam is 0.12MPa. (p2) – Determine • A) the value of n • B) Work done by the steam
- 26. Short Note • = – log = log • P1,P2,t1,m Given – Find V1 • 1 1 = 1 = /M V2 – Final volume The conditions are – Cylinder size : 280 diameter (D) – As the piston moves a distance of 305 mm (h) 2 = ℎ Process in which pV n = C = − − Mass of steam 0.02 Kg (m) P1 = pressure of 0.6MPa (p1) temperature of 200 oC (T1) p2 = 0.12MPa
- 27. Ans A) the value of n log = log
- 28. A) = 1 1 = 2 2 log 1 + log 1 = log 2 + log 2 log 1 − log 2 = log 2 - log 1 log = log Concept poof
- 29. A) the value of n log = log . . Here need v1 and v2 Find V1 – initial volume The initial conditions are Mass of steam 0.02 Kg (m) pressure of 0.6MPa (p1) temperature of 200 oC (T1)
- 30. 1 1 = 1 0.6 1 = 0.02 200 + 273 = = /M − . ⁄ = . . ⁄ − . − Mass of steam 0.02 Kg (m) pressure of 0.6MPa (p1) temperature of 200 oC (T1)
- 31. 1 1 = 1 0.6 1 = 0.02 200 + 273 = /M = . . = ? R = 0.461 KJ/Kg K Mass of steam 0.02 Kg (m) pressure of 0.6MPa (p1) temperature of 200 oC (T1) Gas Molar Weight ( M)Kg/Kmol Air 28.97 Nitrogen 28.01 Oxygen 32 Hydrogen 2.016 Helium 4.004 Carbon dioxide 44.01 Steam 18.02
- 32. 1 1 = 1 0.6 1 = 0.02 0.461 200 + 273 What is the unit of V1 ? 0.6 1 3 = 0.02 0.461 ⁄ 200 + 273 V1 = ? V1 = 0.007268 3 Mass of steam 0.02 Kg (m) pressure of 0.6MPa (p1) temperature of 200 oC (T1) = = 106 106 2 = 2 KJ = ⁄
- 33. Exercise -1 Find V2 – initial volume The conditions are – Cylinder size : 280 diameter (D) – As the piston moves a distance of 305 mm (h) 2 = ℎ 2 = 140 305 mm 2 = 18770920 mm3 2 = ( ) m3 2 = 0.018770920 m3
- 34. A) the value of n 1 1 = 2 2 log 1 2 = log 0.12 0.6 Here need v1 and v2 Find 1 = 0.007268 3 2 = 0.018770920 3 Substituting V1 and V2 log . . = log . . = 1.6963
- 35. pdV-Work - Quasi-Static Processes Process in which pV n = C = − − Process in which pV = Constant = P1V1 OR = P1V1 Constant Volume process (isochoric process) dV = 0 = 0 Constant pressure process (isobaric) = p(v2-v1)
- 36. – Determine B) Work done by the steam • = • Values from given data – P1 = 0. 6 – 2 = 0.12 • Calculated values – = 1.6963 – 1 = 0.007268 3 – 2 = 0.018770920 3 = 0. 6 106 0.007268 3 − 0.12 106 0.018770920 3 1.6963 − = - 3735.67 Nm or J
- 37. – Determine The magnitude and sign of heat transfer = (T2-T1) = * = = −
- 38. Ex - 1 A Cylinder containing 0.4 m3 of gas at 1 bar and 75 ℃. the gas is compressed to 0.15 m3, the final pressure is 4 bar. Take = 1.4, R = 0.2942 KJ/kg ℃ Find 1. Mass of the gas 2. n – (index of compression) 3. Work Done 4. increase in internal energy of gas 5. Heat transfer.
- 39. Ex – 1 - Note 1. Mass of the gas • 1 1 = 1 2. n – (index of compression) – = » log = log 3. Work Done = 4. increase in internal energy of gas = 2 − 1 C = R = 0.2942 KJ/kg ℃ - so use Temperature in ℃ 5. Heat transfer. Q = u +W = (T2-T1) = * = =
- 40. Prob-2 • Air At 0.1 MPa at 25℃, initially occupies a volume of 0.016 m3. it is compressed reversibly and adiabatically to a pressure of 0.6MPa. Find the a) final temperature, b) Final Volume and c)Work Done. Take = 1.4 – for air – Final Temperature • = – Final Volume • = ∆ = = − − Adiabatic process Q = 0
- 41. Prob-3 • 1.3 Kg of liquid having constant specific heat of 2.3 KJ/kgK, is stirred in a well insulated chamber, causing a temperature rise to 15 C. • Calculate the work done and internal energy. • well insulated – No heat transfer or loss Q = 0 Q = u + W Q = 0 W = -u U = m Cv (T2-T1)
- 42. Ex - 2 • 1.3 Kg of liquid having constant specific heat of 2.3 KJ/kgK, is stirred in a well insulated chamber, causing a temperature rise to 15 C. During the process 1.5 KJ of heat is transferred to the system • Calculate the work done and internal energy. • well insulated – heat transfer Q = 1.5KJ Q = u + W U = m Cv (T2-T1) W = Q-u
- 43. Reference • Nag, P. K. 2002. Basic and applied thermodynamics. Tata McGraw-Hill, New Delhi.

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