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Work Transfer
S.Gunabalan
Associate Professor
Mechanical Engineering Department
Bharathiyar College of Engineering & Techn...
Energy interaction in a closed system
• A closed system and its surroundings can
interact in two ways:
– (a) by work trans...
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 J...
Work
• Work - basic modes of energy transfer
Work
• When work is done by a system, it is
arbitrarily taken to be positive, and
• when work is done on a system, taken t...
Displacement Work
Quasi-static process : is an infinitesimal slow process, where each state of the system
is pass through ...
Point-path functions
Work is depends on the path
of the process.
work is a path function
Thermodynamic properties (P,V,T) ...
Cyclic Process
• the initial and final states of the system are
the same.
• The change in any property is zero
• cyclic in...
Thermodynamic Processes
• Constant pressure process (isobaric)
• Pressure constant (p1= p2)
• n = 0
• = ∆u + W (first law ...
Thermodynamic Processes
• Constant Volume process
(isochoric process)
• Volume constant (V1= V2)
• n = ∞
• ∆u = ( 2 − 1)
•...
Thermodynamic Processes
• Constant Temperature process
(isothermal process)
Process in which pV = C
• (T1= T2)
• n = 1
• ∆...
pdV-Work - Quasi-Static Processes
• Process in which pV = C
• 12 = ∫ 	
– PV = P1V1 = C
– P =
P1V1
• 12 = ∫
P1V1
• 12 = P1V...
Thermodynamic Processes
Process in which pV n = C
Polytrophic process
12 = 	
– = 1 1
– P =
12 =
1 1
12 = 1 1
1
= 1 1
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
...
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
...
Thermodynamic Processes
Process in which pV n = C
=
	
=
	
=
∆ = −
= ∆ +
Also
= 	 	
(T2-T1)
=	
	
*
= 	
= 	
Polytrophic proc...
Thermodynamic Processes
Process in which pV = C
=
No heat is transferred to or from the system
It require perfect thermal ...
Thermodynamic Processes
Process in which pV = C
= /
= − +
Hyperbolic process
But not necessarily
T = Constant
Free Expansion
• Expansion of gas against vacuum is called free
expansion
• 	 = 0
• 	 = 0
Heat Transfer
• Heat is defined as the form of energy that is
transferred across a boundary by virtue of a
temperature dif...
Heat Transfer
• At constant pressure
– = 	
– = 	 ( )
– = KJ
• At constant Volume
– = 	
– heat capacity
Heat Transfer is a ...
Heat Transfer
• Heat flow into a system – positive
• heat flow out o f a system - negative
Short Answer
• What is an indicator diagram ?
– An indicator diagram is a trace made by a
recording pressure gauge.
• Defi...
Questions -1
A 280mm diameter cylinder fitted with a frictionless leak
proof piston contains 0.02 Kg of steam at a pressur...
Given data
– Cylinder size : 280 diameter (D)
– Mass of steam 0.02 Kg (m)
– pressure of 0.6MPa (p1)
– temperature of 200 o...
Short Note
• =
– 	 log =		log
• P1,P2,t1,m Given
– Find V1
• 1 1 = 	 1
= /M
V2 – Final volume
The conditions are
– Cylinde...
Ans
A) the value of n
	 log =		log
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
A) the value of n
	 log =		log
. 	
. 	
Here need v1 and v2
Find
V1 – initial volume
The initial conditions are
Mass of ste...
1 1 = 	 1
0.6	 	 	 1 = 	0.02	 	 	 	 	 200 + 273
=
= /M
	− 	 	 . 	⁄
= . 	 . 	⁄
		 − 	 	 .
− 	 	 	 	
Mass of steam 0.02 Kg (...
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....
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	
		
	...
Exercise -1
Find
V2 – initial volume
The conditions are
– Cylinder size : 280 diameter (D)
– As the piston moves a distanc...
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
Subs...
pdV-Work - Quasi-Static Processes
Process in which pV n = C
=
	 −
−
Process in which pV = Constant
= P1V1 OR = P1V1
Consta...
– Determine
B) Work done by the steam
• =
	
• Values from given data
– P1 = 0. 6
– 2 = 	0.12	
• Calculated values
– = 1.69...
– Determine
The magnitude and	sign of heat transfer
= 	 	
(T2-T1)
=	
	
*
=
= −
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 ba...
Ex – 1 - Note
1. Mass of the gas
• 1 1 = 	 1
2. n – (index of compression)
– =
» 	 log =		log
3. Work Done
=
	
4. increase...
Prob-2
• Air At 0.1 MPa at 25℃, initially occupies a volume
of 0.016 m3. it is compressed reversibly and
adiabatically to ...
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 t...
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 t...
Reference
• Nag, P. K. 2002. Basic and applied thermodynamics. Tata McGraw-Hill, New
Delhi.
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02 part3 work heat transfer first law

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Transcript of "02 part3 work heat transfer first law"

  1. 1. Work Transfer S.Gunabalan Associate Professor Mechanical Engineering Department Bharathiyar College of Engineering & Technology Karaikal - 609 609. e-Mail : gunabalans@yahoo.com
  2. 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. 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. 4. Work • Work - basic modes of energy transfer
  5. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 16. Thermodynamic Processes Process in which pV n = C = = = ∆ = − = ∆ + Also = (T2-T1) = * = = Polytrophic process
  17. 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. 18. Thermodynamic Processes Process in which pV = C = / = − + Hyperbolic process But not necessarily T = Constant
  19. 19. Free Expansion • Expansion of gas against vacuum is called free expansion • = 0 • = 0
  20. 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. 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. 22. Heat Transfer • Heat flow into a system – positive • heat flow out o f a system - negative
  23. 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. 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. 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. 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. 27. Ans A) the value of n log = log
  28. 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. 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. 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. 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. 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. 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. 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. 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. 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. 37. – Determine The magnitude and sign of heat transfer = (T2-T1) = * = = −
  38. 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. 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. 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. 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. 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. 43. Reference • Nag, P. K. 2002. Basic and applied thermodynamics. Tata McGraw-Hill, New Delhi.
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