Transcript of "02 part3 work heat transfer first law"
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
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
=
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
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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