5
Temperature
5.1 Definition of Temperature Equality
If 2 bodies are brought into contact, after a period of time, there is no observable change in their physical properties (example: length, electrical resistance, density etc). The bodies are said to be in thermal equilibrium and to be equal in temperature.
The 2 systems are equal in temperature when no change in any property occurs when they are brought into communication.
5.2 The Zeroth Law of Thermodynamics
(This law was formulated after the First Law of Thermodynamics. Since it is more fundamental, it is called the Zeroth Law)
It was observed experimentally that 2 systems that are equal in temperature to a third system are also equal in temperature to each other.
This may appear obvious or trivial but in general 2 systems that behave in the same way to a third system DO NOT necessarily behave in the same way with respect to each other.
For example:
or
5.2
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UEME1112 Goh Sing Yau (May 2014), FES, UTAR
5.3 The use of the Zeroth Law for Temperature measurement.
(1) S3 is brought into contact with S2. After a period of time, the temperature S2 and S3 become equal.
(2) Measure a physical property of S3. (say the length of the mercury After a period of time, the temperatures of S2 and S3 become equal.
(3) S3 is brought into contact with S1. If there is no change in properties of S3, then S1 and S2 are equal in temperature.
If S1 and S2 are at different temperatures, a measure of the change in the physical property
(eg the length of the mercury column) is a measure of the temperature difference.
6. Heat Transfer
6.1: Definition
Heat Transfer is an Interaction between systems that occur as a result of a difference in the
Temperature between the Systems.
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6.2 Positive and Negative Heat Transfer and Work Done
6.3.1 Conduction:
When 2 systems come into contact, Heat Transfer occurs through Conduction.
6.3.2 Convection:
Heat Transfer between a surface and a fluid flow occurs by Convection.
6.3.3 Radiation:
If the systems are not in contact and there is no fluid flow such as in a vacuum, Heat Transfer occurs by Radiation
7.1 The First Law of Thermodynamic
Equivalence between Heat Transfer and Work Done
4.1868 kNm Work Done generates 1 kcal of Heat Transfer or 4.1868 kJ Work Done generates 1 kcal
Heat Transfer
7.2 Cyclic Processes
A cyclic process is one in which the Initial and Final
States are the same.
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7.3
First Law of Thermodynamics states that when a system performs a cyclic process, the sum of Work Done is proportional to the sum of Heat Transfers.
∫ dQ ∝ ∫ dW
∫ dQ = k ∫ dW where k is a constant.
If we choose the units for both dQ and dW to be in kJ, then k = 1
Therefore ∫ dQ = ∫ dW and ∫ (dQ − dW ) = 0
7.4 The First Law as applied to Non cyclic Process
7.4.1 Property of a Property
The change in value of a Property depends only on the End States and Not on the PATH of a
Process.
If 2 Properties can define the State of a system (2 Property Rule), the change in the properties in a process does not depend on the PATH to achieve the initial and end states.
Example: The change in the Longitude and Latitude of a ship in a journey is calculated from the initial and final
Longitudes and Latitudes only and is
NOT dependent of the PATH taken by the ship. The Longitude and Latitude are Properties
But the Length of the journey is NOT a
Property because it depends on the
PATH taken by the ship
7.4.2 Definition of Energy
The increase in ENERGY of a system during a noncyclic process is equal to the
Heat Transfer minus the Work Transfer:
E2 - E1 = Q - W
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UEME1112 Goh Sing Yau (May 2014), FES, UTAR
7.4.2.1 Show that Energy is a Property
∫ (dQ - dW ) = 0
2
1
1A
2C
2
1
∫ (dQ - dW ) + ∫ (dQ - dW ) = 0
∫ (dQ - dW ) + ∫ (dQ - dW ) = 0
1B
2C
2
2
1A
1B
∫ (dQ - dW ) = ∫ (dQ - dW ) = E
2
− E1
The change in Energy E2 - E1 between states 1 & 2 is independent of the PATHS taken either via A or B.
Therefore ENERGY is a PROPERTY.
Case A
Case B
7.4.3 The First Law for Noncyclic Process
Q - W = ∆E
For case A, we consider the system consisting of the ball and the bowl.
The First Law of Thermodynamics :
Q - W = ∆E
Q =0
(No heat transfer)
W = − W12
( Work Done by hand to raise the ball)
∆E =W12
(Change of Energy is Positive)
Energy of system (Ball and Bowl) has increased.
Energy increase is given by hand which is external to the system.
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UEME1112 Goh Sing Yau (May 2014), FES, UTAR
For case B, we consider the system consisting of the ball and the bowl.
The First Law of Thermodynamics :
Q - W = ∆E
Q=0
W =0
(No heat transfer)
(Ball released by hand)
∆E = 0
( No change of Energy)
Energy of system (Ball and Bowl) is conserved.
Potential energy loss is equal to kinetic energy gained by the ball.
7.4.4
Law of Conservation of Energy
This is a special case of the First Law of Thermodynamics where
Q = 0 & W = 0.
Q - W = ∆E
∆E = 0
Energy of the system is conserved if the system is isolated from the surroundings in terms of Heat Transfer and Work Done.
7.5.1 Open and Closed Systems
Until now we have considered Closed Systems that contain all the material undertaking the process. 25
UEME1112 Goh Sing Yau (May 2014), FES, UTAR
Sometimes we encounter situations that are more complex where the processes involve steady flow. We can simplify the analysis if we choose a Control Volume that does not move but allow steady flow across the boundaries.
1
Closed System
"System" Analysis
Shape of System boundary can change.
Open System
"Control Volume" Analysis
Shape of Control Volume cannot change.
2
No mass crosses the System boundary.
Normally mass crosses the
Control Volume boundary at a steady flow rate.
Note that "System" and "Control Volume" are Methods of Analysis. Sometimes, both methods can be used on the same case.
However, one of the methods can provide an easier solution to the problem.
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UEME1112 Goh Sing Yau (May 2014), FES, UTAR
Derivation of The Steady Flow Energy Equation using the System Analysis
7.5.2
Derivation of SFEE (Steady Flow Energy Equation)
(a)
Conservation of Mass
For Steady Flow: dm2 − dm1 = 0 − − − − − − − − − (7.5.2.1)
(b)
The Frist Law of Thermodynamics for Closed System
Q -W = ∆E
W
= W x + ( pdVs )2 − ( pdVs )1
= W x + p 2 v 2 dm2 − p1 v1 dm1
∆E
= E 2 − E1
= (E c2 + e2 dm2 ) − (Ec1 + e1 dm1 )
Q - (Wx + p 2 v2 dm2 − p1 v1 dm1 ) = (Ec2 + e2 dm2 ) − (E c1 + e1 dm1 )
= ( Ec2 − E c1 ) + e2 dm2 − e1 dm1
For Steady Flow:
E c2 = Ec1
dan
dm2 = dm1 = dm
Q -W x = dm( e2 + p 2 v 2 − e1 − p1 v1 ) − − − − − − − − − − − −(7.5.2.2)
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UEME1112 Goh Sing Yau (May 2014), FES, UTAR
For a Pure Substance, the specific energy e is given by : e = u (Internal Energy) + motion, gravity, electric, capillary and magnetic effects
(a)
Energy of a body by virtue of its Motion
Find the Work Done by the Force to bring the body to rest. d(mv) Newton's S econd Law:
F=
dt dv If mass is constant,
=m
dt
2
= ∫ FdL
Work Done by the Force
where 1 and 2 denotes the initial and final states
1
2
= ∫ Fvdt
1
2
=∫m
1
dv vdt dt
2
= ∫ mvdv
1
2
v2
=m
2 1
v 2 v 2
= m 2 − 1
2
2
2
= −m
v1
2
because v1 = v & v 2 = 0
2
Work Done by the body
v1
2
Q − W = E 2 − E1
W = m
2
Q = 0, E 2 = 0:
E1 = W = m
v1
2
2
Kinetic Energy (motion) E k = m
v1
2
2
Specific Kinetic Energy
ek =
v1
− − − − − − − − − − − −(7.5.2.3)
2
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UEME1112 Goh Sing Yau (May 2014), FES, UTAR
(b)
Energy of a body by virtue of its height above sea level as a result of Gravity
Determine the Work Done to bring the body from z metres above sea level to sea level.
Newton's S econd Law:F
For constant mass,
d(mv) dt dv
=m
dt
=
2
Work Done by the Force
= ∫ FdL
1
2
= ∫ mgdL
1
= mg [L ]1
= mg [L2 − L1 ]
2
= − mgz
because L1 = z & L2 = 0
Work Done by the body
= mgz
Q − W = E 2 − E1
Q = 0,
E 2 = 0:
E1 = W = mgz
Potential Energy (height) E p = mgz
Specific Potential Energy e p = gz − − − − − − − − − − − −(7.5.2.4)
e = u (Internal Energy) + motion, gravity, electric, capillary and magnetic effects v2 + gz + (electric, capillary and magnetic effects)
2
In this Thermodynamic course, we deal wi th problems in which electric, capillary and magnetic effects are small and therefore assumed negligible.
e =u +
v2
SFEE: Q -W x = dm ∆ u + pv +
+ gz
2
& &
&
If Q , W x & m is for unit time,
v2 & &
&
Q -W x = m ∆ u + pv +
+ gz
2
&
& where m = dm (mass flow rate across the control volume)
In addition, u + pv = h (Entalpy)
The SFEE (Steady Flow Energy Equation) is given by
v2 & &
&
Q -W x = m ∆ h +
+ gz − − − − − − − − − − − −(7.5.2.5)
2
For multiple flow st reams, the following SFEE can be used
v2 & &
&
Q -W x = ∑ m ∆ h +
+ gz
2
29
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