Download Thermodynamics

Thermodynamics
Introduction and Basic Concepts
by
Asst. Prof. Channarong Asavatesanupap
Mechanical Engineering Department
Faculty of Engineering
Thammasat University
2
What is Thermodynamics?
Thermodynamics is the study that concerns with
the ways energy is stored within a body and how
energy transformations, which involve heat and
work, may take place.
Conservation of energy principle
, one of the most fundamental laws of nature,
simply states that “energy cannot be created or
destroyed” but energy can change from one form to
another during an energy interaction, i.e. the total
amount of energy remains constant.
3
Thermodynamic systems
or simply system, is defined as a quantity of matter or a
region in space chosen for study.
Surroundings are physical space outside the system
boundary.
Surroundings
System
Boundary
Boundary is the surface that separates the system from its
surroundings
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Closed, Open, and Isolated Systems
The systems can be classified into
(1) Closed system consists of a fixed amount of mass and
no mass may cross the system boundary. The closed system
boundary may move.
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(2) Open system (control volume) has mass as well as
energy crossing the boundary, called a control surface.
Examples: pumps, compressors, and water heaters.
6
(3) Isolated system is a general system of fixed mass where
no heat or work may cross the boundaries.
mass
No
energy
No
An isolated system is normally a collection of a main system and its surroundings
that are exchanging mass and energy among themselves and no other system.
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Properties of a system
Any characteristic of a system is called a property. Some familiar
properties are
volume V,
mass m,
density r,
pressure P,
temperature T
and etc.
Density is defined as mass per unit volume
Water@ 20 C , 1 atm
r = 998 kg/m3
The reciprocal of density is the specific volume, which is defined as
Specific Gravity SG is defined as the ratio of the
density of a substance to the density of some standard
substance at a specified temperature (usually water
at 4C).
Temperature (T) is a measure of the average energy of motion, or
kinetic energy, of particles in matter. (or a measure of hotness and coldness)
Temperature scales
Common scale:
Celsius scale

°C
SI unit
Fahrenheit scale

°F
English unit
The Celsius scale is related to the Fahrenheit scale by
Thermodynamic scale(Absolute scale):
Kelvin

K
SI unit
Rankine

R
English unit
Temperature scales
The common scales are related to the absolute scale by
SI unit
English unit
Example: Water boils at 100C at one atmosphere pressure.
At what temperature does water boil in F, K and R.
T(°F) = 100x1.8 + 32 = 212 °F
T(K) = 100 + 273.15 = 373.15 k
T(R) = 212 + 459.67 = 671.67 F
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Pressure(P) is the force per unit area applied in a direction
perpendicular to the surface of an object
P
For English system,
=
F
A
N (Pa)
m2
12
Pressure scales
Absolute scale:
Absolute pressure is the pressure that is measured relative to absolute
zero pressure (absolute vacuum).
Gage scale:
Gage pressure is the pressure that is indicated on a pressure-measuring
device (called a pressure gage). Generally, the device is calibrated to read zero in
the atmosphere.
13
Vacuum pressure
Pressures below atmospheric pressures are called vacuum pressures.
A device that is used to measure vacuum pressure is called a vacuum gage.
Pressure
Symbol
Absolute
Pa
Gage
Pg
Pressure
Eng. unit
Absolute
psia
Gage
psig
Example A pressure gage connected to a valve stem of a truck tire reads 24014
kPa at a location where the atmospheric pressure is 100 kPa. What is the absolute
pressure in the tire, in kPa and in psia?
Pabs  Patm  Pgage
 100 kPa  240 kPa
 340 kPa
The pressure in psia is
Pabs
14.7 psia
 340 kPa
 49.3 psia
1013
. kPa
What is the gage pressure of the air in the tire, in psig?
Pgage  Pabs  Patm
 49.3 psia  14.7 psia
 34.6 psig
15
Intensive and Extensive properties
• Intensive properties are
those that are independent of the
mass of a system.
• Extensive properties are
those whose values depend on the
size—or extent—of the system.
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Internal energy (U)
is defined as the sum of all the microscopic forms
of energy of a system. It is related to the molecular
structure and the degree of molecular activity and
can be viewed as “the sum of the kinetic and
potential energies of the molecules”.
Properties
Extensive
Intensive
Gas:
where
Symbol
U
u=U/m
Unit
J [Joule]
J/kg
U = CV T = mcV T [kJ/kg]
CV = heat capacity at constant volume [kJ/K]
cV = specific heat capacity [kJ/kg-K]
17
Enthalpy (H)
is
a
measure
of
the
total
energy
of
a
thermodynamic system and is defined as “the
summation of the internal energy and the flow
work (PV); H = U + PV”.
Properties
Extensive
Intensive
Gas:
where
Symbol
H
h=H/m
Unit
J [Joule]
J/kg
H = CP T = mcP T [kJ/kg]
CP = heat capacity at constant pressure [kJ/K]
cP = specific heat Capacity [kJ/kg-K]
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Specific heats (c)
is defined as the energy required to raise the
temperature of a unit mass of a substance by one
degree. In thermodynamics, we are interested in
two kinds of specific heats: specific heat at
constant volume cv and specific heat at
constant pressure cp.
1 kg Water
1 kg air
1 kg air
DT = 1°C
DT = 1°C
DT = 1°C
4.18 kJ
0.72 kJ
1.00 kJ
19
How to identify the state of a substance?
1.Equations of state
2.Property tables
3.Property diagrams
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Ideal gas law
PV = nRuT
Pv = RT
where P and T are absolute pressure and temperature, respectively.
R is a gas constant = Ru/M [kJ/kg.K or kPa.m3/(kg.K)]
Ru is a universal gas constant = 8.314 kJ/(kmol.K)
n is the number of moles = m/M
M is Molar mass
21
Example 1
A room of the size 4m x 5m x 6m contains air at P = 100 kPa
and T = 25°C. Determine the mass of air inside the room. Assume
that Ra = 0.287 kPa.m3/kg.K
PV
100kPa  120m 3
m

 140.3kg
3
RaT 0.287kPa  m / kg  K  (25  273) K
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Example 2
The pressure in an automobile tire depends on the temperature of the
air in the tire. When the air temperature is 25°C, the pressure gage reads
210 kPa. If the volume of the tire is 0.025 m3, determine the pressure rise in
the tire when the air temperature in the tire rises to 50°C
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Liquid and vapor phases of a substance
• Compressed liquid (subcooled liquid): A substance that it is not about
to vaporize.
• Saturated liquid: A liquid that is about to vaporize.
• Saturated vapor: A vapor that is about to condense.
• Saturated liquid–vapor mixture: The state at which the liquid and
vapor phases coexist in equilibrium.
• Superheated vapor: A vapor that is not about to condense (i.e., not a
saturated vapor).
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Examples of property tables
.
g = gaseous phase (saturated vapor)
f
=
liquid phase (saturated liquid)
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Examples of property tables
.
26
Examples of property tables
27
Example 3
Find the the density, specific internal energy, and specific enthalpy of
water at T = 25 C and P = Patm.
From table A-4:
1
1
3
rw  

997
.
01
kg
/
m
vw 0.001003m 3 / kg
u w  u f @ 25C  104.83kJ / kg
hw  h f @ 25C  2441.7kJ / kg
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Example 4
Determine the amount of energy required to increase the temperature of
1-kg water in example 2 to 150C.
Phase-change process
A-4:
1. Liquid water at 25C hFrom table
w,1 h f @ 25C  104.83kJ / kg
2. Liquid water at 100C
From table A-4:
hw,2  h f @100C  419.17 kJ / kg
Q1 2  m  Dh
 1  (419.17  104.83)
 314.34kJ
Q2 3  m  Dh
3. Vapor water at 100C
From table A-4:
hw,3  hg @100C  2675.6kJ / kg
 1  (2,675.6  419.17)
 2,256.43kJ
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Example 4
Determine the amount of energy required to increase the temperature of
1-kg water in example 3 to 150C.
Phase-change process
3. Vapor water at 100C
From table A-4:
hw,3  hg @100C  2675.6kJ / kg
4. Super heated water at 150C
From table A-6:
hw,4  hsup erheat @150C  2776.6kJ / kg
Q3 4  m  Dh
 1  (2,776.6  2,675.6)
 101.0kJ
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Example of property diagrams
.
Water
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Psychrometric chart
Air
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Moist air ( Air/Water vapor mixture) properties
• Dry-bulb temperature: The air temperature indicated by a standard
thermometer.
• Wet-bulb temperature: The air temperature indicated by a thermometer
with a wet wick attached to it bulb.
• % Relative Humidity: The amount of water vapor held in the air as a
percent of the maximum amount of water vapor the air can hold at a
specific temperature.
• Enthalpy: The total heat contained in the air.
• Dew point: The air temperature at which condensation begins.
• Humidity ratio: The mass of water vapor held in 1 kilogram of dry air.
33
Example 5
Find the the density, specific enthalpy, humidity ratio and dew-point
temperature of air at T = 25 C and %RH = 50%.
ha  50kJ / kg
a  0.10kg w / kg a
Tdp  14C
ra 
1
1

 1.16kg / m 3
va 0.86
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Example 6
Condensation within the 1 m2 wall
hi = 10 W/(m2.K)
Brick
Fiberglass Gypsum
25C, 50%
35C, 48%
kB= 0.68 W/(m.K)]
kF= 0.038 W/(m.K)]
kG= 0.48 W/(m.K)]
ho = 40 W/(m2.K)
0.1m
0.15m
0.01m
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Example 6 (cont.)
Rwall 
Dx
KA
Rair 
1
hA
Rth,tot  Rtot  Ri  Rb  R f  Rg  Ro
Rth,tot  0.1  0.147  3.947  0.021  0.03  4.24 (m 2  K ) / W
The heat flux is
q 
DT
Rth,tot

(35  25) K
4.24 (m 2 .K ) / W
 2.35 W / m 2
36
Example 6 (cont.)
Fiberglass
2
1
35C, 70%
28C
3
25C, 50%
DT  q  Rth,i
4
DTa  2.35  0.03  0.07C  T1  34.93C
DTg  2.35  0.02  0.05C  T2  34.43C
Condensation forms
within the fiberglass layer.
DT f  2.35  3.947  9.27C  T3  25.15C
37
Example 7
Preventing condensation on cold air ducts.
Ambient air 25C, 50%
Tsurface = 12 C
Cold air
38
Example 7 (cont.)
Preventing condensation on cold air ducts.
Ambient air 25C, 50%
From the example 1, the dp temperature = 16 C
Tsurface < dp temperature
Tsurface = 12 C
Cold air
Water drops form on the cold duct wall where the
surface temperature is below the dewpoint
temperature.
39
Example 7 (cont.)
Preventing condensation on cold air ducts.
Ambient air 25C, 50%
From the example 1, the dp temperature = 16 C

qconv
h0 = 30 W/(m2K)
Cold air
Tsurface = 12 C
Before

qconv
 h(Tambient  Tsurface )
 30(25  12)  390 W / m 2
40
Example 7 (cont.)
Preventing condensation on cold air ducts.
To prevent condensation, Insulation is needed on the duct.
Ambient air 25C, 50%
Ts = ? C
Insulator
b =?
Fiberglass is used,
Cold air
k= 0.038 W/(m.K)]
41
Example 7 (cont.)
Preventing condensation on cold air ducts.
Ambient air 25C, 50%
Ts = ? C
Insulator
h =?
After
(Ts  Tsurface )
q  k
 h(Tambient  Ts )
b
Cold air
0.038
(16  12)
 30(25  16)
b
b  0.00056 m
Heat gain
q  270 W / m 2
????
42
Example 7 (cont.)
Preventing condensation on cold air ducts.
What will happen, if h = 2.5 cm ?
0.038
(Ts  12)
 30(25  Ts )
0.025
Ts  24.37C
Heat gain
(no condensation)
q  18.9 W / m 2
(-93%)
43
First Law of Thermodynamics
is an expression of the conservation of energy principle.
Ein – Eout = DEsys
where DEsys = Efinal – Einit
44
1st Law for closed systems
(1) Energy transferred across the boundary of the closed
system can be divided into 2 forms: Heat and Work
(2) Energy stored in the closed system is represented by the
total internal energy (U)
Therefore, the first law of thermodynamics is written as
(Qin + Win) – (Qout +Wout) = DU+DEp+DEk
DQ – DW = DU+DEp+DEk
45
1st Law for closed systems
If the system does not move with a velocity and has no change in elevation,
the conservation of energy equation reduces to
DQ – DW = U2-U1
dU
Q  W 
dt
[kJ]
where 1 represents initial state
2 represents final state
[kW ]
No motion
(DKE=0)
+
No elevation change
(DPE=0)
46
Example 7
A closed tank has a volume of and is filled with 200 kg of water at the
temperature of 30 C. The water is heated by a 30kW electric heater. How
long does it take for water to reach 45C.
dU


Q W 
dt
Water
200 kg.
20 kW heater
(u2  u1 )
20kW  0 W  mw
Dt
(188.43  125.73)kJ / kg
Dt  200kg
30kW
Dt  418 s
47
Note:
(u2  u1 )
20kW  0 W  mw
Dt
cv (T2  T1 )
 mw
Dt
4.178(45  30)kJ / kg
Dt  200kg
30kW
 417.8 s
48
1st Law for Open systems
Energy and material transfer into or out of the system
boundary.
For steady flow,
 m   m
in
[kg / s ]
out
2
2
V
V
Q  W   m (h 
 gz )   m (h 
 gz ) [kW ]
2
2
out
in
49
Example 8
Determine the amount of energy removed for space cooling if the
infiltration mass flow is 0.35 kg/s
To = 35 C
%RH = 70
Ti = 25 C
%RH = 50
Energy removed
for space cooling,
(Qinf)
50
Example 6 (cont.)
Mass balance:
m in  m out  m inf  0.35 kg / s
Energy balance:
0
0
2
V
Q inf  W  m inf (Dh  D
 Dgz )
2
0
Q inf  m inf (hin  hout )
51
HW#1
1. student living in a 4-m x 6-m x 6-m dormitory room turns on her 150-W fan before she
leaves the room on a summer day, hoping that the room will be cooler when she come
back in the evening. Assume all the doors and windows are tightly closed and
disregarding any heat transfer through the walls and the windows, determine the
temperature in the room when she comes back 10 h later. Use specific heat values at
room temperature, and assume the room to be at 100 kPa and 15 C in the morning when
she leaves.
2. The air in a room is 20C and 50% relative humidity. Will moisture condense on a window
whose surface is 7C? If the room is 4.5m2 and 2.5 m high, how much water is contained in
the room?
3. A chilled-water line carries chilled water at 7C through a room at 21C and 60%RH. How
much fiber glass insulation is needed on the pipe to avoid condensation?