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ESS 2410 Engineering Thermodynamics
Overview and Review
Thermodynamics and Energy
Thermodynamics: the science of “energy”
Energy: the ability to cause change, the capability to do work
Energy can exist in many different forms
thermo (heat) ; dynamics (power)
Thermodynamics: converts heat to power
Thermodynamics: include all aspects of energy,
energy transformation, power generation,
refrigeration, relationship among properties of matter
1
Conservation of energy principle: Energy can change from one
form to another, but the total amount of energy remains constant.
Energy is a thermodynamic property
fundamental law of nature (1st Law of Thermodynamics)
The 2nd Law of Thermodynamics: the energy has quantity
as well as quality
Energy transfer can occur certain direction but not in the
reverse direction.
The actual process occurs in the direction of decreasing
quality of energy.
The ability to do useful work diminishing even though
energy is not.
Thermodynamics: science of energy and entropy.
2
Forms of Energy
Energy can exist in numerous form: thermal, mechanical, kinetic,
potential, electric, magnetic, chemical, nuclear…..
Thermodynamics provides no information about the absolute value
of the total energy. It only deals with the change of total energy.
Microscopic forms of energy: those related to the molecular structure
of a system and the degree of the molecular activities, and they
are independent of outside reference frames.
Completely Random; Disorganized form of Energy
3
Energy
Microscopic
Energy
Macroscopic
Energy
Internal
Energy
Kinetic Energy
Chemical
Bonding Energy
Nuclear
Bonding Energy
Potential Energy
Thermal Energy
Electrical Energy
Chemical
reaction
Nuclear
reaction
Chemical
Energy
Kinetic
Energy
Nuclear
Energy
friction
Kinetic Energy
Sensible Energy
Heat
Generator
Potential
Energy
Mechanical
device
Internal
Thermal Energy
Latent Energy
Internal Thermal Energy
Mechanical
device
Work
Internal Energy: the sum of all the microscopic forms of energy
kinetic energy: sensible energy (sensible heat)
(proportional to temperature)
potential energy: binding force between molecules;
strongest in solid, weakest in gas,
energy required in phase change
latent energy (latent heat)
Chemical Energy and Nuclear Energy
Macroscopic forms of energy: A system possesses as a whole with
respect to some outside reference frame (kinetic and potential energy)
Orderly motion: Organized form of Energy
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Static form of Energy: total energy of a system can be contained or
stored in a system.
Dynamic form of Energy: energy interactions, energy in transit;
defined at system boundary, represent the energy gained or lost by
a system during a process
Heat : energy interaction driving by temperature difference
Organized form of energy is more valuable than the disorganized
form of energy.
Organized energy can be converted to disorganized energy completely.
Only fraction of disorganized energy can be converted into organized
energy by specially built devices called heat engines.
Thermodynamics: the conversions of disorganized energy into
organized energy.
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Application area of thermodynamics:
The science that explains and predicts how much energy we
may extract and how efficiently we may do it under a specific
situation.
Automotive engines; rockets, jet engines, power plant, refrigeration
Deals with various properties of substances and the change in these
properties as results of energy transformation.
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Property of System
Property: Any characteristic of a system,
pressure, temperature, volume, mass, viscosity,
thermal conductivity, modulus of elasticity,
thermal expansion coefficients, color, hardness,…..
Not all properties are independent:
Two properties are independent if one property can be
Varied freely while the other one is constant
Some are defined in terms of others:
density: mass per unit volume
specific properties: property per unit mass
specific volume: volume per unit mass
(1/density),
specific heat, specific power…….
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Intensive Properties: properties are independent of the size of system
pressure, temperature, density……
Specific volume, specific enthalpy, specific internal energy…
Extensive Properties: properties depend on the size-or extent- of the
system
volume, mass, total energy,……
Specific properties are intensive properties
Extensive property per unit mass
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State and Equilibrium
State: the condition of system as determined by its properties
At a given state, all the properties of a system have fixed values
Each of the properties of a substance in a given state has only
one definite value, and these properties always have the same
value for a given state, regardless of how the substance
arrived at that state.
How many?
A state can be specified by a certain number of properties.
Equilibrium: implies a state of balance,
no unbalanced potentials (or driving force)
experiences no changes when it is isolated
from its surroundings.
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Thermodynamics deals with equilibrium system
Thermodynamics
Equilibrium
thermal equilibrium (no temperature difference
mechanical equilibrium (no imbalance force)
phase equilibrium (if phase changes involved)
chemical equilibrium (no chemical changes)
Processes and Cycles
Process: any change that a system undergoes from one equilibrium
state to another equilibrium state.
Path: the series of states through which system passes during
a process.
To describe a process completely: initial state, path, final state
11
Compressed Process P-V Diagram
12
Process Diagram: employ thermodynamics properties as coordinates
to visualize the process, T - P, T - v, P – v, T – s
The prefix – iso is often used to designate a process:
isothermal : constant temperature;
isobaric: constant pressure;
isochoric (isometric): constant volume;
isentropic: constant entropy
Cycles: A system is said to have undergone a cycle if it returns
to its initial state at the end of a series of processes.
13
The State Postulate
no external force
The State Postulate: the state of simple compressible system is
completely specified by two independent, intensive properties
T, VX
T, v
P, T (gas, liquid, not during phase change)
P, v; T, u; P, h; …..
Two properties are independent if one property can be
Varied freely while the other one is constant
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T-v diagram of the heat addition
of water at constant pressure
State 5
State 2
State 3
State 4
State 1
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State 1
water, P = 1 atm, T = 20 ; water exists in liquid phase
compressed liquid or subcooled liquid
Compressed: Pressure is higher than Psat @ T = 20 C
1.013x105 Pa
2.339x103 Pa
Subcooled: Temperature is lower than Tsat @ P = 1 atm
20 C
100 C
Compressed or Subcooled: meaning it is not about to
vaporized
heat addition
temperature of
liquid increases
volume
expansion
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State 2
water reaches 100 C, water is still in liquid form
Tsat @ P = 1 atm
Saturated liquid: a liquid is about to vaporize
Phase change process from liquid to vapor is about to occur
Any heat addition will cause some of the liquid to vaporize
Under saturated condition, temperature and
pressure are not independent.
17
State 3
two- phase mixture, liquid water and water vapor coexist,
temperature is 100 C, pressure is 1 atm
liquid and vapor are at the same temperature
temperature
remains the same
heat
addition
amount of
vapor increases
volume expand
significantly
boiling
vaporization
The vaporization will continue until last drop of water is vaporized
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State 4
entire vessel is filled with saturated water vapor,
T = 100 C, P = 1atm
Any heat loss will cause some of the vapor to condense
Saturated Vapor : a vapor is about to condense
State 5
entire vessel is filled with superheated vapor
T = 300 C, P = 1 atm
Superheated vapor: a vapor that is not about to condense
Temperature is greater than Tsat @ given pressure
Pressure is lower than Psat @ given temperature
heat
addition
temperature of
vapor increases
volume
expansion
19
T-v diagram of constant pressure phase-change process
20
T-v diagram of a pure substance
21
P-v diagram of a pure substance
22
Saturation temperature and saturation pressure
saturation temperature (Tsat): at a given pressure, the temperature
at which a pure substance changes phase
saturation pressure (Psat): at a given temperature, the pressue at
which a pure substance changes phase
During a phase change process, pressure and temperature
are dependent properties.
Tsat  f Psat 
Tsat as Psat
Liquid-vapor saturation curve, Clapeyron equation
Latent heat: energy absorbed or released during a phase change
Table A-2 (760) Saturated Water--temperature table
Table A-3 (762) Saturated Water-- pressure table
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Critical Point: the point at which the saturated liquid and
saturated vapor are identical.
Critical Point Properties: Tcr, Pcr, vcr
Above the critical state, there is no line that separates the
compressed liquid region and superheated vapor region.
At pressure above the critical pressure, there will be no distinct
phase change process. The specific volume of substance will
continually increase. (Figure on page 20)
Above critical state:
Temperature above Tcr: superheated vapor
Temperature below Tcr: compressed liquid
Saturated liquid line: a line that connects the saturated liquid state
Saturated vapor line: a line that connects the saturated vapor state
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Property Table
For most substances, the relationships among thermodynamic
properties are presented in the form of tables,
Property Table of water
Table A-2
Table A-3
Table A-4
Table A-5
Table A-6
Steam Table
Saturated Water – Temperature Table (p.760)
Saturated Water – Pressure Table (p.762)
Superheated Vapor (p.764)
Compressed Liquid Water (p.768)
Saturated Ice-Water vapor (p. 769)
Properties in Property Tables:
Pressure, Temperature, Specific Volume, Internal Energy (specific)
Enthalpy (specific), Entropy (specific)
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1a: Saturated liquid and saturated vapor state
subscripts f : saturated liquid
g : saturated vapor
fg : difference between saturated vapor and
saturated liquid
Need two properties to determine the state
the word “saturated” is one of the condition
1b : saturated liquid-vapor mixture (P and T are dependent property)
Quality (x) : mass fraction of vapor in the mixture;
ratio of mass of vapor to the total mass of the mixture
A property
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Definition of Quality
V  V f  Vg
mtot  ave  m f 
 ave 
mf
mtot

f
f
 mg  g

mg
mtot
g
 ave  1  x  f  x g


 f

 ave
x  ave
 fg
 g  f
f
u ave  1  x  u f  x u g  u f  x u fg
have  1  x  h f  x hg  h f  x h fg
save  1  x  s f  x s g  s f  x s fg
0 x  1
saturated liquid
saturated vapor
quality has no physical meaning for compressed
liquid and superheated vapor
In two-phase region (two phase mixture):
y f  yave  y g
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2 Superheated Vapor
Single phase region: P and T are independent property
lower pressure (P < Psat @ given T)
higher temperature (T > Tsat @ given P)
greater specific volume (v > vg @ given P or T)
higher internal energy (u > ug @ given P or T)
higher enthalpy (h > hg @ given P or T)
greater entropy (s > sg @ given P or T)
3. Compressed Liquid (P and T are independent property)
higher pressure (P > Psat @ given T)
lower temperature (T < Tsat @ given P)
smaller specific volume (v < vf @ given P or T)
lower internal energy (u < uf @ given P or T)
lower enthalpy (h < hf @ given P or T)
smaller entropy (s > sf @ given P or T)
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3. Compressed Liquid (P and T are independent)
Not many data exist, relative independence of compressed liquid
properties from pressure. Treat compressed liquid at the given
temperature.
Given P and T
  f at given T
u  u f at given T
h  h f at given T  f  P  Psat 
Values of u, h, and s can not be measured directly.
Calculated using thermodynamics relations (Chapter 11)
Relation gives changes in thermodynamic properties related
to a reference state.
The values of u and s at reference state is set equal to 0.0
the reference state of water is triple point
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1. water @ P = 150 kPa, T = 111.4 C; find u and v
2. water @ P = 150 kPa, v = 0.0010435 m3/kg; find T and u
3. water @ T = 400 C, v = 0.02641 m3/kg; find P and u
4. water @ T = 120 C, v = 0.44648 m3/kg; find P and u
Answers:
1. v = 0.0010528 m3/kg, u = 466.94 kJ/kg (saturated water)
2. T = 100 C, u = 418.94 kJ/kg (compressed liquid)
3. P = 10 MPa, u = 2832.4 kJ/kg (superheated vapor)
4. P = 198.53 kPa, u = 1516.4 kJ/kg (2-Phase mixture)
30
1.
water @ P = 150 kPa, T = 111.4 C; find u and v
from Table A-3 Tsat @ P = 150 kPa = 111.4 C
T= Tsat @ P = 150 kPa; it is saturated liquid
from Table A-3
vf @ P=150 kPa = 0.0010528 m3/kg
uf @ P=150 kPa = 466.94 kJ/kg
2. water @ P = 150 kPa, v = 0.0010435 m3/kg; find T and u
from Table A-3 vf @ P = 150 kPa = 0.0010528 m3/kg
v < vf; it is compressed liquid
use temperature to determine its properties
from Table A-2 (use v find T)
vf @ T = 100 C = 0.0010435 m3/kg
therefore T = 100 C, and u ~ uf = 418.94 kJ/kg
31
3. water @ T = 400 C, v = 0.02641 m3/kg; find P and u
from Table A-2 Tcritical = 374.14 C
T > Tcritical ; it superheated vapor
use Table A-4 determine its properties
First try P = 2.0 MPa , v@ P=2.0 MPa, T = 400 C = 0.1512 m3/kg
which is greater than 0.02641 m3/kg ;
therefore the Pressure should be higher than 2 MPa.
From Table A-4, @ P = 10 MPa, T= 400 C,
v = 0.02641 m3/kg
Therefore P = 10 MPa, u = 2832.4 kJ/kg
32
4. water @ T = 120 C, v = 0.44648 m3/kg; find P and u
From Table A-2, vf (0.0010603)< v(0.44648)<vg(0.8919)
It is two-phase mixture
Find x (quality)
P = 198.53 kPa
v  vf
0.44648  0.0010603
x

 0.5
vg  v f 0.8919  0.0010603
u  (1  x) u f  x u g  u f  x u fg
 503. 50  0.5  2025.8  1516 . 4 (kJ / kg)
33
Internal Energy, Enthalpy, and Specific Heats of Ideal Gas
For an ideal gas, the internal energy is function of temperature only.
u  u T 
Enthalpy:
u  u (T ), Pv  RT ,  h  u  RT  h  hT 
 u 
 h 
du
dh
Cv  
;Cp  
  Cv 
 Cp 
dT
dT
 T  v
 T  p
du  C T  dT , u  12 C T  dT
dh  CP T dT , h  12 CP T  dT
The specific heat of real gases at low pressures are called
ideal-gas specific heats or zero-pressure specific heats. C p , C
Ideal gas specific heats of monatomic gases (Ar, Ne, He) remain
constant over the entire temperature range.
34
35
Three ways to determine the internal energy and enthalpy change of
ideal gases.
1. Table A-22 (p. 796) Reference point is 0 K
2. Performing integration. (Table A-21, p. 794)
3. Using average specific heats.
u2  u1  Cv ,av T2  T1  Cv ,ave and C p ,ave are evaluated at
T2  T1 
h  h  C T  T 
2
1
P , av
2
1
2
Specific-Heat Relations of Ideal Gases
h  u  RT , dh  du  rdT
C P  Cv  R, C P  Cv  Ru
specific heat ratio:
k  C P Cv
36
The Tds Relations
 Qint rev   Wint rev  d U
 Qint rev  Td S ;  Wint rev  P dV
or
TdS  dU  P dV
Tds  du  P dv
the only possible work mode of
a simple compressible system
Gibbs equation
h  u  Pv  dh  du  pdv  vdp
Tds  du  Pdv  Tds  dh  vdp
du Pdv
dh vdP
ds 

; ds  
T
T
T
T
Tds relations related the entropy change of a system to the change
of other properties.
T, P, u, v, s are properties, therefore, the results obtained are valid
37
for both reversible and irreversible processes.
Entropy Change of Ideal Gas
du Pdv
dh vdP
ds 

;
ds  
T
T
T
T
dT Rdv
dT RdP
ds  Cv

;
ds  C p

T
v
T
P
2
2
dT
v2
dT
P2
s2  s1   Cv
 R ln ; s2  s1   C p
 R ln
1
1
T
v1
T
P1
1. constant specific heat: approximate analysis
T2
v2
T2
P2
s2  s1  Cv,ave ln  R ln ; s2  s1  C p,ave ln  R ln
T1
v1
T1
P1
Good for monoatomic gases, good for most ideal gases
if the temperature change is not very large.
38
2 variable specific heats: exact analysis
dT
P2
s2  s1   C p
 R ln
1
T
P1
2
s T   

T

dT
Cp ; 
T
dT 



C

s
T

s
1 p T 2 2 1 T1 
2
Tabulated value, Table A-22 (air)


P2
s2  s1   
T1   R ln
P1
temperature dependency pressure dependency
of entropy
of entropy
s2
T2  s1
39
Entropy Change of an ideal Gas
Nitrogen gas is compressed from an initial state of 100 kPa and 17℃
to final state of 600 kPa and 57℃. Determine the entropy change of
the nitrogen during this compression process by using (a) property
values from the nitrogen table and (b) average specific heats.
40
State 1: P1 = 100 kPa, T1 = 290 K; State 2: P2 = 600 kPa, T2 = 330 K
(a)
Specific heat is not constant (Table A-23)

P2
P1
 194.459  190.695  8.314 ln
600
  11.133 kJ
kmole  K
800
s2  s1 
(b)

s2  s1  s2 T2   s1 T1   Ru ln
s2  s1  11.133

 0.3974 kJ
kg  K
M
28.013
Specific heat is constant (Table A-20)
T2
P2
s2  s1  C p ,ave ln  R ln
T1
P1
use C p ,ave at 37 C, C p ,ave  1.0394 kJ/kg  K
330
600
s2  s1  1.0394 ln
 0.297  ln
290
100
  0.3978 kJ/kg  K
41
Isentropic Process of Ideal Gas
Constant Specific Heat: Approximate Treatment
T2
v2
s2  s1  Cv ln  R ln  0
T1
v1
T2
v2
Cv ln   R ln
T1
v1
 v1 
T2
 ln  ln 
T1
 v2 
R Cv
R  C p  Cv ; Define k  C p Cv
 T2 
 v1 
 
 
 T1 isentropic  v2 
 T2 
 P2 
 
 
 T1 isentropic  P1 
R Cv
k 1 k
 v1 
 
 v2 
k 1
 P2 
 v1 
;  
 
 P1 isentropic  v2 
ideal gas, isentropic, constant specific heat
k
42
Tv k 1  constant
TP 1k  k  constant
ideal gas, isentropic
constant specific heat
Pv k  constant
Variable Specific Heat: Exact Treatment
s2  s1 

s2
 s1

P2
 R ln  0 
P1



s2
 s2  s1  exp s2 R
P2

 exp
 exp s  R
P1
R


1


 s1

P2
 R ln
P1
dimensionless
function of
Define : exp s R as relative pressure Pr
temperature only
tabulated value
 P2 
 P T  
(Table A-22)
  r2 2 
 
 P1  s constant  Pr1 T1  


43
P1v1 P2v2

T1
T2
Define : T
Pr
v2 T2 P1 T2 Pr1 T2 Pr 2




v1 T1 P2 T1 Pr 2 T1 Pr1
as relative specific volume vr
 v2 
 vr 2 T2  

 

 v1  s constant  vr1 T1  
dimensionless
function of
temperature only
tabulated value
(Table A-22)
44
Isentropic Compression of air in a Car Engine
Air is compressed in a car engine from 22℃ and 95 kPa in a reversible
and adiabatic manner. If the compression ratio V1/V2 of piston-cylinder
device is 8, determine the final temperature of the air.
45
For closed system:
V2
v
1
 2  ;
V1
v1
8
 v2

 v1

 vr 2

 
 isentropic  vr1
 1
 
 8
from Table A - 22; vr1 at T1  295 K is 647.9
vr 2  vr1 / 8  80.99
from Table A - 22

T2  662.7 K
Alternate Solution: specific heat is constant
 T2 
 v1 
 
 
 T1 isentropic  v2 
 v1 
T2  T1  
 v2 
 665.2
k 1
k 1
 29581.3911
K 
The value of k has to be evaluated
at the average of initial and final
temperature. Assume the average
temperature is 450 K.
(665.2+295)/2 = 480.1; need iteration
46
Isentropic Compression of an ideal Gas
Helium gas is compressed in an adiabatic compressor from an initial
state of 14 psia and 50℉ to a final temperature of 320℉ in a
reversible manner. Determine the exit pressure of helium.
Specific heat of helium is constant
P2  T2 
 
P1  T1 
k  k 1
 T2 
P2  P1 
 T1 
k  k 1
1.667 0.667
780 
P2  14  

 510 
 40.5 psia
47
Isentropic Efficiencies of Steady Flow Devices
Isentropic process is a model process for most steady-flow device.
The more closely the actual process approximates the isentropic
process, the better the device will perform.
Isentropic efficiency of turbine:
actual turbine work
isentropic turbine work
wa h 1 h 2 a


ws h 1 h 2 s
T 
48
Isentropic Efficiency of a Steam Turbine
Steam enters an adiabatic turbine steadily at 3 MPa and 400℃ and
leaves at 50 kPa and 100℃. If the power output of the turbine is 2 MW,
determine (a) the isentropic efficiency of the turbine and (b) the mass
flow rate of the steam flowing through the turbine.
49
next
State 1:
P1 = 3 MPa
T1 = 400 C
s1 = 6.9212 kJ/kg K
h1 = 3230.9 kJ/kg
State 2a:
P2a = 50 kPa
T2a = 100 C
h2a = 2682.5 kJ/kg
P2s = 50 kPa
s2s = s1
sf = 1.0910 kJ/kg K
sg = 7.5939 kJ/kg K
State 2s:
(a) s f  s2 s  s g ; x2 s 
s2 s  s f
s fg
h2 s  h f  x2 s h fg  2407.4
 0.897
kJ / kg 
h1  h2 a
T 
 66.7%
h1  h2 s
Figure 6.50
(b) Wa  m h1  h2a ; 2 103  m 3230.9  2682.5
m  3.65 kg / s
50
Isentropic efficiency of compressors and pumps:
isentropic compressor work
actual compressor w ork
w
h h
 s  2s 1
wa h2 a  h1
c 
P 
ws v P2  P1 

wa
h2 a  h1
51
Effect of Efficiency on Compressor Power Input
Air is compressed by an adiabatic compressor from 100kPa and 12℃
to a pressure of 800 kPa at a steady rate of 0.2kg/s. If the isentropic
efficiency of the compressor is 80 percent, determine (a) the exit
temperature of air and (b) the required power input to the compressor.
52
c 
ws h2 s  h1

wa h2a  h1
State 1: T1 = 285 K
(Table A.22)
h1 = 285.14 kJ/kg
Pr1= 1.1584
State 2s: P2 = 800 kPa
Find state 2 s;
isentropic process
 Pr 2   P2 
 P2

  
  Pr 2  Pr1 
 Pr1   P1 
 P1
from Table A.22; Pr 2  9.2672
State 2a
find
h2 a ;
h2 a  575.03

 800 
  1.1584
  9.2672
 100 

 h2 s  517.05
kJ / kg

517.05  285.14
0.80 
kJ / kg;
h2 a  285.14
from Table A.22
T2 a  569.5
K 
53
Isentropic efficiency of nozzle:
N 

actual KE at nozzle exit
isentropic KE at nozzle exit
V22a
V22s
h1  h2a

h1  h2 s
54
Effect of Efficiency on Nozzle Exit velocity
Air at 200 kPa and 950K enters an adiabatic nozzle at low velocity and
is discharged at a pressure of 80 kPa. If the isentropic efficiency of
the nozzle is 92 percent, determine (a) the maximum possible exit
velocity, (b) the exit temperature, and (c) the actual velocity of the air.
Assume constant specific heats for air.
55
P1 = 200 kPa, T1 = 950 K; P2 = 80 kPa
(a) Find T2s (specific heat is constant)
T2 s  P2 s 


T1  P1 
k 1 k
 80 
 950

 200 
 748  K 
0.354 1.354
V22s
V12
h1 
 h2 s 
2
2
V2 s  2h2 s  h1 
 2C p ,ave T2 s  T1 
 666
m / s 
Guess the average temperature is
800 K; to determine k
(748+950)/2 = 849, need iteration
C p,ave T1  T2a 
h

h
1
2
a
(b)  N 

h1  h2 s C p,ave T1  T2 s 

950  T2a
0.92 
 T2a  764
950  748
(C)
N

V22a
V22s
K 
 V2a   NV22s
 639 m/s 
56
Decimal Relationship:
10-12
10-9
Pico(p) nano(n)
pm
nm
奈米
101
102
deka hecto
10-6
micro(  )
m
微米
10-3
10-2
milli(m) centi(c)
mm
cm
毫米
公分
103
106
109
kilo(k) mega(M) giga (G)
km (公里) Mm
Gm
kW
MW
GW
千瓦
百萬瓦
十億瓦
10-1
deci(d)
dm
公寸
1012
tetra(T)
Tm
TW
兆瓦
百: hundred; 千: thousand; 萬: ten thousands; 百萬: million
千萬: ten million; 億: hundred million; 十億: billion
兆 (萬億) : trillion
57