Professor Zhang’s Summary to Chapter 7
Summary: Chapter 7
1. First Law of Thermodynamics
A system possess energy due to (1) the whole system move (translational + rotating)
relative to a reference (i.e., the earth); (2) the whole system possess a height relative to a
reference level (i.e., the ground); and (3) the system’s internal energy possessed by the
mass of the system (sensible kinetic + latent kinetic + chemical bond + nuclear) due to the
microscopic motions and the interactions of the molecules and their constituent particles.
1)
2)
3)
4)
5)
6)
7)
A. Terminology
List the three components (or forms) of the total energy of a system?
Explain equation: E = Ek + Ep + U
Define (in one sentence) kinetic energy, potential energy, and internal energy
What are the two forms of energy transfer between a system and its surroundings
Define Heat
Define Work
Explain the following equations:
Ek = 12 mu 2
E p = mgh
8)
9)
10)
&
&
E k = 12 m u 2
E p = m gh
(1)
(2)
List three components of the internal energy.
List and briefly explain the three energy interactions between an open system and the surroundings
(i.e., due to mass in/out, heat transfer through boundary, work)
List and briefly explain the two energy interactions between a closed system and the surroundings
B. First Law of Thermodynamics
a)
b)
c)
Statement: Energy can neither be created nor destroyed. It can only be transferred from one system to
its surroundings (or vise versa) or be transformed from one form to another.
Energy transfer mechanisms: either an exchange of heat (i.e., heat transfer), the result of a force (i.e.,
work), or both.
Descriptive equation of the Law of Energy Conservation (or the 1st Law of Thermodynamics)
 Rate of
  Energy entering   Heat energy   Energy exiting   Energy used 

 
 
 
 

accumulation
with the
with the

 = 
 +  added to  - 
 -  by the system 
 of energy   input streams   the system   output streams   to perform work 

 
 
 
 

2. Energy Balances
1)
2)
3)
4)
5)
A. Terminology:
Closed system
Open system
Adiabatic process
Isothermal process
Specific properties (specific internal energy, specific kinetic, specific potential energy, specific
volume, specific enthalpy, …): an intensive quantity obtained by dividing an extensive (or its flow rate)
by the total amount (or flow rate) of the process material.
Professor Zhang’s Summary to Chapter 7
6)
Explain H  mHˆ
7)
8)
ˆ = Uˆ + PVˆ
Specific enthalpy: H
What are flow work and shaft work?
and
U  mUˆ , where “m” is the total mass of the process material
9)

Explain equation: W
= W s + W fl
10)
Explain equation: W fl  ( PV )  PoutVout  PinVin
11)
12)
Explain state properties:
What is reference state, give an example
(3)
(4)
(5)
B. Energy Balance for closed systems
There is no material go across the system boundary, therefore the changes in total system energies
(including, kinetic, potential, and internal) between two instants of time (from 1 to 2),
equal to the heat absorbed from the surrounding minus the work performed to the
surrounding. It can be expressed by equation 6 below:
E2  E1  E  Q  W
(6)
Substitute E = U + Ek + Ep into Equation 6, we obtain,
U 2  E k 2  E p 2  U 1  E k 1  E p1  Q  W
(7)
Re-group the terms on the left side of Equation 7 and note ΔU=U2-U1, …, we obtain,
U  Ek  E p  Q  W
(8)
Possible simplifications of Equations 6 to 8:
U = f(chemical composition, state of aggregation, T), independent of P for ideal gas, almost
independent of P for liquids and solids. If no reactions, no phase change, ∆T = 0, ∆P < a few atm,
then ∆U ≈ 0
2.
If system no acceleration, ∆Ek = 0, if no rising or falling, ∆Ep = 0
3.
If system and surrounding has same temperature or system well-insulated, Q = 0, the process is then
called adiabatic
4.
Work done on or by a closed system is accomplished by moment of system boundary against a
resisting force or the passage of an electric current or radiation across the system boundary. Without
these resistances, W = 0
1.
C. Energy Balance for open systems at steady state
There is no accumulation for a steady state process and therefore,
Energy in = Energy out, mathematically this can be expressed as,
Energy in = the total rate of transport of kinetic energy, potential energy, and internal energy by all
process input streams + the rate at which the energy is transferred to the system as heat
Energy out = the total rate of transport of kinetic energy, potential energy, and internal energy by all
process output streams + the rate at which the energy is transferred out of the system as work
Professor Zhang’s Summary to Chapter 7
For a system with “n” streams of input and “k” streams of output, absorbing heat at a rate of Q
and performing shaft work at a rate of W , we have:
s
n
 m i (Uˆ i  PiVˆi 
i 1
k
u2
ui2
 gzi )  Q   m j (Uˆ j  PjVˆ j  j  gz j )  W s (9*)
2
2
j 1
Re-group the above equation and note that H=U+PV, we obtain,
2
2
k
u
u
i
  m i ( Hˆ i   gzi )   m j ( Hˆ j  j  gz j )  Q  W s
2
2
i 1
j 1
n
(9)
Equation 9 can also be written as:
H  E k  E p  Q  W s
where
n:
k:
m i :
m j :
total number of input streams to the system
total number of output streams from the system
mass (or molar) flow rate of input stream “i”
Ĥ i
specific enthalpy in J/g (or in J/mole) of input stream “i”
Ĥ j
specific enthalpy in J/g (or in J/mole) of output stream “j”
ui :
uj :
zi :
zj :
Ws :
Q
average velocity of input stream “i”
average velocity of output stream “j”
elevation of input stream “i” relative to the reference plane
elevation of output stream “j” relative to the reference plane
rate of shaft work at which the system is performing to the surrounding
Û i
specific internal energy in J/g (or in J/mole) of input stream “i”
Û j
specific internal energy in J/g (or in J/mole) of output stream “j”
Pi :
Pj :
Vˆi
Vˆ
pressure of input stream “ i “
pressure of output stream “ j “
specific volume of input stream “i”
j
[J/s]
mass (or molar) flow rate of output stream “j”
rate of heat transfer from the surrounding to the system
specific volume of output stream “j”
For open systems with single input stream, single output stream, at steady state, Equation 9 can be reduced
to:
u u
m ( Hˆ out  Hˆ in  out in  gzout  gzin )  Q  Ws
2
2
2
Professor Zhang’s Summary to Chapter 7
steam turbine,
For
kinetic and potential energy changes can be neglected and the process is close

to adiabatic ( Q =0), therefore,
W s  m ( Hˆ in  Hˆ out )
fluid flow processes
In
, the change in internal energy is essential zero, there is only one
stream in and out, no heat transferred to the fluid, no shaft work, therefore the energy balance equation (9*)
can be reduced to: mechanical energy balance equation:
ΔP
+
ρ

Pressure
change
Δu2
2

Kinetic energy
change
+
gΔz

Potential energy
change
 
 ˆ Q
+  ΔU
-  =
 
m


Friction loss
Exchange with
environment

W
- S

m

Work
(pump/compressor)
(10)
3.
Tables of Thermodynamic Data
a)
b)
c)
What is a state property?
What of a reference state and the purpose of choosing one?
Are the absolute values of specific internal energy and specific enthalpy depends on the reference state
chosen?
Find out regions in a Phase Diagram (T vs. P) for subcooled liquid, saturated liquid, saturated steam,
and superheated steam
d)
4.
General procedures for energy balance
a)
Draw a flowchart and identify all knowns and unknowns. Making sure to show the aggregation state
(liquid, solid, or vapor) of each stream!
Choose a convenient basis of calculation
For a multiple unit process, identify the subsystems for which balances might be written
Analysis for the overall system and each possible subsystem for solubility by comparing the number of
available equations and the number of unknowns
If possible, formulate and simplify material balance equations to determine the flow rates of all stream
components
Write appropriate form of the energy balance and solve it for the desired quantity
If Step 5 not possible, determine as much as possible the energy properties of the streams and combine
energy and material balance equations and solve for the unknowns
b)
c)
d)
e)
f)
g)
NOTE: for a non-reactive system with “n” material components the total
number material balance equations is “n”. However, there is only one energy
balance equation for the system!!!