CH1104 Part II:

Energy Balance
Wang Xin, Associate Professor
Division of Chemical & Biomolecular Engineering
School of Chemical & Biomedical Engineering

N1.2-B1-15
WangXin@ntu.edu.sg
Tel: 6316 8866
CH1104 1
As an chemical engineer designing a process, your job also
include: to determine the overall energy requirement.

Typical problems you may solve using energy

balances:
1. How much power is required to pump 1250 m3/h
of water from a storage vessel to a process unit?
2. How much energy is required to convert 2 ton of
water at 30 ºC to steam at 180 ºC?
3. A highly exothermic reaction A -> B takes place
in a continuous reactor. If 75% conversion is to be
achieved, at what rate must heat be removed to
keep isothermal operation?

CH1104 2
• Textbook:
– “Elementary Principles of Chemical Processes”
by RM Felder & RW Rousseau
• Lecture Notes: (EdveNTUre)
• Summary notes: (EdveNTUre/textbook)
• Video Reference: MIT open Course:
Thermodynamics , Part 1-4.
(http://ocw.mit.edu/OcwWeb/Chemistry/5-
60Spring-2008/VideoLectures/index.htm)

CH1104 3
Content

 Energy and Energy Balances (Chapter 7)

 Balances on Nonreactive Processes (Chapter 8)

 Balances on Reactive Processes (Chapter 9)

 Balances on Transient Processes (Chapter 11)

CH1104 4
7.1. Introduction
• What is Energy ?
– The ability to do WORK, W

Chemical electrical mechanical

CH1104
Sound light thermal 5
 –Work
W = Force × distance, (Unit: Joule)
1 Joule (N.m)
= 1 Newton × 1 Metre
= 107 ergs (dyne·cm)

CH1104 6
–Power (Unit: Watt or Horsepower)
The rate of doing work, P = Work /Time
(1 Watt = Joule/s, 1 Horsepower = 75 kg× 1 m, in 1 second)

75kg
75kg

CH1104 7
1 HP 0.5 HP
• Work & heat: Joule: also an unit for Heat

Heat-Work conversion experiment

Thermo
meter

weight
James P. Joule (1818-1889)

1 Joule = 0.239 cal Tank with Water, Paddle,

1 cal = 4.18 Joule E  Q
CH1104 mgh  mH 2oCP T 8
 Heat capacity (Specific heat capacity or specific
heat, CP): Heat energy required to increase the
Temperature of a Unit quantity of a substance by a
certain Temperature Interval (1°C).
(Unit: Joule/g/K; Joule/g/°C; Joule/mol/K)

CH1104 9
• Law of energy conservation:
( First Law of Thermodynamics)

“Energy can neither be created

nor destroyed (simply converted)”
Perpetual machine

-The law of nature

- The foundation of energy balance

CH1104 10
• Energy issues ?

– Energy shortage drags our economy !

– Energy crisis is looming !!
– Save energy, save the world !!!

 Politically: correct
 Scientifically: wrong ( or misleading)

Why ?
–1st law of thermodynamics: energy can neither
be created nor destroyed. (simply converted)

CH1104 11
• The Quality of Energy
– Shortage of “useful” (High quality) energy
– Plenty of low grade (low quality) energy

“Energy” crisis  “Useful” Energy  Resource

- High quality & low quality energy

(a) High in density (kJ/kg)

(b) Easy to convert to other forms
(High ratio of conversion)
(c) Easy to store, transport
CH1104 (d) Less pollution 12
 Chemical Electricity Mechanical

Thermal
Sound Light Energy

CH1104 13
• Example: Internal Combustion Engine (ICE)
– Otto Engine (Petrol)

Ƞ <30%

Source of energy:
Fuel + jO2  mCO2+ nH2O + Q
CH1104 14
• Why so low efficiency ?

Energy Conversion in ICE:

Chemical  thermal  mechanical

Limited by 2nd law of thermo, Carnot circle,

η=(TH-TL)/TH

• Solutions : Hybrid; Electrical; Fuel cell, etc.

CH1104 15
CH1104 16
 (1). Open system (also called Control volume)
• Mass (and energy) interchange with
surroundings

(2). Closed system

• Only allow energy transfer, no mass
interchange with surroundings

(Isolated system: no mass , no energy

interchange with surroundings, or a closed system
without energy flow)

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 Examples
Open system

Closed system

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CH1104 19
• Three Components of Energy in a System
System energy = summation of these two forms of energy
(1) Kinetic energy: Due to the relative motion

1 2
Ek  mu Unit: kJ
2


 1  2 m : kg / s
Ek  mu 
2 E k : kw
Rate of kinetic
CH1104 energy 20
 (2). Potential, EP, due to the position in a
potential (gravimetric) field

E p  mgh Unit: kJ


 
m : kg / s
E p  m gh 
E p : kw
work ( Joule )
Power ( watt ) 
CH1104 time( s) 21
 (3). Internal U: High T, P

Energies other than kinetics & potential energy of

a system (but excluding the chemical bond energy, e.g. H2
and He at same T,P may have very similar U)

A Summation of Motion/interactions/vibration of
molecules in the system

Generally, a function of Temperature, Pressure,

Components, etc.  f (T, P, x)
CH1104 22
CH1104 23
7.2. First Law of Thermodynamics
Applicability: In a closed system, Energy Transfer
take places only in two forms:
(1) heat, Q
(2) work, W
So that: ΔE = Q + W
where
E (total energy) = U (internal) + EK (kinetic) + EP (potential)

∆E (change in total energy) = EFinal ― EInitial

= ΔU + ΔEK + ΔEP
Note: The sign of W is different from the textbook, to be consistent with other
SCBE subjects.
CH1104 24
•Definition of the Sign of Q and W
Final system Initial system total Net energy
total Energy Energy Transfer

( U + EK + EP )Final ( U + Ek + Ep)Initial Q+W

Q > 0 Heat transfer from surrounding to system

Q < 0 Heat transfer from system to surrounding
W < 0 System does work to surroundings
W > 0 Surroundings does work to system

i.e.: Gain = positive ‘+’ ; lose = negative „-‟

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• Example of applications:

-
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 For a small change in the volume

System gains heat, Q>0

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 for isothermal compression

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 The volumetric work depends on the system path
(how it performs)

V1 P1

V2 P1 V1 P2

V2 P2

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• Simplification of 1st law in some typical systems:

(1) System is Adiabatic (Q = 0, no heat transfer, EK,, EP )

ΔE = Q + W  ΔU = W (physical meaning ?)

(2) System boundary is fixed (and no moving part, W=0,

no volumetric work, no EK,,&, EP )
ΔE = Q + W  ΔU = Q (physical meaning ?)

(3) System is isolated (no heat transfer, no volumetric

work)
ΔE = Q + W  ΔU = 0, Q=0, W=0
CH1104 30
• Energy balance in a closed system
0
Example 7.3-1 (in textbook)
A gas is contained in a cylinder fitted with a movable piston . The initial
temperature is 25C.
The cylinder is placed in boiling water with the piston held in a fixed
position. (1) Heat in the amount of 2.0 kCal is transferred to the gas,
which equilibriates at 100C (and a high pressure). (2) The piston is
then released, and the gas does 100J of work in moving the piston to
its equilibrium position. The final gas temperature is 100C.
Write the energy balance equation for each of the two stages of this
process, and in each case solve for the unknown energy term in the
equation.
In solving this problem, consider the gas in cylinder to be the system,
neglect the change in potential energy of the gas as the piston moves
vertically, and assume the gas is ideal. Express all energy in Joule.

CH1104 31
Solution:
(1) Draw process diagram, mark the knowns
T0=25ºC T1 = 100ºC W = 100 J

2 3
0 1

T3 = 100ºC
T2 = 100ºC
Q = 2 kcal

(2) Assumptions: ideal gas, neglect potential energy,

no friction, no heat loss
(3) Target: to find Energy Balance in stage (0-1), (2-3)
CH1104 32
(4) Identify Tools (theory):
ΔU + ΔEK + ΔEP = Q + W
– Using assumptions,  ΔU = Q + W

Process 0  1,
W=0 (boundary fixed, no volumetric
work done)

U  Q  2 kcal
2  1000cal
  8368 J
0.239 cal J

The gas gained 8368 J of internal energy in

CH1104 process (01) 33
 W =100 J
Process: 23
(Isothermal) 2 3

T2 = 100C T3 = 100C
ΔU + ΔEK + ΔEP = Q + W
=0 !! =0 =0
0=Q+W
Ideal gas
isothermal W= -100J
U= f(T) !!!
Q = 100 J

An additional 100J of heat is transferred to the

CH1104
gas as it expands and re-equilibriated at 100C 34
• First Law in daily life (our body can be treated as
a closed system) Δ U = Q + W

Assumption:
U = Fat/Muscle (accumulated)
Q = Food supply;
W = Exercise;

To reduce U (fat),  ΔU < 0 ,

(1). More heat release & less input, Q < 0
 eat less
(2). Do more work to surroundings, W <0
CH1104  More exercises ! 35
 Summary
• Energy (quality of energy), Heat/work conversion: 1 cal =
4.18Joule
• System, Boundary, Surrounding
• Open system, Closed system, isolated system, adiabatic
system
• E= U+ EK+ EP (U, a function of T only for ideal gas.)
• 1st law for closed system: ΔE =Q+W, Δ E= ΔU+ ΔEK+ ΔEP
• Q>0, Q<0; W>0, W<0
• Volumetric work

CH1104 36
Definition of some important terms
• Steady state process
– System parameters (T,P,V,Q, W, etc. ) do not
change with time.

• Transient state process

– System parameters (T,P,V,Q, W, etc. ) is a
function of time.

m  const

Example: the volume of water in cup

Transient state,V = f(t) before the cup‟s full,
Steady State, V = Const after the cup‟s full
CH1104 37
• Batch/ Continuous operation

Batch process

Continuous
process

Transport of (a) People (b) Oil

• Extensive property of the process material

- A property depends on the size or quantity of
the system, e.g. weight, volume, U, etc.

100KG
100K
G
CH1104 100KG 38
• Intensive Property:
– Independent of size, e.g., Density , T, P, etc.
– Specific property is on the basis of per unit
mass (therefore, intensive properties ):
• Volume/mass specific volume, m3/kg ^
V V W
• Specific kinetic energy (kJ/mol) ^
EK  EK M
• Specific potential energy (kJ/mol) ^
EP  EP M
• Specific internal energy (kJ/mol) ^
U U M

(Specific properties also use the unit: XX /kg)

CH1104 39
 • Why specific properties ?
- Easy to compare /represent the system

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 • Specific properties in life
• GDP , (Gross domestic product) is the market value of
all final goods and services produced within a country
during a given time.

• GDP per capita is the normalized GDP with the

population  specific GDP (intensive property, easy
to compare and more useful)

– For example: Singapore (US), 2010

GDP (extensive prop): 222(14,657) billion USD, ranked 39 (1)
GDP Per Capita (intensive prop): 43K USD, ranked 15 (9)
CH1104 41
7.3. Energy balance for open systems (basics)
•Open system (also called control volume):
-Both Energy & Mass Interchange with surroundings
(cross the boundary)
-Represented by:
a steam turbine
boundary

Steam, Steam,
T1, P1 T2, P2

CH1104 Schematic Diagram of a steam turbine 42

CH1104 43
 (useful)
(useless)

CH1104 44
 = -(Ws+Wfl)

Enthalpy

CH1104 45
• Why H ?
- Calculate the enthalpy change for the expansion of ideal
gas in a closed system (piston-cylinder)
Assuming constant pressure, negligible KE & PE

Solution: for a closed system

ΔU = U2– U1 = Q + W
= Q - P(V2 – V1)  volumetric work
(U2 + P2V2) – ( U1 + P1V1) = Q P

 H2 –H1 = ΔH = Q
^ ^ ^
• Specific Enthalpy: H  U  PV
Q
• Enthalpy rate:
CH1104 46
• Example - Calculation of Enthalpy:
The specific internal energy of helium at 300K and 1 atm is 3800J/mol,
and the specific molar volume at the same temperature is 24.63 L/mol.
Calculate the specific enthalpy of helium at this T and P, and the rate at
which enthalpy is transported by a stream of helium at 300K and 1 atm
with molar flow rate of 250 kmol/h.
Solution:
^ ^ ^
H  U  P V  3800  1 24.63 ( L  atm)
(Conversion factor: 1 L.atm = 101.3 J) , Pls DIY

So that (DIY for details):

^
(1) H  6295 J / mol
  ^
(2) H  n H  6295  250,000  1.57 109 J / h
CH1104 47
 P : N/m2

: Volumetric flow rate
V
m3/s
CH1104 48
• General Equation for E-Balance in an open
system at steady state
    
 H   E k   E p  Q  Ws
 ^ ^
Where: H   m
output
j H j   m j H j
input

 m j u 2j m j u 2j
 Ek  
output 2

input 2

 Ep   m
output
j gz j   m
 j gz j
input

If single working media and no phase changes:

 ^ ^  ^ ^
H   m
output
j H j   m j H j  m( H 2  H 1 )
input
CH1104 ^ ^ ^ ^ ^ 49
H  Ek  E p  Q  Ws
• U&H
– For real gases:
• U& H are function of T & P

– For ideal gases: (PV = nRT)

• U is a function of Temperature only
• H is a function of Temperature only, too .
(Proof: H = U + PV = U + nRT )

– For solid/liquid
• U and specific V is nearly independent of pressure
• H will depend on pressure.

If pressure changes at constant temperature,

how will U and H change?
CH1104 50
• Schematic example of ideal gas and real gas

A hypothetical gas consisting of identical particles of

(1) Negligible volume,
(2) No intermolecular forces.

1. River (ideal gas) 2. Fish tank (real gas)

CH1104 Fish = molecules 51

• Example: E-balance on a turbine
500kg/h steam drive a turbine. The steam enters at 44 atm and 450ºC and a linear
velocity of 60 m/s and leaves at a point 5 m below the turbine inlet at
atmospheric pressure and a velocity of 360 m/s. the turbine delivers a shaftwork
at the rate of 70 kW, and the heat loss from the turbine is estimated to be
10,000 kcal/h. Calculate the specific enthalpy change associated with the
process.

Solutions: 500kg/h
W = 70 kW
44atm
450ºC 5m 500kg/h
60m/s 1 atm
360m/s
Q = -104 kcal/h

^ ^ ^

CH1104
 H  H 2,out  H 1,in  ??? 52
 With the E – Balance of the open system:
    
 H   E k   E p  Q  Ws

 ^ ^
H   m
output
j H j  m
jHj
input

 ^ ^
  ^ Target
 m H out  H in   m  H
 

 Ek  ?


Therefore, we need to work out all Ep  ?


the other parameters. Q ?

Ws ?
CH1104 53
 
Mass Flow m  500 / 3600  0.139 kg / s
(rate)


 E k  u2  u12   8.75 kW


Kinetic E (rate m 2
change) 2

 E p  m z2  z1   6.81  103 kW

Potential E  

(rate change)
Heat Transfer
 kcal 1J 1h 1kW
(rate) Q  104   
h 0.239  103 kcal 3600 s 103 J
  11.6kW

Power output

(rate) W s  70 kW
CH1104 54
Using the first law for open system @ steady
state
    
 H  Q  Ws   E k   E p  90.3 kW


^  ^

 H  m H out  H in 
 

H ^ ^


 H out  H in
m
^  90.3 kJ / s
H   650 kJ / kg
0.139 kg / s
CH1104 55
7.4. Steam Table (Thermodynamic data)
To solve energy balance equation involving water
or saturated steam or superheated steam, etc, we
need ^ ^ ^ ^
-Their physical properties of ( H , U , V , S )

-Their values are Independent of path, but only

depend on the state of the system (T, P, V, c, etc.)

 only depends on ^ ^
 U 21  U 2  U 1
^

P b 2
the initial and ^ ^ ^
final states of the c  H 21  H 2  H 1
system. i.e.
CH1104 1 a 56
T
 – Example: mountain climbing, the change in your
potential energy is independent of the path;

State property

– the work done depends on the path. W= F×S

ΔPE = mgΔh

57
CH1104
• State parameters (T, P, V, c):
– To define H & U of a system, How many
parameters need to be fixed ?
• Ans: Gibbs Phase rule: Freedom, f = C – P + 2

Example: 1. Steam (water vapor)

f= 1–1 +2=2
Therefore, T, P, V, only two are independent, or two state
parameters needed to fix the state of the steam.

Example: 2. Water-Steam in equilibrium (boiling water)

f= 1–2 +2=1
Therefore, T, P, V, only one is independent, or one parameter fix
the state of the steam-water system
CH1104 58
• Reference States
– Impossible to know the absolute value of potential
energy
– Impossible to know the absolute value of
Internal energy: U,
enthalpy H,
entropy S

- Therefore, a reference sate is necessary for

computation purpose.
- Example: Normally, the ground is used as the
reference level (H=0) to calculate potential energy,
Pe = mgΔH

CH1104 59
• Steam Table
^ ^ ^ ^
-Physical properties of ( H , U , V , S ) for

saturated water,
Saturated steam,
Superheated steam, etc.

-Reference point of steam table:

Triple point of water

i.e., @ T = 0.01°C,
P = 0.00611 bar
we set:
^ ^
U 0 H 0
CH1104 60
•Fundamental concepts for Steam Table (review)

(1) Kinetic/dynamic equilibrium:

Suppose the water-vapor system
At a Temperature: the number of molecules escaping
from liquid water equals the number of molecules
condensing into the water
(2) Saturation -- At this point the vapor is said to be
saturated, and the pressure of that vapor (e.g. mmHg)
is called the saturated vapor pressure
(3) Boiling point -- The temperature at which the vapor
pressure is equal to the atmospheric pressure is called
the boiling point.

CH1104 61
• Cooking water in kitchen (@constant pressure)
- with a closed system (piston-cylinder)

P=1atm
P=1atm

P=1atm
P=1atm
30- 100-
100C 100C
20-100C
100C 1000C

Q Q Q Q

CH1104 time 62
• Cooking water under constant pressures in plot

- Saturated liquid (2, f)

- Saturated vapor (4, g)

- Superheated vapor (5, s)

v: specific volume of the system, m3/kg
After one curve, we raise the pressure, e.g. put
more weight on top of the piston, then repeat the
process, again, what we have ???
CH1104 63
 The results:

A multiple pressure T~v diagram

CH1104 64
 •Property Diagrams 1:-- T-v
Bubble point :
• 3 Regions: Table B.5, B.6

– 1. Compressed liquid
– 2. Saturated liquid-vapor
mixture (wet region),

– 3. Superheated vapor,

Dew Point:
Table B.5, B.6

CH1104 65
• Table B.5: Saturated Steam – Temperature

T P
Temperature Reference
State
Corresponding
Sat. pressure
CH1104 Specific volume 66
v
•Table B.6: Saturated Steam –pressure
Reference

Using the linear interpolation for the middle

values, e.g., @ P = 0.019 Tsat = ??
CH1104 67
•Table B.7: Superheated Steam

Superheated
Steam Region

liquid water
Region

Pressure, and the corresponding

Temperature at which the steam is
CH1104 saturated 68
•Example 1: Using the Steam Table
^ ^ ^
Determine U H V of saturated steam at 133.5°C
using Table B.6

CH1104 69
•Answers:
- For Saturated Steam @ 133°C:
^ ^
Psat  3.0bar , U  2543.0 kJ / kg H  2724.7 kJ / kg
^
V  0.606 m 3 / kg
- For Saturated water @ 133°C, what are the above values
??? DIY

Pls Pay attention to unit !!!! (no joking)

Because Many tragedies happened before: such as, J/kg, J/mol, cal/kg,
liter/g, etc.
CH1104 70
•Example 2. Using steam table. Show water at 350C, 10
^ ^ ^
bar is superheated steam, and determine: U V H
Ans: Using Table B.7: P = 10 bar, Tsat = 179.9°C < 350°C
^ ^ ^
V  0.282 m / kg U  2876 kJ / kg H  3159 kJ / kg
3

CH1104 71
•Example 3. Linear interpolation @no phase change
^
Determine P, V for saturated water (liquid) at 9°C:

 Using the values at 8 and 10C

to interpolate the values at 9C

Y
x2,y2

x3,y3
@ x3 , y3 = ?

x1,y1
X
T=9
y3  y1 y2  y1

CH1104 With triangle geometry x3  x1 x2  x1 72
 X(T) Y(P) Y(V)
(x1,y1) = (8, 0.01072)
9 0.011495 113.7
(x2,y2) = (10, 0.01227)
y2  y1
y3  y1  ( x3  x1 )
x2  x1
Checking !!!
CH1104 73
 ^ ^ ^
• Example 4: Double interpolation for V U H
e.g. water @ 3 bar, & 220°C (no phase change)

2865 2968

^
CH1104 Answer: H  2906 kJ / kg 74
• Example 5: Determine the unknown properties of a
steam, with the known 2 state parameters.
– T, P,  find the state of water
• P = 5bar, T= 111C,  compressed liquid

– U, T,  find the state of water

• U = 2883, T = 350C,  Superheated vapor

– H, P,  U, V, T  Superheated vapor
• H = 3052, P = 10 bar, T = 300C, V = 0.258, U = 2794

Note: U, H are state properties, can also be

CH1104
used to determine the system state 75
Self Test: At the constant pressure of 10 bar, water

is heated from 20°C onward, it will boil at the

temperature of____ (°C), with the specific enthalpy of

the vapor phase = _________ ; if temperature is

further increased to 325°C, the steam is then (a)

compressed liquid, (b) overcrowded vapor, or (c) ☺

superheated steam [please tick one only!] with the

specific volume of _______________.

CH1104 76
Example 6: Energy Balance using steam table
Ex 7.5-3@ page 329. Steam at 10 bar absolute with 190C of superheat
is fed to a turbine at a rate of 2000 kg/h. The turbine operation is
adiabatic and effluent is saturated steam at 1bar. Determine: power
output in kw, neglecting kinetic and potential energy changes.

Solution: Draw the diagram & label the knowns

2000 kg/h
1 bar
10 bar, with
Saturated steam
190C superheat

Adiabatic: Q = 0
Ws = ?????
CH1104 77
 Analysis:
1. The system: Open System @ Steady State
2. Without any info of kinetics/potential energy
 ΔEk, ΔEP negligible
3. Turbine is adiabatic

    
 H   E k   E p  Q  Ws
Using Assumptions
and given conditions
   ^ ^
Ws   H  m( H out  H in )
CH1104 78
 ^
-The key is to get the specific enthalpy H of inlet &
outlet steams
^
Inlet P = 10 bar, Tsat = 180 C H
T = 190 + 180 = 370 C 3201 (kJ/kg)
^
outlet P= 1 bar, T =? H
Saturated 2675 (kJ/kg)

 2000kg ( 2675  3201)kJ 1h

Ws  H   
h kg 3600s
  292 kJ / s  292 kW
The turbine delivers 292 kW of work
(“-” : system does work to surroundings )
CH1104 79
•Example 7: E-Balance of one-component process
Ex 7.6-1 @ page 330: two streams of water are mixed to form the
feed to a boiler. Process data are as indicated. The existing steam
emerges from the boiler through a 6-cm ID pipe. Calculate the
required heat input to the boiler in KJ/min if the emerging steam is
saturated at the boiler pressure. Neglect the kinetic energies of the
liquid inlet streams.

Solutions: 1. diagram labeled with knowns and unknowns

Stream 1
Stream 3

Stream 2
CH1104 80
-Analysis:
 E-balance of Open system at steady state

-Procedures:
1. Draw the process diagram, labeled with
knowns & unknowns
2. Determine flow rate of all components via
material balance
3. Get the specific enthalpy of each component
4. Set up energy balance to solve the required
quantity
CH1104 81
1. Draw the flowchart

2. Determine flow rate of all components via

material balance
  
m3  m1  m2  295 kg(v) / min
3.Get the specific enthalpy of each component

• Assuming saturated H2O @ inlet

• Saturated vapor @ outlets (P=17bar)

4.Set up energy balance to solve the required

quantity
CH1104 82
Energy balance of open system @ steady state
    
 H   E k   E p  Q W
  
 H  Ek  Q

1 
 ^ ^ m u 2

H   m
output
j H j   m j H j
input
2

 7.6110 kJ / min 5
V (m / s)  A(m )  u(m / s)
3 2

CH1104 83
 Therefore
   

u V m V m  202 m / s
A A A
0.1166 m3 from Table B.6 at 17
   bar, 204 C
Q   H   Ek
Neglect inlet EK

1  2
  
m u  6.02 10 kJ / min
3

2
Q   H   Ek
 7.61  10  6.02  10  7.67  10 kJ / min
5 3 5

CH1104 84
 •Review: Three rates @ steady state

- (1) Mass flow rate, 

– Constant for steady state m kg / s

- (2)Volumetric flow rate: V m3 / s
– Constant for incompressible liquid

- (3) Linear velocity u m/ s

– Varying with the tube diameter (cross sectional
area)

CH1104 85
 • Review: The relation between the 3 rates

Q: Suppose water (constant density) flowing through the channel


below m3 / s
Volumetric flow rate V

Mass flow rate m kg / s
A1 , u1 A2, u2 Linear velocity u m/s

1 2
A: cross
section area


V (m3 / s)  A1 (m2 )  u(m / s)  A2  u2
 

CH1104
m   (kg / m3 ) V 86
•Example 8: Simultaneous M & E Balance
Ex 7.6-3 @ page 332. Saturated steam at 1atm is discharged from a turbine at a
rate of 1150 kg/h. superheated steam at 300C and 1 atm is needed as a feed to
heat exchanger; to produce it, the turbine discharge stream is mixed with
superheated steam available from a second source at 400C and 1 atm. The
mixing unit operates adiabatically. Calculate the amount of superheated steam
at 300C produced and the required volumetric flow rate of the 400C steam.

Solution: Draw the process diagram, labeled with knowns &

unknowns

Adiabatic Super heated

 
V 1  ?, m2  ?
Super heated
CH1104 87
-Analysis:
 
2 unknowns: m1  ?, m2  ?
2 equations :
(a) Mass Balance:
 
1150  m1  m2 (1)

(b) Energy Balance:

    
 H   E k   E p  Q W s

H  0
CH1104 88
  ^ ^
H   m
output
j H j   m j H j  0
input
 
1150  2676  m1 3278  m 2  3074 (2)

Solving (1) & (2):  

m1  2240 kg / h m2  3390 kg / h

-To convert to volumetric rate of steam @ inlet (1)

^
V 1  3.11 m3 / kg @ 400C, 1 atm(Table B.7)

  ^
V1  m1 (kg / h) V1 (m3 / kg)  6980 m3 / h
CH1104 89
7.5. Mechanical energy balance
• Chemical processes, concerning:
– Q, H, U Normally KE and PE can be neglected
(represented by heat engine, turbine, piston-cylinder
which convert thermal energy to shaft power)

• Mechanical processes (now), concerning:

– Shaft work (Ws)

– Kinetics energy (KE)
– Potential energy (PE)
(represented by hydraulic-turbine, which convert
CH1104 kinetic or potential energy into shaft power) 90
•Example of application: Hydropower plant

CH1104 91
• System analysis: First law for an open system:
    
 H   E k   E p  Q W s
^ ^ u 2j  ^ ^ u 2j   
m j U j  Pj V j   gz j   m j U j  Pj V j   gz j   Q Ws
 2  output  2  input

1/ρ
 
 
P u 2  ^ Q  Ws
  gz    U     
 2   m
 m 
^
Friction loss, F >0
CH1104 92
• First law of open system at steady state for
mechanical energy balance

P u 2 ^ Ws
  gz  F  
 2 m

Assuming:
(1) Steady State flow,
(2) Incompressible fluid
^ ^  
(3) F   U  Q m
CH1104 93
•Bernoulli equation (fluid mechanics)
^ 
If no friction loss, no shaft work F  0 Ws 0

P u 2
  gz  0
 2

Bernoulli equation (ODE)

dy
 Py  Qy n
zy 1n
dx
dz
 (1  n) Pz  (1  n)Q
CH1104
dx 94
•Example 1: Bernoulli equation
Ex 7.7-1 @ page 334. Water flow through the system at the rate
of 20 L/m. Estimate the pressure required at point (1) if friction
losses are negligible

Solution: Draw the process diagram, marked with

knowns and unknowns

-Analysis: The volumetric flow rate

are same @ (1) and (2)

Linear velocity are different for


(1) and (2):
u V
A

Bernoulli equation:  P u 2
  gz  0
CH1104  2 95
 L 1m3 1 10 4 cm 2 1 min
u1  20  3   
min 10 L  (0.25) 2
m 2
60s
 17.0m / s

L 1m3 1 10 4 cm 2 1 min
u2  20  3   
min 10 L  (0.5) 2
m 2
60s
 4.24 m / s
P u 2
With Bernoulli equation   gz  0
 2
P2  P1
3
 135.5 Nm / kg  490 N .m / kg  0
1000kg / m
( P2  1.013 105 N / m 2 )
P1  4.56 105 N / m 2  4.56 105 pa  4.56bar
CH1104 96
• Example 2: Siphoning
Gasoline ρ = 50.0 lbm/ft3 is to be siphoned from a tank. The friction
^
loss in the line is F  0.80 ftlb f / lbm . Estimate how long it will take to
siphon 5.00 gal, neglecting the change in liquid level in the gasoline
tank during this process and assuming that both point (1, at the liquid
surface in the gas tank) and point (2, in the tube just prior to the exit)
are at 1 atm.

Solution: the process diagram with knowns / unknowns

ρ = 50 lbm/ft3

P1=1 atm,
u1 ≈ 0
z1 = 2.5 ft P2=1 atm,
z2 = 0 ft
CH1104 97
-Analysis: Mechanical energy Balance for open system
at steady state: 
P u 2 ^ Ws
  gz  F  
 2 m

For the system: p  p , u1  0, w0
2 1

Therefore : u22 ^
 gz  F  0 u2 = 10.5 ft/s
2
 10.5 ft  (0.125) 2 1 ft 2 3
V  u2  A    2
 3.58  10 ft 3
/s
s 1 144in
volume to be drained
t  3.1 min
volumetric flow rate
CH1104 98
•Example 3: Hydraulic power

Water flows from an elevated reservoir through a conduit to turbine at a

lower level and out of the turbine through a similar conduit. At a point 100m
above the turbine, the pressure is 207 kPa, and at the point 3 m below the
turbine the pressure is 124 kPa. What must be the water flow rate if the
turbine output is 1.00 MW.

Solution: Draw process diagram, mark

knowns. neglect friction loss.
CH1104 99
-Analysis : take the surfaces of water as reference
points, then u1 = u2 = 0. -
Using Bernoulli equation:

P u 2 ^ Ws
  gz  F   
 2 m

  Ws
m
P   gz

CH1104 100
 - Using unit conversion:

CH1104 101
•Applications of Bernoulli equation
P u 2
  gz  0
 2
• Example 1. Flow in a tube. - The relation between
pressure and velocity (incompressible fluid)
A1 , u1 A2, u2 A = Cross-Section
area
1 2
Question: for Δz =0,
what is the relation btw
P and u ??

V ( m 3 / s )  A1 ( m 2 )  u1 ( m / s )  A2  u2

(u22  u12 ) u12 u22

P1  P2  P1   P2 
CH1104 2 2 2 102
• Example 2. Venture meter – measuring
volumetric flow rate
(u22  u12 )
P1  P2 
2
A1
A2
H2O

Hg
P1-P2 = (ρHg-ρH2O)×h

V  A1  u  A2  u2

CH1104 103
• Example application 3: airplane

(u22  u12 )
P1  P2 
2
F = (P1-P2) × A

F = Lifting force
A = Area

CH1104 104
• Example 4. Water spray, a daily life application
(u22  u12 )
P1  P2 
2

• Example 5. Bernoulli‟s ball , How does it work ?

CH1104 105
 Summary of Chapter 7 in PDF
1. Systems
2. First law (open/close/steady)
3. U/H (state property)
4. Steam table
5. Mechanical process

- Tutorial 1

CH1104 106