SOURCE MODELS
Prepared by
Associate Prof. Dr. Mohamad Wijayanuddin Ali
Chemical Engineering Department
Universiti Teknologi Malaysia
page 0
A pipe transporting liquid is shown in Figure 5. A pressure
gradient across the pipe is the driving force for the movement of
liquid. Frictional forces between the liquid and the wall of the pipe
converts kinetic energy into thermal energy. This results in a
decrease in the liquid velocity and a decrease in the liquid
pressure.
Flow of incompressible liquids through pipes is described by
the mechanical energy balance, Equation 1, combined with the
incompressible fluid assumption, Equation 2. The net result is
P
2
ws
u
g


z  F  

2g c
gc
m
(22)
page 1
Figure 5 Liquid flowing through pipe. The frictional flow losses
between the fluid and the pipe wall result in a pressure drop
across the pipe length. Kinetic energy changes are frequently
negligible.
page
2
The friction term, F, is the sum of all of the frictional elements
in the piping system. For a straight pipe, without valves or fitting,
F is given by
2 fLu
F 
gcd
2
(23)
where
ƒ is the Fanning friction factor (no units)
L is the length of the pipe
d is the diameter of the pipe (length)
page 3
The Fanning friction factor, ƒ, is a function of the Reynolds
number, Re, and the roughness of the pipe, e. Table 1 provides
values of e for various types of clean pipe. Figure 6 is a plot of the
Fanning friction factor versus Reynolds number with the pipe
roughness, e/d, as a parameter. Figure 7 presents the data of
Figure 6 in a form useful for certain types of calculations (see
Example 3).
For laminar flow, the Fanning friction factor is given by
f 
16
Re
(24)
For turbulent flow, the data shown in Figure 6 are
represented by the Colebrook equation
 1 e
1.255
 4 log 

 3.7 d
f
Re f

1




(25)
page 4
Table 1 Roughness factor, e, for clean pipes.
Pipe material
e, mm
Riveted steel
1 – 10
Concrete
Cast iron
Galvanized iron
Commercial steel
Wrought iron
Drawn tubing
Glass
Plastic
0.3 – 3
0.26
0.15
0.046
0.046
0.0015
0
0
page 5
Figure 6 Plot of Fanning friction factor, f, versus Reynolds number.
page 6
Figure 7 Plot of 1/ ƒ, versus Re  ƒ. This form is convenient for
page 7
certain types of problems. (see Example 2.)
An alternate from of Equation 25, useful for determining
the Reynolds number the friction factor, ƒ
f  0.25
1

10

Re 1.255 
f

1 e
3.7 d 
(26)
For fully developed turbulent flow in rough pipes, ƒ is
independent of the Reynolds number as shown by the nearly
constant friction factors at high Reynolds number on Figure 6. For
this case Equation 26 is simplified to
d

 4 log  3.7 
e
f

1
(27)
page 8
For smooth pipes, e = 0 and Equation 25 reduces to
Re f
 4 log
1.255
f
1
(28)
Finally, for smooth pipe with the Reynolds number less than
100,000, the following Blasius approximation to Equation 28 is
useful.
f  0.079 Re
1
4
(29)
For piping systems composed of fittings, elbows, valves, and
other assorted hardware, the pipe length is adjusted to
compensate for the additional friction losses due to these fixtures.
The equivalent pipe length is defined as
Lequiv  Lstraight   Lequiv
total
(30)
pipe
page 9
Where the summation is over all of the valves, unions, elbows,
and so on within the piping system. Table 2 provides selected
values for the equivalent lengths. Note that Table 2 includes
corrections for contractions and expansions in the piping system.
For many problems associated with pipe flow the contribution
due to the kinetic energy term in the mechanical energy balance is
negligible and then check the validity of the assumption at the
completion of the calculation.
For problems involving laminar flow, the solution is always
direct. Turbulent flow problems with an unknown pipe diameter,
d, always require a trial-and-error solution. Other types of
turbulent flow problems might be direct or trial-and-error
depending on the work and kinetic energy terms.
page 10
Table 2 Equivalent pipe lengths for various pipe fittings
(Turbulent flow only).
Pipe fitting
Lequiv/d
Globe valve, wide open
~300
Gate valve, wide open
¾ open
½ open
¼ open
90º elbow, standard
45º elbow, standard
Tees
Used as elbow, entering the stem
Used as elbow, entering one of two
sides
Straight through
Pipe connections to vessels
Ordinary, pipe flush with wall
Borda, pipe protruding into vessel
Rounded entrance, union, coupling
~7
~40
~200
~900
30
15
90
60
20
16
30
~0
page 11
Sudden enlargement from d to D
Laminar flow in d:
Re
32

 d ² 
1




D
²



2
Turbulent flow in d:
f ind  
 d ² 
1




4 
 D ² 
Sudden contraction from D to d
(except choked gas flow)
Laminar flow in d
Turbulent flow in d:
Re
160
2

 d ² 
1
.
25




D
²



f ind 
 d ² 
1
.
25



10 
D
²



page 12
Water contaminated with small amounts of hazardous waste
is gravity drained out of a large storage tank through a
straight, commercial steel pipe 100 mm in ID. The pipe is
100 m long with a gate valve near the tank. The entire pipe
assembly is mostly horizontal. If the liquid level in the tank
is 5.8 above the pipe outlet, and the pipe is accidentally
severed 33 m from the tank, compute the flow rate of
material escaping from the pipe.
page 13
The draining operation is shown in Figure 8. Assuming negligible KE
changers, no pressure changes, and no shaft work, the mechanical
energy balance, Equation 22, applied between points 1 and 2, reduces
to
g
z  F  0
gc
For water
  1.0  10 3 kg ms
  1000 kg m 3
The frictional loss term, F, is given by Equation 23
F 
2 fLu ²
gcd
page 14
Figure 8
page 15
The equivalent pipe length is computed from Equation 30. The
equivalent lengths for the pipe connected to the vessel and the gate
valve are required. These are available in Table 2, assuming turbulent
flow in the pipe. For the connection to the vessel, from Table 2,
L
 16  L  (16)(100mm )  1600mm  1.6m
d
For a fully open gate valve, from Table 2
L
 7  L  (7)(100mm )  700mm  0.7 m
d
Then
Lequiv  Lhorizontal   Lequiv
total
and
pipe
 33m  1.6m  0.7 m  35.3m
F 
2 f 35.3m u ²
 706 f u ²
(0.1m)
page 16
Substituting into the mechanical energy balance above and solving
for ū,
gz  .706 f u ²  0
706 f u ²   gz  (9.8 m s ²)( 5.8m)  56.8 m² s²
u ²  0.08051 f
u  0.284
f
The Reynolds number is given by
Re 
d u


0.1m u 1000 kg
 2.84  10 4
1.0  10
3
m3
kg m.s
f  Re
  1.0  10
5
u
f  2.84  10 4
page 17
For commercial steel pipe, from Table 1, e = 0.046 and
e
0.046mm

 0.00046
d
100mm
From Figure 7
1
f
 15.1 
f  0.0662  f  0.00438
The velocity of the fluid is
u  0.284 0.0662  4.29 m s
The cross-sectional area of the pipe is
A
d ²
4

(3.14)( 0.1m)²
 0.00785m ²
4
page 18
The mass flow rate from the release is found by
Q   u A  (1000 kg m 3 )( 4.29 m s)0.00785m ²   33.7 kg s
The kinetic energy term must be checked. Its value is
KE 
4.29 m s ²  9.20
u²

2g c
2
This is compared to the frictional loss term, F
F  706 f u ²  7060.004384.29 m s ²  57.0
Thus, the assumption of negligible KE is not valid for this case. The
liquid has zero initial velocity and is accelerated to a high velocity,
resulting in a significant KE term. For problems involving liquids
flowing in pipes with an initial and final velocity, the KE changes are
usually negligible.
page 19
Start over from the mechanical energy balance, but this time include
the KE term
g
u²
z 
 706 f u ²  0
gc
2gc
Solving for ū
u

gz
1  706 f
2
9.8 m s 

56.84


0.5  706 f  0.5  706 f
The solution to this equation requires a trial and error procedure
since f is a function of ū. the procedure is.
a.
Guess a value for the friction factor, f,
b. Determine average velocity, ū, from above equation,
c.
Determine Reynolds number, Re,
d. Compute f from Colebrook equation, Equation 25, and
e.
Iterate until value of ƒ converges.
page 20
This procedure produces the following results.
ƒ
ū
Re
Computed ƒ
0.004
4.13
4.13 × 105
0.00439
0.00439
3.97
3.97 × 105
0.00438
The last result is close enough. The mass flow rate is


Qm   u A  1000 kg m 3 3.97 m s 0.00785m² 
 31.1 kg s
This represents a significant flow rate. Assuming a 15- minute
emergency response period to stop the release, a total of 28,000
kg of hazardous waste will be spilled. In addition to the material
releases by the flow, the liquid contained within the pipe between
the valve and the rupture will also spill. An alternate system must
be designed to limit the release. This could include a reduction in
the emergency response period, replacement of the piping system
to include additional control valves to stop the flow.
page 21
For flowing liquids the kinetic energy changes are frequently
negligible and the physical properties (particularly the density are
constant. For flowing gases and vapor these assumptions are only
valid for small pressure changes (P1/P2 < 2)and low velocities ( 0.3 ×
speed of sound in gas). Energy contained within the gas or vapor as a
result of its pressure is converted into the kinetic energy as the gas or
vapor escapes and expands through the hole. The density, pressure
and temperature change as the gas or vapor exits through the leak.
Gas and vapor discharges are classified into throttling and free
expansion releases. For throttling releases, the gas issues through a
small crack with large frictional losses; very little of energy inherent
with the gas pressure is converted to kinetic energy. For free
expansion re;eases, most of the pressure energy is converted to
kinetic energy; the assumption of isentropic behavior is usually valid.
Source models for throttling releases require detailed information
on the physical structure of the leak; they will not be considered here.
Free expansion release source models require only the diameter of the
page 22
leak.
A free expansion leak is shown in Figure 9. The mechanical
energy balance, Equation 1, describes the flow of compressible gases
and vapors. Assuming negligible potential energy changes and no
shaft work results in a reduced form of the mechanical energy
balance describing compressible flow through holes.

 u²
 
 2g

c

dP

F 0


(31)
A discharge coefficient, C1, is defined in a similar fashion to the
coefficient defined in the first section.
(32)


dP


 F  C12 



 


Equation 32 is combined with Equation 31 and integrated
between any two convenient points. An initial point (denoted by
subscript o) is selected where the velocity is zero and the pressure is
Po. The integration is carried to any arbitrary final point (denoted
without a subscript). The result is
dP
2
1
C
P
dP
Po



u²
0
2g c
(33)
page 23
Figure 9 A free expansion gas leak. The gas expands isentropically
through the hole. The gas properties (P, T) and velocity change during
the expansion.
page 24
For any ideal gas undergoing an isentropic expansion
Pv g 
P

g
 constant
(34)
where g is the ratio of the heat capacities, g  C P CV .
Substitution of Equation 34 into Equation 33, defining a
new discharge coefficient, Co identical to Equation 5 and
integrating results in an equation representing the velocity of
the fluid at any point during the isentropic expansion
g 1 g


 P 
Po
g
2

1  

u ²  2 g c Co


g  1  o 

 Po 

g 1 g
2 g c C o2 R g To g 
 P 

1  



M
g  1 
 Po 




(35)
page 25
The second form incorporates the ideal gas law for the initial
density, o. Rg is the ideal gas constant and To is the temperature of
the source. Using the continuity equation
(36)
Qm   u A
and the ideal gas law for isentropic expansion in the form
 P 

   o 

 Po 
1
g
(37)
results in an expression for the mass flow rate.
Qm  C o APo
2
g 1 g

 P 
2 g c M g  P  g







R g To g  1  Po 
Po 







(38)
page 26
Equation 38 describes the mass flowrate at any point during
the isentropic expansion.
For many safety studies, the maximum flowrate of vapor
through the hole is required. This is determined by differentiating
Equation 38 with respect to P/Po and setting the derivative equal to
zero. The result is solved for the pressure ratio resulting in the
maximum flow.
Pchoked
 2 




Po
 g  1
g g 1
(39)
The choked pressure is the maximum downstream pressure
resulting in maximum flow through the hole or pipe. For
downstream pressures less than Pchoked the following statements are
valid : (1) the velocity of the fluid at the throat of the leak is the
velocity of sound at the prevailing conditions and (2) the velocity
and mass flowrate cannot be increased further by reducing the
downstream conditions. This type of flow is called choked, critical,
or sonic flow and is illustrated in Figure 10.
page 27
The maximum flow is determined by substituting Equation 39
into Equation 38
Qm choked
 C o APo
gg c M 
R g To
2 

 g  1



g 1 g 1
(40)
where M is the molecular weight of the escaping vapor or gas,
To is the temperature of the source, and Rg is the ideal gas
constant.
For sharp-edged orifices with Reynolds numbers greater than
30,000 (and not choked), a constant discharge coefficient, Co, of
0.61 is indicated. However, for choked flows, the discharge
coefficient increases as the downstream pressure decreases. For
these flows and for situations where Co is uncertain, a
conservative value of 1.0 is recommended.
page 28
A 0.1 inch hole forms in a tank containing nitrogen at
200 psig and 80°F. determine the mass flowrate
through this leak.
page 29
For the diatomic gas nitrogen, g = 1.4. Thus,
Pchoked  0.528200  14.7 psia  113.4psia
An external pressure less than 113.4 psia will result in choked flow
through the leak. Since the external pressure is atmospheric in
this case, choked flow is expected and Equation 40 applies. The
area of the hole is
A
d
4


3.140.1in  1ft 2 144in 2
2
2
4
 5.45 10
5
ft 2
page 30
The discharge coefficient, Co, is assumed to be 1.0. Also,
Po  200  14.7  214.7 psia
T0  80  460  540 o R
g 1 g 1
 2 
 2 




 g 1
 2.4 


Then, using Equation 40
Qm choked
2.4 0.4
6.00
gg c M 
 C o APo
R g To

 0.833
2 

 g 1



 0.335
g 1 g 1

 1.0  5.45  10 5 ft² 214.7 lb f in² 144 in² ft² 

1.4 32.17 ft.lb m
1545 ft.lb
 1.685lb f
Qm choked
f
lb f .s² 28 lb m lb.mole
lb.mole o R 540 o R



0.335
5.064  10  4 lb 2m lb f2 .s 2
 3.79  10  2 lb m s
page 31
Vapor flow through pipes is modelled using two special cases :
adiabatic or isothermal behavior. The adiabatic case corresponds
to rapid vapor flow through an insulated pipe. The isothermal
case corresponds to flow through an uninsulated pipe maintained
at a constant temperature; an underwater pipeline is an excellent
example. Real vapor flows behave somewhere between the
adiabatic and isothermal cases. Unfortunately, the “real” case is
very difficult to model and no generalized and useful equations
are available.
page 32
For both the isothermal and adiabatic cases it is convenient to
define a Mach number as the ratio of the gas velocity to the
velocity of sound in the gas at the prevailing conditions.
u
Ma 
a
(41)
where a is the velocity of sound. The velocity of sound is
determined using the thermodynamic relationship.
a
 P 
gc 
  


S
(42)
which, for an ideal gas, is equivalent to
a
gg c R g T M
(43)
which demonstrates that, for ideal gases the sonic velocity is a
function of temperature only. For air at 20°C the velocity of sound
is 344 m/s (1129 ft/s)
page 33
Adiabatic Flows
An adiabatic pipe containing a flowing vapor is shown in
Figure 11. For this particular case the outlet velocity is less than
the sonic velocity. The flow is driven by a pressure gradient across
the pipe. This expansion leads to an increase in velocity and an
increase in the kinetic energy of the gas. The kinetic energy is
extracted from the thermal energy of the gas; a decrease in
temperature occurs. However, frictional forces are present
between the gas and the pipe wall. These frictional forces increase
the temperature of the gas. Depending on the magnitude of the
kinetic and frictional energy terms either an increase or decrease
in the gas temperature is possible.
page 34
Figure 11 Adiabatic, non-choked flow of gas through a pipe. The gas
temperature might increase or decrease, depending on the magnitude
of the frictional losses.
page 35
The mechanical energy balance, Equation 1, also applies to
adiabatic flows. For this case it is more conveniently written in
the form
dP


W s
ud u
g

dz  dF  
g c
gc
m
(44)
the following assumptions are valid for this case :
g
dz  0
gc
is valid for gases, and
dF 
2 f u ² dL
gcd
From Equation 23, assuming constant f, and
Ws  0
page 36
Since no mechanical linkages are present. An important part of
the frictional loss term is the assumption of a constant Fanning
friction factor, f, across the length of the pipe. This assumption is only
valid at high Reynolds number.
A total energy balance is useful for describing the temperature
changes within the flowing gas. For this open, steady flow process it is
given by
Ws
ud u
g
dh 

dz  q 
g c
gc
m
(45)
Where h is the enthalpy of the gas and q is the heat. The
following assumptions are invoked.
dh  C p dT for an ideal gas,
g
dz  0
gc
q  0
W S  0
is valid for gases,
since the pipe is adiabatic,
since no mechanical linkages are present.
page 37
The above assumptions are applied to Equations 45 and 44.
The equations are combined, integrated (between the initial point
denoted subscript o and any arbitrary final point), and
manipulated to yield, after considerable effort,
T2
Y1
γ 1

where Y1  1 
Ma i2
T1
Y2
2
(46)
P2
Ma1

P1
Ma 2
Y1
Y2
(47)
2
Ma1

1
Ma 2
Y2
Y1
(48)
G   u  Ma1 P1
gg c M
R g T1
 Ma 2 P2
gg c M
R g T2
(49)
page 38
Where G is the mass flux with units of mass/(area time), and
g 1
2
 Ma 22Y1
ln 
 Ma 2Y
1
2

kinetic
energy
  1
1 
 4 fL 






g

0
2 
  Ma 2
Ma 2 
 d 
1
 
compressibility
(50)
pipe
friction
Equation 50 relates the Mach numbers to the frictional losses
in the pipe. The various energy contributions are identified. The
compressibility term accounts for the change in velocity due to
the expansion of the gas.
Equations 49 and 50 are converted to a more convenient and
useful form by replacing the Mach numbers with temperatures
and pressures using Equations 46 through 48.
g  1 P1T2 g  1  P12T22  P22T12

ln

g
P2T1
2g 
T2  T1

G
 1
1



2
 P 2T
P
2 T1
 1 2
2g c M g
T2  T1
R g g  1 T1 P1 2  T2 P2 2
 4 fL(51)

 d 0

(52)
page 39
For most problems the pipe length (L), inside diameter (d),
upstream temperature (T1) and pressure (P1), and downstream
pressure (P2) are known. To compute the mass flux, G, the
procedure is as follows.
1. Determine pipe roughness, e from Table 1. Compute
e/d.
2. Determine the Fanning friction factor, f, from Equation
27. This assumes fully developed turbulent flow at high
Reynolds numbers. This assumption can be checked later,
but is normally valid.
3. Determine T2 from Equation 51.
4. Compute the total mass flux, G, from Equation 52.
For long pipes, or for large pressure differences across the
pipe, he velocity of the gas can approach the sonic velocity. This
case is shown in Figure 12. At the sonic velocity the flow will be
choked. The gas velocity will remain at the sonic velocity,
temperature, and pressure for the remainder of the pipe. For
choked flow, Equations 46 through 50 are simplified by setting
Ma2 = 1.0. The results are :
page 40
(53)
Tchoked
2Y1

T1
g 1
Pchoked
 Ma1
P1
2Y1
g 1
(54)
 choked
g 1
 Ma1
1
2Y1
(55)
Gchoked   u  Ma1 P1
g 1
2
gg c M
R g T1
 Pchoked
gg c M
(56)
R g Tchoked

  1

2Y1
 4 fL 


ln 


1

g

0
2 
2


 d 
 g  1Ma1   Ma1

(57)
Choked flow occurs if the down stream pressure is less than
Pchoked. This is checked using Equation 54.
page 41
Figure 12 Adiabatic, choked flow of gas through a pipe. The
maximum velocity reached is the sonic velocity of the gas.
page 42
For most problems involving choked, adiabatic flows, the
pipe length (L), inside diameter (d), and upstream pressure
(P1) and temperature (T1) are known. To compute the mass
flux, G, the procedure is as follows.
1. Determine the Fanning friction factor, f, using
Equation 27. This assumes fully developed turbulent
flow at high Reynolds number. This assumption can
be checked later, but is normally valid.
2. Determine Ma1, from Equation 57.
3. Determine the mass flux, Gchoked, from Equation 56.
4. Determine Pchoked from Equation 54 to confirm
operation at choked conditions.
page 43