page 0 Prepared by Associate Prof. Dr. Mohamad Wijayanuddin Ali Chemical Engineering Department...

Prepared by

Associate Prof. Dr. Mohamad Wijayanuddin AliChemical Engineering DepartmentUniversiti Teknologi Malaysia

Isothermal flow of gas in a pipe with friction is shown in Figure 13. For this case the gas velocity is assumed to be well below the sonic velocity of the gas. A pressure gradient across the pipe provides the driving force for the gas transport. As the gas expands through the pressure gradient the velocity must increase to maintain the same mass flowrate. The pressure at the end of the pipe is equal to the pressure of the surroundings. The temperature is constant across the entire pipe length.

Isothermal flow is represented by the mechanical energy balance in the form shown in Equation 44. The following assumptions are valid for this case.

is valid for gases, and

form Equation 23, assuming constant f, and

since no mechanical linkages are present. A total energy balance is not required since the temperature is constant.

0≈dzgg

c

dg

dLufdF

c

?2=

0=SWδ

Figure 13 Isothermal, non-choked flow of gas through a pipe.

Applying the above assumptions to Equation 44, and, after considerable manipulation

(58)

(59)

(60)

(61)TR

MgPMauG

Ma

Ma

Ma

Ma

P

P

TT

g

cγρ

ρρ

11

2

1

1

2

2

1

1

2

12

==

=

=

=

where G is the mass flux with units of mass/(area time), and,

(62)

The various energy terms in Equation 62 have been identified.

A more convenient form of Equation 62 is in terms of pressure instead of Mach numbers. This form is achieved by using Equations 58 through 60. The result is

(63)( ) 04

ln2 22

212

2

1 =+−−d

fLPP

TRG

Mg

P

P

g

c

04111

ln222

211

2 =+⎟⎟⎠

⎞⎜⎜⎝

⎛−−

dfL

MaMaMaMa

γkinetic energy compressibility

pipe friction

A typical problem is to determine the mass flux, G, given the pipe length (L), inside diameter (d), and upstream and downstream pressures (P1 and P2). 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 usually valid.

2. Compute the mass flux, G, from Equation 63.

Levenspiel has shown that the maximum velocity possible during the isothermal flow of gas in a pipe is not the sonic velocity as in the adiabatic case. In terms of the Mach number, the maximum velocity is

(64)γ1

choked =Ma

This result is shown by starting with the mechanical energy

balance and rearranging it into the following form.

(65)

the quantity -(dP/dL)--> when Ma --> 1/γ. Thus, for

choked flow in an isothermal pipe, as shown in Figure 14, the

following equations apply.

( ) ⎥⎦

⎤⎢⎣

⎡−

=⎥⎦

⎤⎢⎣

⎡

−=−

?1

1?2

/?1

1?2

Madg

fG

Pgudg

fG

dL

dP

ccc γρρρ

(66)

(67)

(68)

(69)

(70)

where Gchoked is the mass flux with unit of mass/(area/time), and

(71)

TR

MgP

TR

MgPMauuG

Mau

u

Ma

MaP

P

TT

g

c

g

cchoked1111choked

11

choked

11

choked

11

choked

1choked

1

====

=

=

=

γρρ

γ

γρ

ρ

γ

04

111

ln21

21

=+⎟⎟⎠

⎞⎜⎜⎝

⎛−−⎟⎟

⎠

⎞⎜⎜⎝

⎛

dfL

MaMa γγ

Figure 14 Isothermal, choked flow of gas through a pipe. The maximum velocity reached is a/γ.

For most typical problems the pipe length (L), inside diameter

(d), upstream pressure (P1), and temperature (T) are known. The

mass flux, G, is determined using the following procedure.

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 usually valid.

2. Determine Ma1 from Equation 71.

3. Determine the mass flux, G, from Equation 70.

The vapor space above liquid ethylene oxide (EO) in storage

tanks must be purged of oxygen and then padded with 81 psig

nitrogen to prevent explosion. The nitrogen in a particular facility

is supplied from a 200 psig source. It is regulated to 81 psig and

supplied to the storage vessel through 33 feet of commercial

steel pipe with and ID of 1.049 inches.

In the event of a failure of the nitrogen regulator, the vessel will

be exposed to the full 200 psig pressure from the nitrogen

source. This will exceed the pressure rating of the storage vessel.

To prevent rupture of the storage vessel it must be equipped

with a relief device to vent this nitrogen. Determine the required

minimum

mass flow rate of nitrogen through the relief device to prevent

the pressure from rising within the tank in the event of a

regulator failure.

Determine the mass flowrate assuming :

a. an orifice with a throat diameter equal to the pipe

diameter,

b. an adiabatic pipe, and

c. an isothermal pipe.

Decide which result most closely corresponds to the real

situation. Which mass flowrate should be used?

a. The maximum flowrate through the orifice occurs under

choked conditions.

The area of the pipe is

The absolute pressure of the nitrogen source is

The choked pressure from Equation 39 is, for diatomic gas

( )( ) ( )23

222

ft1000.6

4

in144ft1in049.114.3

4

?

−×=

==d

Aπ

2ft

4 ftlb103.09psia7.2147.14200 ×==+=oP

( )( )2

f4

psiachoked

ftlb101.63

psia4.1137.214528.0

×=

==P

Choked flow can be expected since the system will be venting

to atmospheric conditions. Equation 40 provides the maximum

mass flowrate. For nitrogen,γ = 1.4 and

The molecular weight of nitrogen is 28 lbm/lb-mole. Without

any additional information, assume a unit discharge coefficient,

Co = 1.0. Thus,

( ) ( )

335.04.2

2

1

24.04.211

=⎟⎠

⎞⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛+

−+ γγ

γ

( )( )( )( )( )( )

( )( ) ( )

( )slb 16.4

s lblb 1006.5lb 185

335.0R540Rmole-lblbft 1545

mole-lblb 28slblbft 17.324.1

ftlb1009.3ft1000.60.1

m

22f

2m

4f

f

m2

fm

2f

423

=

×=

°°×

××=

−

−

m

m

Q

Q

b. Assume adiabatic, choked flow conditions. For commercial

steel pipe, from Table 1, = 0.046 mm. The diameter of the pipe

in mm is (1.049 in)(25.4 mm/in) = 26.6 mm. Thus,

From Equation 27

00173.0mm 26.6

mm 046.0==

d

( )

00564.0

0751.0

32.1300173.07.3log4

7.3log41

=

=

==

⎟⎠

⎞⎜⎝

⎛=

f

f

d

f ε

For nitrogen, γ = 1.4.

The upstream Mach number is determined from Equation 57.

with Y1 given by Equation 46. Substituting the number provided,

( ) 04

11

1

2ln

2

121

21

1 =⎟⎠

⎞⎜⎝

⎛+⎟⎟⎠

⎞⎜⎜⎝

⎛−−⎥

⎦

⎤⎢⎣

⎡

+

+

d

fL

MaMa

Yγ

γ

γ

( )( )

( )( )( )( )( )

092.1111

4.2

4.02ln2.1

0in 12ft 1in 1.049

ft 3300564.044.1

11

14.1

14.12ln

2

14.1

22

2

22

2

=+⎟⎠

⎞⎜⎝

⎛−−⎟⎟

⎠

⎞⎜⎜⎝

⎛ +

=⎥⎦

⎤⎢⎣

⎡+

⎟⎠

⎞⎜⎝

⎛−−⎥

⎦

⎤⎢⎣

⎡

+

−++

MaMa

Ma

MaMa

Ma

This equation is solved by trial and error for the value of Ma. The

results are tabulated below.

Guessed Ma Value of LHS of equation

0.20 -8.43

0.25 0.043

This last value looks very close. Then from Equation 46

( ) 012.125.02

14.11

2

11 22

1 =−

+=−

+= MaYγ

and from Equations 53 and 54

( )

( )( )

( )

( )( ) 2f

3choked

1

1

choked

choked

1

1

choked

ftlb 107.11psia 4.49psia 7.214230.0

230.0843.025.01

2

R455R46080843.0

843.014.1

012.12

1

2

×===

==+

=

°=°+=

=+

=+

=

P

YMa

P

P

T

Y

T

T

γ

γ

The pipe outlet pressure must be less than 49.4 psia to insure

choked flow. The mass flux is computed using Equation 56

( )

( )( )( )( )( )

( )( )slb81.1

ft 1000.6s ftlb301

s ftlb 301

s lblb1079.1ftlb 1011.7

R455R mole-lblbft 1545

mole-lblb 28s lblbft 17.324.1

ftlb 1011.7

m

232m

2m

22f

2m

32f

3

f

m2

fm

2f

3

chokedchokedchoked

=

×==

=

××=

°°×

×==

−

−

GAQ

TR

MgPG

m

g

cγ

c. For the isothermal case, the upstream Mach number is given

by Equation 71. Substituting the numbers provided

The solution is found by trial and error

052.814.1

1

4.1

1ln

22=+⎟

⎠

⎞⎜⎝

⎛ −−⎥⎦

⎤⎢⎣

⎡MaMa

The choked pressure is, from Equation 67

The mass flowrate is computed using Equation 70.

( )( )2

f3

2f11choked

ftlb108.93psia 0.62

4.1244.0inlb 7.214

×==

== γMaPP

( )( )( )( )

( )( )slb76.1

ft1000.6s ftlb293

s ftlb293

slblb1008.1ftlb1093.8

R540Rmole-lblbft 1545

mole-lblb28s lblbft 17.32

ftlb1093.8

m

232mchoked

2m

22f

2m

32f

3

f

m2

fm

2f

3chokedchoked

=

×==

=

××=

°°×

×==

−

−

AGQ

TR

MgPG

m

g

c

The results are summarized in the following table

Case Pchoked (psia) Qm (lbm/ s)

Orifice 113.4 4.16

Adiabatic pipe 49.4 1.81

Isothermal pipe 62.0 1.76

A standard procedure for these types of problems is to represent

the discharge through the pipe as an orifice. The results show

that this approach results in a large result for this case. The

orifice method will always produce a larger value than the

adiabatic pipe method, insuring a conservative safety design. The

orifice calculation, however, is easier to apply, requiring only the

pipe

diameter and upstream supply pressure and temperature. The

configurational details of the piping are not required, as in the

adiabatic or isothermal pipe methods.

Also note that the choked pressures computed differ for each

case, with a substantial difference between the orifice and

adiabatic/isothermal cases. A choking design based on an orifice

calculation might not be choked in reality due to high

downstream pressures.

Finally,note that the adiabatic and isothermal pipe methods

produce results that are reasonably close. For most real

situations the heat transfer characteristics cannot be easily

determined. Thus, the adiabatic pipe method is the method of

choice; it will always produce the larger number for a

conservative safety design.

Liquids stored under pressure above their normal boiling point

temperature present substantial problems due to flashing. If the

tank, pipe, or other containment device develops a leak, the liquid

will partially flash into vapor, sometimes explosively.

Flashing occurs so rapidly that the process is assumed to be

adiabatic. The excess energy contained in the superheated liquid

vaporizes the liquid and lower the temperature to the new boiling

point. If m is the mass of original liquid, Cp the heat capacity of the

liquid (energy/mass deg), To the temperature of the liquid prior to

depressurization, and Tb the depressurized boiling point of the liquid,

then the excess energy contained in the superheated liquid is given

by

(72)

( )bop TTmCQ −=

This energy vaporizes the liquid. If ΔHv is the heat of vaporization

of the liquid, the mass of liquid vaporized, mv is given by

(73)

The fraction of the liquid vaporized is

(74)

Equation 74 assumes constant physical properties over the

temperature range To to Tb. A more general expression without this

assumption is derived as follows.

( )v

bop

vv

TTmCQm

ΔΗ

−=

ΔΗ=

( )v

bopvv

TTC

m

mf

ΔΗ

−=

The change in liquid mass, m, due to a change in temperature, T

is given by

(75)

Equation 75 is integrated between the initial temperature To (with

liquid mass m) and the final boiling point temperature Tb (with liquid

mass m – mv),

(76)

dTmC

dmv

p

ΔΗ=

∫∫ ΔΗ=

− b

o

v T

Tv

pmm

mdT

C

m

dm

(77)

where and are the mean heat capacity and mean latent

heat of vaporization, respectively, over the temperature range To to

Tb. Solving for the fraction of the liquid vaporized, fv = mv/m,

(78)

( )v

bopvTTC

m

mm

ΔΗ

−−=⎟

⎠

⎞⎜⎝

⎛ −ln

pC vΔΗ

( )[ ]vbopv TTCf ΔΗ−−−= exp1

One lbm of saturated liquid water is contained in a vessel at

350°F. the vessel ruptures and the pressure is reduced to 1 atm.

Compute the fraction of material vaporized using :

a. the steam tables,

b. Equation 74, and

c. Equation 78.

a. The initial state is saturated steam at To = 350°F. from the

steam tables,

The final temperature is the boiling point at 1 atm, or 212°F. At

this temperature, and saturated conditions,

mlbBTU 321.6H

psia 6.134

=

=P

mliquid

mvapor

lbBTU07.180

lbBTU4.1150

=

=

H

H

Since the process occurs adiabatically, Hfinal = Hinitial and the

fraction of vapor (or quality) is computed from,

14.59% of the mass of the original liquid is vaporized.

( )

( )

1459.0

07.1804.115007.1806.321

liquidvaporliquidfinal

=

−+=

−+=

v

v

v

f

f

HHfHH

b. For liquid water at 212°F,

From Equation 74

m

m

lbBTU3.970

FlbBTU01.1

=ΔΗ

°=

v

pC

( ) ( )( )

1436.0

lbBTU3.970

F212350FlbBTU01.1

m

m

=

°−°=

ΔΗ

−=

v

v

bopv

f

TTCf

c. The mean properties for liquid water between To and Tb

are

Substituting into Equation 78

Both expressions work about as well when compared to the

actual value from the steam table.

m

m

lbBTU7.920

FlbBTU04.1

=ΔΗ

°=

v

pC

For flashing liquids composed of any miscible substances,

the flash calculation is complicated considerably, since the

more volatile components will lash preferentially.

Procedures are available to solve this problem.

Flashing liquids escaping through holes and pipes require

very special consideration since two-phase flow conditions

may be present. Several special cases need consideration.

If the fluid path length of the release is very short (through

a hole in a thin-walled container), non-equilibrium

conditions exist, and the liquid does not have time to flash

within the hole; the fluid flashes external to the hole. The

equations describing incompressible fluid flow through

holes apply.

If the fluid path length through the release is greater than

10 cm (through a pipe or thick-walled container),

equilibrium flashing conditions are achieved and the flow is

choked. A good approximation is to assume a choked

pressure equal to liquids stored at a pressure higher than

the saturation vapor pressure. With this assumption the

mass flowrate is given by

(79)

( )sat2 PPgACQ cfom −= ρ

where

A is the area of the release,

Co is the discharge coefficient (unitless),

ρt is the density of the liquid (mass/volume),

P is the pressure within the tank, and

Psat is the saturation vapor pressure of the flashing

liquid at ambient temperature

Liquid ammonia is stored in a tank at 24°C and a pressure

of 1.4 106 Pa. A leak of diameter 0.0945 m forms in the

tank, allowing the flashing ammonia to escape. The

saturation vapor pressure of liquid ammonia at this

temperature is 0.968 106 Pa and its density is 603 kg/m ³.

Determine the mass flowrate through the leak. Equilibrium

flashing conditions can be assumed.

Equation 79 applies for the case of equilibrium flashing

conditions. Assume a discharge coefficient of 0.61.

( )

( )( )( )

( )

skg6.97

m

N10968.0104.1

Ns

kgm1

m

kg6032

4

m 0945.014.361.0

2

266

2

3

2

sat

=

⎟⎠

⎞⎜⎝

⎛×−×

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡⎟⎠

⎞⎜⎝

⎛

⎟⎠

⎞⎜⎝

⎛×

=

−=

m

cfom

Q

PPgACQ ρ

for liquids stored at their saturation vapor pressure, P = Psat,

Equation 79 is no longer valid. For this case the choked,

two-phase mass flowrate is given by

(80)

where v is the specific volume with units if (volume/mass).

The two-phase specific volume is given by

(81)

( )dPdvg

AQ cm −=

fvfg vfvv +=

where

vfg is the difference in specific volume between

vapor and liquid,

vf is the liquid specific volume, and

fv is the mass fraction of vapor.

Differentiating Equation 81 with respect to pressure,

(82)

But, from Equation 74,

(83)

dP

dfv

dP

dv vfg=

dTC

dfv

pv ΔΗ

−=

and, from the Clausius-Clapyron equation, at saturation,

(84)

Substituting Equations 84 and 83 into Equation 82 yields,

(85)

The mass flowrate is determined by combining Equation 85, with

Equation 80.

(86)

fg

v

TvdT

dP ΔΗ=

TCv

dP

dvp

v

fg

2

2

ΔΗ−=

TC

g

v

AQ

p

c

fg

vm

ΔΗ=

Small droplets of liquid also form in a jet of flashing vapor. These

aerosol droplets are readily entrained by the wind and

transported away from the release site. The assumption that the

quantity of droplets formed is equal to the amount of material

flashed is frequently made.

Propylene is stored at 25°C in a tank at its saturation pressure. A

1 cm diameter hole develops in the tank. Estimate the mass

flowrate through the hole. At these conditions, for propylene.

ΔHv = 3.34 x 105 J/kg

vfg = 0.042 m3/kg

Psat = 1.15 106 Pa

Cp = 2.18 103 J/kg K

Equation 86 applies to this case. The area of the leak is

Using Equation 86

( )( ) 25

222

m1085.74

m10114.3

4−

−

×=×

==d

Aπ

( )( )( )

( )( )( )( )

skg774.0

JNm1K298kgKJ1018.2

Nskgm0.1

kgm042.0

m1085.7JNm1kgJ1034.3

3

2

3

255

=

××

××=

ΔΗ=

−

m

p

c

fg

vm

Q

TC

g

v

AQ

The case for evaporation of volatile from a pool of liquid has

already been considered in Chapter 3. The total mass flowrate from

the evaporating pool is given by

where

Qm is the mass vaporization rate (mass/time),

M is the molecular weight of the pure material,

K is the mass transfer coefficient (length/time),

A is the area of exposure,

Psat is the saturation vapor pressure of the liquid,

Rg is the ideal gas constant, and

TL is the temperature of the liquid.

Lgm TR

MKAPQ

sat

=

For liquids boiling from a pool, the boiling rate is limited

by the heat transfer from the surroundings to the liquid in

the pool. Heat is transferred :

1. from the ground by conduction,

2. from the air by conduction and convection, and

3. by radiation from the sun and/or adjacent sources

such as fire.

page 0 Prepared by Associate Prof. Dr. Mohamad Wijayanuddin Ali Chemical Engineering Department...

Documents

Transcript of page 0 Prepared by Associate Prof. Dr. Mohamad Wijayanuddin Ali Chemical Engineering Department...