Full Revised Chapter 7

CHAPTER 7 “CONCEPTS TO PREVENT FIRE AND EXPLOSIONS” Carina Nicole P. Labrague

March 19, 2016

Marvin Arpon

7-1. PROBLEM: Develop a list of steps needed to convert a comm on kitchen in to an XP area. SOLUTION: Assuming ethyl alcohol is the most flammable liquid handled in the kitchen: (Methane and propane are also possible flammables) a.) Change all electrical fixtures and appliances to Class I, Group D and Division 2 b.) Add ventilation, 1 ft3/min-ft2 of floor area c.) Provide ventilation hoses (elephant trunks) for spot ventilation d.) Heat the room with hot water or steam e.) Use high pressure steam or hot oil for the stove f.) Add a sprinkler system which is activated via flame or flame g.) Doors would need to be changed to “ fire doors” h.) Ventilation would need to be adjusted to give higher pressures in adjoining rooms compared to the XP room

7-2 PROBLEM: What bonding and grounding procedures must be followed to transfer a drum of flammable solvent in to a storage tank ? SOLUTION: See Figure 7-18

Main Idea : Bond the drum and storage tank, then ensure that this tandem is grounded.

7-3. PROBLEM: Ethylene oxide is a flammable liquid having a normal boiling temperature below room temperature. Describe a system and a procedure for transferring ethylene oxide from a tank car through a pumping system to a storage tank . Include both inerting and purging as well as bonding and grounding procedures. SOLUTION: N2

N2

Pad vent

Add N2 to : Displace EO Blow out lines Dip Pipe

NOTE: Ethylene oxide vapor is explosive. A positive N2 pad (

35 psig) is required

7-4. PROBLEM: Flammable liquid is being pumped out of a drum into a bucket using a hand pump. Describe an appropriate g rounding and bonding procedure. SOLUTION : Hand Crank Bond Bond

Ground

7-5. PROBLEM:

Using the sweep-through purging method, inert a 100-gal vessel containing 100% air until the oxygen concentration is 1%. What volume of nitrogen is required? Assume nitrogen with no oxygen and a temperature of 77°F.

SOLUTION: C0 = 0 % C1 = 79 % 100 gal

1100 %

C2 = 21 %

air

Qvt is the total volume required

Eq. 7-15 C2−¿ C C1−¿ C ¿ ¿ ¿ Qv t=V ln¿ o

o

fr3 21−0 ¿ ( 100 gal ) ln =¿ (13.4 ft3) (ln21) = 40.7 ft3 Qv t 7.48 gal 1−0

(

7-6. PROBLEM:

) (

)

A 150-f t3 tank containing air is to be inerted to 1% oxgen concentration. Pure nitrogen is available for the job. Because the tank’s maximum allowable working pressure is 150 psia, it is possible to use either the sweep-through or a pressurization technique. For the pressurization technique, multiple pressurization cycles might be necessary , with the tank being returned to atmospheric pressure at the en of each cycle. The temperature is 80°F. a. Determine the volume of nitrogen required for each technique. b . For the pressurization technique, determine the number of cycles required if the pressure purge includes in creasing the pressure to 140 psia with nitrogen and then venting to 0 psig .

SOLUTION: 150 ft3 of air

T = 80oF + 460o = 540o R Sweep through, eq. 7-15 applies; C1 ¿ V ln Qv t C2

( )

;

= 150ln

Pressure purge: 0 psig = 14.7 psia

Use equations 7-1 and 7-2:

Nh =

PhV Rg T

(140 psia ) ( 150 ft 3 ) 3 = ( 10.73 psia−fr )(540 o R) lbmole – o R

= 3.62 lbmole

( 211 )=4

57 ft3

NL =

PLV RgT

==

(14.7 psia ) ( 150 ft 3 ) 3 10.73 psia−fr ( )(540 o R) lbmole – o R

= .381 lbmole

Use equations 7-6 to compute the number of cycles required: NL j ¿ Nh ¿

Yj = Yo

1 21 .381 ln 3.62

ln

ln

J=

=

−3. 044 −2.251

Yj =¿ Yo

( )

NL j ln( N h ¿

Yj Yo N ln L Nh ln

j=

= 1.35 purges

Use 2 purges . Total moles N2 = (2)(3.62 −¿ .381) = 6.48 lbmoles Use equation 7-2 to compute volume N2 required: NL =

V=

PLV RgT

V=

Rg T NL PL

10.73 psia−fr3 ( )(540 o R)(6.48 lbmoles) lbmole – o R ( 14.7 psia )

= 2550 ft3 N2

7-7. PROBLEM: Use a vacuum purging technique to purge oxygen from a 150-f t3 tank containing air. Reduce the oxygen concentration to 1% using pure nitrogen as the inert gas. The temperature is 80°F.

Assume that the vacuum purge goes from atmospheric pressure to 20 mm Hg absolut e. Determine the number of purge cycles required and the total moles of nitrogen used. SOLUTION: Vacuum purge is about 150 ft3 tank from 21% O2 to 1% O2 . The vacuum goes 760 mm to 20mm Hg . T is 540oR. Use equation 7-6 Yo = 0.21

Yj= 0.01

Ph = 760 mm

PL = 20 mm

20 760 ¿ ¿ ¿

0.01 = 0.21

0.01 20 ln ( 0.21 ) = j ln ( 760 )

−3.044 −3.637

j=

= 0.84

(1 cycle)

Total N2 required :

∆

N = (1) ( 1 atm -

20 760

)

150 ft 3 ft 3−atm 0.73 (540 o R) o lbmole R

= 0.37lbmoles = 10.4

lb N2 7-8. PROBLEM: Repeat Problem 7-7 using a combined vacuum and pressure purg e. Use a vacuum of 20 mm Hg absolute and a pressure of 200 psig . SOLUTION: Result depends on whether you pressurize first or evacuate first. 20 PL = ( 760 ¿ ( 14.7 )=0.387 psia

Ph = 200 psig + 1.47 = 214.7 psia Po = 14.7 psia If you pressurize first: Po 14.7 Yo = 0.21 ( P h ¿ = 0.21 ( 214.7 ) = 0.0144 Then use equation 7-6 0.387 0.01 = 0.0144 ( 214.7 ¿¿j

0.01 0.327 ln ( 0.0144 ¿ = j ln ( 214.7 )

−3.646 J = ( −6.318 ¿ = 0.58 ( 1 cycle) Total N2 needed: 150 ft 3 ft 3 psia 10.73 ( 540 ) lbmole o R = 0.0259 [ 200+214.3 ]

[ ( 214.7−14.7 ) +(214.7−0.387) ]

= 157 lbmole N2

= 299 lb N2

If you evacuate first: Yo = 0.21 0.387 0.01 = 0.21 ( 214.7 ¿¿j 0.01 0.21 0.387 ln 214.7 ln

J=

Total N2 needed:

=

−3.04 −6.32 = 0.48 ( 1 cycle)

3

150 ft 3 ft psia 10.73 ( 540 ) o lbmole R

=

[ 214.7−0.387 ]

= 0.0259 [ 214.3 ]

= 5.5 lb

mole = 155 lb N2 Much less then when you pressurize first. 7-9. PROBLEM: Use the sweep-through purging technique to reduce the concentration of toluene from an initial 20% to 1% in a room with a volume of 25,000 f t3 . Assume that the room is purged with air at a rate of 6 room volumes per hour. How long will it take to complete this purge process? SOLUTION : V= 25,000 ft3 Qv = (6) ( 25,000 ft3 / hr) = 150,000 ft3/hr C0 = 0

C1 = .20

C2 = .01 Equation 7-5

Qv t = V ln

T=

V Qv

[ ln

C 1−C0 C 2−C0

[

]

C 1−C0 C 2−C0

T = .499 hr. = 30.0 min.

7-10. PROBLEM:

]

=

25,000 ft 3 150,000 ft 3 /hr

ln

[

.2−0 .01−0

]

Design an inerting system for a pressure vessel to maintain the inert atmosphere at 40 psig. Be sure to account for filling and emptying of the vessel . Indicate the precise location of valves, regulators, pipes, etc.

SOLUTION:

Relief R-1

N2 supply

v-6

vacuum

80 psig v-1

v-2

liquid feed

discharge

R-1

Regulator set @ 40 psig R-2

Regulator relieves @ 45 psig

R-1

Provides primary padding @ 40 psig

R-2

Prevents the pressure from exceeding 45 psig for example when the vessels rises

V1 & V2 By-pass values needed for pressure and vacuum purging PI is the pressure indicator to monitor the results

7-11. PROBLEM:

Design a generalized pressure vessel storage tank for a flammable material . Include the following design features: a. Vacuum and pressure purging . b . Vacuum charging of material from a 55-gal drum . c. Draining the tank contents. Provide precise details on the locaion of valves, regulators, and process lines. SOLUTION: vent nitrogen vacuum vessel

From 55 gallon drum Open T to vent Nitrogen dip leg

drum 1.) First vacuum purge vessel.

`2.) Draw vacuum on vessel. 3.) Sweep through purge drum with nitrogen and add dip leg. 4.) While drawing solvent into vessel (under vacuum), continue to add N2 to the drum to displace the liquid. This procedure always maintains an inerted (non-flammable) atmosphere in the drum and in the vessel. 7-12. PROBLEM: Determine the number of vacuum purges required to reduce a vessel’s oxygen concentration from 21% t o 1% if the nitrogen contains: a. 0 ppm of oxygen . b . 9000 ppm of oxygen SOLUTION; a.) Use equation 7-6 PL Yj = Yo ( Ph ¿ 20 760 ¿ ¿ 1 =¿ 21 20 760 ¿ ln ( ¿ 1 ¿=¿ 21

1 ln ( 21 ¿

=

j ln

20 760

-3.04 = j (-3.64) J = 0.835 purges b.) Use equation 7-12

Yi – Yoxy

PL Ph = ( ¿ Yo – Yoxy) ¿ ¿ j¿

(1-0.9) =

20 760 ¿ ¿ ¿

0.1 ( 20.1 ) = 5.30 =1.5 3.63 20 ln ( ) 760

ln J=

7-13. PROBLEM: Use the system described in Figure 7-14 to determine the voltage developed between the charging nozzle and the grounded tank, and the energy stored in the nozzle. Explain the potential hazard for cases and b from the following table:

Hose length (ft) Hose diameter (in) Flow rate (gpm) Liquid conductivity (mho/cm) Dielectric constant Density (g/cm-3)

Case a 20 02 25 10-8 02.4 00.8

Case b 20 2 25 10-18 19 0.8

SOLUTION :

(part a)

See example 7-14 2 L= (20)(12)(2.54)= 610 cm

A= π r

2 = (3.14) (2.54)

2

= 203

V c = 10−8 mno/ cm

R=

L V

L L = A 10

mno/cm

610 cm 109 203 cm = 3x

Y= 2.55 ft/sec Y= (2.54 ft/sec) (

m ¿ = p. 774 m/s 3.28 ft

ft m D= 2m ( 12 m ) ( 3.28 ft ) = 0.0508m Î = ξr.ξo/y T=

( 2.24 ) (8.55 x 10 sec/0 m) 10 mno/cm −5

T= 2.12 x 10

sec

7-13 continued Eg 7-21

Is =

{

10 x 10 cm m 2(m)2 }{1.7 exp [-L/n]} s

( )

I s = 10x 10−8 cmp −8

V= IR =(1.54x 10

)(3 x 10 )

Assume figures are 3 inch dia and a gap between figures is 0.25 inch and Ex =1.0 for a ft L=(0.25m)( 12 m ) = 0.0208 ft

A=

π π 2 2 02−01 ) = ( 4 4

L=

E1 E 0 A/L

2 2 ( 3 −2 )

1 44

2 = 0.0273 ft

0.0272 −2 −12 C = (1)(27x 10 cont/volt ft) ( 0.0208 ) = 3.5 x 10 Energy between flanges = I = cm/2 2 −12 −4 I= (3.5x 10 ) (46.4) /2 = 3.77x 10

−6 j= 3.7x 10 m¿

This is lower than the MIE for gases (0.1nJ), therefore there is no hazard Solution for problem 7-13 part (b) P1

1 v

8

L/A= 3x 10

10¿8 5 ( 10−18 ) = 2.12x 10 sec

2.12 x I s = 1.546x 10−8 amp{ 1- exp(- 3.7x 105 ( ) 2.12 x )} I s = 1.546x 10−8 amp {1 – exp (- 3.7x 10−13 −13

V=IR =(1.54x 10 −12

C= 3.5x 10

19

)(3x 10

coulomb / volt

)= 4.6x 10

6

volt

J= C V

2

6

−12

/2 = (3.5x 10

)(4.6x 10

J= 5.29 j= 5290 mt >> 0.1 mj

)/2

(Very hazardous)

7-14. PROBLEM: Use the system described in Problem 7-13 part b, to determine the hose diameter required to eliminate the potential hazard resulting from static build up. SOLUTION: 2 A=π r

2 U= 3.11/ j m/s d =inch

−2 d= 2.54x 10 d m d=inch

r= 2x 10

4

hy

d2 d2

14

= 3x 10

14

4/d2= 12x 10

where d 2 = inch I s 10x 10−6 (nd )2 {1

exp(

2.88 x 10 )} n

I s 10x 10−6 (nd )2 {1

exp(

2.88 x 10 )} n

d l =8 = 25x 10−12 coulomb / volt

/d2

d2 = 6 V- IR 2 J= cv /2

Conclusion: This system is hazardous even when the diameter is mereased significantly. The problem is the low conductivity of the fluid. This illustrates the problem with nm conductivity fluids and it illustrates the importance of grounding.

7-15. PROBLEM: Repeat Example 7-2, with a 40,000-gal storage vessel. Assume that the vessel height is equal to the diameter.

SOLUTION: Refer to the example for some values used in this solution

V=

40,000 gal 3 (7.48 gal/ ft )

H=2r Pressure

N=

PV TR

= 5348 ft

3

3

N L =0.342( 5348 ft )= 13.7 1bmoles 133.7 ft 3

n14 = 4.07 ( 5348 ft ) =162.8 1bmoles 133.7 ft J= purges =4.95

iny 1/ y 0 inni/nh

=

¿ .000001/.21 ¿ 13.7/162.8

−12.25 = −2.48

5 pressure purges

∆n = 5(162.8 – 13.7) = 746 1bmoles n2 Vacuum 3 n2 = 0.00901 ( 5348 ft ) = 0.364 1bmole 133.7 ft

3

nh = 0.342 ( 5348 ft ) = 13.68 1bmole 133.7 ft

J=

¿ y 1/ y 0 inn 1/nh

−6/ .21

=

¿(10 ) ¿ (.364 /13.68)

=

−12.25 −3.63

=3.38 purges 4 vacuum purges ∆n= 4(13.68-0.634) = 53.3 1bmoles

7-16.PROBLEM: Review Problem 7-13, part b. what is the most effective way to reduce the hazard of this situation? SOLUTION:

As illustrated in part (b) this is hazardous because the fluid is nonconductive. You reduce the hazards via. a.) b.) c.) d.)

Add a conductive additive Reduce the velocity within the pipe In all cases ground and bound In all cases have your system enerted, and contolled to maintain the low O2

Concentration: O2 < MOC

7-17.PROBLEM: Estimate the charge build up and accumulated energy as a result of pneumatically conveying a dry power through Teflon duct. The powder is collected in an insulated vessel. Repeat the calculation for a transport rate of 50lb/min and 100lb/min for transport times of 1hr and 5 hr. Discuss ways to Improve the safety of this situation. SOLUTION: Kg/s = 0.376

for

50 bb/min

Kg/s = 0.756

for

108 bb/min

−5 Charge rate is 10 coulomb/kg (table 7-3)

3 Determine the capacitance of material assuming a bun density of 30 bb/ ft

Volume1 = (50 bb/min) (60 mm/m) (1m)=3000 bb Volume2 = (5m) = 1500 bb Volume3 = (100) (60) (1) = 600 bb Volume 4 = (100) (60) (5) = 30000 bb

Lb/m 50

By/s

Hr

0.376

1

Lb

ft 3

3000

100

r(ft) 2.9

c 9.8x

Coulomb 1.3x

J 8.6x

10−11 50

0.376

5

15000

500

4.9

1.7x −10

100

0.756

1

6000

200

3.6

0.756

5

30000

1000

6.2

6.8x −2

105 1.4x 2

10

10

10

1.2x

2.7x

2.7x

10−10 100

10−2

2.1x 10−10

10−2 0.136

10−2 4.4x 109

2 V= 4/3 π r

3 v 1 /3 V= ( 4 π )

1/ 3 = (0.239 v )

−12

C= 4π ξr lor = 4n (1) (2.7x 10 −11

C= 3.39x 10 J= Q

2

) cont/volt ft (n)

n

/2c=

Conclusion 1. 2. 3. 4.

All cases are hazardous To improve use metal duct and vessel and ground and bond Inert Note if solids are conductive, then a metal duct and metal vessel plus grounding and bounding would be safe.

7-18. PROBLEM: Compute the accumulated charge and energy for a 100,000-gal vessel being filled with a fluid rate of 200gpm and having a streaming current of 2 x 10-6 amp. Make the calculation for a fluid having a conductivity of 10² mho/cm and a dielectric constant of 2.0. Repeat the calculation for (a) a half full vessel, and (c) a full vessel with an overflow line.

SOLUTION: Is = 2 × 10-6 amps

← 200 gpm

ɤc = 10-18 mho/cm εr

= 2.0

100,000 gal

a.) Full vessel ε Relaxation Time = τ = r ε = ɤc

(

( 2.0 ) 8.85× 10−14

o

−18

10

sec ohm . cm

mho/cm

) =1.77 × 10 sec

F= outlet flow = 0 −t / τ Q = τ I s + ( Q 0−I s τ ) e

sec ⁡ −6 ¿ 2× 10 5 −6 ×10 Q = (1.77 2× 10 coul/¿−¿ 1.77 )( ¿ ¿−t /τ 5 ×10 sec ⁡¿¿ 100,00 gal 60 sec T = ( 200 gal/min ¿ min =30,000 sec

(

)

−1

3.54 × 10 ¿ Q= ¿ 3.54 ×10−1 coul−¿

The capacitance is estimated assuming a spherical geometry:

5

3

ft 4 V = ( 100,00o gal)( 7.48 gal ¿=1.33× 10 ft3 V π 3 4 R=( 1.33× 104 ft 3 (¿) =14.7 ft π 1 3 3 3 ¿ ¿ = √¿ 4

(

)

Capacitance for a sphere (assume ε r =1 for air ¿ C = 4 π ε r εo r −12 −10 = ( 4 ¿ ( π ) ( 1 ) ( 2.7× 10 coul /volt−ft ) ( 14.7 ft )=4.98 ×10 coul/volt

Equation 7-20 volt −10 4.98 ×10 coul/¿ ¿ ¿ (2)¿ J= (5.51× 10−2 coul) ¿ Q2 =¿ 2C 6 3.05 ×10¿ ¿ V= (2)¿ 2J =¿ Q

B.)

1 2

( High enough for ignition)

Fuel Vessel

5

T = 15,000 sec τ =1.77 ×10 sec

(same as part a)

−t / τ Q = τ I s + ( Q 0−I s τ )l

−1

=

J=

V=

3.54 × 10 coul ¿l ¿ −1 3.54 ×10 coul−¿

volt −10 4.98 ×10 coul/¿ ¿ (2)¿ −2 2 Q2 (2.88 ×10 coul) = ¿ 2C

( High enough for ignition)

5 2 J ( 2 ) ( 8.33× 10 J ) 7 = =5.78× 10 volts ( High enough for ignition) −2 Q 2.88× 10 coul

C.) Full Vessel with Overflow

Equation 7-36 applies: Q=

A + Bl−c Is

A=

B=

i /τ +

Q o−

Fn V

t

=

2 ×10−6 1 200 + 177,000 100,000

Is F i/τ + n V

=I s τ

Fn I C = i/τ + V = τ

Q= τ

τ Is ¿l ¿ I s+ ¿

The results will be the same as part a. 7-19. PROBLEM: For Problem7-18, part c, if the inlet flow is stopped, compute the accumulated charge and energy after 5hr and after 20hr. Discuss the consequences of these result. Solution : the inlet flow is stopped after 5 and 20 hours. I s r – ( Qo -

Q=

I s r) e−t /T

Q= Qo −2 −t /177,000 sec = (5.5x 10 cool) e

t 2 = 18,000 sec t 20

= 72,000 sec

Q 2 = (5.51x 10−2 cool) e−18,000/177,000 −2 = 4..98x 10 cool

Q 20 = (5.51x 10−2 cool) e−72,000/177,000 −2 = 3.67x 10 cool

2

(4.98 x 10−2 coul)2 ( 2 ) (4.48 x 10−10 coul/volt )

J5 =

Q 2c

J 20 =

Q 2c

=

25 Q

4.98 x 10−2 coul ¿ = ( 2 ) 2.49 x 106 J ¿

25 Q

3.67 x 10−2 coul ¿ = ( 2 ) (1.35 x 106 J ) ¿

V5 =

=

−2

2

V 20 =

6 = 2.49x 10 J

2

(3.67 x 10 coul) ( 2 ) ( 4.98 x 10−2 coul/volt )

6 = 1.35x 10 J

8

= 1.00x 10 V

7

= 7.36x 10 V

Conclusion a.) As the time constant decreases, the accumulated O, J,Q,V decrease. b.) T increases as Vc decrease. c.) As r decreases, the accumulated Q,J,V decrease. d.) When I s =O, J,Q,V decrease as a function of time

7-20. PROBLEM: Some large storage vessels have a floating head, a flat cover that floats on the liquid surface. As the liquid volume increases and decreases, and decreases, the floating head rises and falls within the cylindrical shell of the vessel. What are the reasons for this design? SOLUTION: The floating head eliminates the vapor space above the liquid. This minimizes explosives vapors. It also eliminates the possibility of free fall filling, which is a condition for generating static charges.

7-21 PROBLEM: Determine the fire water requirements (gpm, number of sprinklers, and pump horsepower) to protect an inside process area of 200 f t2. Assume that the sprinkler nozzles have a 0.5-in orifice, the nozzle pressure is 75 psig , and the rate is 50 gpm .

SOLUTION: Per table 7.7 ,the protection requires 0.50 gpm/ft2 Gpm = 0.5 (2000) = 1000 gpm @ 75 psig No. Of nozzle : 1000/50 = 20

Power =

lb m2 gal min 144 1000 2 2 min 60 sec m ft

( )( 75

Horse Power =

)(

{24, 064 ftseclb } 550

ft lb sec HP

)(

)(

ft 3 =24,064 ( ft lb)/sec 7.48 gal

)

= 437 HP @ 100% efficiency

7-22 PROBLEM: What electrical classification would be specified for an area that has Classes I and II,Groups A and E, and Divisions 1 and 2 motors? SOLUTIONS: The electrical class for areas with these electrical equipment would be : Class I 7-23 PROBLEM:

Group A

Division I

Determine the recommended distance between a process area with toluene and an area with an open flame.Toluene leaks as large as 200 gpm have been recorded. Assume an average wind speed of 5 mph and stability class D.

SOLUTION: 200 gpm Toluene Stability class ‘D’ Wind = 5 mph LEL of Tolouene = 1.2 % Flowrate =

(

=

200 gal min

)(

min 60 sec

)(

ft 3 7.48 gal

8.84 ×10 3 g/ sec

Equation 5-48 applies to this situation: (x,o,o) =

Qm π σ yσzU

¿ c >¿( π )(U ) ¿ Qm σ y σz= ¿ 1.2 % concetration of Tolune = 49.3 g/m3 See example 7-11

u= 2.23 m/sec

)( 62.4ft lb )( 454lb g )( 71 gg ) 3

3

m sec ⁡ 2.23 m/¿ ¿ ¿ 4.93× 10 g / ¿(3.14) ¿ ¿ 8.84 × 103 g /sec σ y σ z= ¿ σz

X 100 10 20 25

13.8 1.10 2.70 3.00

σy 2.84 3.74 6.88 8.38

σ yσz 39.1 4.10 18.6 25.1 ≈ 25.6

Up to 25m may be dangerous 7-24.PROBLEM: Determine the recommended ventilation rate for an inside process area (30,000 f t3) that will handle Class I liquids and gases.

SOLUTION: Assume 12 ft ceilings, 2,500 ft2 floorspace. Ventilation rate : 1 ft

3 air/ min−ft

2

floorspace

= 2500 CFM NOTE: Class I requires spot ventilation to limit flammable concentrations not more than 5 ft from source 7-25. PROBLEM:

For the process area described in Problem 7-24, determine the concentration of propane in th area as a function of time if at t = 0 a 3/4-in propane line breaks (the propane main header is at 100 psig ). The temperature is 80°F. See Chapter 4 for the appropriate source model and Chapter 3 for material balance models. SOLUTION: Assume schedule 40 pipe ID = 0.824 inch 0.824 ¿ ¿ ¿ 2 ¿2 12∈¿ Area =

ft 2 =3.70 ×10−3 ft 2 ¿ ( 3.14 ) (¿¿ 4 ¿)¿ ¿ π d2 =¿ 4

P2 = 14.7 psia P1 will be crirical or choked if P2 < .541 P1 .542 P1 = (.542)(100+14.7) psia = 62.2 psia Flow will be critical Use equation 4-50 to determine the flowrate:

Qm ¿ ¿ 2 0+1 ¿ ¿ ɤ +¿ (ɤ −1) ¿ ¿ ɤ Gc M ¿ RgT o ¿ C o=1∧¿

ɤ =1.32

M = 42 T = 80+ 460 = 540oR Qm ¿ ¿ ¿ lbm −sec 2 lbmole 42lb m / ¿ ¿ ¿ ¿ ¿(1545 ft−lb f /lbmole ° R)(540 ° R) ¿ 2 2.32 32.7 lb ft /¿ ¿ (1.32)¿ ¿ √¿ −2 = ( 61.1 lb f ¿ ( 2.70 ×10 /sec ) =1.65 lb m /sec

P ν

= nRT can be used to solve for the volume of propane being leaked per second

1.65lb m / sec 42lb m /lbmole

N=

(

υ=

0.0393 lbmole sec

= 0.0393 lbmole/sec

)(

10.73 psia ft lbmole° R

3

540° R =15.5 ft /sec )( 14.7 psia ) 3

Do a “concentration” balance υ

dc =Q m−k Q v C dt

min 3 Qv = (2500 cfm)( 60 sec ¿=41.7 ft /sec C = ft

3 PROPANE

/ft

3 TOTAL

3 3 ( 30,000 ft 3 ) dc = 15.5 ft −k 41.7 ft C

dt

sec

(

)

sec

dc =5.17 × 10−4−1.39× 10−3 K C dt dc +1.39 ×10−3 K C=5.17 × 10−4 dt This is a linear differential equation with a solution (C=0 ϱ T =0 ¿ as follows: C=

5.17 × 10−4 +Co e 1.39 ×10 −3 1.39 ×10 K

C o=

C=

−3

Kt

−5.17 × 10−4 −3 1.39 × 10 K

(

−4

)

5.17× 10 (1−e 1.39 ×10 −3 1.39× 10 K

−3

Kt

)

K is a mixing factor between .1 and .5

C=

.372 ( 1−e 1.39 ×10 K

( )

−C

Kt

) −4

For K = .1:

−1.39× 10 t ¿ C = 3.72( e

For K = .5:

−6.95× 10 t ¿ C = .744( e

−4

K = .1 t(sec)

K = .5 ft 3 C( ft /¿ ¿

t(sec)

3

0 10 100 1000 ∞

0 .00517 .0514 .483 3.72

ft 3 C( ft /¿ ¿ 3

0 10 100 1000 ∞

0 .00515 .0500 .373 744

7-27. PROBLEM: Determine the fire water requirements (gpm,number of sprinkler heads, and pump horsepower) to protect an inside process area of 2000 f t2. Assume that the sprinkler nozzles have a 0.5-in orifice and that the nozzle pressure is 75 psig . SOLUTION: Using results of problem 7-25 describe whta safety measures shoul be added to this process area. At t = ∞

C=

Qm Qvk

Asssume the k= 0.5 and the C needs to be below LFL or 21% (see appendix b)

Qv =

Q m ft 3 /min 3 = =ft /min k C ft 3 / ft 3

Qv =

(15.5)ft 3 /min =88,600 cfm ( 0.5 ) ( 0.021 ) ft 3 /ft 3

This is not a practical room for ventilation therefore: a) Add a block valves in the propane ine and block the flow as soon as the concentration 25% of LFL or 0.5%. b) Place local ventilation at all potential leak points. c) Alarm at 10% of LFL. d) Enclose the area where propaneis handled this will increase Qv necessary to keep C < 25% of LFL. e) Decrease the operating pressure that is use 5 psig or as low as possible place the regulator outside of the building. Since this is a storage area, the area should be protected with a closed head system. The entire area will be covered with sprinkler nozzles, but the water supply and supply lines will be based on protecting the remote 3000 ft2 area. The water rate per nozzle is 50 gpm ( table 7-7) The total water supply is based on a 3000 ft2 therefore: (3000 ft2)(0.25 gpm/ft3) = 750 gpm 50 gpm Nozzles for 3000 ft2 = (750 gpm)/ nozzle =15 nozzles

(

Therefore nozzles for 10,000 ft2 10,000 ( 15 nozzles )=50 nozzles 3,000

)

Power = (75 psig)

Horsepower =

(

2

)

144 ¿ min ( 750 gpm ) 2 60 sec ft

(

)(

3

=3.28 HP ( 18048secft −lb )( 550sec−HP ft−lb ) f

f

)

ft =18048 ft −lb f /sec 7.48 gal

(at 100%efficiency)

7-28 PROBLEM: Repeat problem 7-27 assuming the nozzle pressure is 100 psig and the rate is 50 gpm

SOLUTION: gpm = (0.5)(2000) = 1000 gpm at 100 psig nozzles =

1000 58

= 17.2 say 18

1 1 Power = (100)(144)(100)( 60 )( 7.48 ) = 32085 (ft 1bp)/sec µ p=

32085 550

= 58.3 (at 100%efficiency)

7-29 PROBLEM: Determine the water requirement (gpm) and number of nozzles for a deluge system required to protect a 10,000-gal storage tank that has a diameter of 15 f t.Use 0.5-in nozzles with a nozzle pressureof 35 psig , and assume that the vessel contains a reactive solvent. SOLUTION: The surface area of the storage tank must be covered by sprinkler nozzles. (Ignore the bottom surface)

SPRINKLER SYSTEM FOR TANK

(10,000 gal)

15

Solve for the height: H=

v Πr 2

=

(10,000 gal )( ft 3/ 7.48 gal ) 2 ( π )(7.5 ft )

= 7.57 ft

Surface area: 2 A = 2πrh + π r

2 = (2π)(7.5ft) + (π) (7.5 ft ) 2 =356.5 ft

2 + 176.6 ft

2 = 533 ft

Total water requirement: (0.35 gpm/ ft

2

) = 186.6 gpm

Number of nozzles: 186.6 gpm 34 gpm/nozzle

= 7.3 nozzles, or 8 nozzles

These eight nozzles must be placed so that they are less than 8 feet apart.

7-30 PROBLEM: Determine the sprinkler requirements for a chemical process area 150ft by 150ft determine the number of sprinkler heads and the pump specifications for this system (hp and gpm). Assume the friction loss from the last sprinkler from the head to the pump is 50ps and the nozzles (1/2 inch arifice) are at 75 psig SOLUTION: 2 Water requirement = (50ft)(150ft)( 0.5 gpm/ ft )

=11250 gpm Nozzles = (11250gpm)/ 50gpm =225 nozzles Pressure at pump : 75 + 50 = 125psig bb m2 ft 3 Power= (122 m2 )(144 ft 2 )(11250gpm)(min/60sec) ( 748 gal ) =451,200ft 1bf/sec HP = (451,200)/550 = 820 HP (At 100% efficiency)

7-31 PROBLEM: Aceton e (C3H6O) is to be stored in a cylindrical process vessel with a diameter of 5ft and a height of 8 ft . The vessel must be inerted with pure nitrogen before storage of the acetone.A limited supply of pure nitrogen is available at 80 psig and 80°F. A vacuum is available at 30mm Hg absolute pressure. a. Determine the target oxygen concentration for the inerting procedure. b.Decide whether a pressure or vacuum purge, or a combination of both , is the best procedure. c. Determine the number of cycles required for your selected procedure. d.Determine the total amount of nitrogen used.The final pressure in the tank after the inerting procedure is atmospheric.The ambient temperature is 80°F. SOLUTION: a)Stoichiometry

C3 H 6 O + 402 moles 02 moles fuel

→ 3 c o2 + 3 H 2 O

= 4.1

LEL = 2.5% UEL = 13.0% moles 02 moles fuel )= (2.5%)(4) = 10% 02 MOC= LEL¿ B) VACUUM PURGE Y0 = 21% 02 YF = 10% 02

2 VESSEL VOLUME = π r h

=(π)(2.5ft)(8ft) = 157 ft

3

Pl =30mm hg PH = 760 mm Hg Equation 7-6 applies: yj yo

Pl = (P ) h

j

.10 30 j ¿ ( → ( .21 = 760 )

Pressure purge yj Pl j ( ) y o = Ph

pl= 14.7 psig

ph=80 +14.7psig = 94.7 psig

J =.230 purges

21 10 =(

94.7 ¿ 14.7

J = .398 purges

c) Either method will require 1 cycle. Local circumstances will dictate which is better. d)

∆ n= j ( Ph−Pl )

V Rg T

Vacuum: lbmole° R 10.73 ft − psia/¿(540 ° R) ¿ ¿ ( 30 ) ( 14.7 ) (1 ) 14.7− psia (157 ft 3) 760 ∆ n= ¿ 3

(

)

Pressure: lbmole° R 10.73 ft − psia/ ¿(540 ° R) ¿ ¿ ( 30 ) ( 14.7 ) 3 (1 ) 14.7− psia (157 ft ) 760 ∆ n= ¿ 3

(

)

Same for both (assuming 1 cycle per purge method). 7-32 PROBLEM: We are considering the installation of a storage vessel to hold 5000 kg of liquid hydrogen .The hydrogen will be stored in an insulated vessel at 1atm absolute pressure at its normal boiling point of 20K.Physical properties for liquid hydrogen at 20 K:

a. We wish to store the liquid hydrogen in a vertical cylindrical storage tank with an inside diameter of 3m . A vapor volume equal to 10% of

the liquid volume must

also be

included.What is the volume of the liquid and the tank (in m 3)? What height tank (in m)is required? b . A 25-mm schedule 40 pipe (ID: 26.64 mm; OD: 30.02 mm) is connected to the bottom of the tank to drain the liquid hydrogen .If the pipe breaks off , producing a hole with a diamete equal to the OD of the pipe, what is the initial discharge rate of the liquid hydrogen from the hole? Assume the liquid height is at the full 5000 kg level . c. What distance (in m) from the storage will the 3psi side-on overpressure occuring the event of an unconfined vapor cloud explosion involving the entire 5000 kg contents of the vessel? d .We need to develop a procedure to inert the vessel prior to charging it with hydrogen .To what target nitrogen concentration do we need to inert the vessel in order to prevent the formation of a flammable gas mixture during the filling process?

SOLUTION: a.) The density of hydrogen is 70.8 kg/m3 (given) Then V LIQ =

5,000 kg =70.6 m3 3 70.8 kg /m

V TANK =( 1.1 ) ( 70.6 m3 )=77.7 m3 2

V TANK =

πD h=77.7 m3 4

H=

3m ¿ ¿ ¿2 (3.14) ¿ 4 V TANK 4 ( 77.7 m3 ) = ¿ π D2

b.) use equation 4-12 to estimate the discharge rate. For this case,

Pg =0 so the equation

reduces to: Qm= ρA C o √ 2 g h L Assume C o=0.61 for highly turbulent flow.

A=

[

1m (3.14) ( 30.22 mm ) 2 1000 m πD = 4 4

h L=

(

2

)] =7.07 × 10

−4

m2

11.0 m =10.0 m 1.10

Substituting the earlier equation, 9.8 m/s2 ( 10.0 ) ¿ (2)¿ Qm=( 70.8 kg/m3 ) ( 7.07 ×10−4 m2 ) √¿ c.) From appendix B for hydrogen, using the lower heating value for the heat of combustion, ∆ H c =−241.8 kJ /mol Using equation 6-28, with an explosion efficiency of 5%

mol 241.8 kJ /¿ ¿ ( 0.05 ) ( 5,000 kg ) 1 mol ¿ 0.002 kg ɳm ∆ H c m TNT = =¿ E TNT

(

)

For 3 psi overpressure , compute the scale overpressure using equation 6-26 Ps=

Po 3 psi = =0.20 P s 14.7 psi

1 /3 From figure 6-23 , z = 9.2 m/kg

From equation 6-25, mTNT ¿ ¿ R = 6,450 kg = 171 m ¿ ¿ z¿ d.) From table 7-1 for hydrogen , ISOC = 5.7 vol% oxygen. Thus, the nitrogen concentration is 100 – 5.7% = 94.3 nitrogen Could also use equation 7-18 or 7-19, The reaction stoichiometry is:

1 H 2+ O2 → H 2 O thus, z = 0.5 2

From appendix B, for Hydrogen, LFL = 4.0% From table 6-3, for Hydrogen, LOC = 5.0% From Equation 7-18:

ISOC =

z (LFL) (0.5)(0.4) = =2.08 LFL 4.0 oxygen 1− 1− 100 100

( )

( )

Nitrogen = 100-2.08 = 97.9% nitrogen From equation 7-19:

ISOC =

z(LOC ) (0.5)(5.0) = =5.56 LOC 5.0 oxygen Z− 0.5− 100 100

Nitrogen = 100-5.56= 94.4% nitrogen Can also use LOC, but this will give a soewhat higher nitrogen concentration. The LOC for hydrogenis 5% thus, nitrogen = 100-5% = 95% nitrogen. 7-33.PROBLEM: A storage vessel must be prepared for filling with carbon monoxide. The vessel currently contains fresh air. a. What is the target oxygen concentration for this operation in order to prevent the existence of a flammable vapor when the carbon monoxide is added? b . If nitrogen containing 2% by volume oxygen is available at 2 barg, how many pressure cycles are required to inert the vessel properly? SOLUTION: a.) From table 7-1 the ISOC is 7.0 % carbon monoxide. This is the target. This can also be estimated using equation 7-18 or 7-19, but the results are only estimate. b.) Use equation 7-12 to estmate the number of cycles required. PL ( Y j−Y oxy )=(Y o −Y oxy )( P ) h

j

0.070-0.02 = (0.21 - 0.02)

1.031 3.013 ¿ ¿ ¿

0.068 = (0.19)(0.336)j Ln(0.3759) =j ln(0.336) −1.03 J = −1.09 =¿ 0.94 cycles 1 cycle would meet the target, but this is real close to the concentration limit. 2 cycles might be more prudent. 7-34 PROBLEM: Your plant is considering in stalling a 5000 m 3 low-pressure cone-roof storage tank .The tank will store toluene (C7H8). The plant is considering several options: a. A single tank within 10m of the process. b . Multiple,smaller tanks within 10m of the process.This option requires 200m of additional piping plus additional valves. c. A single tank 100m from the process. This requires 150 m of additional piping . d . Multiple, smaller tanks 100m from the process. This requires 1000m of additional piping plus additional valves. Consider each option and list the inherently safer features associated with each . Select the single option that represents the most inherently safe design . Please make sure to provide support for your selection . What additional questions should you ask to improve the inherent safety of this installation ? SOLUTION: Inherent safety includes the folowing method: a. Minimize – reduce quantity

b. Substitute – replace a hazardou material with one less hazardous c. Moderate – reduce temperature and pressure d. Simplify – replace complex equipment / chemistry with something simpler. This problem is related to the two concepts: minimize and simplify. However, they are contradictory here. If we use smaller tanks we minimize the quantity per tank ( but not the total amount) but this requires more piping, valves, level gauges and a lot more maintenance. If we use a single tank we have simplified the process, but we have the entire inventory in one vessel.

The key to solving this problem is to realize that even if we use multiple tanks, the total inventory remains the same. Let’s look at the cases : a. A single tank 10 m to the process : This uses the inherent safety simplify. However, the close proximity to the process will result in severe damage if an explosion occurs. b. Multiple smaller tanks 10 m to the process : the total inventory is the same as configuration a. Unless we are able to separate tanks by a large distance , we have gained little by using smaller tanks . We have also added the complexity by the additional pipes, valves, level gauge, etc. c. A single tank 100 m from the process: This simplifies the process, although it is a bit more complex than configuration a to the additional piping . the inreased distance reduces the impact on the process due to explosion. d. Multiple smaller tanks 100 m from the process: this is the same configuration b, but the distance has been increased . This reduces the impact due to an explosion. Based on the above analysis, option c is the best alternative. The smaller tanks do not reduce over all inventory and adda complexity. But the distance and the smaller tanks reduces the consequences. Other questions to ask:

1. Why are we using so much tolune? 2. Why are we using toluene in the first place? 3. What about inerting / purging and grounding / bonding? 4. What about other protection system?

7-35. PROBLEM: The plant has asked you to consider the consequences of a tank explosion with resulting overpressure for options a and c in Problem 7-34. For both cases assume that the storage tank is drained of all liquid and contains only the saturation vapor pressure for liquid toluene at 25°C and 1 atm total pressure with air. a. What is the volume percent concentration of toluene in the vapor of each tank ? b . If the tank contains air, is this concentration flammable? c. What is the stoichiometric concentration for toluene in air? Is the vapor in the tank fuel rich or fuel lean ? d . If an ignition and explosion occur within the storage tank , estimate the overpressure at the process boundary for each case. e . Which case is acceptable? Discuss. f. What additional design features will you recommend to reduce the probability of an explosion ? SOLUTION: Sngle tank of toluene either 10 m or 100 m from process.

a. From appendix E for toluene (C7H8) 3096.52 sat ln( P ¿=16.0137− T −53.67 T is in deg. K P

sat

is in mm Hg

3096.52 sat ln( P ¿=16.0137− 298−53.67 =3.3402 Psat

= 28.22 mm Hg

Volume % =

28.22 mm H g ×100 = 3.71 % 760 mm H g

b. From appendix B for toluene : LFL = 1.2 % ; UFL = 7.1% It is flammable! c. Stoichiometry : C7H8 + 9 O2 → 7 CO2 + 4 H2O Therefore , z = 9. C st =

100 100 = =2.28 z 9 1+ 1+ 0.21 0.21

The concentration of 3.34 % is fuel rich since is is above stoichiometric concentration. d. For toluene molecular weight = 92 gm/ gm-mole = 92 kg / mol At 298 K , 1 atm for toluene vapor:

mol 92 kg/¿ ¿ (1 atm)¿ PM ρ= =¿ RgT Toluene is only 3.71% of the total volume: 3 M = (0.0371)(5000 m3)( 3.76 kg /m ¿ = 697 kg

Use Equation 6-98 . Use an explosion efficiency of 100% since it is enclosed. From appendix B , the energy of explosion for toluene is th lower heating value for ombustion or 3733.9 kJ/ mol = 40,586 kJ/kg kg 40,586 kJ /¿ ¿ TNT! (1.0)(697 kg) ¿ M TNT =¿ For 100 m from process: r Z=

=

1 /3

m

TNT

100 m =5.49 ( 6037 Kg )1/ 3

From figure 6-23 or equation 6-27, Ps = 0.4 The overpressure is then : po = pspa = 0.4 atm = 5.9 psi For 10 m from the process r Z=

=

1 /3

m

TNT

From figure 6-23 , po =70 ps

10 m =0.55 ( 6037 Kg )1/ 3

Po = 70 atm = 1029 psi = 7,09 kPa!!! e. The overpressure for both cases is huge , resulting in complete desctruction of the plant. f. Need inerting / purging plus grounding and bonding.

7-36 PROBLEM: A 1000 m3 storage vessel contains liquid methyl alcohol (CH4O). The vessel is padded with a gas mix obtained from a membrane separation unit.The gas from the membrane unit contains 98% nitrogen (plus 2% oxygen ). The vessel is padded to a total pressure of 10 mm Hg gauge. We must prepare the vessel for entry for the annual inspection of the inside of the vessel .The liquid is first drained from the tank prior to this operation, and then the empty tank must be inerted using a sweep purging method prior to opening the vessel and allowing air to enter. Assume an ambient temperature of 25°C and 1 atm . a. What is the concentration of gas (in vol. %) within the tank after draining the liquid and prior to inerting ? b . Use a triangle diagram to estimate the target fuel concentration (in vol.%) for the inerting operation . c. If we use a sweep purging inerting procedure, using the 98% nitrogen sweep gas from the membrane unit, how much total sweep gas(in m3 at 25°C and 1 atm) is required to achieve the desired target concentration ? d .If the gas from the membrane unit is supplied at the rate of 5 kg/min, how long(in min)will it take to achieve the desired target concentration ?

SOLUTION: Methanol: CH4O, molecular weight = 32

Temperature = 25oC = 298 k Pressure =1 atm Padding gas : 98% nitrogen , 2% oxygen Padding pressure : 10 mm Hg a. From appendix E for methanol: 3626.55 sat ln( P ¿=18.5875− T −34.29 =4.83 Psat =¿ 125.9 mm Hg The mole fraction of methanol in the vapor is : Y=

P sat 125.9 mm Hg = =0.1635 PTOT 10 mm Hg+360 mm Hg

The volume percent methanol in the vapor is 16.3%. The remaining pad gas is 100 – 16.3 = 83.6% Nitrogen in vapor gas = (0.98)(83.6%) = 82.0% Oxygen vapor in gas = (0.02)(83.6%) = 1.7% The gas composition in the vapor in space of the vessel has the following composition: Nitrogen :

82.0%

Methanol :

16.3%

Oxygen:

1.7%

Total : 100.0 b. For methanol from appendix B: LFL = 7.5 % ; UFL = 36.0%

From table 6-3 , LOC = 10%

The stoichiometric combustion is given by; CH4O+3/2 O2 → CO2 + 2 H2O Therefore z = 1.5 z 1.5 = =0.60 ⇒ 60 oxygen 1+z 2.5 See the attached flammability diagram. The target concentration is 13% fuel from the intersection of the two lines. c. Use equation7-15; C2−¿ C C1−¿ C ¿ ¿ ¿ Qv t=V ln¿ o

o

The concentrations are in terms of fuel , not oxygen. Therefore, no fuel C o=0 ( ¿inert ) C1 = 16.3 % C2 = 13.0 % ∴Q v t=( 1000 m3 ) ln

=226 m of inert gas ( 16.3 13.0 ) 3

d. Molar volume of gas :

(

3

)(

22.4 liters m gm−mole 1000 liters 5 kg 0.764 m3 min kg

( )(

)

298 K =0.764 )( 1 gm−mole )( 0.032 kg 273 K )

3 = 3.82 m / min

m3 / kg

T=

226 m 3 =59.2 min 3.82 m 3 /min

LOC

0

100 Target conc. 13% fuel

20

80

40

60

Oxygen

Fuel

60

40

UFL=36

STOICH

82.0 % N2

80

20

16. 3%M2OH 1.7%O2 LFL= 7.5%

100

0 0

20

40 N i t r o g en

60

80

100 98%

N2 , 2 % O 2

7-37 PROBLEM: A propane storage tank with a volume of 10,000 liters is being taken out of service for maintenance.The tank must be drained of its liquid propane, depressurized to atmospheric pressure, and then inerted with nitrogen prior to opening the tank to air. The temperature is 25°C and the ambient pressure is 1 atm . a. Determine the required target fuel cocentration in the tank prior to opening the tank . b . A sweep purge will be used for the inerting procedure. If pure nitrogen is available at a delivery rate of 0.5 kg /min , what is the minimum time (in min) required to redu ce t h e f u el con cen t rat ion t o t h e t arg et valu e? c. Wh at is the total amount (in kg ) of nitrog en required to do the job? SOLUTION: a. We need to calculate the out-of-service fuel concentration from equation 7-16. From appendix B for propane, LFL = 2.1 % fuel in air. The stoichiometry combustion equation is : C3 H 8 +5 O2 → 3CO 2 +4 H 2 o This z=5 Substituing into equation 7-16:

OSFC =

LFL 2.1 = =4.2 LFL 2.1 1− 1−5 21 21

( )

( )

We will need to reduce the fuel concentration to below 4.2 % Our starting condition are : Propane Fuel: 100% Our target concentration for inerting is : Propane fuel : 4.2% ; Nitrogen :95.8%

b. Sweep purge. Use equation 7-15. C o=0 C1 = 16.3 % C2 = 13.0 % ∴Q v t=V ln

C1 100 =( 10,000 liters ) ln =31,700 L of inert gas C2 4.2

( )

( )

At these conditions the molar volume is: 22.4 L K =¿ ( gm−mole )( 298 273 K )

24.4 L/gm-mole

At a flow rate of 0.5 kg/min: 24.4 L =437 L/min ( 0.5minkg )( 1000kggm )( 1 gm−mole )( 28 gm gm−mole ) The time required is: 31,700 L T = 437 L/min =72.5 min The total nitrogen required: 31,700 L nitrogen

28 gm 1 kg =¿ ( 1 gm−mole )( )( 24.4 L gm−mole 1000 gm )

36.4 kg of nitrogen