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 :

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

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 %

1 C2 = 21 %

Qvt is the total volume required Eq. 7-15 C_2−¿Co C1−¿C_o ¿ ¿ ¿ Q_vt=V ln¿ Qv t ¿(100 gal)

(

fr 3 7.48 gal

)

ln

(

21−0 1−0

)

=¿ (13.4 ft3) (ln21) = 40.7 ft3 7-6. PROBLEM: 100 gal 100 % air

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:

T = 80o_{F + 460}o _{= 540}o _R Sweep through, eq. 7-15 applies;

Qv t ¿V

(

ln C₁ C₂

)

= 150ln

(

21 1

)

=4 57 ft3 ;_{Pressure purge:} 0 psig = 14.7 psia

Use equations 7-1 and 7-2:

Nh = P_hV RgT = (140 psia)

(

150 ft3

)

(10.73 psia−fr 3 lbmole – o_R )(540 o R) = 3.62 lbmole 150 ft3 _of air

NL = P_RLV gT = = (14.7 psia)

(

150 ft3

₎

(10.73 psia−fr 3 lbmole – o_R )(540 o_{R) = .381 lbmole}

Use equations 7-6 to compute the number of cycles required:

Yj = Yo N_L Nh ¿j ¿ ln

(

Yj Y_o

)

=¿ j ln( N_L Nh ¿ _{j =} lnYj Yo ln NL Nh J = ln 1 21 ln.381 3.62 = −3. 044_{−2.251 = 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 =} PLV RgT V = R_gT N_L PL V = ( 10.73 psia−fr3 lbmole – oR )(540o_{R)(6.48 lbmoles)} (14.7 psia) = 2550 ft 3 _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 540o_{R. Use equation 7-6} Yo = 0.21 Yj= 0.01 Ph = 760 mm PL = 20 mm 0.01 = 0.21 20 760 ¿ ¿ ¿ ln (0.01_0.21) = j ln (₇₆₀20 ) j = −3.044_{−3.637 = 0.84 (1 cycle)} Total N2 required : ∆ N = (1) ( 1 atm - _{760 )} 20 150 ft3 0.73 ft3−atm lbmoleoR(540 o_{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.

Ph = 200 psig + 1.47 = 214.7 psia Po = 14.7 psia

If you pressurize first:

Yo = 0.21 (

P_o Ph

¿ _{= 0.21 (} 14.7

214.7 ) = 0.0144

Then use equation 7-6

0.01 = 0.0144 ( 0.387 214.7 ¿ ¿j ln ( 0.01 0.0144¿ = j ln ( 0.327 214.7 ) J = ( −3.646_−6.318¿ = 0.58 ( 1 cycle) Total N2 needed: 150 ft3 10.73 ft 3 psia lbmoleo_R(540)

[

(214.7−14.7 )+(214.7−0.387)

]

= 0.0259

[

200+214.3

]

= 157 lbmole N2 = 299 lb N2 If you evacuate first:

Yo = 0.21 0.01 = 0.21 ( 0.387 214.7 ¿ ¿j J = ln0.01 0.21 ln0.387 214.7 = −3.04_{−6.32 = 0.48 ( 1 cycle)} Total N2 needed:

= 150 ft3 10.73 ft 3 psia lbmoleoR(540)

[

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 ft}3 _{/ hr) = 150,000 ft}3_/hr C0 = 0 C1 = .20 C2 = .01 Equation 7-5 Qv t = V ln

[

C₁−C₀ C₂−C₀

]

T = _QV_v ln

[

C₁−C₀ C₂−C₀

]

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

[

.2−0 .01−0

]

T = .499 hr. = 30.0 min. 7-10. PROBLEM:

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

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

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

Yj = Yo ( P_L Ph ¿ 20 760 ¿ ¿ 1 21=¿ ln ( 20 760 ¿ ¿ 1 21¿=¿

ln ( ₂₁1 ¿ = j ln ₇₆₀20 -3.04 = j (-3.64) J = 0.835 purges b.) Use equation 7-12 Yi – Yoxy = ( P_L Ph ¿ ¿ ¿j¿ Yo – Yoxy) (1-0.9) = 20 760 ¿ ¿ ¿ J = ln

(

0.1 20.1

)

ln

(

20 760

)

=5.30 3.63=1.5 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:

Case a Case b

Hose length (ft) 20 20

Hose diameter (in) 02 2

Flow rate (gpm) 25 25

Liquid conductivity (mho/cm) 10-8 ₁₀-18

Dielectric constant 02.4 19

SOLUTION : (part a) See example 7-14 L= (20)(12)(2.54)= 610 cm2 A= π r2 = (3.14) (2.54)2 = 203 V_{c = 10}−8 mno/ cm R = _V L _AL=_{10 mno/cm} L 610 cm_{203 cm = 3x 10}9 Y= 2.55 ft/sec Y= (2.54 ft/sec) ( _{3.28 ft}m ¿ = p. 774 m/s D= 2m ( _{12 m )}ft (_{3.28 ft}m ) = 0.0508m Î = ξr.ξo/y T = (2.24 )(8.55 x 10 sec/0 m)_{10 mno/cm} T= 2.12 x 10−5 sec 7-13 continued Eg 7-21 Is ₌ {10 x 10 cm

(

m s

)

2(m) 2 _{}{1.7 exp [-L/n]}}

I_{s = 10x 10}−8 cmp

V= IR =(1.54x 10−8 )(3 x 10 )

Assume figures are 3 inch dia and a gap between figures is 0.25 inch and Ex =1.0 for a

L=(0.25m)( _{12 m ) = 0.0208 ft}ft A= π_{4 (} 02 2 −01 2 ) = π_{4 ( 3}2−22 ) _{44 = 0.0273 ft}1 2 L= E1E0 A/L C = (1)(27x 10−2 cont/volt ft) ( 0.0272_{0.0208 ) = 3.5 x 10}−12

Energy between flanges = I = cm/2

I= (3.5x 10−12 ) (46.4)2 /2 = 3.77x 10−4 j= 3.7x 10−6m¿

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 L/A= 3x 10}8 ₍ 10 ¿8 10−18 ) = 2.12x 10 5 sec I_{s = 1.546x 10}−8 amp{ 1- exp(- 3.7x 105(2.12 x_{2.12 x}) )} I_{s = 1.546x 10}−8 amp {1 – exp (- 3.7x 10−13 V=IR =(1.54x 10−13 )(3x 1019 )= 4.6x 106 volt C= 3.5x 10−12 coulomb / volt

J= C V2 /2 = (3.5x 10−12 )(4.6x 106 )/2 J= 5.29 j= 5290 mt >> 0.1 mj (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: A = π r2 U= 3.11/ j2 m/s d =inch d= 2.54x 10−2 d m d=inch r= 2x 104 hy d 2_{d 2 = 3x 10}14 4/d2= 12x 1014 /d2 where d2 = 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 )}} dl =8 = 25x 10−12 coulomb / volt

d_{2 = 6}

V- IR J= cv2 /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= _{(7.48 gal/ ft}40,000 gal3 ) = 5348 ft 3 H=2r Pressure N= PV_TR

N_{L =0.342(} 5348 ft3

133.7 ft )= 13.7 1bmoles

n_{14 = 4.07 (} 5348 ft3

133.7 ft ) =162.8 1bmoles

J= purges iny 1/ y 0_{inni/nh =} ¿_¿.000001/.21_{13.7/162.8 =} −12.25_−2.48 =4.95 5 pressure purges ∆n = 5(162.8 – 13.7) = 746 1bmoles n2 Vacuum n2 _{= 0.00901 (} 5348 ft 3 133.7 ft ) = 0.364 1bmole n_{h = 0.342 (} 5348 ft3 133.7 ft ) = 13.68 1bmole J = ¿_{inn 1/nh =} y 1/ y 0 ¿(10 −6/ .21 ) ¿(.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?

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

a.) Add a conductive additive

b.) Reduce the velocity within the pipe c.) In all cases ground and bound

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

Determine the capacitance of material assuming a bun density of 30 bb/ ft3 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 By/s Hr Lb _ft3 r(ft) c Coulomb J 50 0.376 1 3000 100 2.9 9.8x 1.3x 8.6x

10−11 10−2 105 50 0.376 5 15000 500 4.9 1.7x 10−10 6.8x 10−2 1.4x 102 100 0.756 1 6000 200 3.6 1.2x 10−10 2.7x 10−2 2.7x 10−2 100 0.756 5 30000 1000 6.2 2.1x 10−10 0.136 4.4x 109 V= 4/3 π r2 V= (_{4 π}3 v) 1 /3 = (0.239 v )1/ 3 C= 4π ξr lor = 4n (1) (2.7x 10−12 ) cont/volt ft (n) C= 3.39x 10−11 n J = Q2 /2c= Conclusion

1. All cases are hazardous

2. To improve use metal duct and vessel and ground and bond 3. Inert

4. 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:

← 200 gpm Is = 2 × 10-6 _amps ɤc = 10-18 _mho/cm

εr = 2.0

100,000 gal

a.) Full vessel

Relaxation Time = τ =εr εo ɤ_c = (2.0)

(

8.85× 10−14 sec ohm . cm

)

10−18mho/cm =1.77 × 10 5_sec F= outlet flow = 0 Q = τ Is+

(

Q0−Isτ

)

e−t / τ Q = (1.77 sec ⁡ ¿ 2× 10−6coul/¿−¿ ×105sec ⁡¿¿ 1.77 ×105 )( 2× 10 −6 ¿ ¿−t /τ

T = ( _{200 gal/min}100,00 gal ¿

(

60 sec_min

)

=30,000 sec

Q =

3.54 × 10−1 ¿ ¿

3.54 ×10−1coul−¿

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

(

1.33× 10 4_ft3 π

)

=14.7 ft 3 4¿ ¿ 1 3₌3 √¿

Capacitance for a sphere (assume εr=1 for air¿

C = 4 π εrεor

= ( 4¿(π ) (1)

(

2.7× 10−12coul /volt−ft

)

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

( High enough for ignition)

T = 15,000 sec τ =1.77 ×105sec (same as part a) Q = τ Is+

(

Q0−Isτ

)

l −t / τ = 3.54 × 10−1_coul ¿l ¿ 3.54 ×10−1 coul−¿ J = volt 4.98 ×10−10coul/¿ ¿ (2)¿ Q2 2 C= (2.88 ×10−2_coul)2 ¿

( High enough for ignition)

V = 2 J_Q =(2)

(

8.33× 10 5

J

)

2.88× 10−2coul=5.78× 10

7

volts ( High enough for ignition)

C.) Full Vessel with Overflow

Equation 7-36 applies: Q = A +Bl −ct A = I_s i/τ +Fn V = 2 ×10 −6 1 177,000+ 200 100,000 B = Qo− Is i/τ +Fn V =I_sτ

C = i/τ +F_Vn=I_τ Q = τ τ I_s ¿l ¿ I_s+¿

The results will be the same as part a. 7-19. PROBLEM:

ForProblem7-18, 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. Q= Is r – ( Qo - Is r) e−t /T Q= Qo = (5.5x 10−2 cool) e−t /177,000 sec t_{2 = 18,000 sec} t_{20 = 72,000 sec} Q_{2 = (5.51x 10}−2 cool) e−18,000/177,000 = 4..98x 10−2 cool Q20 = (5.51x 10−2 cool) e−72,000/177,000 = 3.67x 10−2 cool

J5 ₌ Q 2 2 c = (4.98 x 10−2_coul)2 (2)(4.48 x 10−10_{coul/volt ) = 2.49x} 106J J_{20 =} Q2 2 c = (3.67 x 10−2coul)2 (2)(4.98 x 10−2coul/volt ) = 1.35x 10 6 J V_{5 =} 25 Q = 4.98 x 10−2_coul ¿ (2) 2.49 x 106 J ¿ = 1.00x 108V V_{20 =} 25 Q = 3.67 x 10−2coul ¿ (2)(1.35 x 106 J) ¿ = 7.36x 107V 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 Is =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 =

(

75_mlb2

)

(

144 m2 ft2

)

(

1000 gal min

)(

min 60 sec

)

(

ft3 7.48 gal

)

=24,064 (ft lb)/sec Horse Power =

{

24, 064ft lb sec

}

550 ft lb sec HP = 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 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

)(

_{60 sec}min

)

(

ft 3 7.48 gal

)(

62.4 lb ft3

)

(

454 g lb

)(

7 g 1 g

)

= 8.84 ×103g/ sec

Equation 5-48 applies to this situation:

<(> (x,o,o) = _{π σ}Qm_yσzU

¿c >¿(π )(U )

¿

σ_yσ_z=Qm_¿

1.2 % concetration of Tolune = 49.3 g/m3 See example 7-11 u= 2.23 m/sec

m3 sec ⁡ 2.23 m/¿ ¿ ¿ 4.93× 10 g /¿(3.14)¿ ¿ σyσz= 8.84 × 103_{g /sec} ¿ X _σz σy σyσz 100 13.8 2.84 39.1 10 1.10 3.74 4.10 20 2.70 6.88 18.6 25 3.00 8.38 _{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 ft3air/ min−ft2 floorspace

= 2500 CFM

NOTE: Class I requires spot ventilation to limit flammable concentrations not more than 5 ft from source

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 Area = 0.824 ¿ ¿ ¿2¿2 12∈¿ ft ¿ 2 =3.70 ×10−3ft2 (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

Q_m ¿ ¿ 2 0+1 ¿ ¿ ɤ+¿(ɤ−1) ¿ ¿ ɤ G_cM RgTo ¿ ¿ Co=1∧¿ ɤ =1.32 M = 42 T = 80+ 460 = 540o_R Q_m ¿ ¿ ¿ lbm−sec 2 lbmole 42lb_m/¿ ¿ ¿ ¿ ¿(1545 ft−lbf/lbmole ° R)(540 ° R) ¿ 2 2.32 32.7 lb_ft/¿ ¿ (1.32)¿ ¿ √¿ = ( 61.1 lbf¿

(

2.70 ×10−2/sec

)

=1.65 lbm/sec

N = 1.65lbm /sec

42lbm/lbmole = 0.0393 lbmole/sec

υ=

(

0.0393 lbmole_sec

)

(

10.73 psia ft 3 lbmole° R

)

(

540° R 14.7 psia

)

=15.5 ft 3 /sec Do a “concentration” balance υdc_dt=Qm−k QvC

Qv = (2500 cfm)( _{60 sec}min ¿=41.7 ft3/sec

C = ft3PROPANE/ft3TOTAL

(

30,000 ft3

₎

dc dt= 15.5 ft3 sec −k

(

41.7 ft3 sec

)

C dc dt=5.17 × 10 −4_{−1.39× 10}−3_{K C} dc_dt+1.39 ×10−3K C=5.17 × 10−4

This is a linear differential equation with a solution (C=0 ϱ T =0¿ as follows:

C = 5.17 × 10 −4 1.39 ×10−3K+Coe 1.39 ×10−3 Kt Co= −5.17 × 10−4 1.39 × 10−3_K C =

(

5.17× 10 −4 1.39× 10−3K

)

(1−e 1.39 ×10−3 Kt )

C=

(

.372_K

)

(

1−e1.39 ×10 −C_Kt

)

For K = .1: C = 3.72( e−1.39× 10 −4 t ¿ For K = .5: C = .744( e−6.95× 10 −4 t ¿ K = .1 K = .5 t(sec) C( ft 3 ft3/¿ ¿ t(sec) C( ft 3 ft3/¿ ¿ 0 0 0 0 10 .00517 10 .00515 100 .0514 100 .0500 1000 .483 1000 .373 ∞ 3.72 ∞ 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 = _QQm vk

Qv= Q_m k C= ft3 /min ft3/ft3 =ft 3 /_min Qv= (15.5)ft3 /min (0.5) (0.021) ft3 /ft3=88,600 cfm

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/ft}3_{) = 750 gpm}

Nozzles for 3000 ft2 = (750 gpm)/

(

50 gpm

nozzle

)

=15 nozzles

Therefore nozzles for 10,000 ft2

Power = (75 psig)

(

144¿ 2 ft2

)

(750 gpm)

(

min 60 sec

)

(

ft3 7.48 gal

)

=18048 ft −lbf/sec Horsepower =

(

18048 ft −lb_f sec

)(

sec−HP 550 ft−lb_f

)

=3.28 HP (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 = 1000₅₈ = 17.2 say 18 Power = (100)(144)(100)( _{60 )(}1 _{7.48 )}1 = 32085 (ft 1bp)/sec µ p = 32085₅₅₀ = 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)

(10,000 gal)

15

Solve for the height: H = _Πrv2 = (10,000 gal)(ft3/ 7.48 gal) (π )(7.5 ft )2 = 7.57 ft Surface area: A = 2πrh + π r2 = (2π)(7.5ft) + (π) (7.5 ft )2 =356.5 ft2 + 176.6 ft2 = 533 ft2

Total water requirement: (0.35 gpm/ ft2 ) = 186.6 gpm Number of nozzles:

186.6 gpm

34 gpm/nozzle = 7.3 nozzles, or 8 nozzles

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:

Water requirement = (50ft)(150ft)( 0.5 gpm/ ft2 ) =11250 gpm

Nozzles = (11250gpm)/ 50gpm =225 nozzles Pressure at pump : 75 + 50 = 125psig

Power= (122 bb_m2 )(144 m2 ft2 )(11250gpm)(min/60sec) ( ft3 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

C₃H_{6O + 402 → 3 c o2 + 3 H2}O moles 0₂ moles fuel = 4.1 LEL = 2.5% UEL = 13.0% moles 0₂ moles fuel MOC= LEL¿ )= (2.5%)(4) = 10% 0 2 B) VACUUM PURGE Y0 = 21% 02 YF = 10% 02 VESSEL VOLUME = π r2h =(π)(2.5ft)(8ft) = 157 ft3 Pl =30mm hg PH = 760 mm Hg Equation 7-6 applies: y_j yo = ( Pl P_h) j → ( .10_.21¿ = (₇₆₀30 ) j J =.230 purges Pressure purge y_j yo = ( Pl Ph ) j pl= 14.7 psig ph=80 +14.7psig = 94.7 psig

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 R_gT Vacuum: lbmole° R 10.73 ft3 −psia/¿(540 ° R) ¿ ¿ ∆ n= (1)

(

14.7−(30) (14.7) 760 psia

)

(157 ft 3 ) ¿ Pressure: lbmole° R 10.73 ft3 −psia/¿(540 ° R) ¿ ¿ ∆ n= (1)

(

14.7−(30) (14.7) 760 psia

)

(157 ft 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 m3_{)? 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 VLIQ= 5,000 kg 70.8 kg /m3=70.6 m 3 VTANK=(1.1)

(

70.6 m3

)

=77.7 m3 VTANK= π D2 4 h=77.7 m 3

H = 3 m ¿ ¿ ¿2 (3.14)¿ 4 V_TANK π D2 = 4

(

77.7 m3

)

¿

b.) use equation 4-12 to estimate the discharge rate. For this case, Pg=0 so the equation reduces to:

Qm=ρACo

√

2 g hL

Assume Co=0.61 for highly turbulent flow.

A= π D2 4 = (3.14)

[

(30.22 mm)

(

1m 1000 m

)

]

2 4 =7.07 × 10 −4 m2 hL= 11.0m 1.10 =10.0 m

Substituting the earlier equation, 9.8 m/s2(10.0)

¿ (2)¿

Qm=

(

70.8 kg/m3

) (

7.07 ×10−4m2

)

√¿

c.) From appendix B for hydrogen, using the lower heating value for the heat of combustion, ∆ Hc=−241.8 kJ /mol

mol 241.8 kJ /¿ ¿ (0.05) (5,000 kg)

(

1 mol 0.002 kg

)

¿ mTNT= m ɳ ∆ H_c E_TNT =¿

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

Ps= P_o Ps = 3 psi 14.7 psi=0.20 From figure 6-23 , z = 9.2 m/kg1 /3 From equation 6-25, R = m_TNT ¿ ¿ 6,450 kg ¿ ¿ z¿ = 171 m

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: H2+ 1

2O2→ H2O thus, z = 0.5 From appendix B, for Hydrogen, LFL = 4.0%

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

ISOC = z (LFL) 1−

(

LFL 100

)

=(0.5)(0.4) 1−

(

4.0 100

)

=2.08 oxygen Nitrogen = 100-2.08 = 97.9% nitrogen From equation 7-19: ISOC = z(LOC ) Z−LOC 100 =(0.5)(5.0) 0.5−5.0 100 =5.56 oxygen 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.

(

Yj−Yoxy

)

=(Yo−Yoxy)(

PL

P_h)

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) J = −1.03_−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 m3 _{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.

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:

a. From appendix E for toluene (C7H8) ln( Psat¿=16.0137−_{T −53.67}3096.52 T is in deg. K Psat _{is in mm Hg} ln( Psat¿=16.0137−_298−53.673096.52 =3.3402 Psat _{= 28.22 mm Hg} Volume % = 28.22 mm H g_{760 mm H g} ×100 = 3.71 %

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

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

mol 92 kg/¿ ¿ (1 atm)¿ ρ= PM RgT =¿

Toluene is only 3.71% of the total volume:

M = (0.0371)(5000 m3_{)( 3.76 kg /m}3¿ _{= 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 /¿ ¿ (1.0)(697 kg)¿ MTNT=¿ TNT!

For 100 m from process:

Z = r m1 /3 TNT = 100 m (6037 Kg)1/ 3=5.49

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

Z = _m1 /3r TNT = 10 m (6037 Kg)1/ 3=0.55 From figure 6-23 , p_o ps =70

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:

Temperature = 25o_{C = 298 k} Pressure =1 atm

Padding gas : 98% nitrogen , 2% oxygen Padding pressure : 10 mm Hg

a. From appendix E for methanol:

ln( Psat¿=18.5875−_{T −34.29}3626.55 =4.83

Psat

=¿ _{125.9 mm Hg}

The mole fraction of methanol in the vapor is :

Y = P sat

PTOT

= 125.9 mm Hg

10 mm Hg+360 mm Hg=0.1635

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%

The stoichiometric combustion is given by; CH4O+3/2 O2 → CO2 + 2 H2O Therefore z = 1.5 z 1+z= 1.5 2.5=0.60 ⇒ 60 oxygen See the attached flammability diagram.

The target concentration is 13% fuel from the intersection of the two lines. c. Use equation7-15; C_2−¿Co C1−¿Co ¿ ¿ ¿ Q_vt=V ln¿

The concentrations are in terms of fuel , not oxygen. Therefore,

no fuel C_o=0(_¿inert) C_{1 = 16.3 %} C_{2 = 13.0 %} ∴Q_vt=

(

1000 m3

₎

_ln

(

16.3 13.0

)

=226 m 3_{of inert gas}

d. Molar volume of gas :

(

22.4 liters gm−mole

)

(

m3 1000 liters

)

(

1 gm−mole 0.032 kg

)(

298 K 273 K

)

=0.764 m 3 / kg

(

5 kg min

)

(

0.764 m3 kg

)

= 3.82 m 3 / min

T = 226 m 3

3.82 m3/min=59.2 min

LOC 0 100

Target conc. 13% fuel

20 80 40 60 O x y g e n F u e l 60 40 UFL=36 STOICH 82.0 % N2 80 20 16. 3%M2OH 1.7%O2 LFL= 7.5% 100 0 0 20 40 60 80 100 N i t r o g en 98% N2 , 2 % O2

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 : C3H8+5 O2→ 3CO2+4 H2o This z=5

Substituing into equation 7-16:

OSFC = LFL 1−

(

LFL 21

)

= 2.1 1−5

(

2.1 21

)

=4.2

We will need to reduce the fuel concentration to below 4.2 % Our starting condition are : Propane Fuel: 100%

b. Sweep purge. Use equation 7-15. Co=0 C_{1 = 16.3 %} C_{2 = 13.0 %} ∴Q_vt=V ln

(

C1 C₂

)

=(10,000 liters) ln

(

100 4.2

)

=31,700 L of inert gas At these conditions the molar volume is:

(

22.4 L

gm−mole

)(

298 K

273 K

)

=¿ 24.4 L/gm-mole At a flow rate of 0.5 kg/min:

(

0.5 kg min

)(

1000 gm kg

)(

1 gm−mole 28 gm

)(

24.4 L gm−mole

)

=437 L/min

The time required is:

T = _{437 L/min}31,700 L =72.5 min

The total nitrogen required:

31,700 L nitrogen

(

1 gm−mole 24.4 L

)(

28 gm gm−mole

)(

1 kg 1000 gm

)

=¿ 36.4 kg of nitrogen