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CORK INSTITUTE OF TECHNOLOGY INSTITIÚID TEICNEOLAÍOCHTA CHORCAÍ

Semester 2 Examinations 2009/10

Module Title: Process Plant Equipment

Module Code: MECH 8017

School: Mechanical and Process Engineering Programme Title: Bachelor of Science (Honours) in Process Plant Technology Programme Code: EPPTE_8_Y4 External Examiner(s): Mr. N. Kingston, Mr. J. Phelan Internal Examiner(s): Dr. F. Murphy Instructions: Answer THREE Questions Duration: Two Hours Sitting: Summer 2010 Requirements for this examination: Graph paper and Mathematical Tables to be provided Note to Candidates: Please check the Programme Title and the Module Title to ensure that you have received the correct examination paper. If in doubt please contact an Invigilator.

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Q1. (a) The increasing use of pneumatic conveying systems for fine particle transportation is

leading to the wider use of fabric filters to ensure almost complete gas - solids separation.

Describe some of the more important features of fabric filters. (10 marks)

(b) A conveying system is designed to transport Portland cement through a 100 mm diameter

(D) pipe. It is supplied with air at a rate (V&) of 11 m3/min Free Air Delivery. The cement is

conveyed on a batch basis using a bottom discharge blow vessel. The conveying rate (G&) is

60 t/h. The ambient air density ( ρ ) is 1.2 kg/m3. The mean particle size ( d) is 15 µm.

Assume a system separation efficiency (η) of 98%.

Using the following formulae and Figure (Question 1), recommend a filter size in terms of

the filter area. (24 marks)

Figure (Question 1):

FORMULAE:

The mass flow ratio, QG&&

=µ Dust collector efficiency, 100∗

∗∗ −=

α

α

ρρρ

ηc

cwcc

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ρcα∗ = dust concentration (dust-laden gas) ρcw

∗ = dust concentration (cleaned gas)

Air rate, ρ×=VQ && Filter area,γVAF&

=

γ = gas/cloth ratio from Figure (Question 1).

Q2. A 600 mm man way (bolted flange connection) is to have the following design conditions:

Design pressure (P) = 4 MPa Allowable flange stress (Sa) = 135 MPa

Allowable bolt stress (Sb) = 172 MPa Flange inside diameter = 590 mm

Flange outside diameter = 900 mm Bolt circle diameter = 800 mm

24 bolts each of 944 mm2 root area are to be utilised.

The gasket is spiral wound and graphite filled with an outer ring.

Gasket outside diameter (God) = 690 mm Gasket inside diameter (Gid) = 630 mm

Gasket factor (m) = 3.0 Gasket seating stress (y) = 69 MPa

Gasket effective width (b) = 10 mm.

Calculate the required cover thickness. (20 marks)

Figure (Question 2) (a schematic drawing of the cover flange)

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FORMULAE:

The minimum required bolt load for gasket seating: Wm2 = πbGy

The minimum required bolt load during operation: Wm1 =π4G 2P + 2bπGmP

Bolt load to protect from over tightening and flange overloading: W =Ab + Am2

2Sa

Cover plate thickness: t =G0.3PS

+1.9WhGSG 3

SYMBOLS:

Am1 = the required bolt area for operating case

Am2 = the required bolt area for gasket seating

Ab = the actual bolt area

S = stress at the centre of the plate.

Q3. Determine the reinforcement requirements for a 300 mm diameter opening in a cylindrical

pressure vessel 1 m in diameter subjected to an internal pressure of 5 MPa. The shell and

nozzle allowable stress is 120 MPa. The shell and nozzle thickness are 25 mm and 32 mm,

respectively. The reinforcement scheme is shown in Figure (Question 3). (34 marks)

FORMULAE: (based on the ASME design philosophy)

The minimum required shell thickness is given by: trs =PRs

S − 0.6P

Reinforcement limit parallel to shell surface is the larger of ts+tn+0.5d or d.

Reinforcement limit normal to the shell surface is the smaller of 2.5ts or 2.5tn

SYMBOLS:

P = Design Pressure Rs = Shell Radius S = Allowable Stress

d = opening diameter

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Figure (Question 3):

Q4. (a) Describe how ventilating air can influence the point at which a flammable gas or vapour

enters its explosion range and reach a nonhazardous concentration. (14 marks)

(b) A process takes place in an enclosed chamber operating at 168°F. The process emits 30

gallons of flammable liquid per hour. From the 30 gallons of liquid, 80% will evaporate in

60 minutes. Dilution ventilation is required for removing and reducing the flammable

vapour concentration to below 25% of the lower explosive limit (LEL). Assume that the

explosion limits of the flammable vapour in air range from 1.25-10% at 21°C. Assume that

a ratio of 98.75% of air and 1.25% of vapour is too lean to produce an explosion. The

specific gravity of the liquid is 0.72, the vapour density is 8 and 1 US gallon = 3.8 x 10-3 m3.

Determine the volume of air required for diluting the flammable vapour from the process.

(20 marks)

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FORMULAE:

Va =3.78 × Sp.Gr.

1.2 ×VD× LEL ×C

where Va = vapour released in cubic metres per gallon of solvent per hour

3.78 = the equivalent mass of 1 gallon of water in kilogrammes

1.2 = the mass of 1 cubic metre of air in kilogrammes at 21°C ambient temperature

VD = vapour density of the solvent

Sp.Gr. = specific gravity of the solvent

LEL = lower explosive limit

C = correction factors for the LEL of the solvent vapour at elevated temperature (C = 1 for

temperatures up to 250°F and C = 0.7 for above 250°F)

Correction factor for air to compensate for higher operating temperatures is given by:

AirDensityCorrectionFactor = 273+ T2273+ T1

T1 is 21°C ambient temperature

T2 is the actual dilution air temperature