Primary Reformer Mechanical Design Assignment Help

In this section a mechanical design for the primary reformer tubes inside the furnace will be carried out. From the chemical design section we assumed each tube is 9 m long and has an internal diameter of 0.12 m.

Mechanical Engineering Assignment Help Order Now

§ Design Pressure:

The pressure inside the vessels is 30 bar. 10% above operating pressure was considered for safety:

= (30-1) x 1.1


=32 bar

= 3.19 N/mm2

§ Design Temperature:

Design temperature = 800oC (1073 K.)

§ Thickness of tube:

The minimum thickness required to resist internal pressure is :

primary reformer mechanical design

Primary Reformer Mechanical Design Assignment Help By Online Tutoring and Guided Sessions from AssignmentHelp.Net

e = the minimum thickness required.(mm)

Pi= Internal pressure. (kN/mm2)

Di=Internal diameter. (mm)

f= Design stress factor.

e = (120x3.19)/((2x115)-3.19)

e= 1.7 mm

We add 1mm for corrosion e= 1.7+1 =2.7 mm.

§ Heads and closures:

The 2 ends of each tube are designed to be closed by a head. There are three types of heads:

1. Hemispherical heads.

2. Ellipsoidal heads.

3. Torispherical heads.

Ø For a standard Torispherical head:

§ Crown radius Rc = Di = 0.12m.

§ Knuckle radius= 6% of Rk =7.2x10-3m.

§ A head of this size would be formed by pressing no sides, therefore j= 1.

Text Box: Cs = ¼ (3+ (Rc/Rk)1/2

Cs = Stress concentration factor for Torispherical heads.

Cs = ¼ (3 + ((0.12/7.2-3))-1/2)

Cs= 1.77

mechanical help

e = (3.2 x 120 x 1.77)/ ((2x115) + 3.2 (1.77-0.2))

Thickness of head e = 2.9 mm.

Ø For a standard ellipsoidal head:

online mechanical helpMost standard ellipsoidal heads are manufactured with a major and minor axis ratio of 2:1

e =(3.2x 120)/ ((2x115)- (0.2x3.2)

e = 1.7 mm =2.77mm (1 mm corrosion allowance)

Ellipsoidal head would probably be the most economical choice as it needs approximately half the thickness of a torisphrical head.

n Calculation of reinforcement required:

The presence of an opening weakens the shell and gives rise to stress concentration around the openings. To compensate the effect of an opening the thickness is increased in the area adjacent to the openings. There are two openings in each tube, one for the feed at the top and the other for the exiting product at the bottom (fig 4.13)

Mechanical Engineering Assignment Help

A welded pad will be used:

dh = 0.03 m

dr = 2 x dh

= 2x 0.03 = 0.06m

dh = diameter of hole (opening) m

dr = diameter of pad + opening adjacent surface.

( A factor of 1.5 to 2 is usually used in this equation)

n Design of loadings:

Primary Stress:

There are three types of stress due to the internal pressure.

  1. Horizontal (circumferential) stress:

Mechanical Engineering Homework Help

Stress h = (3.2x120) /(2x2.7)

= 71.11 N/mm2

P= Internal pressure. N/mm2

Di= Internal diameter.(mm)

t = shell (tube) thickness. (mm)

  1. Longitudinal stress:
Mechanical Assignment Help

Stress L = (3.2x120) / (4x2.7)

= 35.5 N/mm2

P= Internal pressure. N/mm2

Di= Internal diameter.(mm)

t = shell (tube) thickness. (mm)

  1. Compressive stress:

For stainless steel E= 200,000 N/mm2

Mechanical Homework Help

Stress C = 2x104 x(2.7/122.7)

= 440 N/mm2

Stress C= critical bulking stress

t = thickness

Do = outside diameter.

The maximum compressive stress should not exceed 440 N/mm2 in the vessel or the vessel may fail by elastic instability.

  1. Weight Loads:

The major sources of dead weight loads are:

Ø The vessel shell.

Ø The vessel filling ( e.g. nozzles).

Ø Internal fittings.

Ø External fittings.

Ø Insulations.

Wv = 240 x Cv x Dm (Hv + 0.8 Dm) t

Wv = Total weight of shell, excluding internal fittings.

Cv = a factor to account for the weight of nozzles, internal supports...etc.

( a factor of 1.08 is used with only few internal fittings)

Hv = length of cylindrical section.(m)

Dm= mean diameter of vessel ( Di + t) in (m).

t= shell thickness (mm).

Wv = 240 x 1.08 x 0.1277 (9 + 0.8 x 0.1227 ) x 2.7

= 813 N

  1. Earthquake loading:

Sites in the United Kingdom do not carry out seismic stress analysis in preliminary design of vessels. Since this mechanical design is for a plant in the U.K the seismic stress analysis will neglected. (page 837)

Fs = ae (w/g)

g=the acceleration due to gravity.

ae=the acceleration of the vessel due to earthquake.

W= the weight of the vessel.

  1. Dead weight of the vessel.

W= Wv ( = 781 kN) + catalyst weight

Catalyst used= Nickel

Density = 43.5kg.m3

Mass = 43.5x 0.407 = 17.7m3

Weight = mass x g

= 173.5 N

Assignment Help

Dead weight = 1044.5N

  1. Wind Loading:

Fw = wind pressure x Main diameter

Fw = 1280 N/mm2 x 0.12

Fw = 153.6 N/m

  1. Bending moment:

Mx =(Fw/Di) x L2

Mx = (153.6 / 0.12) x 92

Mx = 103680 N/m

§ Analysis Stress:

Ø Pressure Stress

Stress σ L = 35.5 N/mm2

Stress σ h = 71. N/mm2

Ø Dead Weight stress

Stress σ w = Wv /( 3.14 x (Di +t) x t

Stress w = 1044.5 x (3.14 x (120 + 2.7 ) 2.7

Stress w = 1.0 N/mm2 (compressive)

Ø Bending stress:

Do = 122.7 mm (outside diameter)

Iv = (3.14/ 64) x ( Do4 – Di4 )

Iv = the second moment of area of the vessel about the plane of bending.

Iv = (3.14 / 64) x ( 122.7x104 - 120x104)

Iv = 9.4x105

Stress σ b = +/- (M/Iv) x ((Di/2)+t)

Bending stress=

Bending stress=+/- (103680 /9.4x105) x ((120/2)+ 2.7)

Bending stress= 69.15 N/mm2


Ø The resultant longitudinal stress:

Stress σ z = stress l + stress w +/- stress b

(Stress σ w is compressive therefore –ve)

Stress σ z = 35.5 – 1 + 69.15 = 33.65 N/mm2

Stress σ z = 35.5 - 1 - 69.15 = -34.65 N/mm2

Text Box: 34.6
Homework Help
Online Mechnical Help
Mechnical Tutor
Text Box: 33.6
Text Box: 71.1
Text Box: 71.1 Text Box: 71.1 Text Box: 71.1
Text Box: Up wind
Text Box: Down wind

The great difference between the principal stresses will be on the down wind side:

(71.1 – (-34.6)) = 107.5 N/mm2

The vessel will stand the internal stress since the difference between the principal stresses is below the design stress.


§ type of support: Bracket support (Single plate)

Text Box: Fbs = 60L x tc

1) for the back row tubes :

Fbs = 60 x 150 x 20 =18x 104 N

2)For the front row tubes:

Fbs = 60 x 250 x20 = 30x104 N

L= length of bracket support (mm)

T = thickness of support (mm)


Chemical Engineering design, Coulson and Richardson, vol.6, 3rd Edition, section 13.

Vessel Data Sheet

Description (Furnace):Primary Reformer

Type of vessel

Reactor Tubes

No of tubes Required:



0.105 m3


Methane and Steam



Nickel catalyst


53.5 kg/m3






Max working Pressure



Design Pressure


Max working Temperature



Design Temperature



Estimation of Temp inducing creep




Stainless Steel 314

Joint Factor


Corrosion Allowance

1 mm


2.7 mm

End Type

Ellipsoidal head


2.9 mm

Joint Factor


End Type

Ellipsoidal head


2.9 mm

Joint Factor


Type of support



20 mm



Wind Load Design

6.65 kN/m



Dead Weight

1044 N

Weight of catalyst

170 N

Bending Stress

15.7 N/mm2

Dead weight Stress

0.42 N/mm2

Max design load of bracket support

372 k/N

Project title: Synthesis of Ammonia from Natural Gas

Company : Zeta Process Engineering

Designed By:

Course: Beng Chemical Engineering

Talal Omar

Project: Design Project


Dr S.Larkai

London South Bank University


Dr R.Best

Assignment Help | Mechanical Engineering Assignment Help | Mechanical Engineering Homework Help | Online Tutoring