# 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.

§ __Design Pressure:__

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

= (30-1) x 1.1

=31.9

=32 bar

= __3.19 N/mm ^{2}__

§ __Design Temperature:__

Design temperature = __800 ^{o}C (1073 K__.)

§ __Thickness of tube:__

The minimum thickness required to resist internal pressure is :

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

*e = the minimum thickness required.(mm)*

*P _{i}= Internal pressure. (kN/mm^{2})*

*D _{i}=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 R_{c} = D_{i} = 0.12m.

§ Knuckle radius= 6% of R_{k} =7.2x10^{-3}m.

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

*C _{s} = Stress concentration factor for Torispherical heads.*

C_{s} = ¼ (3 + ((0.12/7.2^{-3}))^{-1/2})

C_{s}= 1.77

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:

Most 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)

*A welded pad will be used:*

*d _{h}* = 0.03 m

*d _{r}* = 2 x d

_{h}

= 2x 0.03 = __0.06m__

*d _{h} = diameter of hole (opening) m*

*d _{r} = 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.

**Horizontal (circumferential) stress**:

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

= __71.11 N/mm ^{2}__

*P= Internal pressure. N/mm ^{2}*

*Di= Internal diameter.(mm)*

*t = shell (tube) thickness. (mm*)

**Longitudinal stress:**

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

= __35.5 N/mm ^{2}__

* P= Internal pressure. N/mm ^{2}*

*Di= Internal diameter.(mm)*

*t = shell (tube) thickness. (mm)*

**Compressive stress:**

For stainless steel E= 200,000 N/mm^{2}

Stress C = 2x10^{4} x(2.7/122.7)

= __440 N/mm ^{2}__

*Stress C= critical bulking stress*

* t = thickness*

*D _{o} = outside diameter.*

*The maximum compressive stress should not exceed 440 N/mm ^{2} in the vessel or the vessel may fail by elastic instability.*

**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__

**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 = a_{e} (w/g)

*g=the acceleration due to gravity.*

*a _{e}=the acceleration of the vessel due to earthquake.*

*W= the weight of the vessel.*

**Dead weight of the vessel**.

W= Wv ( = 781 kN) + catalyst weight

Catalyst used= Nickel

Density = 43.5kg.m3

Mass = 43.5x 0.407 = 17.7m^{3}

Weight = mass x g

= __173.5 N__

__Dead weight = 1044.5N__

- Wind Loading:

Fw = wind pressure x Main diameter

Fw = 1280 N/mm^{2} x 0.12

Fw __= 153.6 N/m__

- Bending moment:

M_{x} =(Fw/Di) x L^{2}

M_{x} = (153.6 / 0.12) x 9^{2}

M_{x} = __103680 N/m__

§ __Analysis Stress:__

Ø Pressure Stress

Stress σ L = __35.5 N/mm ^{2}__

Stress σ h = __71. N/mm ^{2}__

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/mm ^{2} (compressive)__

Ø Bending stress:

D_{o} = 122.7 mm (outside diameter)

Iv = (3.14/ 64) x ( D_{o}^{4} – D_{i}^{4} )

*I _{v} = the second moment of area of the vessel about the plane of bending.*

Iv = (3.14 / 64) x ( 122.7x10^{4} - 120x10^{4})

**Iv** = __9.4x10 ^{5}__

Stress σ b = +/- (M/Iv) x ((Di/2)+t)

Bending stress=

Bending stress=+/- (103680 /9.4x10^{5}) x ((120/2)+ 2.7)

** Bending stress**=

**69.15 N/mm**^{2}σ

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/mm^{2}

Stress σ z = 35.5 - 1 - 69.15 = -34.65 N/mm^{2}

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

(71.1 – (-34.6)) = 107.5 N/mm^{2}

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

__Support:__

§ *type of support: Bracket support (Single plate)*

1) for the back row tubes :

F_{bs} = 60 x 150 x 20 =18x 10^{4} N

2)For the front row tubes:

F_{bs} = 60 x 250 x20 = 30x10^{4} N

L= length of bracket support (mm)

T = thickness of support (mm)

__Reference:__

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

**Vessel Data Sheet**

**Description (Furnace):Primary Reformer**

**Type of vessel**

Reactor Tubes

**No of tubes Required:**

614

**Capacity**

0.105 m^{3}

**Contents**

Methane and Steam

Premixed

**Catalyst**

Nickel catalyst

**Density**

53.5 kg/m^{3}

**Diameter**

0.12

m

**Length**

m

**Max working Pressure**

38

bar

**Design Pressure**

bar

**Max working Temperature**

1200

K

**Design Temperature**

1073

K

**Estimation of Temp inducing creep**

1315

^{o}C

**Materials**

Stainless Steel 314

**Joint Factor**

0.70

**Corrosion Allowance**

1 mm

**Thickness**

2.7 mm

**End Type**

Ellipsoidal head

**Thickness**

2.9 mm

**Joint Factor**

0.70

**End Type**

Ellipsoidal head

**Thickness**

2.9 mm

**Joint Factor**

0.70

**Type of support**

Brackets

**Thickness**

20 mm

**Material**

Steel

**Wind Load Design**

6.65 kN/m

**Radiography**

None

**Dead Weight**

1044 N

**Weight of catalyst**

170 N

**Bending Stress**

15.7 N/mm^{2}

**Dead weight Stress**

0.42 N/mm^{2}

**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**

**Supervisor**

**Dr S.Larkai**

**London** **South Bank University**

**Approved:**

**Dr R.Best**

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