Equipment Design: Mechanical Aspects

Prof. Shabina Khanam

Department of Chemical Engineering
Indian Institute of Technology – Roorkee

Lecture 08
Design of Heads

Welcome to the third lecture of week 2 and here we are discussing design of heads. In lecture 1
and lecture 2, we have covered the theory part related to this and in this lecture we will discuss a
few examples to make you understand how design of different types of heads will be carried out.
So, let’s start with example 1.
(Refer Slide Time: 00:48)

In example 1, a process vessel is to be operated at maximum operating pressure of 2 MN/m2

(gauge). Vessel has diameter of 3 m and length of 5 m. This vessel is made of steel having
allowable design stress value of 120 MN/m2 at design temperature, corrosion allowance is 3 mm
and joint efficiency factor as 1.

And in this problem, we need to determine the thickness of shell to nearest integer, so this we have
to consider nearest integer and second part we need to determine thickness of flat head having
plates welded to the end of the shell. No inside welding. So, in this A and B we have to design
the shell as well as flat head and in part C, we have to determine the thickness of conical head
when apex angle is 120oC.
(Refer Slide Time: 01:45)

No knuckle part is there and we have to find out thickness both at and away from the junction. In
the similar line, part 2 is there we apex angle is 100 and here we have ψ value which is different
than the α and here I am having the knuckle part also. So, considering these variations, we will
demonstrate how the thickness of conical head will be carried out. Further, in part D, we need to
determine the thickness of plate required to fabricate a dished head for the vessel.

Specifications of the head are Ri is 1.1Do, ri is 0.07Do and Sf 70 mm, and finally we have to
determine the blank diameter of plate for forming the dished heads. So, in all these 5 sections, we
need to determine different types of heads, so let’s start with part A where we need to find out
thickness of shell.
(Refer Slide Time: 02:57)
Now, to find out thickness of shell, we have to calculate the design pressure. Maximum working
pressure in gauge is given as 2 MN/m2, design pressure would be 5% extra to that.

Allowable stress value, J value, corrosion allowance and outer diameter, all these values are
𝑜𝑃𝐷
known, and here this is the expression for minimum thickness of shell, 2𝑓𝐽+𝑃 , so putting all these

values in this, we can find out 26.022 mm minimum thickness of shell. We will add corrosion
allowance into this to get the thickness as 29.022 and as it is mentioned in part A. that we need to
find out thickness for nearest integer, we can consider 30 mm as next integer value to this. So, in
this way we compute thickness of shell as we have already demonstrated in fifth lecture of week
1.

Further, in part B, we have to find out thickness of flat head, so let’s discuss that. So, this is the
𝑝
expression for calculation of thickness of flat head where C, D, E and √𝑓, all values we need to

extract first. p and f we already know. De and C will depend on type of attachment and type of
attachment is flat head having plates welded to the end of the shell, no inside welding. So, this
type of connection is there where D = Di, that is inner diameter of shell and C in this case is 0.7.

So, De how I can find out, that is 3 m is the outer diameter of shell minus 2 x 30 mm, 30 mm is the
standard thickness of shell. So, considering these two, 2.94 is the value of De. C value, as I have
discussed, it is 0.7, so tmin we can find out as 272.248 mm and further we will not add any corrosion
allowance at tmin is greater than 30 mm. So, if you remember the terminology lecture, that is lecture
4 of week 1, we have discussed that when thickness is greater than 30 mm, we will not add
corrosion allowance into this. Therefore, in this case also, it will not be added.

Standard value will be next integer value to this value. So, final thickness value for flat head is
273 mm, which is quite high.
(Refer Slide Time: 05:50)

Now part C I am having where 120o apex angle with sharp edge that is no knuckle part is there
and we have to find out thickness at and away from the junction, so at the junction, apex angle is
120, so α in that case would be 60. This is the expression to find out the thickness where P is the
design pressure, De is the outer diameter of head, but here we can consider for shell because it will
be almost equal, so we can assume the equality between the two.

C we need to find out through table 3.6. For C value, I have to find value of SI and SI would be
equal to alpha which is 60 in this case, okay. Now, if you focus on this table, this is table 3.6
where SI is given as 60o, okay. As no knuckle part is there, we have discussed in lecture 2 that if
no knuckle part is there, then ri/De should be equal to 0.01. So, corresponding to 60o, we can find
out C value as 3.2 as already mentioned over here.
So, considering these values, we can find out thickness of conical section at the junction and as it
is greater than 30 mm, we will not add corrosion allowance into this. So, final thickness we can
consider as 84 mm. So, in this way we find out thickness at the junction when no knuckle part is
there. Let us continue for the thickness away from the junction.
(Refer Slide Time: 07:34)

So, thickness away from the junction that can be found out L distance away, so in this case L
𝐷 𝑡
would be 0.5√cos𝑒 Ψ. So, De is the outer diameter of shell as we have discussed that would be 3 m

and t is the actual thickness at the junction and here we have computed the thickness as 84 mm.
3∗0.085
So, in this case 0.5√ 𝐶𝑜𝑠60 .

So, considering all these values we can find out L as 0.354965 m. Putting this value in expression,
we can find out value of Dk and that would be 2.325183 m and here in this Dk expression, this t s
will be nothing but the standard thickness of shell, which we have found as 30 mm. So that value
we have kept over here. Considering value of Dk, P f J and α, we can find out minimum thickness
of conical section away from the junction and that comes out as 41.049 mm, again it is greater than
30 mm, so we will not add corrosion allowance.

So, final t in this case will be 45, so here we can consider the thickness of conical section at the
junction and away from the junction. And in next part, we will discuss thickness of conical section
at the junction and away from the junction when knuckle part will be there. In first part, we have
not considered any knuckle part.
(Refer Slide Time: 09:18)

So, in this part, 100o apex angle is given where ri is 0.5De, when knuckle part will not be there, it
is 0.01 * De as we have considered in last slide and ψ is equal to 25. So, in this case if you consider
α would be 50 and ψ would be 25, so these both angles are different. So, 2α is 100, α is 50, then
ri is 0.5De and ψ 25. So, for this combination, I have to find out the value of C.
(Refer Slide Time: 10:00)
 𝑟𝑖
Further, if you focus on table 3.6, is given as 0.05, which is falling between 0.04 and 0.06,
𝐷𝑒

okay, and ψ will be given as 25, which is falling between 20 and 30. So, at the junction, we need
𝑟
to find out value of C for a particular combination that is 𝐷𝑖 that is 0.05 and ψ 25.
𝑒

𝑟𝑖
So, for that case, we will consider as 0.04 and value of C at 20o ψ is 0.8, and further, if I am
𝐷𝑒
𝑟
having 𝐷𝑖 0.04 and ψ value is 30, so C in that case would be 1. So, what I am finding is value of
𝑒
𝑟
C for 0.04 and 0.06 at Si equal to 25. So, 𝐷𝑖 0.04 if I am considering and ψ I am taking as 25, so
𝑒

value will lie in between these 2, so that value we can consider as 0.9. Otherwise, if ψ value is 28
o
, in that case, C value can be obtained by interpolation. In this case, it is directly in between or
middle value of these 2, because ψ is lying in between 20 and 30.

So, ψ is 25, so directly we can take value of C at middle point of 0.8 and 1 and then this value of
𝑟
C becomes 0.9, okay. In the similar line, I need to find out C value for 𝐷𝑖 0.6 and at 20oC, so that
𝑒
𝑟𝑖
value comes at 0.7 and again for 30 oC, I need to find value of 𝐷 at 0.06, so at 0.06 at 20o ψ value
𝑒

of C comes as 0.7, and at 30 , we have value of C as 0.9. So, at 0.06 and 25o, value of C I can
o

consider as 0.8. So, what I have found, value of C here and value of C here.

Now, we have to find value of C at 0.05, so that value will be middle point of these 2 values. So,
𝑟
considering that, 𝐷𝑖 0.05 and ψ 25, we can find C value as 0.85, which is in between this and this.
𝑒

So, in this way we find the value of C and then putting that value of C in this expression, we can
compute as t as 22.3125 mm, further because it is less than 30 mm, we had to add corrosion
allowance. So, here I am getting the value after adding corrosion allowance as 25.3125 and further
we have this value t – that is 26.83 mm.

How this 26.83 mm comes? Because I have to add 6% in this value, okay? If you remember the
lecture 2, there we have discussed that wherever I am having forming section, I need to add 6%
extra, okay. So, that because it is with knuckle at the junction, here I need to add 6% extra, so 1.06
x 25.3125 gives this value that is 26.83 mm and then we can take standard value than this from
table B1. So, in this way we calculate thickness at the junction, now for away from the junction,
we will follow the usual method, which we have observed in previous part.
(Refer Slide Time: 14:21)

So, away from the junction, we will find out L value, we will find out Dk, and putting all these
values in the expression of minimum thickness and then it comes as 37.1715 mm and then we can
see standard thickness from table B1 which comes out as 40 mm. Here, we have not added any
corrosion allowance, not 6% extra because away from the junction all parts are straight parts, so
there is no need to add 6% extra. So, in this way, we calculate thickness at the junction and away
from the junction in conical head when knuckle part is there and when knuckle part is not there.
(Refer Slide Time: 15:09)
Now, next part of the problem is, to determine the thickness of plate required to fabricate dished
head for the vessel and specifications are Ri is given as 1.1 Do, ri 0.07 Do and Sf as 70 mm. So,
considering this we have Ri and ri, Sf value 70 mm I am having and this is the expression to find
out thickness of dished head where main concern is about factor C, okay. Now, if you remember,
ℎ𝐸
how we can find out factor C? For this, first of all we need to find out value of .
𝐷𝑜

Do I already now, to calculate he, we have 3 different expressions and whatever would be the
minimum value among that, that we will select as he. So, to compute 3 values of he, we assume
that ro is equal to ri which is 3.3 m, because Ri is 1.1 Do, so that should be 3.3 m, and similarly we
will assume ro equal to ri which is equal to 0.21 m. Now, why I am assuming ri is equal to ro
because thickness of head should be added to ri to calculate ro, but at present I am not having
thickness of head, and therefore for computation purpose, I am taking that ri and ro both are equal
and further we will resolve that.

Considering this, we can find out ho based on this expression, which comes out as 0.49215 and
𝐷𝑜2 𝐷𝑜 𝑟𝑜
we can compute as 0.681818 and similarly √ 2
is 0.56125. In that way, we can find out all
4𝑅𝑜

3 values and hE is minimum of above three and that should be equal to ho, so here it should be hE.
So, hE should be equal to minimum of above three and which we can find for ho and which comes
out as 0.4921538, so in this way we can find out value of hE.
(Refer Slide Time: 17:51)
 ℎ𝐸
Once I am having value of hE, I can find out as 0.164 and I am assuming opening as
𝐷𝑜
𝑡
compensated, okay. So, in that case I can use values of , so this is the whole expression and if
𝐷𝑜
𝑡
I am having , I can having P value, f value, J value, all these values are constant and which gives
𝐷𝑜
𝑡 ℎ𝐸
collectively 8.75 times power minus 3 x C. So, I need to select and see corresponding to
𝐷𝑜 𝐷𝑜

0.164 in such a way so that this expression should be satisfied and the corresponding value of C I
am getting as 1.85, that I have found through interpolation, and how I can find this, let’s discuss
that.
(Refer Slide Time: 18:48)
 ℎ𝐸
To find out that value, I will consider table 4.1A from B. C. Bhattacharya book and here is
𝐷𝑜
ℎ𝐸
0.164. If you consider this table, here I am having 0.15 and 0.2 and t/Do have different values
𝐷𝑜

which varies from 0.002 to 0.04, okay. So, corresponding to 0.164 value, I have to find C factor
𝑡
for all t/Do values, okay. So, for example, 𝐷 is 0.002, corresponding to 0.15 value comes as 4.55
𝑜

and corresponding to 2, value comes as 2.3.

So, corresponding to 0.164, value will lie in between these 2 and that value we can find through
interpolation and that comes out as 3.92 and similarly for 0.005 value, we can find C value in
between these two, which comes out as 2.3912 and in the same line we can find value of C
corresponding to 0.164 for 0.01, 0.02 and 0.04 and these values are summarized or mentioned over
ℎ𝐸
here. Now, to find out value of C for , I will consider all these values, okay. Now, how I need
𝐷𝑜
𝑡
to compute this, if I am having 𝐷 as 0.00875 that we have already found in the last slide and to
𝑜𝐶

𝑡
start computation, I am starting with 𝐷 as 0.005, okay?
𝑜

𝑡
So, corresponding to 0.005, C value comes as 2.3912, which is mentioned over here.
𝐷𝑜
𝑡
Considering this value, we can find out value of this expression 𝐷 as 0.00209. Now, here you
𝑜𝐶

need to see that what I have to find was 0.00875 and I am getting very less value than this. It
𝑡
means what? It means that if you consider the expression , C is coming in denominator. So,
𝐷𝑜 𝐶
𝑡
expression comes as 0.00209 and I want to find the value equal to 0.00875. What is the
𝐷𝑜 𝐶

meaning of that? This expression has to increase its value from 002 to 0.00875, okay? How can
I increase the value, to reduce the value of C, okay. So, further in which direction I have to move?
𝑡
That we can decide while comparing value of .
𝐷𝑜 𝐶

So, what we can see that, now I have to deal with lesser value of C and if you consider this table,
𝑡
what I can observe that, as I move towards higher value of , I am getting lesser value of C. It
𝐷𝑜

means I have started from here, so correct direction would be this, not this, okay. So, next value
 𝑡
of I am considering as 0.01 and corresponding to this value of C, I am getting as 1.954 and
𝐷𝑜
𝑡
putting this value in the expression, I am getting 𝐷 as 0.005, which is coming close to this, but
𝑜𝐶

still it is very far. So, further I need to reduce the value of C and therefore I need to increase the
𝑡
value of .
𝐷𝑜
𝑡
So, further I am considering as 0.02, corresponding value of C is 1.7876 and corresponding
𝐷𝑜
𝑡 𝑡
value of 𝐷 is 0.01118, okay. So what is happening over here, that now for 𝐷 , whatever value I
𝑜 𝐶 𝑜

𝑡
am getting for 𝐷 𝐶, it is increasing than whatever I was expected to get. It means further I need to
𝑜

𝑡
increase the value of C. So, what is the main point over here, that when I am having value of
𝐷𝑜
𝑡
as 0.01, I am getting lesser value of 𝐷 𝐶, but if I am having t/Do as 0.02, I am getting higher value
𝑜

𝑡 𝑡
of 𝐷 𝐶. So, exact value of C will lie between 𝐷 0.01 to 0.02. So, next value I am considering as
𝑜 𝑜

0.015 and C value I am getting 1.8708 through interpolation and then

𝑡 𝑡
I am getting as 0.008, and further it has to increase, so I am considering as 0.016,
𝐷𝑜 𝐶 𝐷𝑜
𝑡
corresponding value of C 1.85416 through interpolation and further value of is coming as
𝐷𝑜 𝐶
𝑡
0.0086, which is increased, so further I need to take another value of , which is I am taking as
𝐷𝑜
𝑡
0.0162 and finally I can get 𝐷 as 0.00875 at C value 1.85.
𝑜𝐶

(Refer Slide Time: 24:57)

So, this 1.85 I have taken over here and then corresponding value of t I can find as 0.04856, which
is 48.56 mm. I have to add 6% extra to this and the value comes as 51.476 and next standard value
coming as 56 mm. So, in this way, I can find thickness of dished heads.
(Refer Slide Time: 25:25)

ℎ𝐸
Further, you can find out value of C from this graph also which is figure 3.7 in code. Here, 𝐷𝑜
𝑡
0.164, so you can draw a line of 0.164 and then you need to choose different value of 𝐷 and C in
𝑜

𝑡
such a way so that it satisfies the expression of 𝐷 as we have explained through table 4.1.
𝑜𝐶

(Refer Slide Time: 25:54)

Now, here, further you need to consider that when we have started the computation of dished
heads, we have assumed that Ro should be equal to Ri , and while assuming this, I have considered
t should be 0. Then only it is possible. So, when t equal to 0, Ri should be Ro, okay, and next
value of t I am getting as 56. So, if difference of these two consecutive values will be greater than
10 mm, I have to repeat the whole calculation.

So, in this case, repetition is required, so for that purpose Ro is considered as Ri + ts that is 56 will
be added over here and the value comes as 3.356 and similarly r o we can find out. So, for revised
value of Ro and ro, ho value, this expression value and this expression values are found out and
ℎ ℎ
then 𝐷𝐸 we can consider as 0.174. Now, 𝐷𝐸 corresponding to 0.174, I have to find out value of C,
𝑜 𝑜

which comes out as 1.7 based on the method which I have explained in last slide.

So corresponding to C 1.7, you can find out tmin 44.625, 6% will be added to this and then final
value would be 50. So, previous thickness was 56, now thickness is 50, so that difference is less
than 10 mm, hence no extra calculation will be required, so final thickness of dished head can be
considered as 50 mm.
(Refer Slide Time: 27:53)
Now, here I am having the final part of this example where blank diameter needs to be determined.
Since thickness is greater than 25 mm, we will use this expression and blank diameter we can find
as 3.4014 m.
(Refer Slide Time: 28:07)

Here, I am having example 2. In this example, a vessel is being operated at a maximum pressure
of 2 MN/m2, outer diameter 2.5 m and length is 6 m we have. Allowable stress and corrosion
allowance are given like this and it is made of class 2 where double welded butt joint with full
penetration is used for all joints. What we need to find is, thickness of plate used to fabricate the
torispherical head for this vessel and here we have fully compensated openings.
So, compensated openings are provided and we need to consider Ri as 1.2 Do and ri as this and I
have to carry out only 1 iteration, okay, that is part 1. In part B, what is the maximum diameter of
uncompensated opening can be made if 2:1 ellipsoidal head of same thickness as computed in part
A is used and finally we need to find out flat formed head thickness where this connection is made.
(Refer Slide Time: 29:21)

So, part 1 is calculation of torispherical heads, so it is computed based on same method as we have
explained in the last example. So, for that purpose, we need to calculate Ri and ri and then hE we
can find out using the method explained in previous example, which comes out as 0.15.
(Refer Slide Time: 29:48)
 ℎ𝐸
Following the table 4.1A, we can calculate value of C corresponding to as 0.15 and which
𝐷𝑜

comes as 2.016 and corresponding to this we can calculate thickness as 0.4156, 6% extra I have
taken over here and then we can take that standard from table B1. So, in this way we calculate the
thickness of torispherical heads.
(Refer Slide Time: 30:19)

Now, main point over here is part B where I have to find out the maximum diameter of the
ℎ𝐸
uncompensated opening for 2:1 ellipsoidal head where is 0.25, okay, of same thickness as
𝐷𝑜

computed in part A. So, this is the repression for ellipsoidal head, the same we have used for
torispherical head and thickness we have found out as 45 standard, so here corresponding
minimum thickness I have to put, so once this t will be same, it means C will also be same.

ℎ𝐸
So, C I am taking as 2.016 and comes as 0.25. How it comes as 0.25? Because if you see
𝐷𝑜

torispherical head, okay, this is 2 and this is 1 ratio. So, if am considering this particular section,
ℎ𝐸
that should be 0.5 divided by 2, so it is 0.25, so corresponding to 0.25 and C value, I have to
𝐷𝑜

find value of this expression.

This is table 4.1E. Here it is 0.25 and value is lying between 1.6 and 2.015. If you see value 2.016
is lying between 2. So while interpolating value of C, we can find out value of this expression
between this, which comes out as 2.9244 and here D would be uncompensated opening diameter,
which comes out as 0.94. So, in this way we calculate the maximum diameter of uncompensated
opening.
(Refer Slide Time: 32:10)

Part C is very simple that is flat formed head where C for this type of connection is 0.45, D is equal
to Di, how I can find this? it is Do that is 2.5 minus 2ts. ts we have found as 25 mm. Considering
this value, we can find out t as 130.45 mm and then I have to add 6% extra in this because it is
formed section, so final value comes as 139 mm. So, that is for example 2.
(Refer Slide Time: 32:50)

And here I am having some of the books which you can refer to study this topic.
(Refer Slide Time: 32:56)
And now we have the summary of video, and here I am combing summary of lecture 1, lecture 2
and lecture 3 as different types of heads used for pressure vessel are discussed. Selection of these
heads is discussed and design equation along with detailed method to design these heads are
described and to design different heads, a few worked examples are illustrated with detail solution.
And that is all for now, thank you.