11 replies to this topic

[rishabh29288]

Q.1) A pressure vessel is to be designed for some chemical process.Total volume of the vessel required is 10 m^3.The vessel operates at a pressure of 4 kg/cm^2 and material used for fabrication have an allowable stress of 1000 kg/cm^2.Welded joint efficiency is 85% and corrosion allowance is 3mm. Weight of vessel with its content is 7500 kg.The torque exerted over the vessel is 75 kgcm.Bending moment induced is negligible.The vessel is closed by two elliptical heads of major to minor axis ratio 2 :1. Estimate-
a)Optimum proportion of vessel
b)Minimum thickness required for the vessel.

Q.2) For a pressure vessel for same operating condition and same diameter, it is desired to choose either a hemisherical head or elliptical head. Discuss which one is better and why?

[Art Montemayor]

rishabh29288:

If you are a student at a university and this is a homework problem assigned to you, it is logical that this assignment is a result of you attending at least one or several lectures involving the fabrication of pressure vessels. This is an easy problem, related to the mechanical equations on the fabrication of pressure vessels, their wall thickness and othe criteria. You should follow your lecture notes (if you went, and stayed awake). If you don't have any lectures on this subject, tell me and I will give you a reference book where you will find all the relationships on how to solve this problem. (Hint: a sphere is the strongest, natural geometric object to withstand both external and internal pressures)

Now, as a favor to me, please tell me what university you attend and whether this course is a chemical engineering course or a required course, outside of the chemical engineering department. Also tell me what year of study you are in. I am curious to know.

[rishabh29288]

no, actually this question came in my final semester exams and there has been a lot of argument about the question....the first one.I know the answer of the second one....just wanted to have a discussion about it.Can you please help with the first question as it was of 12 marks and lot of it is at stake whether i get it right or wrong.Though the question might look easy on the face value, there are loop holes which i shall tell u once u tell how to go about it.
well, i am a final year student pursuing bachelors in chemical engineering at school of chemical tech, IP university main campus,delhi

[rishabh29288]

anyone has any answer to the above?plzz try and help if possible....

[pavanayi]

Rishabh,


Don't expect anyone to solve any sums for you. Plain and simple.
If you want to know/calculate how you might be scored in the subject, the best approach is to talk to your professor about the way you have answered it.
If you are trying to solve a chemical engineering question, and are stuck somewhere, there might be someone here who can guide you in the right way. But I don't think anyone would spend their time in solving your question and give back to you on a platter.

If you have attempted the question, as you claim you have, upload and describe how you have solved it, and also explaining your concerns about the same. Then expect someone to reply.

I hope this guidance helps you now and also in your future professional life.

[rishabh29288]

ok i will first explain how i went about it and then kindly let me know if i am correct or not.

i used the shear strain energy criterion for design of tall vessels...
the equivalent stress will have to be less than - f*J*1.3

where f=pressure of 1000kg/cm^2 and J=0.85

resultant longitudinal stress would comprise only of the stress by pressure i.e pD^2/4t(D + t)

D + t- total width of shell, p is the operating pressure of 4kg/cm^2

since the bending moments are negligible, hence no stress due to dead load, wind or seismic forces etc....
and then i calculated the hoop stress p(D +t)/2t

and the shearing stress 2T/(pi)t(D+t)D
T is the torque......
equated this in sheri strain energy criterion........and got the diameter
now for a 2:1 ellipsoidal head.........the ht. of the head will be reated to diameter of vessel by

H=D/4............and then summing up got the total ht of the vessel

now plz.....can we have a discussion abt this?
a

[Art Montemayor]

rishabh29288:

First you present a problem with 2 questions. Later, you state that you know the answer to the second question. Then, why ask it??

You also followed up and say that you “just wanted to have a discussion about it”. OK, lets just discuss it. But why do you now change the story again, and ask: “plzz try and help if possible”.

I think you are either confused or don’t know what you seek. We certainly don’t know now whether you have a question or just want a discussion.

I can discuss the following about your so-called quiz problem:

As any third-year mechanical engineer can tell you, a cylindrical pressure vessel’s wall is determined by:

tw = DP x R / [(JE x SA) + 0.4 x DP)]

where,

tw = wall thickness, inches
DP = design pressure, psig
R = vessel’s cylindrical radius, inches
JE = joint efficiency, fraction
SA = Allowable stress of the material, psi

The thickness of a 2:1 ellipsoidal head is:

tw = PLM / (2SE – 0.2P)

where,

tw = wall thickness, inches
P = design pressure, psig
L = inside crown radius, inches
M = value from ASME table VIII, App 1-4.1
S = allowable stress, psi
E = joint efficiency, fraction

With the above equations and the equation for the volume of a cylinder, you can find the volume of the vessel – except for the volume of the ellipsoid. That equation you have to obtain as well. This is assuming that you can come up with the diameter and the tangential length of the vessel. More on this later.

Knowing the thickness of the cylinder you can safely assume that the ellipsoidal heads will have the same thickness and you can calculate the weight of the cylinder – but you will need a miracle to come up with the method to calculate the weight of the ellipsoidal heads.

NOW, let me tell you that you can’t determine the “optimum” size of the vessel. The reason is that the term “optimum” is based on economics – something that goes well beyond this simply-stated problem. Whoever wrote this problem is living in a dream world or a fantasy if he/she believes that you can come up with an “optimum proportion of vessel”. You need much more information than what has been given – and a lot of time.

Is that what you wanted to “discuss” – or for me to tell you how to respond to such a problem? Academic problems are fine. They are supposed to teach students a lesson. This problem goes beyond what a university can furnish as basic data – the economics involved in fabricating a real world pressure vessel.

Now that I have told you how to go about it, I don’t need for you to tell me where the “loop holes” are in this simple problem. What I need is for you to tell us all why you – and your university – are involved in the study of fabricating pressure vessels in the school of Chemical Engineering without having any strength of materials courses or mechanical engineering courses. Are you supposed to be studying Chemical Engineering or how to fabricate pressure vessels? Studying and knowing about pressure vessel design (and fabrication) is something that I wish all universities should teach Chemical Engineers. I have been advocating this on our Forums for years. However, before taking that subject, you should also take courses in strength of materials and other pre-requisites before discussing “optimum” vessel sizes. That's the point I want to make here.

[rishabh29288]

sir, point or rather points well taken....but what hapens to the torque?how do we use that here?and i really couldn't get an idea what kind of torque?handling torque on the vessel i suppose....but its confusing, there can be a torque due to the eccentricity of the vessel or its discontinuities....whats ur take about this?
i know the formulas u just stated here sir,can u plzz say where my method in two posts above is wrong?

i guess i am too small an entity to question the way education system is being run here,papers being set in this country etc etc....but trust me,our college has one of the better situations in India.As a part of our course we are made to go throu an exam which has all the questions like this which is surprising u.you wont appreciate the other 7 questions tht came in tht exam of mine if thts the case coz its all along the same lines.but i cant change all this and have to just study harder and try and do better in whatever education system i have.
thanx sir.

[Art Montemayor]

You still have not told us the total background regarding this problem appearing on your final exam. WHAT final exam? Is this exam related to a pressure vessel design course? What was the nature of the exam? Surely, you were versed or had an opportunity to study the related material for this exam. Please give us all the details if you would like to "discuss". Otherwise, if you retain the facts and details to yourself, there can be no discussion - only a preaching or a lecture coming from you. Kindly answer our questions, as we have answered yours.

I don't know what the term "torque" in the problem is related to. What I personally know from field experience is that you don't require any "torque" values to design the wall thickness of a pressure vessel - UNLESS the vessel is subject to forces applied in the field that result in a torque. But what kind of torque? --and in what position? I can give that information no importance unless it is fully defined and explained. Without a meaniful explanation, I can only give it the same importance as to what color the vessel will be painted in - no importance.

I can't tell you where your "method in two posts above is wrong" because you have not given us any method on how you answered this problem. In fact, you haven't given us very much information except a verbatim copy of the problem as it was given to you -- am I correct? I am the only one who has quoted the actual equations on how a pressure vessel is calculated for wall thickness - and ultimately for weight. Am I not correct?

[rishabh29288]

no sir.....the subject was "process eqipment designing' and we were not having any access to any material or tables or anything except the data i giave u in the "verbatim" copy of the question. The course had topics on design preliminaries, designing of thick vessels, thin vessels, design of heads,flanges,gaskets,blinds etc etc....hope u got a general idea........but no tables or ASME data or anything but just the question, the mind and the scientific calculator.
sir....my underlying anxiety in asking this question is becase it carried 12 marks and i believe it will be the difference between my higher end marks or the lower end marks....a make or break ques.
well if torque has no importance why go in depth about shear strain energy criterion while designing of pressure vessel.....this is getting all the more baffling!!

i dnt know about the "optimum" proportion being related to economics but will tell u how i went abt it....

i wrote the volume of the vessel in terms of the volume of the shell and the head........the differentiated it with respect to the height using the criteria that for an ellipsoidal head, the height will be one fourth the diameter of the vessel
i got the minimum thickness.........then found an expression by putting this value and obtaining a general expression for thickness.
then using the method i wrote to calculate from which i get the other dimension of diameter, i put it here to calculate the "optimum thickness"

any take?

[Art Montemayor]

The English word "Optimum" means the best, most desirable answer. The best and most desirable is the vessel that will do the job for the least cost. The least cost (assuming simplicity and ideality) is the vessel that requires less steel - which involves the wall thickness and the length of the vessel that are the variables in the equation for the volume of the vessel - a given. Again, this is assuming additional simplicity and ideality because the ellipsoidal heads are not subject to the cylindrical realationships. But, you could assume that the ultimate volume and weight of the heads will be a minimal factor.

As the diameter is reduced, the wall thickness is also reduced; but the cylindrical tangential length is also increased. The optimum lies where these two factors "cross" each other. In other words, there is an ideal minimum reached if you plot the weight of the cylinder versus the diameter (holding the total volume constant. If you differentiate such a relationship and equate the result to zero, it should yield the lowest point in the curve - the "optimum".

[kkala]

I quite agree with Art Montemayor that Strength of Materials and some "Mechanical Design of Equipment" should be taught in Chemical Engineering Faculties, as it happened in the past. This is particularly true for not industrialized countries, where specialization may be useless, while ability to understand (and if possible do) numerous simpler tasks is useful. Coulson and Richardson's "Chemical Engineering, Vol 6, Chapter of "Mechanical Design of Process Equipment" seems (among others) a good introduction to the subject. Without being an expert, following comments (subject to criticism) may be helpful.
1. As pointed out by Art Montemayor in the last post, "optimum" was probably considered by the instructor as minimum steel weight necessary for the vessel; this simplified view should have been clarified in writing. Same "optimization" of steel in Alumina digestion autoclaves was used by Russian Engineers in a feasibility study of 1982.
2. Calculations should be based on design pressure, being about 4+2=6 kgf/cm2 g.
3. Torque of 75kgf*cm=0.75kgf*m (trick?) is quite a small value, so it can be neglected.
4. If D=x m and cylider H=y m, vessel volume is π*x2*y/4, thus π*x2*y/4=10 or y=40/(π*x2) (neglecting head volumes).
5. Necessary wall thickness is 6*x*1000/(2*0.85*1000) + 3 = 3.53x+3 mm
6. Vessel surface is π*x*y+0.5π*x2 (considering head surfaces as flat)
7. Steel weight of vessel is (π*x*y+0.5π*x2)*8*(3.53x+3)kg.
8. Considering (4) above, steel weight gets the form W(x)=(40/x+0.5πx2)*(28.2x+24) kg. This gets minimum W~1340, when x~2.33 and y~2.35, as a plot in excel can indicate (differentiation can be another method). That is "optimum" tank would have D ~ H ~ 2.33 m (~ means approx equal).
9. Consideration of elipsoid head volumes and surfaces (equal thickness as cylinder) would give more precise results (at the expense of more labor).