Im designing a couple of tanks with a cone bottom, I'm using API 620 for this. Im not sure If I got my numbers right. Can you help me?
This is the information for the tank
Wshell = 3345 lb
Wimpeller = 486 lb
Wliqshell = 52,712 lb
Wtop = 638 lb
Wcone = 445 lb
P at 20ft = 8.66PSI
Density = 1
Shell height = 20ft
Shell diameter = 44in
Cone height = 10"
Cone diameter = 44in
Cone angle = 12.8 degrees
It has a skirt right where the cone touches the cylindrical shell, the skirt is supporting the cone.
I used Structural Analysis and Design of Process Equipment 3rd. Edition to do the calculations.
If I look at the point where the cylinder touches the cone above the cone I get these numbers
Sum Fy = 0
Wt + V - P = 0
57,181 + 88*pi*V - 8.66*pi*R^2 = 0
V = -16.31 lb/in, using V to get T1, i get that T1 = V = -16.31, the shell is under compression
T2 = PR = 8.66 * 44
T2 = 381 lb/in
Im using a t = 3/16"
My actual compressive stress = 16.31/0.1875 = 86.98 PSI
Actual tensile stress = 381/0.1875 = 2032 PSI
Using 1.8x10^6 (t/R) to get allowable compressive stress, i get Sca = 7,670 PSI
M = 86.98/7670 = 0.01
N = 2032/16000 = 0.12
N^2 + N*M + M^2 <1
0.0157 < 1 so using a 3/16" for the course (cylinder) should be ok. correct?
Now, if I look below the line under consideration, this is where I get confused, I don't know how to treat the skirt
these are my calculations
Since the V calculated before is connecting to the cone, V should be the same for the cone area due to continuity, correct?
if that is the case V = 16.31 lb/in
My T1 = V/sin12.8
T1 = 74.73 lb/in
T2 = PR/sin12.8
T2 = (6.66)(44)/0.22
T2 = 1,732 lb/in
Both forces are in tension.
t = T1/SE , t = T2/SE
T2 > T1 so I'm using T2 to calculate my thickness.
t = 1,732/16000 (0.85) = 0.127", I'm using a 1/4" material for the cone, so i should be ok, correct?
If I move my line halfway into the cone (6" down) and consider what's above my line.
I get the W of the initial calculations + Whalf cone + Win the cone
That gives me a new Wt = 57,826 lb
If I do my FBD I get the following
Sum Fy = 0
Wt + V - P = 0
Wt + 44*pi*V - 8.84*pi*R^2
V = -321.09 lb/in
using V to calculate T1 i get:
T1 = V/sin12.8
T1 = -1,459.5 lb/in
T2 = 855.7 lb/in
T1 is under compression and T2 is under tension.
I do the same procedure and i get the following
t = 1/4" for the cone
Sc = 1459.5/0.25 = 5,838 PSI
St = 855.7/0.25 = 3,423 PSI
Sca = 1.8x10^6 (t/R)
Sca = 20,454 PSI
N = 0.2139
M = 0.2854
0.19 < 1
Using a 1/4" for the cone should be ok?
am I missing something else?
I did not use the formulas on the standard because I could not figure out the direction of (Wt + F)/At on both cylinders and cones. Also, I was not sure which At I needed to use since Wshell and Wcone is acting only on the thickness and Wliq is acting on the area of the cyl/cone
Thank you for your help.