7.2 Vessel Volume, Surface Area, Circumference & Head Layout Computations

7.2 Vessel Volume, Surface Area, Circumference & Head Layout Computations

Core Trade Concept: Boilermakers regularly fabricate, assemble, and rig massive cylindrical pressure vessels, steam drums, deaerators, and storage tanks. Mastering vessel layout mathematics—including Mean Diameter plate stretchouts, shell surface areas, volumetric fluid capacities, hydrostatic fill weights, and formed head blank geometry—is critical to ensure precise roll fit-up, code-compliant weld seams, and safe rigging capacities.

1. Circumference, Neutral Axis & Mean Diameter in Plate Rolling

When flat steel plate is rolled into a cylindrical cylinder on hydraulic pyramid or pinch rolls, the metal undergoes severe plastic deformation:

• The outer fibers of the plate are placed in tension and stretch (lengthen).
• The inner fibers of the plate are placed in compression and shrink (shorten).
• Exactly midway through the plate thickness lies the Neutral Axis, where the metal experiences zero elongation and zero compression.

                      NEUTRAL AXIS IN ROLLED PLATE
                      
         <------------------- OD (Tension - Stretches) ------------------->
       +-------------------------------------------------------------------+
       |                                                                   |
       | - - - - - - - - - - NEUTRAL AXIS (Dm) - - - - - - - - - - - - - - | t (Thk)
       |                                                                   |
       +-------------------------------------------------------------------+
         <------------------ ID (Compression - Shortens) ----------------->

Mean Diameter ($D_m$) Formulation

To ensure that a rolled cylinder closes to the exact design diameter with zero gap or overlap at the longitudinal weld seam, the flat plate cut length (Stretchout) must be calculated using the Mean Diameter ($D_m$) (the diameter measured to the neutral axis):

$D_m = \text{Inside Diameter (ID)} + \text{Plate Thickness } (t)$

$D_m = \text{Outside Diameter (OD)} - \text{Plate Thickness } (t)$

Plate Stretchout Length ($L$)

$\text{Stretchout Length } (L) = \pi \times D_m = 3.14159 \times D_m$

Critical Trade Error Warning:

• Calculating stretchout using the ID results in a shell that is too short, causing a large gap at the longitudinal seam that cannot be pulled together.
• Calculating stretchout using the OD results in a shell that is too long, causing the plate edges to overlap and requiring torch-cutting and beveling.

Worked Example: Heavy Plate Rolling Stretchout

A boilermaker must cut a flat plate of SA-516 Grade 70 carbon steel with a thickness of $1.500\text{ inches}$ to roll a vessel shell with an Inside Diameter (ID) of $96.000\text{ inches}$ ($8\text{ feet}$ ID):

$\text{Mean Diameter } (D_m) = \text{ID} + t = 96.000\text{ in.} + 1.500\text{ in.} = 97.500\text{ inches}$

$\text{Stretchout Length } (L) = \pi \times D_m = 3.14159 \times 97.500\text{ in.} = 306.305\text{ inches} \approx 306\text{--}\frac{5}{16}\text{ inches}$

(Note: If the boilermaker had mistakenly used the ID, the plate would be $301.593\text{ in.}$, resulting in a disastrous $4.712\text{ in.}$ gap at the joint).

2. Cylindrical Shell Surface Area & Volumetric Computations

                         CYLINDRICAL VESSEL GEOMETRY
                         
                                  +-------+
                                 /         \  Head Area (A_head)
                                +-----------+
                                |           |
                                |           |
                                |     V     |  Shell Height / Length (H)
                                |  Volume   |
                                |           |
                                |           |
                                +-----------+
                                 \         /  Diameter (D)
                                  +-------+

Cylindrical Surface Area

The lateral surface area of a cylindrical shell (excluding the two heads) is the stretchout circumference multiplied by the shell length or height ($H$):

$A_{shell} = \pi \times D \times H$

Trade Application: Surface area is used to calculate the required square footage of insulation, exterior protective lagging, refractory fireproofing, and epoxy coating systems (where paint coverage is rated in $\text{sq ft/gallon}$).

Cylindrical Volume Formulas

The internal volumetric capacity of a cylindrical vessel (exclusive of dished heads) is calculated using radius ($r$) or diameter ($D$):

$V = \pi \times r^2 \times H = \frac{\pi}{4} \times D^2 \times H \approx 0.7854 \times D^2 \times H$

Essential Trade Conversion Factors Table

| Measurement Parameter | Conversion Relationship | Trade Multiplication Factor | | :--- | :--- | :--- | :--- | | Cubic Inches to Cubic Feet | $1\text{ cu ft} = 1{,}728\text{ cu in.}$ ($12 \times 12 \times 12$) | $\text{Volume (cu ft)} = \frac{\text{Volume (cu in.)}}{1{,}728}$ | | Cubic Feet to Gallons | $1\text{ cu ft} = 7.4805\text{ U.S. gallons}$ | $\text{Gallons} = \text{cu ft} \times 7.4805$ | | Gallons to Water Weight | $1\text{ U.S. gallon of water} = 8.33\text{ lbs}$ (at $60^\circ\text{F}$) | $\text{Weight (lbs)} = \text{Gallons} \times 8.33$ | | Cubic Feet to Water Weight | $1\text{ cu ft of water} = 62.4\text{ lbs}$ | $\text{Weight (lbs)} = \text{cu ft} \times 62.4$ | | Carbon Steel Plate Weight | $1\text{ sq ft of } 1\text{-inch thick steel} = 40.8\text{ lbs}$ | $\text{Steel Weight (lbs)} = \text{Area (sq ft)} \times t\text{ (in.)} \times 40.8$ | | Density of Carbon Steel | $\rho_{steel} = 0.2833\text{ lbs/cu in.} = 490\text{ lbs/cu ft}$ | $\text{Weight (lbs)} = \text{Volume (cu in.)} \times 0.2833$ |

3. Hydrostatic Fill Weight & Rigging/Foundation Load Analysis

During pre-commissioning of boilers and pressure vessels, ASME Section I and Section VIII mandate a hydrostatic pressure test (typically $1.3\text{ to }1.5 \times \text{MAWP}$) by completely filling the vessel with water and venting all trapped air.

Because water is extremely dense ($62.4\text{ lbs/cu ft}$), the water fill weight frequently dwarfs the dry metal weight of the vessel. Boilermakers and rigging supervisors must calculate the combined Total Hydrostatic Load to prevent crane overload during shop hydro, catastrophic structural collapse of temporary rigging cribbing, or foundation settlement.

$\text{Total Hydrostatic Load} = \text{Dry Shell Weight} + \text{Water Fill Weight} + \text{Internal Hardware Weight}$

                    HYDROSTATIC LOAD BALANCE PROFILE
                    
       +-------------------------------------------------------+
       | [Dry Vessel Shell Steel]  ~25% of Total Weight         |
       +-------------------------------------------------------+
       | [Internal Hydrostatic Water Fill]  ~75% of Total Weight|
       | ===================================================== |
       | TOTAL RIGGING & FOUNDATION LOAD = STEEL + WATER       |
       +-------------------------------------------------------+

Comprehensive Worked Hydrostatic Load Problem

Scenario: A vertical deaerator storage tank has an inside diameter of $8.00\text{ feet}$ ($96\text{ inches}$, $r = 4.00\text{ ft}$) and a straight shell length of $20.00\text{ feet}$. The shell is fabricated from $1.00\text{ in.}$ thick carbon steel plate.

Step 1: Calculate Shell Surface Area & Dry Shell Metal Weight: $\text{Shell Mean Diameter } (D_m) = 8.00\text{ ft} + \left(\frac{1.00}{12}\text{ ft}\right) = 8.0833\text{ ft}$ $\text{Shell Lateral Area} = \pi \times D_m \times H = 3.14159 \times 8.0833\text{ ft} \times 20.00\text{ ft} = 507.88\text{ sq ft}$ $\text{Dry Shell Weight} = 507.88\text{ sq ft} \times 1.00\text{ in. (thk)} \times 40.8\text{ lbs/sq ft/in.} = 20{,}721\text{ lbs}$ (Adding approx. $4{,}279\text{ lbs}$ for two dished heads and nozzles yields a total dry vessel weight of $\approx 25{,}000\text{ lbs}$).

Step 2: Calculate Internal Cylindrical Volume: $V = \pi \times r^2 \times H = 3.14159 \times (4.00\text{ ft})^2 \times 20.00\text{ ft} = 3.14159 \times 16 \times 20 = 1{,}005.31\text{ cu ft}$

Step 3: Convert Volume to Gallons: $\text{Water Volume} = 1{,}005.31\text{ cu ft} \times 7.4805\text{ gal/cu ft} = 7{,}520.2\text{ U.S. gallons}$

Step 4: Calculate Total Water Fill Weight: $\text{Water Weight} = 1{,}005.31\text{ cu ft} \times 62.4\text{ lbs/cu ft} = 62{,}731\text{ lbs}$ $(\text{Alternatively: } 7{,}520.2\text{ gal} \times 8.33\text{ lbs/gal} = 62{,}643\text{ lbs})$

Step 5: Calculate Total Hydrostatic Test Load: $\text{Total Load} = 25{,}000\text{ lbs (Dry Steel)} + 62{,}731\text{ lbs (Water)} = 87{,}731\text{ lbs} \approx 43.87\text{ tons}$

Key Takeaway: The water adds over $31\text{ tons}$ to the dry vessel weight, representing over $71%$ of the total load on the support saddles!

4. Dished Pressure Vessel Head Geometries & Blank Layout

Pressure vessels are enclosed with formed dished heads. The geometric shape of the head governs its pressure-containing capability, internal volume, and the flat blank diameter required for pressing or spinning.

                      PRESSURE VESSEL HEAD PROFILES
                      
     HEMISPHERICAL (R = D/2)          2:1 ELLIPSOIDAL (h = D/4) 
         +-------------+                   +-------------+
       /                 \               /                 \
      |                   |             |                   |
      |                   |             +-------------------+
      +-------------------+             ASME TORISPHERICAL (F&D)
      (Thinnest Wall;                   (Crown R = OD; Knuckle r = 6% OD)
       Highest Pressure)                (Economical Standard Industrial)

Formed Head Specifications & ASME Section VIII Characteristics

(Note: $\text{SF} =$ Straight Flange or cylindrical skirt length, typically $1.5\text{ to }3.0\text{ inches}$, provided for butt-welding the head to the cylindrical shell away from high knuckle bending stresses).

Worked Example: ASME Torispherical (F&D) Head Blank Sizing

A boilermaker must cut a circular plate blank to press an ASME Torispherical F&D head for an air receiver tank with an Outside Diameter (OD) of $48.00\text{ inches}$ and a Straight Flange (SF) of $2.00\text{ inches}$:

$D_{blank} \approx (1.14 \times \text{OD}) + (2 \times \text{SF}) = (1.14 \times 48.00\text{ in.}) + (2 \times 2.00\text{ in.}) = 54.72 + 4.00 = 58.72\text{ inches} \approx 58\text{--}\frac{3}{4}\text{ in.}$