Horizontal Tank Volumev1.0.0Method: Circular Segment

Horizontal Tank Volume Calculator

Liquid volume, fill percentage, ullage and mass for horizontal tanks with flat, spherical or 2:1 elliptical ends.

Cylinder

Partially-filled horizontal cylinder with flat ends

SIIMP

Tank Dimensions

Diameter (D)

Length (L)

Liquid Height (h)0 – 2 m

EmptyHalfFull

Fluid Density

kg/m³

Fluid preset

Cross-section

50.0% full

Tank Profile

Results

Inventory result

Horizontal Tank Inventory

Tank configurationFlat Ends Profile

Liquid Volume50% Full

7,854.0 L

= 7.854 m³

Total Capacity (100%)

15,708.0 L

15.708 m³

Remaining (Ullage)

7,854.0 L

7.854 m³

Liquid Mass

7,854.0 kg

7.854 t (at 1,000 kg/m³)

Fill Percentage

50.0 %

By total capacity

Volume vs Liquid Height

Strapping / Dip Table

Height (m)	Volume (L)	Volume (m³)	Fill %
0	0	0	0
0.1	293.6	0.294	1.9
0.2	817.5	0.818	5.2
0.3	1,477.5	1.477	9.4
0.4	2,236.5	2.236	14.2
0.5	3,070.9	3.071	19.6
0.6	3,963.4	3.963	25.2
0.7	4,899.6	4.9	31.2
0.8	5,867.4	5.867	37.4
0.9	6,855.7	6.856	43.6
1	7,854	7.854	50
1.1	8,852.3	8.852	56.4
1.2	9,840.6	9.841	62.6
1.3	10,808.4	10.808	68.8
1.4	11,744.6	11.745	74.8
1.5	12,637	12.637	80.4
1.6	13,471.5	13.471	85.8
1.7	14,230.5	14.23	90.6
1.8	14,890.5	14.89	94.8
1.9	15,414.3	15.414	98.1
2	15,708	15.708	100

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Calculation Breakdown

Worked in metres and cubic metres; displayed results are converted to the selected units.

Governing Equations & FormulasFormulas

Mathematical foundations and engineering design equations

Circular Segment Area (Filled Shell)

A_seg = r² · arccos((r − h)/r) − (r − h) · √(2·r·h − h²)

Wetted cross-sectional area of horizontal cylinder at liquid depth h

Cylindrical Shell Volume

V_shell = A_seg · L

Liquid volume contained along straight cylindrical body length L

Liquid Mass

m = ρ · V

Stored fluid mass computed from density at operating temperature

Step-by-Step CalculationWorking Steps

Sequential evaluation with parameter substitution and unit tracking

Step & Parameter	Expression & Substitution	Result
1Tank radius	r = D / 2 = 2 / 2	= 1 m
2Circular segment area	A_seg = r² cos⁻¹((r − h)/r) − (r − h)√(2rh − h²) = 1² × cos⁻¹((1 − 1)/1) − (1 − 1) × √(2 × 1 × 1 − 1²)	= 1.570796 m²
3Straight-shell liquid volume	V_shell = A_seg · L = 1.570796 × 5	= 7.853982 m³
4Flat-end head volume	V_heads = 0	= 0 m³
5Total liquid volume	V = V_shell + V_heads = 7.853982 + 0	= 7.853982 m³
6Liquid mass	m = ρ · V = 1,000 × 7.853982	= 7,853.982 kg

Variable DefinitionsNomenclature

Mathematical symbols, units, and definitions used in calculations

DInside diameter of the tank

LStraight cylindrical shell length between the ends

hMeasured liquid height from the bottom of the tank

rInside radius of the tank (D/2)

A_segLiquid-filled circular-segment area of the cylindrical shell

m²

V_shellLiquid volume in the straight cylindrical shell

m³

V_headsCombined liquid volume in both end heads

m³

VTotal liquid volume in the tank

m³

ρStored-fluid density at the operating temperature

kg/m³

mLiquid mass (ρ × V)

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Engineering Tips

Computes the circular-segment area of a horizontal cylinder and multiplies it by the tank length.

V = L · [r² · cos⁻¹((r − h) / r) − (r − h) · √(2rh − h²)]

Valid for flat-ended cylinders where 0 ≤ h ≤ D.
Use the internal straight length for L.