--- title: "Total pressure and centre on tank face AB" subject: "Fluid Mechanics" chapter: 3 key: "semester_3/fluid_mechanics:oil-tank-pressure-on-ab" source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/oil-tank-pressure-on-ab site: BCT Engineering --- # Total pressure and centre on tank face AB Fluid Mechanics · Chapter 3: Hydrostatics - Asked in: 2076 Chaitra, 6+2 marks - Video: [Inclined Surface Submerged in a Liquid](https://www.youtube.com/watch?v=EBSD2DZ5eBM) — TutorialsPoint - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/oil-tank-pressure-on-ab - Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-3 ## Question A tank full of oil (S = 0.8) as shown in figure. Determine total pressure and centre of pressure on surface AB of the tank. Check your result with pressure diagram also. Take length of the tank 6 m. [Figure: tank with pressure gauge reading 19.62 kPa at top; surface AB inclined from A (bottom) to B (top), vertical height 5 m; horizontal dimensions 4 m and 2 m] ## Short answer The total pressure force onAB is1268 kN, acting3.14 m along the slope from top B (vertical depth2.92 m below B). The centre of pressure is obtained from the trapezoidal pressure diagram. ## Answer **Reading of the figure:** AB is the inclined side wall, vertical height 5 m and horizontal run 2 m, so its length is $L=\sqrt{5^{2}+2^{2}}=5.385$ m; width of tank $b=6$ m; A at the bottom, B at the top. The oil surface (level of B) is under a gauge pressure $p_0=19.62$ kPa. Oil $S=0.8$, $\gamma=0.8(9.81)=7.848$ kN/m³. ```text B p0 = 19.62 kPa \ \ AB, L = 5.385 m \ A \ 5 m vertical ``` Equivalent oil head of the gauge pressure: $\dfrac{19.62}{7.848}=2.5$ m of oil above B. **Total pressure** (centroid is 2.5 m below B): $$ A_{AB}=5.385\times6=32.31\ \text{m}^2 $$ $$ p_c=19.62+7.848(2.5)=39.24\ \text{kPa} $$ $$ F=p_cA_{AB}=39.24\times32.31=1267.9\ \text{kN} $$ **Centre of pressure.** Distance of the centroid from the equivalent free surface along the slope, with $\sin\theta=5/5.385$: $$ \bar y=\frac{2.5+2.5}{\sin\theta}=5.385\ \text{m} $$ $$ \bar y_{cp}-\bar y=\frac{L^{2}}{12\,\bar y}=\frac{29}{12(5.385)}=0.449\ \text{m} $$ The centre of pressure is $L/2+0.449=3.141$ m from B along AB (vertical depth below B $=3.141\sin\theta=2.917$ m). **Check with the pressure diagram.** Pressures: at B, 19.62 kPa; at A, $19.62+7.848(5)=58.86$ kPa (trapezium): $$ F=\frac{19.62+58.86}{2}\times32.31=1267.9\ \text{kN} $$ $$ x_{B}=\frac{L}{3}\cdot\frac{2(58.86)+19.62}{58.86+19.62}=\frac{5.385}{3}\times1.7500=3.141\ \text{m from B} $$ **Answer: $F\approx1268$ kN, acting 3.14 m from B along the slope (2.92 m vertically below B).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the 19.62 kPa gauge reading handled? A. Converted to an equivalent oil head 19.62/7.848 = 2.5 m above B B. Added to the pressure only at A C. Ignored because oil is incompressible D. Converted using the density of water to 2 m Answer: A. It is replaced by an imaginary free surface 2.5 m of oil above B. (Key point: Gauge pressure as oil head) 2. What is the total force on AB? A. About 633 kN B. About 1268 kN, from p_c x area = 39.24 x 32.31 C. About 2535 kN D. About 39.24 kN Answer: B. Centroid is 2.5 m below B so p_c = 19.62 + 7.848(2.5) = 39.24 kPa; area = 5.385 x 6. (Key point: Total force) 3. Where does the force act? A. At the centroid, 2.69 m from B B. At 2/3 of the way from A C. About 3.14 m from B along AB, below the centroid by L^2/(12 ȳ) D. At A Answer: C. The centre of pressure is always below the centroid: 2.69 + 0.449 = 3.14 m. (Key point: Centre of pressure) 4. What does the pressure diagram on AB look like and what gives the force? A. A trapezium, 19.62 kPa at B to 58.86 kPa at A; F = mean pressure x area B. A triangle starting at zero at B C. A rectangle of 39.24 kPa only D. A triangle starting at zero at A Answer: A. The gauge pressure makes the diagram a trapezium; (19.62 + 58.86)/2 x 32.31 = 1268 kN. (Key point: Pressure diagram check) ## Related in Fluid Mechanics - [Gauge pressure in cleaning fluid tank](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Acleaning-fluid-tank-gauge-pressure) — Fluid Mechanics, chapter 2 - [Pressure diagram in rotating closed tank](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Aclosed-rotating-tank-pressure-diagram) — Fluid Mechanics, chapter 2 - [Absolute and gauge pressure, mercury manometer](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Aabsolute-gauge-pressure-vapour-mercury) — Fluid Mechanics, chapter 2 - [Force and position on inclined door](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Adoor-force-and-position) — Fluid Mechanics, chapter 3 - [Resultant force on curved surface BCDEF](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Acurved-surface-bcdef-force) — Fluid Mechanics, chapter 3 - [Accelerated oil tank, spilled volume](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Aaccelerated-cube-tank-spilling) — Fluid Mechanics, chapter 3 ## Related in other subjects - [Pressure to drive oil up inclined pipe](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Aviscous-fluid-upslope-pressure) — Hydraulics, chapter 1 - [Total pressure and centre of pressure derivation](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Atotal-pressure-centre-pressure) — Hydro Power, chapter 1, asked 2× - [Surcharge 15 kPa, stresses at clay centre](https://www.bctengineer.com/question.md?key=semester_4%2Fsoil_mechanics%3Asurcharge-15kpa-stresses-clay-center) — Soil Mechanics, chapter 9 - [Two-compartment vessel gauges: reading of gauge C](https://www.bctengineer.com/question.md?key=semester_2%2Ffundamental_of_thermodynamics_and_heat_transfer%3Atwo-compartment-gauges-a4bar-b2bar) — Fundamental of Thermodynamics & Heat Transfer, chapter 1