Manometers

Fluid Mechanics · FE Reference Handbook section

Fluid Mechanics

27 formulas

10 exam-style examples

~60 min

All Fluid Mechanics lectures

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

Bober, W., and R.A. Kenyon, Fluid Mechanics, Wiley, 1980. Diagrams reprinted by permission of William Bober and Richard A. Kenyon.
Note that the difference between the two densities is used.
Another device that works on the same principle as the manometer is the simple barometer.
Bober, W., and R.A. Kenyon, Fluid Mechanics, Wiley, 1980. Diagrams reprinted by permission of William Bober and Richard A. Kenyon.
Forces on Submerged Surfaces and the Center of Pressure
LIQUID h Patm z PLANAR VIEW FROM ABOVE
The pressure on a point at a vertical distance h below the surface is:
If atmospheric pressure acts above the liquid surface and on the non-wetted side of the submerged surface:

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Manometer pressure relation — solve for pressure at point 2 — Manometers

a mercury manometer measuring pipeline pressure Given pressure at point 1 (p_1) = 66,300 Pa; specific weight fluid 1 (gamma_1) = 9,010 N/m^3; height fluid 1 (h_1) = 0.6300 m; specific weight fluid 2 (mercury) (gamma_2) = 130,200 N/m^3; height fluid 2 (h_2) = 0.2600 m, determine the pressure at point 2 (p_2) in Pa.

Given

$pressure at point 1 (p_1) = 66,300 Pa$
$specific weight fluid 1 (gamma_1) = 9,010 N/m^3$
$height fluid 1 (h_1) = 0.6300 m$
$specific weight fluid 2 (mercury) (gamma_2) = 130,200 N/m^3$
$height fluid 2 (h_2) = 0.2600 m$

Find

pressure at point 2 (p_2), in Pa

Start with the thinking

The governing relation printed in this handbook section is Manometer pressure relation.
Everything except p_2 is given, so isolate p_2 symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
A U-tube manometer connected to a pipeline is read to find an unknown pressure.

Figure 1 — schematic for Manometer pressure relation — solve for pressure at point 2 — Manometers

Step-by-step solution

Step 1 — State the governing relation:
$p_1 + \gamma_1 h_1 - \gamma_2 h_2 = p_2$
Step 2 — Rearrange the relation so that p_2 stands alone on the left-hand side.
Step 3 — List the givens: pressure at point 1 (p_1) = 66,300 Pa, specific weight fluid 1 (gamma_1) = 9,010 N/m^3, height fluid 1 (h_1) = 0.6300 m, specific weight fluid 2 (mercury) (gamma_2) = 130,200 N/m^3, height fluid 2 (h_2) = 0.2600 m.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$p_{2} = 38124\ \text{Pa}$
Step 6 — Check: returning p_2 = 38,124 Pa to
$p_1 + \gamma_1 h_1 - \gamma_2 h_2 = p_2$
reproduces the given quantities, and both sides carry the same units.

Answer:

p_{2} = 38124\ \text{Pa}

Why the other options are there

76,249 — kept a factor of two that cancels in the correct rearrangement.
19,062 — dropped that same factor in the other direction.
41,937 — rounded an intermediate value before the final step.

Reference: FE Handbook — Manometers

Example 2

Differential manometer reading — solve for pressure difference — Manometers (2)

a water manometer across a filter bed Given manometer fluid density (\rho) = 11,000 kg/m^3; gravitational acceleration (g) = 9.8100 m/s^2; manometer deflection (h) = 1.0000 m, determine the pressure difference (\Delta p) in Pa.

Given

$manometer fluid density (\rho) = 11,000 kg/m^3$
$gravitational acceleration (g) = 9.8100 m/s^2$
$manometer deflection (h) = 1.0000 m$

Find

pressure difference (\Delta p), in Pa

Start with the thinking

The governing relation printed in this handbook section is Differential manometer reading.
Everything except \Delta p is given, so isolate \Delta p symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
A U-tube manometer measures the pressure difference across a device in a pipeline.

Figure 2 — schematic for Differential manometer reading — solve for pressure difference — Manometers (2)

Step-by-step solution

Step 1 — State the governing relation:
$\Delta p = \rho g h$
Step 2 — Rearrange symbolically for \Delta p:
$\Delta p = \rho g h$
Step 3 — List the givens: manometer fluid density (\rho) = 11,000 kg/m^3, gravitational acceleration (g) = 9.8100 m/s^2, manometer deflection (h) = 1.0000 m.
Step 4 — Substitute the given values:
$\Delta p = 11000 9.8100 1.0000$
Step 5 — Evaluate:
$\Delta p = 107910\ \text{Pa}$
Step 6 — Check: returning \Delta p = 107,910 Pa to
$\Delta p = \rho g h$
reproduces the given quantities, and both sides carry the same units.

Answer:

\Delta p = 107910\ \text{Pa}

Why the other options are there

215,820 — kept a factor of two that cancels in the correct rearrangement.
53,955 — dropped that same factor in the other direction.
118,701 — rounded an intermediate value before the final step.

Reference: FE Handbook — Manometers

Example 3

Manometer pressure relation — solve for pressure at point 1 — Manometers (3)

a differential manometer across a pump Given specific weight fluid 1 (gamma_1) = 9,750 N/m^3; height fluid 1 (h_1) = 0.9200 m; specific weight fluid 2 (mercury) (gamma_2) = 131,400 N/m^3; height fluid 2 (h_2) = 0.4000 m; pressure at point 2 (p_2) = 191,700 Pa, determine the pressure at point 1 (p_1) in Pa.

Given

$specific weight fluid 1 (gamma_1) = 9,750 N/m^3$
$height fluid 1 (h_1) = 0.9200 m$
$specific weight fluid 2 (mercury) (gamma_2) = 131,400 N/m^3$
$height fluid 2 (h_2) = 0.4000 m$
$pressure at point 2 (p_2) = 191,700 Pa$

Find

pressure at point 1 (p_1), in Pa

Start with the thinking

The governing relation printed in this handbook section is Manometer pressure relation.
Everything except p_1 is given, so isolate p_1 symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
A U-tube manometer connected to a pipeline is read to find an unknown pressure.

Figure 3 — schematic for Manometer pressure relation — solve for pressure at point 1 — Manometers (3)

Step-by-step solution

Step 1 — State the governing relation:
$p_1 + \gamma_1 h_1 - \gamma_2 h_2 = p_2$
Step 2 — Rearrange the relation so that p_1 stands alone on the left-hand side.
Step 3 — List the givens: specific weight fluid 1 (gamma_1) = 9,750 N/m^3, height fluid 1 (h_1) = 0.9200 m, specific weight fluid 2 (mercury) (gamma_2) = 131,400 N/m^3, height fluid 2 (h_2) = 0.4000 m, pressure at point 2 (p_2) = 191,700 Pa.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$p_{1} = 235290\ \text{Pa}$
Step 6 — Check: returning p_1 = 235,290 Pa to
$p_1 + \gamma_1 h_1 - \gamma_2 h_2 = p_2$
reproduces the given quantities, and both sides carry the same units.

Answer:

p_{1} = 235290\ \text{Pa}

Why the other options are there

470,580 — kept a factor of two that cancels in the correct rearrangement.
117,645 — dropped that same factor in the other direction.
258,819 — rounded an intermediate value before the final step.

Reference: FE Handbook — Manometers

Example 4

Differential manometer reading — solve for manometer deflection — Manometers (4)

a manometer reading the head loss across a venturi Given pressure difference (\Delta p) = 52,932 Pa; manometer fluid density (\rho) = 5,930 kg/m^3; gravitational acceleration (g) = 9.8100 m/s^2, determine the manometer deflection (h) in m.

Given

$pressure difference (\Delta p) = 52,932 Pa$
$manometer fluid density (\rho) = 5,930 kg/m^3$
$gravitational acceleration (g) = 9.8100 m/s^2$

Find

manometer deflection (h), in m

Start with the thinking

The governing relation printed in this handbook section is Differential manometer reading.
Everything except h is given, so isolate h symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
A U-tube manometer measures the pressure difference across a device in a pipeline.

Figure 4 — schematic for Differential manometer reading — solve for manometer deflection — Manometers (4)

Step-by-step solution

Step 1 — State the governing relation:
$\Delta p = \rho g h$
Step 2 — Rearrange symbolically for h:
$h = \dfrac{\Delta p}{\rho g}$
Step 3 — List the givens: pressure difference (\Delta p) = 52,932 Pa, manometer fluid density (\rho) = 5,930 kg/m^3, gravitational acceleration (g) = 9.8100 m/s^2.
Step 4 — Substitute the given values:
$h = \dfrac{52932}{5930 9.8100}$
Step 5 — Evaluate:
$h = 0.9099\ \text{m}$
Step 6 — Check: returning h = 0.9099 m to
$\Delta p = \rho g h$
reproduces the given quantities, and both sides carry the same units.

Answer:

h = 0.9099\ \text{m}

Why the other options are there

1.8198 — kept a factor of two that cancels in the correct rearrangement.
0.4550 — dropped that same factor in the other direction.
1.0009 — rounded an intermediate value before the final step.

Reference: FE Handbook — Manometers

Example 5

Manometer pressure relation — solve for height fluid 2 — Manometers (5)

a manometer used to calibrate a pressure gage Given pressure at point 1 (p_1) = 95,900 Pa; specific weight fluid 1 (gamma_1) = 9,340 N/m^3; height fluid 1 (h_1) = 0.7400 m; specific weight fluid 2 (mercury) (gamma_2) = 132,400 N/m^3; pressure at point 2 (p_2) = 187,100 Pa, determine the height fluid 2 (h_2) in m.

Given

$pressure at point 1 (p_1) = 95,900 Pa$
$specific weight fluid 1 (gamma_1) = 9,340 N/m^3$
$height fluid 1 (h_1) = 0.7400 m$
$specific weight fluid 2 (mercury) (gamma_2) = 132,400 N/m^3$
$pressure at point 2 (p_2) = 187,100 Pa$

Find

height fluid 2 (h_2), in m

Start with the thinking

The governing relation printed in this handbook section is Manometer pressure relation.
Everything except h_2 is given, so isolate h_2 symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
A U-tube manometer connected to a pipeline is read to find an unknown pressure.

Figure 5 — schematic for Manometer pressure relation — solve for height fluid 2 — Manometers (5)

Step-by-step solution

Step 1 — State the governing relation:
$p_1 + \gamma_1 h_1 - \gamma_2 h_2 = p_2$
Step 2 — Rearrange the relation so that h_2 stands alone on the left-hand side.
Step 3 — List the givens: pressure at point 1 (p_1) = 95,900 Pa, specific weight fluid 1 (gamma_1) = 9,340 N/m^3, height fluid 1 (h_1) = 0.7400 m, specific weight fluid 2 (mercury) (gamma_2) = 132,400 N/m^3, pressure at point 2 (p_2) = 187,100 Pa.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$h_{2} = -0.6366\ \text{m}$
Step 6 — Check: returning h_2 = -0.6366 m to
$p_1 + \gamma_1 h_1 - \gamma_2 h_2 = p_2$
reproduces the given quantities, and both sides carry the same units.

Answer:

h_{2} = -0.6366\ \text{m}

Why the other options are there

-1.2732 — kept a factor of two that cancels in the correct rearrangement.
-0.3183 — dropped that same factor in the other direction.
-0.7003 — rounded an intermediate value before the final step.

Reference: FE Handbook — Manometers

Example 6

Differential manometer reading — solve for manometer fluid density — Manometers (6)

a mercury manometer across an orifice plate Given pressure difference (\Delta p) = 1,001 Pa; gravitational acceleration (g) = 9.8100 m/s^2; manometer deflection (h) = 0.7700 m, determine the manometer fluid density (\rho) in kg/m^3.

Given

$pressure difference (\Delta p) = 1,001 Pa$
$gravitational acceleration (g) = 9.8100 m/s^2$
$manometer deflection (h) = 0.7700 m$

Find

manometer fluid density (\rho), in kg/m^3

Start with the thinking

The governing relation printed in this handbook section is Differential manometer reading.
Everything except \rho is given, so isolate \rho symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
A U-tube manometer measures the pressure difference across a device in a pipeline.

Figure 6 — schematic for Differential manometer reading — solve for manometer fluid density — Manometers (6)

Step-by-step solution

Step 1 — State the governing relation:
$\Delta p = \rho g h$
Step 2 — Rearrange symbolically for \rho:
$\rho = \dfrac{\Delta p}{g h}$
Step 3 — List the givens: pressure difference (\Delta p) = 1,001 Pa, gravitational acceleration (g) = 9.8100 m/s^2, manometer deflection (h) = 0.7700 m.
Step 4 — Substitute the given values:
$\rho = \dfrac{1001}{9.8100 0.7700}$
Step 5 — Evaluate:
$\rho = 132.5\ \text{kg/m^3}$
Step 6 — Check: returning \rho = 132.5 kg/m^3 to
$\Delta p = \rho g h$
reproduces the given quantities, and both sides carry the same units.

Answer:

\rho = 132.5\ \text{kg/m^3}

Why the other options are there

265.0 — kept a factor of two that cancels in the correct rearrangement.
66.2589 — dropped that same factor in the other direction.
145.8 — rounded an intermediate value before the final step.

Reference: FE Handbook — Manometers

Example 7

Manometer pressure relation — solve for pressure at point 2 (case 2) — Manometers (7)

a mercury manometer measuring pipeline pressure Given pressure at point 1 (p_1) = 101,300 Pa; specific weight fluid 1 (gamma_1) = 9,150 N/m^3; height fluid 1 (h_1) = 1.1300 m; specific weight fluid 2 (mercury) (gamma_2) = 132,400 N/m^3; height fluid 2 (h_2) = 0.3800 m, determine the pressure at point 2 (p_2) in Pa.

Given

$pressure at point 1 (p_1) = 101,300 Pa$
$specific weight fluid 1 (gamma_1) = 9,150 N/m^3$
$height fluid 1 (h_1) = 1.1300 m$
$specific weight fluid 2 (mercury) (gamma_2) = 132,400 N/m^3$
$height fluid 2 (h_2) = 0.3800 m$

Find

pressure at point 2 (p_2), in Pa

Start with the thinking

The governing relation printed in this handbook section is Manometer pressure relation.
Everything except p_2 is given, so isolate p_2 symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
A U-tube manometer connected to a pipeline is read to find an unknown pressure.

Figure 7 — schematic for Manometer pressure relation — solve for pressure at point 2 (case 2) — Manometers (7)

Step-by-step solution

Step 1 — State the governing relation:
$p_1 + \gamma_1 h_1 - \gamma_2 h_2 = p_2$
Step 2 — Rearrange the relation so that p_2 stands alone on the left-hand side.
Step 3 — List the givens: pressure at point 1 (p_1) = 101,300 Pa, specific weight fluid 1 (gamma_1) = 9,150 N/m^3, height fluid 1 (h_1) = 1.1300 m, specific weight fluid 2 (mercury) (gamma_2) = 132,400 N/m^3, height fluid 2 (h_2) = 0.3800 m.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$p_{2} = 61328\ \text{Pa}$
Step 6 — Check: returning p_2 = 61,328 Pa to
$p_1 + \gamma_1 h_1 - \gamma_2 h_2 = p_2$
reproduces the given quantities, and both sides carry the same units.

Answer:

p_{2} = 61328\ \text{Pa}

Why the other options are there

122,655 — kept a factor of two that cancels in the correct rearrangement.
30,664 — dropped that same factor in the other direction.
67,460 — rounded an intermediate value before the final step.

Reference: FE Handbook — Manometers

Example 8

Differential manometer reading — solve for pressure difference (case 2) — Manometers (8)

a water manometer across a filter bed Given manometer fluid density (\rho) = 4,080 kg/m^3; gravitational acceleration (g) = 9.8100 m/s^2; manometer deflection (h) = 0.2800 m, determine the pressure difference (\Delta p) in Pa.

Given

$manometer fluid density (\rho) = 4,080 kg/m^3$
$gravitational acceleration (g) = 9.8100 m/s^2$
$manometer deflection (h) = 0.2800 m$

Find

pressure difference (\Delta p), in Pa

Start with the thinking

The governing relation printed in this handbook section is Differential manometer reading.
Everything except \Delta p is given, so isolate \Delta p symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
A U-tube manometer measures the pressure difference across a device in a pipeline.

Figure 8 — schematic for Differential manometer reading — solve for pressure difference (case 2) — Manometers (8)

Step-by-step solution

Step 1 — State the governing relation:
$\Delta p = \rho g h$
Step 2 — Rearrange symbolically for \Delta p:
$\Delta p = \rho g h$
Step 3 — List the givens: manometer fluid density (\rho) = 4,080 kg/m^3, gravitational acceleration (g) = 9.8100 m/s^2, manometer deflection (h) = 0.2800 m.
Step 4 — Substitute the given values:
$\Delta p = 4080 9.8100 0.2800$
Step 5 — Evaluate:
$\Delta p = 11207\ \text{Pa}$
Step 6 — Check: returning \Delta p = 11,207 Pa to
$\Delta p = \rho g h$
reproduces the given quantities, and both sides carry the same units.

Answer:

\Delta p = 11207\ \text{Pa}

Why the other options are there

22,414 — kept a factor of two that cancels in the correct rearrangement.
5,603 — dropped that same factor in the other direction.
12,328 — rounded an intermediate value before the final step.

Reference: FE Handbook — Manometers

Example 9

Manometer pressure relation — solve for pressure at point 1 (case 2) — Manometers (9)

a differential manometer across a pump Given specific weight fluid 1 (gamma_1) = 9,230 N/m^3; height fluid 1 (h_1) = 1.2500 m; specific weight fluid 2 (mercury) (gamma_2) = 131,200 N/m^3; height fluid 2 (h_2) = 0.2100 m; pressure at point 2 (p_2) = 192,900 Pa, determine the pressure at point 1 (p_1) in Pa.

Given

$specific weight fluid 1 (gamma_1) = 9,230 N/m^3$
$height fluid 1 (h_1) = 1.2500 m$
$specific weight fluid 2 (mercury) (gamma_2) = 131,200 N/m^3$
$height fluid 2 (h_2) = 0.2100 m$
$pressure at point 2 (p_2) = 192,900 Pa$

Find

pressure at point 1 (p_1), in Pa

Start with the thinking

The governing relation printed in this handbook section is Manometer pressure relation.
Everything except p_1 is given, so isolate p_1 symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
A U-tube manometer connected to a pipeline is read to find an unknown pressure.

Figure 9 — schematic for Manometer pressure relation — solve for pressure at point 1 (case 2) — Manometers (9)

Step-by-step solution

Step 1 — State the governing relation:
$p_1 + \gamma_1 h_1 - \gamma_2 h_2 = p_2$
Step 2 — Rearrange the relation so that p_1 stands alone on the left-hand side.
Step 3 — List the givens: specific weight fluid 1 (gamma_1) = 9,230 N/m^3, height fluid 1 (h_1) = 1.2500 m, specific weight fluid 2 (mercury) (gamma_2) = 131,200 N/m^3, height fluid 2 (h_2) = 0.2100 m, pressure at point 2 (p_2) = 192,900 Pa.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$p_{1} = 208915\ \text{Pa}$
Step 6 — Check: returning p_1 = 208,915 Pa to
$p_1 + \gamma_1 h_1 - \gamma_2 h_2 = p_2$
reproduces the given quantities, and both sides carry the same units.

Answer:

p_{1} = 208915\ \text{Pa}

Why the other options are there

417,829 — kept a factor of two that cancels in the correct rearrangement.
104,457 — dropped that same factor in the other direction.
229,806 — rounded an intermediate value before the final step.

Reference: FE Handbook — Manometers

Example 10

Differential manometer reading — solve for manometer deflection (case 2) — Manometers (10)

a manometer reading the head loss across a venturi Given pressure difference (\Delta p) = 84,979 Pa; manometer fluid density (\rho) = 2,850 kg/m^3; gravitational acceleration (g) = 9.8100 m/s^2, determine the manometer deflection (h) in m.

Given

$pressure difference (\Delta p) = 84,979 Pa$
$manometer fluid density (\rho) = 2,850 kg/m^3$
$gravitational acceleration (g) = 9.8100 m/s^2$

Find

manometer deflection (h), in m

Start with the thinking

The governing relation printed in this handbook section is Differential manometer reading.
Everything except h is given, so isolate h symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
A U-tube manometer measures the pressure difference across a device in a pipeline.

Figure 10 — schematic for Differential manometer reading — solve for manometer deflection (case 2) — Manometers (10)

Step-by-step solution

Step 1 — State the governing relation:
$\Delta p = \rho g h$
Step 2 — Rearrange symbolically for h:
$h = \dfrac{\Delta p}{\rho g}$
Step 3 — List the givens: pressure difference (\Delta p) = 84,979 Pa, manometer fluid density (\rho) = 2,850 kg/m^3, gravitational acceleration (g) = 9.8100 m/s^2.
Step 4 — Substitute the given values:
$h = \dfrac{84979}{2850 9.8100}$
Step 5 — Evaluate:
$h = 3.0395\ \text{m}$
Step 6 — Check: returning h = 3.0395 m to
$\Delta p = \rho g h$
reproduces the given quantities, and both sides carry the same units.

Answer:

h = 3.0395\ \text{m}

Why the other options are there

6.0789 — kept a factor of two that cancels in the correct rearrangement.
1.5197 — dropped that same factor in the other direction.
3.3434 — rounded an intermediate value before the final step.

Reference: FE Handbook — Manometers

Manometers

Handbook notes for this section

Core formulas for this FE topic

P0=P2+γ2h2−γ1h1=P2+g(ρ2h2−ρ1h1)P_{0} = P_{2} + \gamma_{2}h_{2} - \gamma_{1}h_{1} = P_{2} + g (\rho_{2} h_{2}- \rho_{1} h_{1})P0​=P2​+γ2​h2​−γ1​h1​=P2​+g(ρ2​h2​−ρ1​h1​)

Ifh1=h2=hIf h_{1} = h_{2} = hIfh1​=h2​=h

P0=P2+(γ2−γ1)h=P2+(ρ2−ρ1)ghP_{0} = P_{2} + (\gamma_{2} - \gamma_{1})h = P_{2} + (\rho_{2} - \rho_{1})ghP0​=P2​+(γ2​−γ1​)h=P2​+(ρ2​−ρ1​)gh

P=pressureP = pressureP=pressure

γ=specificweightoffluid\gamma = specific weight of fluidγ=specificweightoffluid

h=heighth = heighth=height

g=accelerationofgravityg = acceleration of gravityg=accelerationofgravity

ρ=fluiddensity\rho = fluid densityρ=fluiddensity

Patm = PA = Pv + γh = PB + γh = PB + ρghNeeds technical review

Pv=vaporpressureofthebarometerfluidPv = vapor pressure of the barometer fluidPv=vaporpressureofthebarometerfluid

h = y sin θ h θyNeeds technical review

P = Patm + ρgh, for h ≥ 0Needs technical review

P=pressureP = pressureP=pressure

Patm=atmosphericpressurePatm = atmospheric pressurePatm=atmosphericpressure

PC=pressureatthecentroidofareaPC = pressure at the centroid of areaPC=pressureatthecentroidofarea

PCP=pressureatcenterofpressurePCP = pressure at center of pressurePCP=pressureatcenterofpressure

yC=slantdistancefromliquidsurfacetothecentroidofareayC = slant distance from liquid surface to the centroid of areayC=slantdistancefromliquidsurfacetothecentroidofarea

yC=hC/sin⁡θyC = hC/\sin \thetayC=hC/sinθ

hC=verticaldistancefromliquidsurfacetocentroidofareahC = vertical distance from liquid surface to centroid of areahC=verticaldistancefromliquidsurfacetocentroidofarea

yCP=slantdistancefromliquidsurfacetocenterofpressureyCP = slant distance from liquid surface to center of pressureyCP=slantdistancefromliquidsurfacetocenterofpressure

hCP=verticaldistancefromliquidsurfacetocenterofpressurehCP = vertical distance from liquid surface to center of pressurehCP=verticaldistancefromliquidsurfacetocenterofpressure

θ=anglebetweenliquidsurfaceandedgeofsubmergedsurface\theta = angle between liquid surface and edge of submerged surfaceθ=anglebetweenliquidsurfaceandedgeofsubmergedsurface

IxC=momentofinertiaaboutthecentroidalx−axisIxC = moment of inertia about the centroidal x-axisIxC=momentofinertiaaboutthecentroidalx−axis

yCP=yC+IxC/yCAyCP = yC + IxC/yCAyCP=yC+IxC/yCA

yCP = yC + ρg sin θ IxC/PCANeeds technical review

Wetted side: FR = (Patm+ ρgyC sin θ)ANeeds technical review

Patm acting both sides: FR = (ρgyC sin θ)ANeeds technical review

Worked exam-style examples