Difference between revisions of "Fluid Pressure"
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<math>h</math> = [[Depth]] in the [[fluid]] | <math>h</math> = [[Depth]] in the [[fluid]] | ||
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| + | ===Calculating Pressure in a Fluid=== | ||
| + | {| class="wikitable" | ||
| + | | style="height:20px; width:200px; text-align:center;" |A whale dives to a depth of 1300m below sea level. Given the density of [[water]] is 1000kg/m<sup>3</sup>, calculate the [[pressure]] on the whale due to the water correct to two [[Significant Figures|significant figures]]. | ||
| + | | style="height:20px; width:200px; text-align:center;" |Normal atmospheric pressure at sea level is 101kPa. A miner works in a mine shaft 550m below the surface of the Earth. Given the density of the [[Earth's Atmosphere|atmosphere]] at this level is 1.2kg/m<sup>3</sup>, calculate the [[pressure]] on the miner correct to two [[Significant Figures|significant figures]]. | ||
| + | | style="height:20px; width:200px; text-align:center;" |A piece of [[Lithium]] is submersed 4.3cm deep in oil prevent it from oxidising. The oil has a [[density]] of 932kg/m<sup>3</sup>, calculate the [[pressure]] on the [[Lithium]] due to the oil correct to two [[Significant Figures|significant figures]]. | ||
| + | |- | ||
| + | | style="height:20px; width:200px; text-align:left;" |'''1. State the known quantities''' | ||
| + | |||
| + | [[Density]] = 1000kg/m<sup>3</sup> | ||
| + | |||
| + | [[Gravitational Field Strength]] = 9.8N/kg | ||
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| + | [[Depth]] = 1300 | ||
| + | | style="height:20px; width:200px; text-align:left;" |'''1. State the known quantities''' | ||
| + | |||
| + | [[Density]] = 1.2kg/m<sup>3</sup> | ||
| + | |||
| + | [[Gravitational Field Strength]] = 9.8N/kg | ||
| + | |||
| + | [[Depth]] = 550m | ||
| + | | style="height:20px; width:200px; text-align:left;" |'''1. State the known quantities''' | ||
| + | |||
| + | [[Density]] = 932kg/m<sup>3</sup> | ||
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| + | [[Gravitational Field Strength]] = 9.8N/kg | ||
| + | |||
| + | [[Depth]] = 4.3cm = 0.043m | ||
| + | |- | ||
| + | | style="height:20px; width:200px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers into the [[equation]] and [[Solve (Maths)|solve]].''' | ||
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| + | <math> P = \rho g h</math> | ||
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| + | <math> P = 1000 \times 9.8 \times 1300</math> | ||
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| + | <math> P = 12740000Pa</math> | ||
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| + | <math> P \approx 13000000Pa</math> | ||
| + | |||
| + | | style="height:20px; width:200px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers into the [[equation]] and [[Solve (Maths)|solve]].''' | ||
| + | |||
| + | <math> P = \rho g h</math> | ||
| + | |||
| + | <math> P = 1.2 \times 9.8 \times 550</math> | ||
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| + | <math> P = 6468Pa</math> | ||
| + | |||
| + | Since the [[miner]] also has normal atmospheric pressure 101kPa is added. | ||
| + | |||
| + | <math> P_{Total} = 6468 + 101000</math> | ||
| + | |||
| + | |||
| + | <math> P_{Total} = 107468Pa</math> | ||
| + | |||
| + | <math> P_{Total} \approx 110000Pa</math> | ||
| + | | style="height:20px; width:200px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers into the [[equation]] and [[Solve (Maths)|solve]].''' | ||
| + | |||
| + | <math> P = \rho g h</math> | ||
| + | |||
| + | <math> P = 932 \times 9.8 \times 0.043</math> | ||
| + | |||
| + | <math> P = 392.7448Pa</math> | ||
| + | |||
| + | <math> P \approx 390Pa</math> | ||
| + | |} | ||
Revision as of 14:26, 11 February 2019
Contents
Key Stage 4
Meaning
Fluid Pressure is the force per unit area on a surface due to particles colliding with that surface.
About Fluid Pressure
- The force responsible for pressure in a fluid is the weight of all the particles above the object.
- The pressure in a fluid decreases with height and increases with depth. The deeper into a fluid the greater the pressure.
- Since pressure increases with depth then the bottom of an object will experience more pressure than the top of the object. This causes a force called upthrust.
Equation
NB: You do not need to remember this equation.
Pressure = (Density of the Fluid) x (Gravitational Field Strength) x (Depth in the fluid)
\( P = \rho g h\)
Where\[ P\] = Pressure at that depth.
\( \rho\) = Density of the fluid (This symbol is a 'rho' not a 'p')
\( g \) = Gravitational Field Strength
Calculating Pressure in a Fluid
| A whale dives to a depth of 1300m below sea level. Given the density of water is 1000kg/m3, calculate the pressure on the whale due to the water correct to two significant figures. | Normal atmospheric pressure at sea level is 101kPa. A miner works in a mine shaft 550m below the surface of the Earth. Given the density of the atmosphere at this level is 1.2kg/m3, calculate the pressure on the miner correct to two significant figures. | A piece of Lithium is submersed 4.3cm deep in oil prevent it from oxidising. The oil has a density of 932kg/m3, calculate the pressure on the Lithium due to the oil correct to two significant figures. |
| 1. State the known quantities
Density = 1000kg/m3 Gravitational Field Strength = 9.8N/kg Depth = 1300 |
1. State the known quantities
Density = 1.2kg/m3 Gravitational Field Strength = 9.8N/kg Depth = 550m |
1. State the known quantities
Density = 932kg/m3 Gravitational Field Strength = 9.8N/kg Depth = 4.3cm = 0.043m |
| 2. Substitute the numbers into the equation and solve.
\( P = \rho g h\) \( P = 1000 \times 9.8 \times 1300\) \( P = 12740000Pa\) \( P \approx 13000000Pa\) |
2. Substitute the numbers into the equation and solve.
\( P = \rho g h\) \( P = 1.2 \times 9.8 \times 550\) \( P = 6468Pa\) Since the miner also has normal atmospheric pressure 101kPa is added. \( P_{Total} = 6468 + 101000\)
\( P_{Total} \approx 110000Pa\) |
2. Substitute the numbers into the equation and solve.
\( P = \rho g h\) \( P = 932 \times 9.8 \times 0.043\) \( P = 392.7448Pa\) \( P \approx 390Pa\) |