PHYS110 Chapter 11: Static Fluids Comprehensive Study Guide

Fundamental Concepts of Static Fluids and Pressure

• Definition of Static Fluid Pressure:

  • Pressure ($P$) is defined as the normal force ($F$) exerted per unit surface area ($A$):     $P = \frac{F}{A}$
  • Pressure is a scalar quantity, having magnitude but no direction.

• Proportionality Relationships for Pressure:

  • Pressure is directly proportional to the applied force ($F$).
  • Pressure is inversely proportional to the contact surface area ($A$).
  • For circular contact surfaces, the area is calculated using radius ($r$) or diameter ($d$):     $A = \pi r^2$$r = \frac{d}{2}$

• Weight Force Equation:

  • The force produced by gravity on a mass ($m$) is its weight ($w$):     $F = w = m \times g$
  • Gravitational acceleration ($g$) is standardly taken as $9.81\,m/s^2$ or $9.8\,m/s^2$.

• Fluid Density:

  • Density ($\rho$) is defined as mass per unit volume ($V$):     $\rho = \frac{m}{V}$
  • Density is an intrinsic physical property dependent on the specific type of material.
  • Density of fresh water: $\rho_{\text{water}} = 1000\,kg\,m^{-3} = 1\,g\,cm^{-3}$.
  • Density of liquid mercury: $\rho_{\text{mercury}} = 13600\,kg\,m^{-3} = 13.6\,g\,cm^{-3}$.

• Compressibility Classifications:

  • Incompressible Fluid: A fluid whose density remains constant ($\rho = \text{constant}$) regardless of applied pressure.
  • Liquids are physically categorized as incompressible fluids.
  • Gases are categorized as compressible fluids because gas density varies significantly with pressure.
  • An ideal stationary fluid in static equilibrium is defined as an incompressible fluid.

Mechanical Equilibrium of Fluid Elements

• Forces Acting on a Static Fluid Element:

  • A fluid element stationary inside a fluid container experiences three primary vertical forces:
  1. Weight of the fluid element ($\mathbf{w}$): Directed downwards due to gravity.
  2. Upward contact force ($\mathbf{F}_{\text{up}}$): Pushed upward by the fluid directly below the element.
  3. Downward contact force ($\mathbf{F}_{\text{down}}$): Pushed downward by the fluid directly above the element.

• Equilibrium Condition:

  • Because the fluid element is in static mechanical equilibrium, it neither rises nor sinks.
  • The net vertical force equals zero:     $\sum F_{\text{net},y} = 0$$+F_{\text{up}} - w - F_{\text{down}} = 0$

Pascal's Law and Hydrostatic Pressure Variation

• General Form of Pascal's Law:

  • The differential form of pressure change across a vertical height change ($\Delta y$) in a uniform fluid is:     $\Delta P = -\rho \times g \times \Delta y$$P_2 - P_1 = -\rho \times g \times (y_2 - y_1)$
  • Statement of General Pascal's Law: The difference in pressure between two vertical positions in a fluid of constant density is directly proportional to the vertical distance between those positions. The proportionality factor is the product of fluid density ($\rho$) and acceleration due to gravity ($g$).
  • Primary Application: Used in its general differential form when the open surface of the fluid cannot be identified or exposed to the environment (such as blood circulating inside the closed human cardiovascular system).

• Pressure in Liquids with a Free/Visible Surface:

  • When a liquid has an open surface exposed to atmosphere at depth $d$:     $\Delta P = \rho \times g \times d$$P = P_{\text{atm}} + \rho \times g \times d$

• Key Characteristics and Limitations of Pascal's Law:

  • Does not apply to pressure variations across the atmosphere because atmospheric air is compressible and its density decreases with altitude.
  • Does not apply to gases because gases are compressible.
  • Assumes fluid density remains strictly constant as pressure increases.
  • Contains no information regarding the shape or geometry of the container; hydrostatic pressure depends purely on depth ($d$), not container width or volume.
  • In liquid water, absolute pressure increases by approximately $1\,atm$ for every $10\,m$ of depth.

Pressure Classifications and Measurement Devices

• Absolute Pressure ($P_{\text{absolute}}$):

  • The pressure relative to absolute zero pressure (perfect vacuum).
  • Absolute pressure is always positive ($P_{\text{absolute}} > 0$).
  • Relationship formula:     $P_{\text{absolute}} = P_{\text{gauge}} + P_{\text{atm}}$

• Atmospheric Pressure ($P_{\text{atm}}$):

  • The total pressure exerted by the cumulative weight of all atmospheric gases above ground level.
  • Atmospheric pressure decreases as elevation above sea level increases.
  • Always positive. Measured using a instrument called a barometer.
  • Standard values of atmospheric pressure at sea level:     $P_{\text{atm}} = 1.013 \times 10^5\,Pa$$P_{\text{atm}} = 101.3\,kPa \approx 100\,kPa$$P_{\text{atm}} = 1\,atm$$P_{\text{atm}} = 760\,mmHg$

• Gauge Pressure ($P_{\text{gauge}}$):

  • Pressure measured relative to local atmospheric pressure ($P_{\text{atm}}$).
  • Relationship formula:     $P_{\text{gauge}} = P_{\text{absolute}} - P_{\text{atm}}$
  • Sign behavior:
  • Positive ($P_{\text{gauge}} > 0$) when absolute pressure is greater than atmospheric pressure ($P_{\text{absolute}} > P_{\text{atm}}$).
  • Negative ($P_{\text{gauge}} < 0$) when absolute pressure is less than atmospheric pressure ($P_{\text{absolute}} < P_{\text{atm}}$).
  • Zero ($P_{\text{gauge}} = 0$) when absolute pressure equals atmospheric pressure ($P_{\text{absolute}} = P_{\text{atm}}$).

Medical Applications: Blood Pressure and Cardiovascular Physics

• Clinical Blood Pressure Definition:

  • Human blood pressure measured by medical personnel represents the gauge pressure of blood inside major systemic arteries at the level/height of the heart.
  • Measured in units of millimeters of mercury ($mmHg$).

• Cardiovascular System Divisions:

  • High-Pressure System:
  • Comprises the aorta, systemic arteries, arterioles, and systemic capillaries.
  • Arterial pressure fluctuates between:
    • Lower pressure limit (Diastolic pressure): $10.7\,kPa = 80\,mmHg$.
    • Upper pressure limit (Systolic pressure): $16.0\,kPa = 120\,mmHg$.
  • Low-Pressure System:
  • Comprises systemic veins and the entire pulmonary circulation.
  • Pressure fluctuates within a low range of $1.3\,kPa$ to $3.3\,kPa$ ($10\,mmHg$ to $25\,mmHg$).

• Hydrostatic Effect in Standing Human Body:

  • Standard adult male reference dimensions:
  • Total standing height: $173\,cm = 1.73\,m$
  • Vertical distance from heart to brain: $51\,cm = 0.51\,m$
  • Vertical distance from heart to feet: $122\,cm = 1.22\,m$
  • Density of human blood: $\rho_{\text{blood}} = 1.06\,g\,cm^{-3} = 1060\,kg\,m^{-3}$
  • Hydrostatic pressure differences occur across body columns according to Pascal's law.
  • Negative gauge pressures can develop in blood vessels located in head regions elevated above heart level while standing upright.
  • Systolic blood pressure typically increases with advancing age.

Pascal's Principle and Hydraulic Systems

• Pascal's Principle (Transmission of Fluid Pressure):

  • Statement: A pressure change applied anywhere to an enclosed, confined, ideal stationary fluid is transmitted undiminished (equally) throughout the entire fluid and to the walls of its container.

• Mathematical Formulation:$P_{\text{in}} = P_{\text{out}}$$\frac{F_{\text{in}}}{A_{\text{in}}} = \frac{F_{\text{out}}}{A_{\text{out}}}$

• Hydraulic Lift Operations:

  • Area relationship: The output piston area is significantly larger than the input piston area ($A_{\text{out}} \gg A_{\text{in}}$).
  • Force amplification relationship: Consequently, the output force is magnified significantly relative to input force ($F_{\text{out}} \gg F_{\text{in}}$).
  • Mechanism: Applying a small effort force ($F_{\text{in}}$) to a small circular piston area ($A_{\text{in}}$) creates an elevated fluid pressure that exerts a massive upward lifting force ($F_{\text{out}}$) on a large heavy-duty piston area ($A_{\text{out}}$).

Complete Worked Problems and Numerical Solutions

• Example: Force Exerted by Atmospheric Pressure on Human Body Area

  • Problem Statement: Find the force exerted on $1\,cm^2$ of a human body surface by standard atmospheric pressure ($P_{\text{atm}} = 1.013 \times 10^5\,Pa$).
  • Given Data: $A = 1\,cm^2 = 1 \times 10^{-4}\,m^2$, $P_{\text{atm}} = 1.013 \times 10^5\,Pa$.
  • Step-by-Step Calculation:$F = P_{\text{atm}} \times A$$F = (1.013 \times 10^5\,Pa) \times (1 \times 10^{-4}\,m^2)$$F = 10.13\,N$
  • Answer Options: a. $100\,N$ | b. $1013\,N$ | c. $9.81\,N$ | d. $10.13\,N$
  • Correct Option: d ($10.13\,N$).

• Problem 11.4: Pressure Increase in Syringe Fluid

  • Problem Statement: What is the pressure increase in the fluid in a syringe when a force of $50.0\,N$ is applied to its circular plunger of radius $1.25\,cm$?
  • Given Data: $F = 50.0\,N$, $r = 1.25\,cm = 0.0125\,m$.
  • Step-by-Step Calculation:$A = \pi \times r^2 = \pi \times (0.0125\,m)^2 = 4.9087 \times 10^{-4}\,m^2$$\Delta P = \frac{F}{A} = \frac{50.0\,N}{4.9087 \times 10^{-4}\,m^2} = 1.0185 \times 10^5\,Pa = 101.85\,kPa \approx 1.02 \times 10^5\,Pa = 102.04\,kPa$
  • Answer Options: a. $1.02 \times 10^5\,Pa$ | b. $102.04\,kPa$ | c. all of these | d. none
  • Correct Option: c (all of these, as $1.02 \times 10^5\,Pa$ equals $102\,kPa$).

• Example 11.2(c): Floor Pressure Under Standing Man

  • Problem Statement: Determine the pressure a standard man ($70\,kg$) exerts on the floor when standing, given his feet contact area is $0.04\,m^2$.
  • Given Data: $m = 70\,kg$, $A = 0.04\,m^2$, g = 9.81\,m/s^2$.\n - *Step-by-Step Calculation:*\n    w = m \times g = 70\,kg \times 9.81\,m/s^2 = 686.7\,N\n    P = \frac{w}{A} = \frac{686.7\,N}{0.04\,m^2} = 17167.5\,Pa \approx 1.7 \times 10^4\,Pa\n - *Answer Options:* a. 17\,Pa$| b.$180\,Pa$| c.$3000\,Pa$| d.$1.7 \times 10^4\,Pa\n - *Correct Option:* d (1.7 \times 10^4\,Pa).\n\n- **Example 11.1: Absolute Pressure at 10.0 m Depth in Lake**\n - *Problem Statement:* What is the pressure 10.0\,m$below the surface of a lake? Density of water$\rho = 1000\,kg/m^3$,$P_{\text{atm}} = 1.013 \times 10^5\,Pa$.
  • Given Data: $d = 10.0\,m$, $\rho = 1000\,kg/m^3$, P_{\text{atm}} = 1.013 \times 10^5\,Pa$.\n - *Step-by-Step Calculation:*\n    P_{\text{gauge}} = \rho \times g \times d = 1000\,kg/m^3 \times 9.81\,m/s^2 \times 10.0\,m = 9.81 \times 10^4\,Pa\n    P_{\text{absolute}} = P_{\text{atm}} + P_{\text{gauge}} = 1.013 \times 10^5\,Pa + 0.981 \times 10^5\,Pa = 1.994 \times 10^5\,Pa \approx 1.99 \times 10^5\,Pa \approx 2\,atm\n - *Answer Options:* a. 1.99 \times 10^5\,Pa$| b.$2\,atm | c. all of these | d. none\n - *Correct Option:* c (all of these).\n\n- **Concept Question 11.1: Nautilus Night Feeding Depth Pressure**\n - *Problem Statement:* The nautilus rises to a depth of 60\,m below the surface of the Pacific Ocean at night. What is the water pressure at that depth?\n - *Calculation:* Every 10\,m$of water depth adds approximately$1\,atm$of hydrostatic pressure. At$60\,m$,$P_{\text{gauge}} = 6\,atm$, making absolute water pressure$7\,atm.\n - *Answer Options:* (a) 3\,atm$| (b)$4\,atm$| (c)$5\,atm$| (d)$6\,atm$| (e)$7\,atm\n - *Correct Answer:* (e) 7\,atm$(absolute pressure) or (d)$6\,atm (gauge water pressure).\n\n- **Concept Question 11.2: Nautilus Daytime Migration Pressure Variation**\n - *Problem Statement:* The nautilus dives to depths up to 420\,m during daytime. What pressure variation occurs during this daily vertical migration?\n - *Calculation:*\n    \Delta P = \frac{420\,m}{10\,m/atm} = 42\,atm\n - *Answer Options:* (a) Less than 30\,atm$| (b)$32\,atm$| (c)$34\,atm$| (d)$36\,atm$| (e)$38\,atm$| (f)$40\,atm$| (g) More than$40\,atm\n - *Correct Option:* (g) More than 40\,atm$($42\,atm).\n\n- **Example 11.2(a): Column Height of Mercury Barometer**\n - *Problem Statement:* Determine height (h$) for a mercury barometer when atmospheric pressure is$1.013 \times 10^5\,Pa$, density$\rho_{\text{Hg}} = 13.6\,g/cm^3 = 13600\,kg/m^3$.
  • Step-by-Step Calculation:$P_{\text{atm}} = \rho_{\text{Hg}} \times g \times h$$h = \frac{P_{\text{atm}}}{\rho_{\text{Hg}} \times g} = \frac{1.013 \times 10^5\,Pa}{13600\,kg/m^3 \times 9.81\,m/s^2} = 0.760\,m = 760\,mm$
  • Answer Options: a. $760\,mm$ | b. $7\,mm$ | c. $60\,mm$ | d. none of these
  • Correct Option: a ($760\,mm$).

• Example 11.2(b): Column Height of Water Barometer

  • Problem Statement: Determine height ($h$) for a water barometer when atmospheric pressure is $1.013 \times 10^5\,Pa$, density \rho_{\text{water}} = 1.0\,g/cm^3 = 1000\,kg/m^3$.\n - *Step-by-Step Calculation:*\n    h = \frac{P_{\text{atm}}}{\rho_{\text{water}} \times g} = \frac{1.013 \times 10^5\,Pa}{1000\,kg/m^3 \times 9.81\,m/s^2} = 10.33\,m\n - *Answer Options:* a. 760\,m$| b.$103.3\,m$| c.$10.33\,m$| d.$200\,m\n - *Correct Option:* c (10.33\,m).\n\n- **Example: Pressure on Table Under Cylinder Container**\n - *Problem Statement:* A cylindrical container on a table is filled with water. According to Pascal's law, hydrostatic pressure on the table depends on which parameter of water?\n - *Answer Options:* a. Volume | b. Area | c. Velocity | d. Height\n - *Correct Option:* d (Height, as P = \rho \times g \times h).\n\n- **Example 11.3(a): Brain-to-Feet Pressure Difference in Standing Standard Man**\n - *Problem Statement:* Calculate the pressure difference between brain and feet in a standing standard man (173\,cm$tall), blood density$\rho_{\text{blood}} = 1.06\,g/cm^3 = 1060\,kg/m^3$.
  • Step-by-Step Calculation:$h = 173\,cm = 1.73\,m$$\Delta P = \rho \times g \times h = 1060\,kg/m^3 \times 9.81\,m/s^2 \times 1.73\,m = 17989.86\,Pa$$\Delta P = \frac{17989.86\,Pa}{133.322\,Pa/mmHg} = 134.93\,mmHg \approx 134.8\,mmHg$
  • Answer Options: a. $200\,mmHg$ | b. $134.8\,mmHg$ | c. $300\,mmHg$ | d. $100\,mmHg$
  • Correct Option: b ($134.8\,mmHg$).

• Example: Gauge Pressure Under Sea Level Given Absolute Pressure

  • Problem Statement: Find the gauge pressure under sea level if absolute pressure at that depth is $250\,kPa$.
  • Step-by-Step Calculation:$P_{\text{gauge}} = P_{\text{absolute}} - P_{\text{atm}} = 250\,kPa - 100\,kPa = 150\,kPa$
  • Answer Options: a. $250\,kPa$ | b. $100\,kPa$ | c. $150\,kPa$ | d. none
  • Correct Option: c ($150\,kPa$).

• Example: Fish Swimming Depth Pressure Change

  • Problem Statement: Find the change in pressure if a fish swims from depth $10\,m$ below water surface to depth $50\,m$.
  • Step-by-Step Calculation:$\Delta d = 50\,m - 10\,m = 40\,m$$\Delta P = \frac{40\,m}{10\,m/atm} = 4\,atm \approx 3.9\,atm$
  • Answer Options: a. $3.9\,atm$ | b. $40\,atm$ | c. $2\,atm$ | d. $5\,atm$
  • Correct Option: a ($3.9\,atm$).

• Concept Question 11.3: Oceanographer Gauge Pressure Reporting

  • (a) What value reported for surface water?
  • Calculation: At surface, P_{\text{absolute}} = P_{\text{atm}} \implies P_{\text{gauge}} = P_{\text{absolute}} - P_{\text{atm}} = 0$.\n - Options: a. positive | b. negative | c. zero | d. none\n - Correct Option: c (zero).\n - *(b) When would they report negative values underwater?*\n - Calculation: Underwater, pressure increases continuously with depth (P > P_{\text{atm}}), so gauge pressure underwater is strictly positive.\n - Options: a. at large depth | b. at surface | c. nowhere | d. everywhere\n - Correct Option: c (nowhere).\n\n- **Problem 11.7: Hydraulic Lift Car Lift Force Calculation**\n - *Problem Statement:* In a hydraulic lift, the diameter of the larger piston is 0.30\,m$, and the diameter of the small piston is$0.030\,m$. Determine the force needed on the small piston to lift a car of mass$1200\,kg.\n - *Given Data:* D_{\text{out}} = 0.30\,m$,$D_{\text{in}} = 0.030\,m$,$m = 1200\,kg.\n - *Step-by-Step Calculation:*\n    F_{\text{out}} = m \times g = 1200\,kg \times 9.8\,m/s^2 = 11760\,N\n    \frac{A_{\text{in}}}{A_{\text{out}}} = \left(\frac{D_{\text{in}}}{D_{\text{out}}}\right)^2 = \left(\frac{0.030\,m}{0.30\,m}\right)^2 = (0.1)^2 = 0.01\n    F_{\text{in}} = F_{\text{out}} \times \left(\frac{A_{\text{in}}}{A_{\text{out}}}\right) = 11760\,N \times 0.01 = 117.6\,N\n - *Answer Options:* a. 117.6\,N$| b.$500\,N$| c.$20\,N$| d.$300\,N\n - *Correct Option:* a (117.6\,N).\n\n# Physics Unit Conversions Reference\n\n- **Mass Conversions:**\n - 1\,kg = 1000\,g = 10^3\,g\n - 1\,g = 10^{-3}\,kg\n\n- **Volume Conversions:**\n - 1\,m^3 = 1000\,L = 10^6\,cm^3\n - 1\,L = 10^{-3}\,m^3 = 1000\,cm^3\n - 1\,cm^3 = 10^{-6}\,m^3\n\n- **Length Conversions:**\n - 1\,m = 100\,cm = 1000\,mm\n - 1\,cm = 10^{-2}\,m\n - 1\,mm = 10^{-3}\,m\n\n- **Area Conversions:**\n - 1\,m^2 = 10^4\,cm^2 = 10^6\,mm^2\n - 1\,cm^2 = 10^{-4}\,m^2\n - 1\,mm^2 = 10^{-6}\,m^2\n\n- **Force Conversions:**\n - 1\,kN = 1000\,N = 10^3\,N\n\n- **Pressure Conversions:**\n - 1\,Pa = 1\,N/m^2\n - 1\,kPa = 1000\,Pa = 10^3\,Pa\n - 1\,atm = 1.013 \times 10^5\,Pa = 101.3\,kPa \approx 760\,mmHg\n - 1\,mmHg = 133.322\,Pa$$