Density and Pressure

Fluid density, pressure, Pascal's principle, and pressure variation with depth

💧 Density and Pressure

Introduction

Fluid mechanics begins with understanding the fundamental properties of fluids: density and pressure. These concepts form the foundation for analyzing fluid behavior in both static and dynamic situations.

Density

Density is mass per unit volume, measuring how tightly matter is packed:

ρ=mV\rho = \frac{m}{V}

where:

  • ρ\rho (rho) = density (kg/m³)
  • mm = mass (kg)
  • VV = volume (m³)

Common Densities:

| Material | Density (kg/m³) | |----------|----------------| | Water | 1000 | | Ice | 917 | | Air (STP)| 1.29 | | Mercury | 13,600 | | Aluminum | 2700 | | Gold | 19,300 |

Specific Gravity

Specific Gravity (SG) is the ratio of a substance's density to water's density:

SG=ρsubstanceρwaterSG = \frac{\rho_{substance}}{\rho_{water}}

  • Dimensionless (no units)
  • SG < 1: floats in water
  • SG > 1: sinks in water

Pressure

Pressure is force per unit area:

P=FAP = \frac{F}{A}

where:

  • PP = pressure (Pa or N/m²)
  • FF = force perpendicular to surface (N)
  • AA = area (m²)

Common Units:

  • Pascal (Pa): 1 Pa = 1 N/m²
  • Atmosphere (atm): 1 atm = 101,325 Pa ≈ 101 kPa
  • Bar: 1 bar = 100,000 Pa
  • mmHg: 760 mmHg = 1 atm
  • psi: 14.7 psi = 1 atm

Pressure in Fluids

Unlike solids, fluids cannot sustain shear stress - they flow. This creates unique pressure properties:

  1. Pressure acts perpendicular to surfaces - fluid pressure always pushes normal to any surface

  2. Pressure is isotropic - equal in all directions at a point

  3. Pressure increases with depth

Pressure vs. Depth

For a fluid at rest, pressure increases linearly with depth:

P=P0+ρghP = P_0 + \rho g h

where:

  • PP = pressure at depth hh
  • P0P_0 = pressure at surface (usually atmospheric)
  • ρ\rho = fluid density (kg/m³)
  • gg = 9.8 m/s²
  • hh = depth below surface (m)

💡 Key Insight: Pressure depends only on vertical depth, not container shape!

Gauge vs. Absolute Pressure

  • Absolute pressure: Total pressure including atmosphere Pabsolute=Patm+PgaugeP_{absolute} = P_{atm} + P_{gauge}

  • Gauge pressure: Pressure relative to atmosphere Pgauge=ρghP_{gauge} = \rho g h

Most pressure gauges (like tire gauges) read gauge pressure.

Pascal's Principle

Pascal's Principle: A change in pressure applied to an enclosed fluid is transmitted undiminished to every point in the fluid.

Hydraulic Systems

This principle enables hydraulic lifts:

F1A1=F2A2\frac{F_1}{A_1} = \frac{F_2}{A_2}

Force multiplication: F2=F1A2A1F_2 = F_1 \frac{A_2}{A_1}

Trade-off: Volume conservation means: d1A1=d2A2d_1 A_1 = d_2 A_2

Small piston moves far, large piston moves little distance.

Atmospheric Pressure

At sea level:

Patm=101,325 Pa=101.3 kPa=1 atmP_{atm} = 101,325 \text{ Pa} = 101.3 \text{ kPa} = 1 \text{ atm}

This is equivalent to:

  • 10.3 m column of water
  • 760 mm column of mercury

Atmospheric pressure decreases with altitude.

Problem-Solving Strategy

  1. Identify pressure type needed (gauge or absolute)
  2. Choose reference point (usually surface)
  3. Apply P=P0+ρghP = P_0 + \rho g h
  4. Remember hh is vertical depth only
  5. Check units (convert to SI)

Common Mistakes

❌ Forgetting atmospheric pressure in absolute calculations ❌ Using horizontal distance for hh ❌ Mixing gauge and absolute pressure ❌ Unit confusion (1 atm ≠ 1 Pa) ❌ Thinking pressure depends on container shape

📚 Practice Problems

1Problem 1easy

❓ Question:

A cube of aluminum has sides of length 5.0 cm. The density of aluminum is 2700 kg/m³. What is the mass of the cube?

💡 Show Solution

Given:

  • Side length: L=5.0L = 5.0 cm =0.050= 0.050 m
  • Density: ρAl=2700\rho_{Al} = 2700 kg/m³

Find: Mass mm

Solution:

Step 1: Calculate volume of cube. V=L3=(0.050)3=1.25×10−4 m3V = L^3 = (0.050)^3 = 1.25 \times 10^{-4} \text{ m}^3

Step 2: Use density formula. m=ρV=(2700)(1.25×10−4)=0.338 kgm = \rho V = (2700)(1.25 \times 10^{-4}) = 0.338 \text{ kg}

Answer: 338 g

2Problem 2easy

❓ Question:

A cube of aluminum has sides of length 5.0 cm. The density of aluminum is 2700 kg/m³. What is the mass of the cube?

💡 Show Solution

Given:

  • Side length: L=5.0L = 5.0 cm =0.050= 0.050 m
  • Density: ρAl=2700\rho_{Al} = 2700 kg/m³

Find: Mass mm

Solution:

Step 1: Calculate volume of cube. V=L3=(0.050)3=1.25×10−4 m3V = L^3 = (0.050)^3 = 1.25 \times 10^{-4} \text{ m}^3

Step 2: Use density formula. m=ρV=(2700)(1.25×10−4)=0.338 kgm = \rho V = (2700)(1.25 \times 10^{-4}) = 0.338 \text{ kg}

Answer: 338 g

3Problem 3medium

❓ Question:

A submarine is at a depth of 200 m below the ocean surface. The density of seawater is 1025 kg/m³. What is the absolute pressure at this depth? What is the gauge pressure?

💡 Show Solution

Given:

  • Depth: h=200h = 200 m
  • Seawater density: ρ=1025\rho = 1025 kg/m³
  • g=9.8g = 9.8 m/s²
  • Patm=1.01×105P_{atm} = 1.01 \times 10^5 Pa

Find: Absolute and gauge pressure

Solution:

Step 1: Calculate gauge pressure (water only). Pgauge=ρgh=(1025)(9.8)(200)P_{gauge} = \rho g h = (1025)(9.8)(200) Pgauge=2.01×106 Pa=2.01 MPaP_{gauge} = 2.01 \times 10^6 \text{ Pa} = 2.01 \text{ MPa}

Step 2: Calculate absolute pressure (total). Pabsolute=Patm+PgaugeP_{absolute} = P_{atm} + P_{gauge} Pabsolute=1.01×105+2.01×106P_{absolute} = 1.01 \times 10^5 + 2.01 \times 10^6 Pabsolute=2.11×106 Pa=2.11 MPaP_{absolute} = 2.11 \times 10^6 \text{ Pa} = 2.11 \text{ MPa}

Step 3: Convert to atmospheres. Pabsolute=2.11×1061.01×105≈20.9 atmP_{absolute} = \frac{2.11 \times 10^6}{1.01 \times 10^5} \approx 20.9 \text{ atm}

Answer:

  • Gauge: 2.01 MPa
  • Absolute: 2.11 MPa or 20.9 atm

4Problem 4medium

❓ Question:

A submarine is at a depth of 200 m below the ocean surface. The density of seawater is 1025 kg/m³. What is the absolute pressure at this depth? What is the gauge pressure?

💡 Show Solution

Given:

  • Depth: h=200h = 200 m
  • Seawater density: ρ=1025\rho = 1025 kg/m³
  • g=9.8g = 9.8 m/s²
  • Patm=1.01×105P_{atm} = 1.01 \times 10^5 Pa

Find: Absolute and gauge pressure

Solution:

Step 1: Calculate gauge pressure (water only). Pgauge=ρgh=(1025)(9.8)(200)P_{gauge} = \rho g h = (1025)(9.8)(200) Pgauge=2.01×106 Pa=2.01 MPaP_{gauge} = 2.01 \times 10^6 \text{ Pa} = 2.01 \text{ MPa}

Step 2: Calculate absolute pressure (total). Pabsolute=Patm+PgaugeP_{absolute} = P_{atm} + P_{gauge} Pabsolute=1.01×105+2.01×106P_{absolute} = 1.01 \times 10^5 + 2.01 \times 10^6 Pabsolute=2.11×106 Pa=2.11 MPaP_{absolute} = 2.11 \times 10^6 \text{ Pa} = 2.11 \text{ MPa}

Step 3: Convert to atmospheres. Pabsolute=2.11×1061.01×105≈20.9 atmP_{absolute} = \frac{2.11 \times 10^6}{1.01 \times 10^5} \approx 20.9 \text{ atm}

Answer:

  • Gauge: 2.01 MPa
  • Absolute: 2.11 MPa or 20.9 atm

5Problem 5medium

❓ Question:

A cube of metal with sides of 5.0 cm has a mass of 1.5 kg. (a) What is its density? (b) Will it float in water (ρ_water = 1000 kg/m³)? (c) What is the pressure at the bottom of the cube when it rests on a table?

💡 Show Solution

Solution:

Given: side = 5.0 cm = 0.050 m, m = 1.5 kg

(a) Density: Volume = (0.050)³ = 1.25 × 10⁻⁴ m³ ρ = m/V = 1.5/(1.25 × 10⁻⁴) = 12,000 kg/m³

(b) Will it float? ρ_metal = 12,000 kg/m³ > ρ_water = 1000 kg/m³ No, it will sink (denser than water)

(c) Pressure at bottom: Weight: W = mg = 1.5 × 10 = 15 N Area: A = (0.050)² = 2.5 × 10⁻³ m² P = F/A = 15/(2.5 × 10⁻³) = 6,000 Pa or 6.0 kPa

6Problem 6medium

❓ Question:

A cube of metal with sides of 5.0 cm has a mass of 1.5 kg. (a) What is its density? (b) Will it float in water (ρ_water = 1000 kg/m³)? (c) What is the pressure at the bottom of the cube when it rests on a table?

💡 Show Solution

Solution:

Given: side = 5.0 cm = 0.050 m, m = 1.5 kg

(a) Density: Volume = (0.050)³ = 1.25 × 10⁻⁴ m³ ρ = m/V = 1.5/(1.25 × 10⁻⁴) = 12,000 kg/m³

(b) Will it float? ρ_metal = 12,000 kg/m³ > ρ_water = 1000 kg/m³ No, it will sink (denser than water)

(c) Pressure at bottom: Weight: W = mg = 1.5 × 10 = 15 N Area: A = (0.050)² = 2.5 × 10⁻³ m² P = F/A = 15/(2.5 × 10⁻³) = 6,000 Pa or 6.0 kPa

7Problem 7hard

❓ Question:

A swimming pool is 2.5 m deep. (a) What is the absolute pressure at the bottom? (b) What is the gauge pressure at the bottom? (c) What is the force on a 2.0 m × 1.0 m rectangular section at the bottom? Use ρ_water = 1000 kg/m³, P_atm = 1.01 × 10⁵ Pa, g = 10 m/s².

💡 Show Solution

Solution:

Given: h = 2.5 m, ρ = 1000 kg/m³, P_atm = 1.01 × 10⁵ Pa

(a) Absolute pressure: P = P_atm + ρgh P = 1.01 × 10⁵ + (1000)(10)(2.5) P = 1.01 × 10⁵ + 2.5 × 10⁴ P = 1.26 × 10⁵ Pa or 126 kPa

(b) Gauge pressure: P_gauge = ρgh = (1000)(10)(2.5) = 2.5 × 10⁴ Pa or 25 kPa

(c) Force on bottom: Area = 2.0 × 1.0 = 2.0 m² F = PA = (1.26 × 10⁵)(2.0) F = 2.52 × 10⁵ N or 252 kN

Note: If using gauge pressure, F = (2.5 × 10⁴)(2.0) = 5.0 × 10⁴ N (force due to water only)

8Problem 8hard

❓ Question:

A swimming pool is 2.5 m deep. (a) What is the absolute pressure at the bottom? (b) What is the gauge pressure at the bottom? (c) What is the force on a 2.0 m × 1.0 m rectangular section at the bottom? Use ρ_water = 1000 kg/m³, P_atm = 1.01 × 10⁵ Pa, g = 10 m/s².

💡 Show Solution

Solution:

Given: h = 2.5 m, ρ = 1000 kg/m³, P_atm = 1.01 × 10⁵ Pa

(a) Absolute pressure: P = P_atm + ρgh P = 1.01 × 10⁵ + (1000)(10)(2.5) P = 1.01 × 10⁵ + 2.5 × 10⁴ P = 1.26 × 10⁵ Pa or 126 kPa

(b) Gauge pressure: P_gauge = ρgh = (1000)(10)(2.5) = 2.5 × 10⁴ Pa or 25 kPa

(c) Force on bottom: Area = 2.0 × 1.0 = 2.0 m² F = PA = (1.26 × 10⁵)(2.0) F = 2.52 × 10⁵ N or 252 kN

Note: If using gauge pressure, F = (2.5 × 10⁴)(2.0) = 5.0 × 10⁴ N (force due to water only)

9Problem 9hard

❓ Question:

A hydraulic lift has an input piston (diameter 5.0 cm) and output piston (diameter 30 cm). (a) What force must be applied to lift a 2000 kg car? (b) If the input piston is pushed down 20 cm, how far does the car rise?

💡 Show Solution

Given:

  • d1=5.0d_1 = 5.0 cm =0.050= 0.050 m, d2=30d_2 = 30 cm =0.30= 0.30 m
  • Car mass: m=2000m = 2000 kg

Solution:

Part (a): Find input force

Step 1: Calculate areas. A1=πr12=π(0.025)2=1.96×10−3 m2A_1 = \pi r_1^2 = \pi (0.025)^2 = 1.96 \times 10^{-3} \text{ m}^2 A2=πr22=π(0.15)2=7.07×10−2 m2A_2 = \pi r_2^2 = \pi (0.15)^2 = 7.07 \times 10^{-2} \text{ m}^2

Step 2: Find output force (car weight). F2=mg=(2000)(9.8)=19,600 NF_2 = mg = (2000)(9.8) = 19,600 \text{ N}

Step 3: Apply Pascal's principle. F1=F2A1A2=(19,600)1.96×10−37.07×10−2F_1 = F_2 \frac{A_1}{A_2} = (19,600) \frac{1.96 \times 10^{-3}}{7.07 \times 10^{-2}} F1=543 NF_1 = 543 \text{ N}

Part (b): Find output displacement

Step 4: Use volume conservation. A1d1=A2d2A_1 d_1 = A_2 d_2 d2=d1A1A2=(0.20)1.96×10−37.07×10−2d_2 = d_1 \frac{A_1}{A_2} = (0.20) \frac{1.96 \times 10^{-3}}{7.07 \times 10^{-2}} d2=0.0055 m=5.5 mmd_2 = 0.0055 \text{ m} = 5.5 \text{ mm}

Answer:

  • (a) Input force: 543 N
  • (b) Car rises: 5.5 mm

Mechanical advantage = 36, but distance disadvantage = 36. Energy conserved!

10Problem 10hard

❓ Question:

A hydraulic lift has an input piston (diameter 5.0 cm) and output piston (diameter 30 cm). (a) What force must be applied to lift a 2000 kg car? (b) If the input piston is pushed down 20 cm, how far does the car rise?

💡 Show Solution

Given:

  • d1=5.0d_1 = 5.0 cm =0.050= 0.050 m, d2=30d_2 = 30 cm =0.30= 0.30 m
  • Car mass: m=2000m = 2000 kg

Solution:

Part (a): Find input force

Step 1: Calculate areas. A1=πr12=π(0.025)2=1.96×10−3 m2A_1 = \pi r_1^2 = \pi (0.025)^2 = 1.96 \times 10^{-3} \text{ m}^2 A2=πr22=π(0.15)2=7.07×10−2 m2A_2 = \pi r_2^2 = \pi (0.15)^2 = 7.07 \times 10^{-2} \text{ m}^2

Step 2: Find output force (car weight). F2=mg=(2000)(9.8)=19,600 NF_2 = mg = (2000)(9.8) = 19,600 \text{ N}

Step 3: Apply Pascal's principle. F1=F2A1A2=(19,600)1.96×10−37.07×10−2F_1 = F_2 \frac{A_1}{A_2} = (19,600) \frac{1.96 \times 10^{-3}}{7.07 \times 10^{-2}} F1=543 NF_1 = 543 \text{ N}

Part (b): Find output displacement

Step 4: Use volume conservation. A1d1=A2d2A_1 d_1 = A_2 d_2 d2=d1A1A2=(0.20)1.96×10−37.07×10−2d_2 = d_1 \frac{A_1}{A_2} = (0.20) \frac{1.96 \times 10^{-3}}{7.07 \times 10^{-2}} d2=0.0055 m=5.5 mmd_2 = 0.0055 \text{ m} = 5.5 \text{ mm}

Answer:

  • (a) Input force: 543 N
  • (b) Car rises: 5.5 mm

Mechanical advantage = 36, but distance disadvantage = 36. Energy conserved!

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