CFR.7 CVS1 - Static Fluids and Fluid Pressure Notes
Static Fluids and Fluid Pressure
Fluids
A fluid conforms to the shape of its container (liquids and gases).
Study of fluids:
Hydrostatics: Fluids at rest (Pascal’s Principle, Archimedes' Principle/Buoyancy).
Hydrodynamics: Fluids in motion (Continuity equation, Bernoulli equation - the physics of the cardiovascular system).
Hydrostatics (Fluids at Rest)
Density: Mass per unit volume.
Density = \frac{mass}{volume}
Unit: kg/m^3
Water density: 1000 kg/m^3
Blood density: 1060 kg/m^3
Pressure
Pressure: Force per unit area.
Pressure = \frac{Force}{Area}
Unit: Pascal (Pa) or N/m^2
Pascal’s Principle
Any change in pressure applied to a completely enclosed fluid is transmitted undiminished to all parts of the fluid and the enclosing walls.
Example: Hydraulic lift to raise a car.
Hydraulic Lift Example
A 100 N force is applied to a 1 cm² end of a hydraulic lift.
How much force is applied to the 100 cm² end?
What mass of car would this lift?
Solution
P1 = P2
\frac{F1}{A1} = \frac{F2}{A2}
\frac{100 N}{1 cm^2} = \frac{F_2}{100 cm^2}
F_2 = 10000 N
F = mg
m = \frac{F}{g} = \frac{10000 N}{9.8 m/s^2} = 1020 kg
Clinical Applications of Pascal's Principle
Onset of Glaucoma: Approximately 5 cm³ of aqueous humor is produced daily; normal intra-ocular pressure is 12-24 mm Hg. In glaucoma, fluid cannot escape, increasing intra-ocular pressure (up to ~85 mm Hg), leading to optic nerve damage and potential blindness. Lasers can create exit channels to reduce pressure.
Protection of the Unborn Baby: Amniotic fluid diffuses the impact of sudden forces applied to a pregnant woman's abdomen.
Screening for Brain Tumors: CSF surrounds the brain and spinal cord (volume ~150 cm³). A brain tumor reduces CSF volume, increasing fluid pressure. A spinal tap connected to a water manometer between the 3rd and 4th lumbar vertebrae can measure CSF pressure. Elevated pressure may indicate a brain tumor.
Decubitus Ulcers (Bedsores): Pressure ulcers caused by prolonged periods in the same position, restricting blood flow around bony regions (elbows, heels, hips, etc.).
It is important to be able to explain how the above examples relate to Pascal's Principle.
Laplace’s Law
The larger the vessel radius, the larger the wall tension required to withstand a given internal fluid pressure.
T = P \cdot r
Where:
T is wall tension.
P is internal fluid pressure.
r is the radius of the vessel.
Laplace’s Law and Bladder Dynamics
Passive Filling: Filling the bladder with fluid exerts pressure on the wall, generating stress.
Muscular Contraction: Contraction generates pressure on the fluid within the lumen. The magnitude depends on the lumen's dimensions.
P = \frac{T}{r}
Pressure and volume law compliance: V \sim r
Application of Laplace’s Law on the Circular System
Heart Wall: The left ventricle is thicker and has a smaller radius than the right ventricle. The pressure in the left ventricle is several times larger than in the right ventricle.
Aneurysms: Ballooning of a vessel increases its radius, increasing tension on the already weakened wall, until pressure equalizes or the vessel bursts.
Enlarged Heart (Congestive Heart Failure): As the radius increases, the heart must exert a larger force to create sufficient blood pressure, resulting in damage.
Worked Example: Wall Tension in Artery
Calculate wall tension (T) in an artery of radius 0.5 cm with a blood pressure of 110 mmHg.
Given: Density of mercury (m) = 13.6 \times 10^3 kg/m^3, g = 9.8 m/s^2
Convert pressure to Pascals (Pa) using P = \rho gh
P = (13.6 \times 10^3 kg/m^3)(9.8 m/s^2)(110 \times 10^{-3} m) = 1.5 \times 10^4 Pa
Convert cm to m: 0.5 cm = 0.5 \times 10^{-2} m
Use T = P \cdot r
T = (1.5 \times 10^4 Pa)(0.5 \times 10^{-2} m) = 75 N/m
Homework: What would T be if an aneurysm formed with a radius of 3 cm?
Atmospheric Pressure
Caused by the weight of air around the Earth.
1 atm = 1.013 \times 10^5 Pa
Pressure at Depth
Consider a container (open at the top) containing a fluid of density \rho.
Pressure at level A within the fluid is the sum of atmospheric pressure (H) and the pressure due to the column of liquid above that level.
P_A = H + \rho gh
The units of pressure must be consistent when adding.
Pressure and Diving
We breathe air at atmospheric pressure, so there is equilibrium between the air outside and inside the body.
If we breathe normal atmospheric air and dive to 5 meters underwater:
Outward pressure from the body = 1 atm
External pressure = 1 atm (atmosphere) + 0.5 atm (fluid above, since 10 m of water ≈ 1 atm of pressure).
Pressure in a Tank
Tank of height h and cross-sectional area A filled with water of density \rho .
Volume: V = Ah
Mass: m = \rho V = \rho Ah
Weight: W = mg = \rho gAh
Gauge pressure (pressure due to water): P_{water} = \rho gh
Total pressure at depth h: P{total} = P{external} + \rho gh
Example: Pressure and Depth
At what depth in water will the liquid pressure equal atmospheric pressure?
\rho gh = P_{atm}
h = \frac{P_{atm}}{\rho g} = \frac{1.013 \times 10^5 Pa}{(1000 kg/m^3)(9.8 m/s^2)} \approx 10 m
At 10 m depth, the pressure is twice the atmospheric pressure.
Lung volume is reduced to ½ of its original volume at this depth.
Worked Example: Force on Aneurysm
Calculate the maximum force exerted by blood on an aneurysm with a max blood pressure of 140 mmHg and an area of 25 cm².
Convert pressure to Pascals:
P = \rho gh = (0.14 m)(13.6 \times 10^3 kg/m^3)(9.8 m/s^2) = 1.87 \times 10^4 Pa
F = PA = (1.87 \times 10^4 Pa)(25 \times 10^{-4} m^2) = 46.6 N
The large force means the aneurysm may burst.
Measurement of Pressure
Many devices used: open tube manometer (U-shaped tube filled with Hg) is simplest.
P{absolute} = P{atm} + \rho gh
P_{absolute}: Total pressure
P_{atm}: Atmospheric pressure
\rho gh: Gauge pressure
\rho: Density
h: Height of mercury (Hg).
Excess pressure above P_{atm} is gauge pressure.
Barometer and sphygmomanometer are also examples.
Blood Pressure Variation in the Body
Blood pressure varies due to \rho gh factor.
h: measured relative to heart (+h below/ -h above).
Low P{head} and high P{leg} can cause fainting and fluid build-up (oedema) through capillary walls.
Standing requires more work.
Acceleration Causing Blackout
Upward acceleration (a) augments gravity thus we have effective gravity = a+g
Pressure difference: \Delta P = \rho (a+g)h
Pressure at head reduced.
Example: a = 3g
\Delta P_{heart-head} = 1.04 \times 10^3 \times 4g \times 0.4 = 16 kPa
If pressure from the heart is 13.3 kPa, the head receives no blood, causing a blackout.
0.4 m is the approximate distance from heart to head.
Density of blood ≈ 1.04 \times 10^3 kg/m^3
Distance heart-head ≈ 0.4 m, heart-feet ≈ 1.4 m.
Homework Problems
If saline solution in a central venous pressure measurement rises to 10 cm above the right atrium level, what is the central venous pressure in mm-Hg? (Density of saline is 1.01 g/cm³).
Answer: 7.4 mm-Hg
If the heart muscle can produce a maximum wall tension in the left ventricle of 1.2 \times 10^3 N/m, calculate the pressure (in N/m^2 and mm-Hg) that the 7.0 cm radius ventricle can produce.
Answer: 3.43 \times 10^4 N/m^2, 257 mm Hg
If the distance from the heart to the head is 40 cm and from the heart to the feet is 1 m, find the pressure at the head and feet for a person in a standing position (Heart pressure = 100 mm-Hg, density of blood = 1.06 \times 10^3 kg/m^3).
Answer: approx. 69 mm-Hg (head), 178 mm-Hg (feet)
A 70 kg adult has 5 liters of blood. If the density of the blood is 1075 kg/m³, find the mass of the blood.
The bladder can hold approximately 600 ml when completely filled. If the density of urine is 1032 kg/m³, find the mass of the urine held in the bladder.
Stroke volume (SV) is the volume of blood pumped from one ventricle of the heart with each beat. It is calculated by subtracting the volume of blood in the ventricle at the end of a beat (end-systolic volume ESV) from the volume of blood just prior to the beat (end of diastolic volume EDV).
If: end-diastolic volume (EDV) = 120 ml, end-systolic volume (ESV) = 50 ml, density of blood = 1075 kg/m³
Find the mass of the blood pumped in each heartbeat.
Note: 1 ml = 1 cm³ (cc), 1 L = 1000 cm³ = 0.001 m³