Question

MCAT Tutorial: Introduction to Fluid Dynamics

Answer 1

\({\bf{Definitions:}}\)
- density: mass/volume, also represented by ρ (rho, not a lowercase p, be careful w/ your handwriting)
- specific gravity: density of a substance/density of water
- buoyancy: the force exerted by displaced fluid; force that causes objects to float
- Archimedes' Principle: an object in a fluid experiences a buoyant force equal to the weight of displaced fluid

Buoyant Force = ρgV, units are Newtons, can sometimes be expressed as a mass differential ρV

\({\bf{Example~Problem~1:}}\) an object has a mass of [a] grams. Submerged in water, it has an apparent mass of [b] grams. What is the buoyant force?

Take the mass differential to be [a] - [b]. Then convert the mass differential to the appropriate volume of water, and plug into Buoyant Force = ρgV

\({\bf{Example~Problem~2:}}\) Devise an experiment to determine the density of an object with known mass/weight.

1. Find the weight of the object when submerged.
2. Find the difference between the submerged and dry weight.
3. Set that difference equal to the bouyant force, using the density of water and the volume of displaced water in the equation, to solve for displaced water.
4.. Assume the object has the same V obtained in step 3.
5. Divide mass/V to find the density.

Answer 2

\({\bf{Pressure:}}\) F/A, units are pascals; atmospheric pressure = 1.01e5 Pascals

The pressure at depth h in an open, fluid-filled container is:

P_o + ρgh

where P_o is the pressure at the surface

- gauge pressure: difference between internal and external pressure
- Pascal's Principle: for a closed container, pressure applied on one part of the fluid gets applied to the rest of the fluid
P1 = P2
F1/A1 = F2/A2

- manometer: measures the atmospheric pressure via a column of mercury

Answer 3

\({\bf{Fluid~Dynamics~Equation~Dump:}}\)

mass flow, based on conservation of mass
Av = volume flow rate
A1v1 = A2v2
be careful with your calculations, make sure you are using the surface area for your calculations and not some other given quantity like radius

surface tension: gamma = F/L for a film in a loop
for a loop with length L and loop-expanding force F

Bernoulli Equation:

p1 + ρgh + (1/2)ρv^2

Poiseuille's Law: gives flow rate in termes of viscosity
viscosity has units of poise

Answer 4

\({\bf{Example~Problem~3:}}\)

Water is leaking from a tank above the ground. The tank is [a] meters deep and the hole is [b] meters from the top of the tank. The open surface area of the tank is [d] m^2. The leak is coming from a hole of [c] cm. Calculate the speed of the water and flow rate from the leak

First: A1v1 = A2v2
solve for v1 in terms of v2

use the surface area of the open surface for A1 and the area of the hole for A2.

Then plug these into the bernoulli equation

for practical purposes the pressure at the top and at the leak are equivalent and the velocity at the top of the tank is near 0 for the bernoulli equation.

Therefore

rho * g * h + (1/2)(rho)v1^2 = rho * g * h2

h2 - h1 is just the height difference between the top and the leak

so from there just plug in the v1 and solve for v2

Answer 5

Anyway, that's the end of my tutorial, I hope it was a helpful resource. Source material is the 2nd Edition Barron's Prep book for the new MCAT