Fluids and solids

Question 1

Give the formula for density. Is this definition true for all states of matter?

Answer 1

Density = mass/volume

True for solids, fluids and gases (all phases).

Question 2

What is specific gravity (SG) and how is it calculated?

Answer 2

SG = (density of a substance)/(density of water)

Often the same as its density due to the fact that the density of water is about 1 g/mL

SG is dimensionless (no units) and therefore a good standard measure of relative density

Question 3

Define pressure

Answer 3

Pressure is defined as the force (F) per unit area (A)

P = F/A

The force (F) is the normal (perpendicular) force to the area.

The SI unit for pressure is the pascal (1 Pa = 1 N/m^2)

Question 4

Give the formulations of pressure for potential energy/unit volume (5)

Answer 4

P = F/A = mg/A = (mg/a)/(h/h) = mgh/v = pgh

p = density
h = depth below surface

Question 5

What are 6 characteristics of force and pressure on incompressible liquid fluids?

Answer 5

1. Forces exerted by fluids are always perpendicular to the surface of the container
2. The fluid pressure (hydrostatic) is directly proportional to the depth of the fluid and to its density
3. At any particular depth, the pressure of the fluid is the same in all directions
4. Fluid pressure is independent of the shape or area of its container
5. An external pressure applied to an enclosed fluid is transmitted uniformly throughout the volume of the liquid (Pascal’s law)
6. An object which is completely or partially submerged in a fluid experiences an upward force equal to the weight of the fluid displaced (Archimedes’ principle)

Question 6

How can Archimedes principle be used to calculate specific gravity? If SG = 0.90, how much of the object is below the surface of water?

Answer 6

Specific gravity is equivalent to the fraction of the height of a buoyant object below the surface of the fluid.

Thus, if SG = 0.90, then 90% of the height of the object would be immersed in water.

Therefore, less dense objects float

Question 7

Give the units and equivalencies for pressure in SI, atm, bars, mmHg torr and imperial

Answer 7

SI: 1 Pa = 1 N/m^2

1.00 atm = 1.01x10^5 Pa = 1.01 bar = 760 mmHg = 760 torr = 14.7 lb/in^2

Question 8

What two equations describe fluids in motion? Describe fluids in motion in terms of streamline and turbulent flow.

Answer 8

• The continuity equation
• Bernoulli’s equation

Fluids in motion are assumed to have streamline (laminar flow) which means that the motion of every particle in the fluid follows the same path. Turbulent flow occurs when that definition cannot be applied (resulting in energy dissipation from collisions and increase in frictional drag)

Question 9

What is viscosity (η)? What is it measuring?

Answer 9

Analogous to friction between moving solids. It is resistance to flow (for layers of a fluid in streamline flow) past each other.

Viscosity (η) results in dissipation of mechanical energy. As one layer flows over another, the second is set into motion as energy is transmitted to it (transfer of momentum)

The greater the transfer of this momentum from one layer to another, the more energy that is lost and the slower th layers move.

η is the measure of the efficiency of transfer of this momentum.

Question 10

What does a high viscosity coefficient (η) indicate about a fluid?

Answer 10

Higher transfer of momentum and greater loss of mechanical energy (and thus loss of velocity)

Question 11

How can the streamline or turbidity of a fluid in motion be measured?

Answer 11

Reynold’s Number (R)

R = vdρ / n

v: velocity of flow
d: diameter of the tube
ρ: density of the fluid
η: viscosity coefficent

laminar flow occurs when R < 2300 and turbulent flow occurs when R > 4000. In the interval between 2300 and 4000, laminar and turbulent flows are possible and are called “transition” flows, depending on other factors, such as pipe roughness and flow uniformity.. As v, d, ρ and/or η increases, the flow becomes more turbulent.

Question 12

Describe cohesive and adhesive forces of fluids

Answer 12

Cohesive forces: fluid molecules attract each other

Adhesive forces: attractive forces between fluid and surface

Question 13

If a liquid is lining an object, cohesive forces cause the molecules to retract to the smallest possible surface area. What is the consequence of this?

(give two formulas for surface tension)

Answer 13

This creates a potential energy present in the surface.

This is proportional to the surface area (A)

Ep = γA

γ: surface tension = Ep/A = joules/m^2
or 
γ = F/I
F: force of contraction of the surface
I: length along surface

Question 14

If a tube is inserted into a body of fluid, what will happen if:

cohesive forces > adhesive forces
cohesive forces < adhesive forces

Answer 14

cohesive forces > adhesive forces
- Liquid is lower in tube relative to surface of body of fluid

adhesive forces < cohesive forces
- Liquid is higher in tube relative to surface of body of fluid

Question 15

What is stress on solid objects defined as? Give the definition of the modulus of elasticity (ME)

Answer 15

The ration of the force to the area over which it acts.

ME = Stress/strain

Strain is defined as the relative change in dimensions or shape of an object caused by stress

Question 16

List three types of stress on solid objects

• Tensile stress
• Compressive stress
• Shearing stress

Answer 16

• Tensile stress
• Compressive stress
• Shearing stress

Question 17

Give the two commonly used moduli of elasticity for solid objects

Answer 17

1. Young’s Modulus (Y) for compressive or tensile stress

Y = (longitudinal stress)/(longitudinal strain)
Y = (F/A)/(ΔI/I) = (FxI)/(AΔI)

Where I = length

2. Shear modulus (S) or the modulus of rigidity

S = shearing stress/shearing strain 
S = (F/A)/tanΦ

Question 18

There are two liquids in two containers. The second liquid is twice as dense as the first one.

Pressure measurements at different depths are taken for the first liquid:

• 5 cm depth = 250 N/m^2 pressure
• 10 cm depth = 450 N/m^2 pressure

What is the pressure of the second liquid at a depth of 10 cm?

Answer 18

Because the density is twice as much, the pressure due to gravitational forces of the liquid is twice is much. However, the data implies an extra pressure of 50 N/m^2 at zero depth, so this must be added (400 x 2 + 50 = 850).

850 N/m^2 is the answer.

Question 19

An object with 15 grams mass is immersed in benzene and suffers an apparent loss of mass of 5 grams. What is the approximate specific gravity of the object? (Specific gravity of benzene = .7)

Answer 19

2.1

The buoyant force on an object immersed in a fluid is equal to the weight of the fluid displaced by the object (Archimedes’s principle).

There were 5g of liquid displaced, thus the ratio of object mass to fluid mass is 15/5 = 3. The specific gravity of the object (mass per unit volume compared to water) is three times the specific gravity of bense (3 x .7 = 2.1) because the volumes of object and displaced liquid are equal.

Question 20

As a balloon rises in water, the temperature of the gas inside falls. Why is this?

Answer 20

As the pressure in the balloon decreases, the gas expands and does work on the balloon/water, thereby decreasing its internal energy.

Question 21

A large cylindrical concrete tank is filled with water and has a horizontal facing near the bottom. If the valve is opened to drain the tank, where is the speed of the flowing water the greatest?

Answer 21

At the narrowest end of the pipe.

For a given volume flow rate, the speed of fluid flow is inversely proportional to the cross-sectional area through which the fluid flows.

Continuity equation:

A1v1 = A2v2

Question 22

Give Archmide’s principle.

In physics problems, this principle allows you to substitute m and F for ___, ___ and ____?

Answer 22

If an object is floating on or immersed in a fluid, then the fluid exerts an upward buoyant force given by:

F = ρVg

v = volume of displaced fluid

m = ρV
F = PA
F(buoyant) = ρVg

Question 23

When you are presented with a problem where something is floating in a fluid, how can you determine it’s specific gravity?

Answer 23

1. Write a force equation (F = mg, which becomes 0 = F - mg)
2. Replace m with ρV of object and replace buoyant force with ρVg of water)

0 = ρ(water)V(displaced)g - ρVg

3. Express amount of water displaced as fraction (eg. 2/3 for 2/3 submerged) and cancel out V and g

0 = ρ(water)(fraction) - ρ(object)

ρ(object)/ρ(water) = fraction

ρ(object)/ρ(H20) = specific gravity

Question 24

What is gauge pressure?

Answer 24

Most pressure that isn’t measuring the atmospheric pressure in some way is measured as gauge pressure. This includes things like blood pressure and the pressure in a column of water.

Eg. The blood pressure of a woman with a systolic blood pressure of 110 torr at standard conditions is actually 870 torr (110 + 760)

P(gauge) = P - P(atm)

Question 25

The molecules at the surface of a fluid experience a cohesive force directed into the fluid. If the surface becomes bent for some reason, there is a restoring force, making the surface smooth or flat. The larger the distortion, the larger the force, up to a maximum.

How is this maximum force calculated?

Answer 25

Fmax = γL

Where γ is the surface tension (a function of the fluid) and L is the length of the edge of the object in contact with the fluid (eg. circumference of a stick).

Question 26

Give the formula concerning velocity and area for figuring out flow rate

Answer 26

f = av

flow rate = (cross sectional area)(velocity)

It’s that simple! That is, for incompressible fluids (liquids) and compressible fluids (eg. air) if they are not compressed).

Question 27

In Reynold’s equation, η (viscosity kg/m*s) is in the denominator, what is the consequence of this?

Answer 27

Higher viscosity reduces turbulence. The flow starts getting turbulent around 40 Re and is turbulent around 20,000 Re

Question 28

How can pressure at one point in a body of fluid be related to another for vertical separation? Horizontal separation?

Answer 28

Vertical separation: Pressure 2 = Pressure 1 + ρgh

Horizontal separation: Pascal’s law (same throughout)

Question 29

An object is floating in a fluid with a specific gravity (given) and is subject to a buoyant force (given).

You are asked to approximate the specific gravity of an unknown liquid where the same object has a given buoyant force. How?

Answer 29

1. Form a ratio of buoyant forces
   eg. (Fb in unknown liquid)/(Fb in known liquid)
2. Calculate for:

(Fb in unknown liquid)/(Fb in known liquid) = (density of unknown liquid)/(density of the known liquid)

Specific gravity of the unknown is then just:

density of unknown/density of water (1)

So basically you just have to find the density of the unknown with the equation above.. Simple!

Question 30

Two blocks of the same density are completely submerged in water. One block has a mass equal to m and volume equal to V. The other has a mass equal to 2m. What is the ratio of the first block’s apparent weight to the second block’s apparent weight?

Answer 30

1:2

There are two forces acting on a block when it is completely submerged in water: its weight, mg, acting downward. And the buoyant force acting upward. The apparent weight in water is equal to the actual weight of the block in air, mg, minus the buoyant force.

The buoyant force exerted on a block when it is submerged in water is equal to the wedight of water that the block displaces. The weight of water displaced is equal to mwg, where mw is the mass of the water displaced and g is the acceleration due to gravity. Using the equation m = ρV, we can express the mass of water displaced as being mw = ρV. We are given that the mass and volume of the first block are ma and V, respectively. We are told that the second block has a mass of 2m and the same density as the first block. The density of the first block is just m/v, so m/v must equal the mass of the second block over the volume of the second block. The mass of the second block is 2m, so the volume of the second block must equal 2V. Thus, the second block displaces twice as much water, and the buoyant force on the second block is twice as great.

Now we can express the apparent weights of the two blocks when submerged in water. The block of mass m has an apparent weight of mg - mwg. And the block of mass 2m has an apparent weight of 2mg - 2mwg. So the apparent weight of block one is one half the apparent weight of block two, the correct ratio is 1 to 2.

Question 31

Pay attention, this is deceptively simple. When you have objects of different densities (and therefore suspended in a fluid at different heights), how do you calculate the change in pressure between them.

For good measure, give the formula for calculating absolute pressure at a depth h below the surface of a fluid.

Answer 31

Density of the objects is irrelevant in this case, you just need the density of water (1000 kg/m^3)

Δp = ρgΔh

Absolute pressure below surface = p(atmospheric) + ρgh

Easy peasy

Remember, density in Bernoulli’s equation always refers to the density of the fluid!!!

Question 32

How do you determine the volume flow rate of a fluid in a pipe (or blood in an artery!)?

Answer 32

(fluid velocity) x (cross sectional area)

A = πr^2

if diameter is given, be sure to use

A = π(D/2)^2 = π(D^2)/4

Question 33

An unknown solid weighs 31.6 N. When submerged in water, its apparent weight is 19.8 N. What is the specific gravity of the unknown sample?

Answer 33

The buoyant force on the solid is the difference between its weight (31.6 N) and its apparent weight in water (19.8 N).

This force equals the weight of the water displaced, which in turn, equals the product of the volume of the solid and the density of the water.

The specific gravity of the solid (ie. the ratio of its weight to the weight of an equal volume of water), is then:

31.6/(31.6 - 19.8) = 2.68