General Physics I (aka PHYS 2013)vvanchur/2015PHYS2013/Slides4.pdfequilibrium condition might not be...

CHAPTER 12 CHAPTER 19 REVIEW

General Physics I(aka PHYS 2013)

PROF. VANCHURIN(AKA VITALY)

University of Minnesota, Duluth(aka UMD)


OUTLINE

CHAPTER 12

CHAPTER 19

REVIEW


CHAPTER 12: FLUID MECHANICS

I Section 12.1: DensityI Section 12.2: Pressure in a FluidI Section 12.3: BuoyancyI Section 12.4: Fluid FlowI Section 12.5: Bernoulli’s EquationI Section 12.6: Viscosity and Turbulence


SECTION 12.1: DENSITY

I We have already seen that the local destiny of a material can bedefined as

ρ =dmdV

. (1)

I When the object has uniform (i.e. position independent) density,then the local density is the same as average density defined as

ρ =mV. (2)

I For the same substance this number does not change even if themass and volume might be different.

I For example both a steel wrench and a steel nail have the samedensity which the density of steel.



I In SI the units of density are given by kilogram per cubic meter

1 kg/m3 =1 kg1 m3 (3)

but gram per unit centimeter are also widely used

1 g/cm3 = 1000 kg/m3. (4)

I Another useful (but dimensionless) measure of density isspecific density also known as relative density. It is defined as aratio of density of a given substance to density of water (attemperature 4.0◦C),

specific density =ρsubstance

ρwater. (5)



Densities of common substances:



Example 12.2. Find the mass and weight of the air (at 1 atm and 20◦C) in aliving room with 4.0m× 5.0m floor and a ceiling 3.0m high, and the massand weight of an equal volume of water.



Example. Rank the following objects in order from highest to lowest averagedensity:(i) mass 4.00 kg, volume 1.60× 10−3 m3;(ii) mass 8.00 kg, volume 1.60× 10−3 m3;(iii) mass 8.00 kg, volume 3.20× 10−3 m3;(iv) mass 2560 kg, volume 0.640 m3;(v) mass 2560 kg, volume 1.28 m3.


SECTION 12.2:PRESSURE IN A FLUIDPressure.

I In fluids pressure might change from one place to another andthus it is convenient to define a local pressure as

p =dF⊥dA

(6)

which reduces to average pressure introduced earlier

p =F⊥A

(7)

for uniform (i.e. position independent) pressures.

I Units of pressure were already introduced in the previouschapter,

1 Pa ≡ 1 N/m2

1 atm ≈ 1.013× 105 Pa1 psi ≈ 6900 Pa. (8)


SECTION 12.2:PRESSURE IN A FLUID

Example 12.2. In the living room with 4.0m× 5.0m floor what is the totaldownward force on the floor due to air pressure of 1.00 atm?


SECTION 12.2:PRESSURE IN A FLUIDPressure with depth.

I Consider an infinitesimal volume element of fluid dV = dxdydz,where y-axis points upwards.

I In the equilibrium all forces acting on the object must add up tozero and thus along y-axis we have

− (ρdV) g− (p + dp) dA + pdA = 0−ρ (dxdydz) g− (p + dp) (dxdz) + p (dxdz) = 0

−ρgdy− (p + dp) + p = 0dpdy

= −ρg (9)

and thus the pressure must change linearly with y.I The above equation can be solved by direct integration, i.e.∫ p(y2)

p(y1)

dp = −∫ y2

y1

ρgdy

p(y2)− p(y1) = ρg (y1 − y2) . (10)



I In terms of depthd = y2 − y1 (11)

and reference pressurep0 = p(y2) (12)

pressure at arbitrary depth is given by

p = p0 + ρgd. (13)

I Thus if p0 (atmospheric pressure) and p(pressure at the bottomof liquid) is the same than d must be the same:



I If we change pressure p at the surface of fluid, the pressure willchange by the same amount everywhere in the fluid.

I Hydraulic lift uses the same principle to measure weights

p =F1

A1=

F2

A2⇒ F2 =

A2

A1F1. (14)

I Pascal’s law: Pressure applied to an enclosed fluid is transmittedundiminished to every portion of the fluid and the walls of thecontaining vessel.



Gauge pressure. It often useful to measure relative pressurecompared to atmospheric pressure,

patm = 1 atm ≈ 14.7 psi ≈ 1.01× 105 Pa. (15)

For example, if the absolute pressure of a car tire is

p = 47 psi (16)

then it is often said that the gauge pressure is

pgauge = p− patm = 32 psi. (17)



Example 12.3. Water stands 12.0 m deep in a storage tank whose top isopen to the atmosphere. What are the absolute and gauge pressure at thebottom of the tank?



I To measure the gauge pressure directly, one can use anopen-tube manometer,

pgauge = p− patm = ρg(y2 − y1) (18)

I To measure air-pressure one can use a barometer where thedifference in heights tells you what the atmospheric is

patm = p0 + ρg(y2 − y1) = ρg(y2 − y1). (19)



Example 12.4. A manometer tube is partially filled with water. Oil (whichdoes not mix with water) is poured into the left arm of the tube until theoil-water interface is at the midpoint of the tube as shown. Both arms of thetube are open to the air. Find a relationship between the heights hoil andhwater.


SECTION 12.3:BUOYANCY

I Any object placed in a fluid experiences a force (buoyant force)arising due to changes of the pressure inside fluid.

I To prove it consider an arbitrary element of fluid:

I If the fluid is in equilibrium then the sum of all forces mustbalance the force of gravity

B = Fgravity or B = Vρfluidg (20)



I Now if we fill the shape with some other material, then theequilibrium condition might not be satisfied, but the buoyantforce due to water pressure would not change.

B = Vρfluidg (21)

I Archimedes’s principle: When a body is completely or partiallyimmersed in a fluid, the fluid exerts an upward force on the body equalto the weight of the fluid displaced by the body.



Example 12.5. A 15.0− kg solid gold statue is raised from the sea bottom.What is the tension in the hosting cable (assumed massless) when thestatues isa) at rest and completely underwater.b) at rest and completely out of water.



Example. You place a container of seawater on a scale and note reading onthe scale. You now suspend the statue of Example 12.5 in the water. Howdoes the scale reading change?(i) it increases by 7.84 N;(ii) it decreases by 7.84 N;(iii) it remains the same;(iv) none of these.


SECTION 12.4: FLUID FLOW

I Consider a simple model of fluid which is incompressible(density is constant) and inviscid (vanishing internal friction).

I Then one can follow trajectories (or flow lines) of small elementsof water (or particles) which will flow along these trajectories.

I The flow is steady if the flow line do no change with time. Thisdoesn’t mean that velocities of particles do not change with time.

I When the flow becomes irregular the small scales and large scaleinteract with each other giving rise to turbulence.



I Consider a flow of fluid through a pipe

I If the fluid is incompressible (i.e. constant density), then

ρA1ds1 = ρA2ds2 or A1v1dt = A2v2dt (22)



I Which give us the (1D) continuity equation fornon-compressible fluid

A1v1 = A2v2. (23)

I The continuity equation equates the volume flow rate acrossdifferent cross-sectional areas

dVdt

= Av (24)

I and can be easily generalized to the case when densities dochange

ρ1A1v1 = ρ2A2v2. (25)



Example 12.6. Incompressible oil of density 850 kg/m3 is pumped through acylindrical pipe at a rate of 9.5 liters per second.(a) The first section of the pipe has a diameter of 8.0 cm. What is the flowspeed of the oil? What is the mass flow rate?(b) The second section of the pipe has a diameter of 4.0 cm. What is the flowspeed and the mass flow rate in that section?



Example. A maintenance crew is working on a section of a three-lanehighway, leaving only one lane open to traffic. The result is much slowertraffic flow (a traffic jam.) Do cars on a highway behave like:(i) the molecules of an incompressible fluid or(ii) the molecules of compressible fluid?


SECTION 12.5: BERNOULLI’S EQUATION

I As fluid moves through pipe external forces such asgravitational force can do work on the fluid.

I Bernoulli’s equation:

p1 + ρgy1 +12ρv2

1 = p2 + ρgy2 +12ρv2

2 (26)


SECTION 12.5: BERNOULLI’S EQUATIONExample 12.7. Water enters a house through a pipe with an inside diameterof 2.0 cm at an absolute pressure of 4.0× 105 Pa. A 1.0 cm diameter pipeleads to the second-floor bathroom 5.0 m above. When the flow speed at theinlet pipe is 1.5 m/s, find the flow speed, pressure and volume flow in thebathroom.



Example 12.8. A gasoline storage tank with cross-sectional area A1, filled toa depth h. The space above the gasoline contains air at pressure p0, and thegasoline flows out the bottom of the tank through a short pipe withcross-sectional area A2. Derive expression for the flow speed in the pipe andthe volume flow rate.



Example 12.9. Venturi meter is used to measure flow speed in a pipe.Derive an expression for the flow speed v1 in terms of the cross-sectionalareas A1 and A2 and the difference in height h of the liquid levels in the twovertical tubes.



Example. Lift of an airplane.


CHAPTER 19: THE FIRST LAW OF THERMODYNAMICS

I Section 19.0: PreliminariesI Section 19.1: Thermodynamic systemsI Section 19.2: Work done by a systemI Section 19.3: Paths between statesI Section 19.4: Internal EnergyI Section 19.5: Kinds of processesI Section 19.6: Internal Energy of ideal gasI Section 19.7: Heat CapacitiesI Section 19.8: Adiabatic Processes


SECTION 19.0: PRELIMINARIES

I Our main objective is to understand the behavior of systemswith a very large number of degrees of freedom N � 1 (e.g.NA = 6.02214× 1023 molecules in a box).

I For N . 10 analytical methods may be useful; for N . 1010

computer may work; for N ∼ 1 googol = 10100 statistical physicsmay be the only tool.

I There are two standard ways to study the large N limit:I phenomenological (e.g. thermodynamics) andI fundamental (e.g. statistical mechanics).

We will only discuss a phenomenological approach.I The main idea is that only a small number of measurable (or

thermodynamics) parameters:I volume V,I pressure p,I temperature T, etc.

should be sufficient for describing so-called equilibrium states.


SECTION 19.0: PRELIMINARIES

I Thermodynamics parameters (i.e. p,V,T) are related to eachother by the so-called equation of state,

f (p,V,T) = 0. (27)

I The equation of state for the so-called ideal gas is given by

pV = nRT (28)

and of the van de Waals gas(p +

an2

V2

)(V − nb) = nRT (29)

where n is the amount of substance described in moles,

R = 8.31 J/mol·K (30)

and a and b constant which (roughly speaking) represent theattractive intermolecular force and the size of moleculesrespectively.


SECTION 19.1: THERMODYNAMICS STATES

I As thermodynamic systems evolves from one thermodynamicstate to another the macroscopic parameters might only changein in such a way that the equation of state remains invariant.

I There are two different mechanisms that allow to change athermodynamic state of the system from the outside:

I Heat:

Q > 0 when the system is heatedQ < 0 when the system is cooled (31)

I Work

W > 0 when the system does workW < 0 when work is done on the system. (32)

I Note that the sign convention for work is different in mechanics.


SECTION 19.2: WORK DONE BY SYSTEM

I The work done by a system can be calculated by consideringtransfer of energy by gas molecules when the piston is moving

dW = Fdx = pAdx = pdV (33)

I By integrating both sides we obtain

W =

∫ V2

V1

p(V) dV. (34)

I On pV diagram the work equals to area under p(V):



I When pressure does not change with volume an expression forwork can be obtained by integrating Eq (34)

W = p(V2 − V1). (35)

I Expansion corresponds to positive dV and thus positive WI Contraction corresponds to negative dV and thus negative W



Example 19.1. As an ideal gas undergoes an isothermal (constanttemperature) expansion at temperature T, its volume changes from V1 to V2.How much work does the gas do?


SECTION 19.3: PATHS BETWEEN STATES

I Work done by a thermodynamic system depends on the path ittakes in (p,V,T) space.

I Speaking mathematically dW is not an exact differential and forthis reason sometimes written as δW.

I For example, consider the following processes:


SECTION 19.3: PATHS BETWEEN STATES

I Similarly the heat transferred to a system Q depends on the pathit takes and thus dQ (or perhaps δQ) is not an exact differential.

I Isothermal vs. free expansion:


SECTION 19.4: INTERNAL ENERGY

The First Law of Thermodynamics

I There is, however, a thermodynamic quality whose change doesnot depend on the path it takes known as internal energy, U.

I In thermodynamics internal energy is defined not an absolutesense, but as a differential

dU = dQ− dW (36)

after integration

∆U = U2 −U1 = Q−W (37)

I Note that in mechanics potential energy corresponding toconservative forces is another physical quantity of the same typethat we have already discussed in some details.

I Equation (37) is also known as the first law of thermodynamics.


SECTION 19.4: INTERNAL ENERGYThree Laws of Thermodynamics

I We have already seen that W = pdV and dQ = SdT, where S isanother thermodynamic quantity known as entropy whichmeasures the amount of disorder in the system.

I Then the first law of thermodynamics can be written as

dU = SdT − pdV (38)

I It is an important experimental fact that entropy almost alwaysgrows with time

dSdt≥ 0 (39)

which is also known as the second law of thermodynamics.

I There is also a third law of thermodynamics stating that there isan absolute zero temperature T = 0 K at which entropy is zero

S(0 K) = 0. (40)



I There are two special kinds of the thermodynamic processes.I The first one is a cyclic process

U2 = U1 ⇒ Q = W (41)

I The second one is for a system in isolation

W = Q = 0 ⇒ U2 = U1. (42)

I Cyclic process of human body:



Example 19.4. Consider the following pV- diagram.

In process a→ b, 150 J of heat is added to the system; in process b→ d,600 J of heat is added to the system. Find(a) the internal energy change in process a→ b(b) the internal energy change in process a→ b→ d(c) the total hear added in process a→ c→ d.


SECTION 19.5: KINDS OF PROCESSES

I Adiabatic: Q = 0 ⇒ ∆U = −WI Isochoric: dV = 0 ⇒ dW = pdV = 0 ⇒ ∆U = QI Isobaric: dp = 0 ⇒ dW = pdV ⇒ ∆U = Q− p (V2 − V1)

I Isothermal: dT = 0 ⇒ dU = 0 ⇒ Q = W.


SECTION 19.6: HEAT CAPACITIES

I How much heat should be added to a system to change itstemperature?

I This is described bydQdT

= nC (43)

where n is mass per unit mole of the material and C is a molarheat capacity (or simply heat capacity).

I The heat capacity depends on the material, but moreover itdepends on the process by which the heat is added.

I If we consider isobaric and isochoric process, then thecorresponding heat capacities are notated by Cp and CVrespectively.


SECTION 19.6: HEAT CAPACITIES OF AN IDEAL GAS

I These two constants can be related to each other by applying thefirst law to the following transformation

1. isochoric decrease in temperature and decrease in pressure2. isobaric increase in volume and increase in temperature

I Then

∆U = Q−W0 = nCV (T1 − T2) + nCV (T2 − T1)− p1 (V2 − V1) (44)

and for ideal gas

0 = nCV (T1 − T2) + nCp (T2 − T1)− nR (T2 − T1) (45)

orCp = CV + R. (46)



I Using the so-called kinetic theory of gases one can show that forthe ideal monatomic gases

CV =32

R Cp =52

R (47)

and thus

γ ≡Cp

CV=

53≈ 1.67 (48)

but from example for diatomic molecules

γ ≡Cp

CV=

75

= 1.4. (49)



Example 19.6. A typical dorm room or bedroom contains about 2500 molesof air. Find the change in the internal energy of this much air when it iscooled from 35.0◦C to 20.0◦C at constant pressure of 1.00 atm.


SECTION 19.7: INTERNAL ENERGY OF IDEAL GAS

I The internal energy of an ideal gas depends only on itstemperature, not on its pressure of volume.

I For example, when partition is broken to start a free expansionof gas into the vacuum region, both the volume and pressure arechanged in such a way that temperature remains constant.

I It is easy to see by considering isochoric process that

dU = dQ− dW = dQ = nCVdT (50)


SECTION 19.8: ADIABATIC PROCESS OF IDEAL GAS

I During adiabatic process

dQ = 0 ⇒ dU = −dW = −pdV. (51)


SECTION 19.8: ADIABATIC PROCESS OF IDEAL GAS

I For an ideal gas the equation of state is

pV = nRT (52)

anddU = nCVdT (53)

which impliesTVγ−1 = constant (54)

I By substituting back into equation of state we get

pVT

= pVγ = constant . (55)

I It is also straightforward to calculate work

W =1

γ − 1(p1V1 − p2V2) . (56)


SECTION 19.8: ADIABATIC PROCESS OF IDEAL GASExample 19.7. The compression ratio of a diesel engine is 15.0 to 1; that is,air in a cylinder is compressed to 1

1.50 of its initial volume.(a) If the initial pressure is 1.01× 105 Pa and the initial temperature is 27◦C(300 K), find the final pressure and the temperature after adiabaticcompression.(b) How much work does the gas do during the compression if the initialvolume of the cylinder is 1.00 L = 1.00× 10−3m3?(Use the value CV = 20.8 J/mol·K and γ = 1.400 for air.)


CHAPTER 12: FLUID MECHANICS

I Density and pressure

I Average density and pressureI Specific density (relative to water)I Gauge pressure (compared to air)I Calculating pressure with depth

I Buoyancy

I Archimedes’s principleI Calculating buoyancy forceI Reaction force to buoyancy force

I Fluid flow

I Understanding incompressible flowI Applying continuity equationI Applying Bernoulli’s equation


CHAPTER 19: THE FIRST LAW OF THERMODYNAMICS

I ThermodynamicsI Thermodynamic statesI Equation of state of ideal gasI Work done by a systemI The first law of thermodynamicsI Heat capacities at const. pressure and const. volume

I ProcessesI Adiabatic (heat and first law of thermodynamics)I Isochoric (work and first law of thermodynamics)I Isobaric (work and first law of thermodynamics)I Isothermal (int. energy and first law of thermodynamics)I PV diagrams of these processes

I Ideal gasI Relations between Cp and CV capacitiesI Internal energy of ideal gasI Solving for adiabatic process of ideal gas

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