t. k. ng, hkust lecture i: (1) introduction to fluids. (2) zeroth and first law of thermodynamics

T. K. Ng, HKUSTT. K. Ng, HKUST

Lecture I:

(1) Introduction to fluids.

(2) Zeroth and first law of thermodynamics

Introduction to fluids

A fluid is a substance that can flow and conform to the boundaries of any container in which we put them.

e.g. water, air, glass.

Basic properties of fluids

AFP /

Vm /Density (mass per unit volume) -

Pressure (force per unit area) -

Basic properties of fluids

AFP /

Pressure (force per unit area) -

Notice that from definition, pressure may depend on direction. However, this is not the case for static fluids.

(why?).

Basic properties of fluids

AFP /

Pressure (force per unit area) -

Unit of pressure:

1 pascal (Pa) = 1 Newton per square meter.

1 atm. = 1.01 x 105 Pa

Fluids at rest

Pressure increases when we go “deeper” into water – why?

)(

,,

,

21

2211

12

yyAm

ApFApF

mgFF

Fluids at rest

Pressure of a fluid in static equilibrium depends on depth only

ghpp

or

yygpp

0

2112 ),(

Example

Which one of the four container + fluid has highest pressure at depth h?

How about if (d) is move up (down) by distance h?

Pascal’s Princple

• A change in the pressure applied to an enclosed incompressible fluid is transmitted undiminished to every portion of the fluid and to the walls of the container as a direct consequence of Newton’s Law.

Example: Hydraulic level

• Applied force Fi change in pressure p=Fi/Ai=Fo/Ao.

• Therefore output force is Fo=FiAo/Ai.

• Therefore

• Fo > Fi if Ao > Ai

• How about work done?

Archimedes’ Principle

• Buoyant force – upward force in liquid because of increasing pressure in liquid as one goes down below the surface.

• (a) a hole in water. Notice that the hole is in static equilibrium if it is filled with water.

Archimedes’ Principle

• (a) a hole in water. Notice that the hole is in static equilibrium if it is filled with water.

• Therefore the upward force = mfg, mf = mass of displaced water.

Archimedes’ Principle

• (b) The hole in water is replaced by a solid with the same shape.

• Since nothing changes in water, therefore the upward force = mfg, mf = mass of displaced water = buoyant force

Archimedes’ Principle

• (c) The solid in water is replaced by a piece of wood with mw < mf..

• In this case the wood float on the surface with Fb=mwg.

Archimedes’ Principle

• When a body is fully or partially submerged in a fluid, a buoyant force Fb from the surrounding fluid acts on the body. The force is directed upward and has a magnitude equal to the weight mfg of the fluid that has been displaced by the body.

Archimedes’ Principle

• Question: Imagine a large sphere of water floating in outer space. The sphere of water is formed under its own gravity. Is there any buoyant force if an object enters this sphere of fluid?

Flowing liquids• The continuity equation – conservation of

mass in a incompressible liquid flow.

2211

2211

vAvA

or

tvAtvAV

v = velocity of fluid flowing through area

A in the tube

Example• What is the volume flow rate of water if

Ao=1.2cm2, A=0.35cm2 and h=45mm.

./34

./6.282

2

300

220

2

0

20

2

00

scmvAR

scmAA

ghAv

ghvv

AvvA

V

Bernoulli’s Equation

• Bernoulli’s Equation is a consequence of conservation of energy in steady flow.

)(2

12

1

2

1

;

21

22

21

22

vvV

mvmvK

KW

Bernoulli’s Equation

• Bernoulli’s Equation is a consequence of conservation of energy in steady flow.

VpVpW

yygVW

WWW

p

g

Pg

12

12 )()(

;

Bernoulli’s Equation

• Adding together, we obtain

cgyvp

or

gyvpgyvp

2

22221

211

2

1

2

1

2

1

(Bernoulli’s Equation)

Example

• What is the speed v of the water emerging from the hole?

• Show that v2=2gh (same as free fall)

Thermodynamics (I)

• Temperature & equilibrium

• Temperature

• - something we can all feel (hot/cold)

• - measured in Kelvin (SI unit)

• - there exists a lower limit (= 0 K) but apparently no upper limit. (room temperature ~ 300 K)

Thermodynamics (I)

• Physics of Temperature

• Fundamental question: Under what physical condition the temperatures of 2 objects are equal? (Notice this is independent of the scale we used to measure temperature)

Thermodynamics (I)

• Physics of Temperature

• We assume (based on daily life experience) is that if two objects (in a closed environment) are in contact with each other for long enough time, they will reach thermal equilibrium (stop changing) same temperature

Thermodynamics (I)

• Physics of Temperature

• Furthermore (zeroth law of thermodynamics): If bodies A and B are each in thermal equilibrium with a third body T, then they are in thermal equilibrium with each other.

Thermodynamics (I)

Zeroth law of thermodynamics

Thermodynamics (I)

Temperature Scales:

The Celsius and Fahrenheit Scales

-can relate to the Kelvin scale using the following rules:

(Celsius)

(Frhrenheit)o

cF

oc

TT

TT

32)5/9(

,15.273

Heat and Temperature

• Question: What changes physically when the temperature of an object changes?

Heat and Temperature

• We call the change “heat”(Q)

• Unit of heat calorie(cal) = amount of heat needed to raise the temperature of 1g of water from 14.5°C to 15.5°C, i.e. any physical process that can make the above change is said to provide 1 calorie of heat.

Heat and Temperature

• Unit of heat calorie(cal) = amount of heat needed to raise the temperature of 1g of water from 14.5°C to 15.5°C, i.e. any physical process that can make the above change is said to provide 1 calorie of heat.

• Notice that nowadays we know heat is a form of energy and 1 cal. = 4.186 Joule.

Heat and Temperature

• Heat Capacity (C) of an object - is the proportionality constant between an amount of heat and the change in temperature that it produces in the object.

• Usually; Q = C(Tf-Ti), Ti, Tf are initial and final Temperatures of the object.

Heat and Temperature

• Specific Heat (c)

• = Heat Capacity per unit mass of a material

• Unit: cal/g.K or J/kg.K

• Molar Specific Heat (c)

• = Heat Capacity per mole of material (1 mole = 6.02x1023 elementary units)

• Units: J/mol.K

Heat and Temperature

Almost constant

Heat and Temperature

In general we have to specify the conditions under which heat transfer occurs to define specific heat, since different conditions may lead to different values of specific heat. For example, cV and cP are specific heats measured at constant volume and pressure, respectively. For gases, the difference between the two is large.

Heat and Temperature

When heat is absorbed by a solid or liquid, the temperature may not rise. Instead, heat may simulate a transformation of the sample from one state, or phase, to another. For example, ice to water or water to water vapor. The temperature of the sample remains unchanged in the process. The amount of heat per unit mass that must be transferred when a system undergoes a phase change is called Heat of Transformation L. The total heat transferred by a sample of mass m is Q=Lm

Heat as a form of energy

Question: How do you know that heat is a form of energy as used in mechanics?

Ans: If heat is a form of energy, we expect that there should be a way to use heat to do work (heat engine).

A simple Heat-Work System

• Work- controlled by lead shots which exert pressure on the system.

• Heat - controlled by thermal reservoir which changes temperature.

• System - characterized by Pressure (P), Temperature (T) and Volume (V).

A simple Heat-Work System

• Work- adding or removing lead shots piston displaced by amount ds.

PdVPAdssdFdw

.

dV=change in volume

A simple Heat-Work System

• Total Work done in changing volume from Vi to Vf

f

i

V

V

PdVW

Notice: temperature remains fixed.

A simple Heat-Work System

• Notice we can also change the volume by heating the system. This is an example of converting heat energy (raising temperature) to work done (change in volume)

P-V diagram• To find out what is the work done in a general

thermodynamics process we use P-V diagram. • E.g.

P-V diagram• Total Work done = area under curve.

• (a) W> 0, Q (Heat) unknown?

• (b) (ia) W > 0, Q > 0 (system heats up),

• (af) W = 0, by adjusting Pressure and Temperature together

• (c) = (b) in reverse order

• (d) Wighf > Wicdf

• (e)W < 0, when volume decreases, but keeping pressure >0.

• (f) Thermodynamic cycle, net work done > 0.

• Problem: How about Q?

First Law of Thermodynamics

• When a system changes from a given initial state to a final state, the work done W and the heat Q are different for different path of changes. However, the quantity Q -W always remains the same for all paths. It depends only on what are the initial and final states.

First Law of Thermodynamics

• The change in (internal) energy of the system is given by

WQEEE if int,int,int

or in differential form,

dWdQdE int

Indicating that there are two ways to change the (internal) energy of a system.

First Law of Thermodynamics

• Example: The figure at the right shows four paths on a p-V diagram along which a gas can be taken from state i to state f. Rank the paths according to (a)the change ΔEint, (b)The work done W by the gas, and (c) the magnitude of the heat transfer Q, greatest first.

First Law of Thermodynamics

• Example: Ideal gases.• For all gases, when the densities of the gases

are low, they all tend to obey the same law, PV=nRT (Ideal gas law) where n is the number of moles of gas present, and R is a universal constant, called gas constant, with value R = 8.31J/mol.K.

• the number of elementary units (atoms/molecules) in one mole of the material = 6.023 x 1023 (Avogadro’s number).

Work done by Ideal gases (examples)

(1)Isothermal processes (Temperature kept constant) P=nRT/V

f

i

f

i

V

V i

fV

V V

VnRT

V

dVnRTPdV ln

(2) Constant Pressure processes

)()( ifif

V

V

TTnRVVPPdVf

i

Example

Initially, containers A and B are kept at temperatures and pressures PA,PB and TA,TB, respectively, with

VB=4VA. What will be the final pressure of the system if the valve is opened with TA and TB kept the same?

Example

BBAB

AAAA

RTnVP

RTnVP

)4(,

Initially,

After valve opened

)(

)4(

,

fB

fABA

Bf

BA

AfAA

nnnn

RTnVP

RTnPV

Example

Using the Ideal Gas Law and Archimedes’ Principle, estimate what is the minimum temperature needed to heat up a hot air balloon with volume 1m3 and mass 2.0g such that the balloon rises at atmospheric pressure and room temperature (300K). (R=8.31J/mole.K. Take molar mass of air to be 29g/mole ). How about if the volume is (0.1m)3?

Example

2

1

122211 T

TnnRTnRTnPV

buoyant force=mmolar(n1-n2)g

=0.002g

End