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Aceasta fituica rezuma Thermodinamics.
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Domeniu: Mecanica

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1. intro in engineering thermodynamics

1.1.Energy – measures the quantity of motion; in e.t. a great interest

is how to transform the energy from a form to another

1.1.1 Heat and Work – both are energy transfer forms; occur in an interaction;

Heat is the energy transfer form due to the interaction bet 2 bodies at dif temp.;

the driving force is Work is the energy transfer form due to the mechanical

interaction bet 2 bodiesin mechanics: displacement of the body; in thermod: p

1.2.Thermody. system – it’s an assembly of bodies that interact bet them and

the surroundings by heat, work and substance transfer Q,W,m; types: opened sys:

the exchange of Q,W,m occurs; closed sys:just Q,W; isolat adiab: W,m?, isol: none;

1.3. Equil state – the sys is in equil when there is the same pressure and temp;

thermal equil(same T);mechanical(same p);internal(inside);external (at the border)

1.4. Thermo. Parameters – characterize the energetic state of a sys in equil;

a) intensive param: don’t dep on the mass of the sys; p,T –can be measured;

b) extensive p: dep on the mass – can be measured; volume V[m3]=m[kg]v[m3/kg];

internal energy U[J]=m[kg]u[J/kg];enthalpy H[J]=m[kg]h[J/kg];

entropy S[J/K]=m[kg]s[J/kgK]1.5. Thermal eq of state – a relation bet several param;

gen we have 2 deg of freedom f(p,V,T)=0

1.6. Thermod process – the passing of the sys from several intermediary equil

states (initial -> final state); if final st=init st => closed process or cyclic one; if not =>

opened process; types: isometric process (V=ct); isobaric(p=ct) isothermal (T=ct);

adiab (´Q=0) w.r.t. keeping a param const; wrt the time: steady state regim

Law of the perfect gases

Boyle – Mariotte

At T=ct. the volume of perfect gas varies inverse prop

with its pabsolute ; T=ct. ópV=ct. or pv=ct. for m=1kg;

v=V/m [m3/kg] v=1/Á[kg/m3] isometric –

Gay - Lussac law

Isobaric p=ct. V of the p. gass

Varies prop with its T


=> dilatation coef

other forms:

c=a+bt; c-mass specific heat

if v=0 => t=-1/±=-273.15 C;

T= [K]

V/Vo=T/To => V/T=ct. for m=1kg => v/T=ct. specific vol.

The normal state

-the technical state: po=760mmHg to=20C

-the physical state: pN=760mmHg TN=273.15K

def: 1 normal cubic meter=the quantity of gas which

occupies V=1m3 in normal state cond (1m3N)

Avocadro Law: same v of diff p.g., in same p,T cond.

Contain same no of molecules. NA=NB

Chemistry knowledge: 1.molecular mass of a subst:

Nondimensional no. which shows how many times is

the molecule’s mass wrt the 12th part of molecule

mass C12 ex: MO2=32[-]; 2.kilomol – a quantity of

subst. ginven in kg= molecular mass; ex: 1kmol O2=

32 kg O2 3.molecular mass=the equiv o a kmol in kg


The conseq of Avocadro: C1: molar vol of gases

Let A,B gases: nA=NA*¼A nB=NB*¼B - ¼A,¼B-molecule mas

nA/nB =NA*¼A/NB*¼B=¼A/¼B=(¼A/(¼C12/2))/(¼B/(¼C12/2))=

MA/MB=(mA/VA)/(mB/VB)=ÁA/ÁB=vB/vA=>MA vA =MBvB ó

At same p,T - Mv is the same for all perfect gases

M[kg/Kmol]v[m3/kg]=Mv[m3/Kmol]=VM- molar volume

óVMA=VMB if pA=pB TA=TB=>at same p,T - VM is the same

for all p.g.; C2: No. of Avogadro – in one kmol of gas the

no. of molecules is the same NA6.023*1026molec/kmol

C3: universal constant of gases; let A,B gases => at

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