me 201 firstlaw

First Law of Thermodynamics

Dr. Md. Zahurul Haq

ProfessorDepartment of Mechanical Engineering

Bangladesh University of Engineering & Technology (BUET)Dhaka-1000, Bangladesh

[email protected]://teacher.buet.ac.bd/zahurul/

ME 201: Basic Thermodynamics

c

Dr. Md. Zahurul Haq (BUET) First Law of Thermodynamics ME 201 (2013) 1 / 17

First Law of Thermodynamics Closed (CM) System

Conservation of Energy for a Closed System

First Law of Thermodynamics

When a system undergoes a cyclic change, the net heat to or from the

system is equal to the net work from or to the system.

J

Q =

W

Mechanical equivalent of heat,J =

{4.1868 kJ/kcal

1.0 in SI unit

c



First Law of Thermodynamics for a Change in State

T363

W = J Q Q = W [J = 1.0 in SI unit] 21 QA + 12 QC = 21 WA + 12 WC 1 21 QB + 12 QC = 21 WB + 12 WC 2

1

2 :2

1 QA2

1 QB =2

1 WA2

1 WB 21 QA 21 WA = 21 QB 21 WB21(Q W )A =

21(Q W )B =

21(Q W ) is independent of the path and dependent only on the

initial and final states; hence, it has the characteristics of a property

and this property is denoted by energy, E.

Q W = dE Q12 W12 = E

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Physical significance of the property E: it represents all the energyof the system in the given state. This energy might be present in avariety of forms, such as:

Kinetic Energy (KE): energy of a system associated with motion.

Potential Energy (PE): energy associated with a mass that is

located at a specified position in a force field.

Internal Energy (U): some forms of energy, e.g., chemical, nuclear,

magnetic, electrical, and thermal depend in some way on the

molecular structure of the substance that is being considered, and

these energies are grouped as the internal energy of a system, U.

KE & PE are external forms of energy as these are independent of

the molecular structure of matter. These are associated with the

selected coordinate frame and can be specified by the macroscopic

parameters of mass, velocity & elevation.

Internal energy, like kinetic and potential energy, has no natural

zero value. Therefore, it is necessary to arbitrarily define the

specific internal energy of a substance to be zero at some state

that is referred to as the reference state.c



Internal Energy (U): A Thermodynamic Property

T133

Various forms of microscopic

energies that make up sensible

energy.

T134

Internal energy of a system is the

sum of all forms of the microscopic

energies.c



E = U + KE + PE +

Q W = dU + d(KE) + d(PE) + Q

dt W

dt= dU

dt+

d(KE )dt

+d(PE )

dt+ = dEcm

dt

dEcmdt

= _Q _W

dU = 21 dU = U2 U1 = m(u2 u1) d(KE) = mVdV = 21 d(KE) = 12m(V22 V21) d(PE) = mgdZ = 21 d(PE) = mg(Z2 Z1) = mghQ12 W12 =

[(U2 U1) +

12m(V

22 V

21) + mg(Z2 Z1)

]

q12 w12 =[(u2 u1) +

12(V

22 V

21) + g(Z2 Z1)

] (u2 u1)

c



T129

T130 T131 T132

c



Enthalpy (H)): A Thermodynamic Property

H U + PV h u + Pv

T117

Q12 W12 = E Q12 W12 = U if KE 0, PE 0 W12 = 21 PdV = P(V2 V1) Q12 = U2 U1 + P(V2 V1) Q12 = (U2 + P2V2) (U1 + P1V1) Q12 = H2 H1The heat transfer in a constant-pressure quasi-equilibrium process is

equal to the change in enthalpy, which includes both the change in

internal energy and the work for this particular process.

c



Joules Free Expansion Experiment

T139

Valve is opened and allowed to

equilibrate.

No change in water temperature.

So, no heat transfer takes place.

1st Law: Q12 = 0, W12 = 0, = U = 0.P & V changed during this process, but without any change in U.

So, U 6= f (P ,V ) = U = f (T ) for ideal gas.h = u + Pv = u + RT 7 H = f (T ) for ideal gas.

c



Specific Heats

C 1m

QT

CV 1m

(QT

)V

CP 1m

(QT

)P

q = du + w = du + Pdv qT

= uT

+ P dvT cV ( qT )

V=(uT

+ P dvT

)V

=(UT

)V+ 0

= cV ( qT )V

= dudT

as u = f (T ) for ideal gas

q = du + w = d(h Pv) + Pdv = dh vdP qT

= hT

v dPT cP ( qT )

P=(hT

v dPT

)P=(hT

)P 0

= cP ( qT )P= dh

dTas h = f (T ) for ideal gas

h = u+Pdv = u+RT dh = du+RdT cPdT = cV dT+RdTSpecific heat ratio, k cP

cV

= cP cV = R cV = Rk1 cP = kRk1c



Ideal gas models:

R = RuM

cP (T ) cV (T ) = R

cV (T ) =R

k (T )1 : k(T ) =cP (T )cV (T )

du = cV dT u2 u1 = 21 cV (T )dTdh = cPdT h2 h1 = 21 cP (T )dT T128

T126 T127c



T115

First Law of Thermodynamics for closed system

c


First Law of Thermodynamics Open (CV) System

Conservation of Energy for CV System

T099

c



T100

dEcv

dt= _Q _W + _miei _meee

= _Q _W + _mi

(ui +

V2i

2+ gzi

) _me

(ue +

V2e

2+ gze

)

= _Q _Wcv + _mi

(hi +

V2i

2+ gzi

) _me

(he +

V2e

2+ gze

)

_Wcv _Ws + _Wb

_Wf = P( _Vi _Ve) = P( _mivi _meve)

h u + Pv

c



First Law of Thermodynamics (FLT) for CV System

dEcvdt

= _Q _Wcv +

i

_mi

(hi +

V2i

2+ gzi

)e

_me

(he +

V2e

2+ gze

)

Closed System: 7 _mi = _me = 0.dEcvdt

= _Q _Wnet

Steady-State-Steady Flow (SSSF) System:dmcvdt

= 0 =i _mi =e _me : dEcvdt = 0One-inlet, One-exit & Steady-state: 7 _mi = _me = _m .

0 = _Q _Wcv + _m[(h1 h2) +

(V

21V

22

2

)+ g(z1 z2)

]

c



Mass Continuity Equation

T122

_m = VndAc

T123

Vavg =1

Ac

Ac

VndAc

T124

_m = _V = AV = AVv

c



Example: Feedwater heater at steady-state. Determine _m2 & V2. Assume,

v2 vf (T2).

T125

dmcvdt

=

i _mi

e _me

dmcv/dt = 0 i _mi = _m1 + _m2 e _me = _m3

_m = AV

_m3 = 3(AV)3

2 = (T = T2,P = P2) 3 = (x = 0.0,P = P3)= _m2 = 14.15 kg/s, V2 = 5.7 m/s .

c


First Law of ThermodynamicsClosed (CM) SystemOpen (CV) System

me 201 firstlaw

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