First

Law

ofT

hermodynam

ics

Dr.

Md.

Zah

uru

lH

aq

Pro

fessorD

epartm

ent

ofM

echanica

lEngin

eering

Bangla

desh

University

ofEngin

eering

&Tech

nolo

gy

(BU

ET

)D

haka

-1000,Bangla

desh

http

://za

huru

l.buet.a

c.bd/

ME

6101:

Cla

ssicalT

herm

odynam

ics

http://zahurul.buet.ac.bd/ME6101/

©D

r.M

d.

Zahuru

lH

aq

(BU

ET

)First

Law

ofT

herm

odynam

ics

ME6101

(2021)

1/

27

Overview

1First

Law

ofT

herm

odyn

amics

&Energy

Closed

(CM

)System

Open

(CV)

System

2Tran

sient

Pro

cesses:First

Law

ofT

herm

odyn

amics

©D

r.M

d.

Zahuru

lH

aq

(BU

ET

)First

Law

ofT

herm

odynam

ics

ME6101

(2021)

2/

27

First

Law

ofT

herm

odynam

ics

&Energ

yClo

sed

(CM

)Syste

m

Conservation

ofEnergy

fora

CM

System

First

Law

ofT

herm

odyn

amics

(FLT

)

When

asystem

undergo

esa

cyclicch

ange,

the

net

heat

to/fromth

esystem

iseq

ual

toth

enet

work

from/to

the

system.

J

∮δQ

=

∮δW

Mech

anica

leq

uiv

alen

tof

hea

t,J=

{4.1868

kJ/kca

l

1.0

inSI

unit

©D

r.M

d.

Zahuru

lH

aq

(BU

ET

)First

Law

ofT

herm

odynam

ics

ME6101

(2021)

3/

27

First

Law

ofT

herm

odynam

ics

&Energ

yClo

sed

(CM

)Syste

m

First

Law

ofT

hermodynam

icsfor

aChange

inState

T363

⇒∮δW

=J∮δQ

⇒∮δQ

=∮δW

[J=

1.0

inSI

unit]

⇒∫

21δQ

A+∫

12δQ

C=

∫21δW

A+∫

12δW

C1○

⇒∫

21δQ

B+∫

12δQ

C=

∫21δW

B+∫

12δW

C2○

•1○−

2○: ∫

21δQ

A−∫

21δQ

B=

∫21δW

A−∫

21δW

B

⇒∫

21δQ

A−∫

21δW

A=

∫21δQ

B−∫

21δW

B∫

21(δQ

−δW

)A=

∫21(δQ

−δW

)B=

···∫

21 (δQ

−δW

)is

indep

enden

tof

path

and

dep

enden

ton

lyon

the

initial

and

final

states;hen

ce,it

has

the

characteristics

ofa

property

and

this

property

isden

otedby

energy,

E.

δQ

−δW

=dE

⇛Q

12−W

12=

∆E

©D

r.M

d.

Zahuru

lH

aq

(BU

ET

)First

Law

ofT

herm

odynam

ics

ME6101

(2021)

4/

27

First

Law

ofT

herm

odynam

ics

&Energ

yClo

sed

(CM

)Syste

m

•Energ

y(E

):represen

tsall

forms

energy

ofth

esystem

inth

egiven

state.It

migh

tbe

present

ina

varietyof

forms,

such

as:•

Kin

etic

Energ

y(K

E):

energ

yofa

systemasso

ciatedw

ithm

otio

n.

•Pote

ntia

lEnerg

y(P

E):

energ

yasso

ciatedw

itha

mass

that

islo

cated

ata

specifi

edpositio

nin

aforce

field

.•

Inte

rnalEnerg

y(U

):so

me

forms

ofen

ergy,

e.g.,

chem

ical,nuclear,

mag

netic,

electrical,an

dth

ermal

dep

endin

gin

som

eway

on

the

molecu

larstru

cture

ofth

esu

bstan

ceth

atis

bein

gco

nsid

ered,an

d

these

energ

iesare

gro

uped

asth

ein

ternal

energ

yofa

system,U

.

•K

E&

PE

areextern

alform

sof

energy

asth

eseare

indep

enden

tof

the

molecu

larstru

cture

ofm

atter.T

hese

areasso

ciatedw

iththe

selected

coord

inate

frame

and

canbe

specifi

edby

the

macroscop

icparam

eters

ofm

ass,velo

city&

elevation.

•In

ternal

energy,

likekin

etican

dpoten

tialen

ergy,has

no

natural

zero

value.

So,

intern

alen

ergyof

asu

bstan

ceis

arbitrarily

defi

ned

tobe

zeroat

some

state,kn

own

asRefe

rence

Sta

te.

©D

r.M

d.

Zahuru

lH

aq

(BU

ET

)First

Law

ofT

herm

odynam

ics

ME6101

(2021)

5/

27

First

Law

ofT

herm

odynam

ics

&Energ

yClo

sed

(CM

)Syste

m

InternalEnergy

(U):

AT

hermodynam

icProp

erty

T133

Vario

us

form

sofm

icrosco

pic

energ

ies

makin

gup

sensib

leen

ergy.

T134

Intern

alen

ergy

isth

esu

mofall

form

sof

the

micro

scopic

energ

ies.

©D

r.M

d.

Zahuru

lH

aq

(BU

ET

)First

Law

ofT

herm

odynam

ics

ME6101

(2021)

6/

27

First

Law

ofT

herm

odynam

ics

&Energ

yClo

sed

(CM

)Syste

m

•E=

U+KE+PE+···

⇒δQ

−δW

=dE=

dU+d(K

E)+d(P

E)+···

⇒δQdt−

δWdt=

dECM

dt

=dUdt+

d(KE)

dt

+d(PE)

dt

+···

•dECM

dt

=dQdt−

dWdt=

Q−W

⇒dU⇒

∫21dU

=U

2−U

1=

m(u

2−u1 )

⇒d(K

E)=

mVdV⇒

∫21d(K

E)=

12m(V

22−V

21 )

⇒d(P

E)=

mgdZ⇒

∫21d(P

E)=

mg(Z

2−Z

1 )=

mgh

Q12−W

12=

[(U2−U

1 )+

12m(V

22−V

21 )+mg(Z

2−Z

1 )+···

]≃(U

2−U

1 )

q12−w

12=

[(u2−u1 )

+12(V

22−V

21 )+g(Z

2−Z

1 )+···

]≃(u

2−u1 )

©D

r.M

d.

Zahuru

lH

aq

(BU

ET

)First

Law

ofT

herm

odynam

ics

ME6101

(2021)

7/

27

First

Law

ofT

herm

odynam

ics

&Energ

yClo

sed

(CM

)Syste

m

T129

T130

T131

T132

©D

r.M

d.

Zahuru

lH

aq

(BU

ET

)First

Law

ofT

herm

odynam

ics

ME6101

(2021)

8/

27

First

Law

ofT

herm

odynam

ics

&Energ

yClo

sed

(CM

)Syste

m

������

����

�

�

�

� � �

����

�

��

��� ��

�

��

�

�

��

�����������������������������

� ���������������������������

T1367

First

Law

ofT

herm

odyn

am

icsfo

rclo

sedsystem

©D

r.M

d.

Zahuru

lH

aq

(BU

ET

)First

Law

ofT

herm

odynam

ics

ME6101

(2021)

9/

27

First

Law

ofT

herm

odynam

ics

&Energ

yClo

sed

(CM

)Syste

m

Mora

nEx.

2.3

⊲T

he

pisto

n-cylin

der

material

isa

ceramic

com

posite

and

thus

a

good

insu

lator.D

etermin

eth

eheat

transfer

from

the

resistorto

the

air,in

kJ,for

asystem

consistin

gof(a)

the

airalo

ne,

(b)

the

airan

dth

episto

n.

T1071

(a):PApisto

n=

PatmApisto

n+mg⇒

P=X

•W

12=

P(V

2−V

1 )=X

,Q

12=

m(u

2−u

1 )+W

12=X

(b):

W12=

Patm(V

2−V

1 )=X

,∆PE=

mpisto

n gh=

mpisto

n g(

V2−V

1

Apisto

n

)

=X

Q12=

m(u

2−u

1 )+∆PE+W

12=X

©D

r.M

d.

Zahuru

lH

aq

(BU

ET

)First

Law

ofT

herm

odynam

ics

ME6101

(2021)

10

/27

First

Law

ofT

herm

odynam

ics

&Energ

yClo

sed

(CM

)Syste

m

Mora

nEx.

3.4

⊲Estim

atenet

heat

transfer

and

work.

T1072•

States

1,2

&3:

fixed

•wnet=

∫21Pdv+∫

32P✚ ✚❃

0

dv=

P(v

2−v1 )

•∆u=

(u3−u

1 )⇒

qnet=

∆u+wnet=X

©D

r.M

d.

Zahuru

lH

aq

(BU

ET

)First

Law

ofT

herm

odynam

ics

ME6101

(2021)

11

/27

First

Law

ofT

herm

odynam

ics

&Energ

yClo

sed

(CM

)Syste

m

Enthalpy

(H):

AT

hermodynam

icProp

erty

H≡

U+PV

⇛h≡

u+Pv

T117

⇒Q

12−W

12=

∆E

⇒Q

12−W

12=

∆U

ifKE

→0,PE→

0

•W

12=

∫21PdV

=P(V

2−V

1 )

⇒Q

12=

U2−U

1+P(V

2−V

1 )

⇒Q

12=

(U2+P

2 V2 )

−(U

1+P

1 V1 )

⇒Q

12=

H2−H

1

Heat

transfer

ina

constan

t-pressure

quasi-eq

uilibriu

mpro

cessis

equal

to

the

chan

gein

enth

alpy,w

hich

inclu

des

both

the

chan

gein

intern

alen

ergy

and

the

work

forth

isparticu

larpro

cess.

©D

r.M

d.

Zahuru

lH

aq

(BU

ET

)First

Law

ofT

herm

odynam

ics

ME6101

(2021)

12

/27

First

Law

ofT

herm

odynam

ics

&Energ

yClo

sed

(CM

)Syste

m

CengelEx.

4-5

:⊲

Electric

Heatin

gofGas

atConstan

tPressu

re:T

2=

?

T118

⇒m

=0.0

25kg

⇒Q

net=

−3.7

kJ

⇒W

e=

−0.2(1

20)(3

00)/

1000

kJ

⇒h

1=

h(3

00

kPa,x

=1)

⇒h

2=

h(3

00

kPa,T

2 )

⇒Q

net−W

net=

∆E≃

∆U

⇒Q

net−(W

e+W

b )=

∆E≃

∆U

⇒Q

net−W

e−P(V

2−V

1 )=

U2−U

1

⇒Q

net−W

e=

m(h

2−h

1 )→

h2=X

⇒h

2=

h(3

00kPa,T

2 )⇒T

2=

200oC⊳

©D

r.M

d.

Zahuru

lH

aq

(BU

ET

)First

Law

ofT

herm

odynam

ics

ME6101

(2021)

13

/27

First

Law

ofT

herm

odynam

ics

&Energ

yO

pen

(CV

)Syste

m

T1073

Contro

lVolu

me,

enclo

sing

atree.

©D

r.M

d.

Zahuru

lH

aq

(BU

ET

)First

Law

ofT

herm

odynam

ics

ME6101

(2021)

14

/27

First

Law

ofT

herm

odynam

ics

&Energ

yO

pen

(CV

)Syste

m

Conservation

ofEnergy

forCV

System

T099

©D

r.M

d.

Zahuru

lH

aq

(BU

ET

)First

Law

ofT

herm

odynam

ics

ME6101

(2021)

15

/27

First

Law

ofT

herm

odynam

ics

&Energ

yO

pen

(CV

)Syste

m

T100

dEcv

dt

=Q

−W

+m

i ei−m

e ee

=Q

−W

+m

i

(

ui+

V2i

2+gzi

)

−m

e

(

ue+

V2e

2+gze

)

=Q

−W

cv+m

i

(

hi+

V2i

2+gzi

)

−m

e

(

he+

V2e

2+gze

)

•W

=W

s+W

b+W

f=

Wcv+W

f

•W

f=

−P(V

i−Ve )

=−P(m

i vi−m

e ve )

•h≡

u+Pv

©D

r.M

d.

Zahuru

lH

aq

(BU

ET

)First

Law

ofT

herm

odynam

ics

ME6101

(2021)

16

/27

First

Law

ofT

herm

odynam

ics

&Energ

yO

pen

(CV

)Syste

m

First

Law

ofT

hermodynam

ics(F

LT)

forCV

System

dECV

dt

=Q

−W

cv+∑

i

mi

(

hi+

V2i

2+gzi

)

−∑

e

me

(

he+

V2e

2+gze

)

•Clo

sed

Syste

m:7→

mi=

me=

0.dECM

dt

=Q

−W

net

•Clo

sed

&Adia

batic

(Isola

ted)

Syste

m:7→

mi=

me=

0,Q

=0.

dECM

dt

=−W

net ⇒

∆ECM

=−W

ad

•Ste

ady-S

tate

-Ste

ady

Flo

w(S

SSF)

Syste

m:

dm

CV

dt

=0⇒

∑im

i=

∑em

e:

dEcv

dt

=0

•O

ne-in

let,

One-e

xit

&Ste

ady-sta

te:7→

mi=

me=

m.

0=

Q−W

CV+m[

(h1−h2 )

+(

V21−V

222

)

+g(z

1−z2 )]

©D

r.M

d.

Zahuru

lH

aq

(BU

ET

)First

Law

ofT

herm

odynam

ics

ME6101

(2021)

17

/27

First

Law

ofT

herm

odynam

ics

&Energ

yO

pen

(CV

)Syste

m

[Mora

nEx.

4.5

]:⊲

Air

Com

pressorPow

er:D

etermin

epow

erreq

uired

,W

cv .

T147

⇒Stead

ystate

⇒dEcv/dt=

0

⇒Z

2=

Z1

⇒Q

cv=

−180kJ/min

=−

3.0

kW

•m

=ρAV

•ρ=

P/RT

dECV

dt

=Q

−W

cv+∑

i

mi

(

hi+

V2i

2+gzi

)

−∑

e

me

(

he+

V2e

2+gze

)

0=

Q−W

cv+m[

(h1−h

2 )+(

V21−V

22

2

)]

⇒h

1−h

2=

Cp(T

1−T

2 )=

−160.8

kJ/kg

⇒ρ

1=

P1 /RT

1=

1.2

0m

3/kg,m

=ρ

1 A1V

1=

0.7

2kg/s

⇒W

cv=

−119.4

kW(w

ork

input

required

)⊳

©D

r.M

d.

Zahuru

lH

aq

(BU

ET

)First

Law

ofT

herm

odynam

ics

ME6101

(2021)

18

/27

First

Law

ofT

herm

odynam

ics

&Energ

yO

pen

(CV

)Syste

mBernoulli’s

Equation

•h=

u+Pv

→dh=

du+Pdv+vdP

,so

for

isotherm

alpro

cess(du=

0)an

din

compressib

lefluid

(dv=

0):

⇒dh=

vdP

h2−h1=

v(P

2−P

1 )=

P2−P

1ρ

•For

astead

ystate

flow

device

if∆PE

6=0,

∆KE6=

0,W

cv

=0

andQ

cv

=0:

⇒0=

0−

0+m[

(h1−h2 )

+(

V21−V

222

)

+g(z

1−z2 )]

P1

ρg+

V21

2g+z1=

P2

ρg+

V22

2g+z2

Pρg

:pressu

rehead

V2

2g

:velo

cityhead

z:

elevationhead

©D

r.M

d.

Zahuru

lH

aq

(BU

ET

)First

Law

ofT

herm

odynam

ics

ME6101

(2021)

19

/27

Tra

nsie

nt

Pro

cesse

s:First

Law

ofT

herm

odynam

ics

First

Law

ofT

hermodynam

ics:Transient

Analysis

dECV=

δQ

−δW

cv+∑

in

(

h+

V2

2+gz

)

i

dm

i−∑out

(

h+

V2

2+gz

)

e

dm

e

∫t0

(

dECV

dt

)

dt=

∫t0 (Q

−W

cv)dt+∑

in

∫t0

mi e

idt−∑out

∫t0

me e

edt

e≡

h+

V2

2+gz

⇒If

CV

isfixed

insp

ace,Ecv=

Ucv

⇒∫(

dEcv

dt

)

dt=

∫dUcv=

∫d(m

u)=

m2 u

2−m

1 u1=

∆Ucv

⇒∫dUcv=

∫d(m

u)=

∫(m

du+udm):

alternative

form.

⇒∫dUcv=

Q−W

cv+∫(

h+

V22+gz)

idm

i−∫(

h+

V22+gz)

edm

e

©D

r.M

d.

Zahuru

lH

aq

(BU

ET

)First

Law

ofT

herm

odynam

ics

ME6101

(2021)

20

/27

Tra

nsie

nt

Pro

cesse

s:First

Law

ofT

herm

odynam

ics

Inth

ean

alysisof

transien

tflow

system,tw

om

odels

arew

idely

used

:

1U

nifo

rmSta

te:

statew

ithin

CV

atan

yin

stant

isuniform

throu

ghou

t

the

CV.H

owever,

the

statew

ithin

CV

may

chan

gew

ithtim

e.It

implies

rapid

orin

stantan

eous

approach

toeq

uilibriu

mat

alltim

esfor

the

mass

with

inCV.

2U

nifo

rmFlo

w:

stateof

mass

crossing

aCS

isin

variant

with

time.

How

ever,m

assflow

rateacross

that

particu

larCS

may

varyw

ithtim

e.

This

condition

isfreq

uen

tlym

etw

hen

the

flow

into

transien

tsystem

is

supplied

froma

verylarge

reservoir.

∆m

CV=

m2−m

1=

mi−m

e

∆ECV=

m2 u

2−m

1 u1=

Q−W

CV+m

i ei−m

e ee

©D

r.M

d.

Zahuru

lH

aq

(BU

ET

)First

Law

ofT

herm

odynam

ics

ME6101

(2021)

21

/27

Tra

nsie

nt

Pro

cesse

s:First

Law

ofT

herm

odynam

ics

Charging

ofEvacuated

Vessel

Mass

enters

only

aton

esection

ofCS

and

no

efflux

ofm

atter.

T119

Assu

mption

s:

•U

niform

inlet

flow

•U

niform

statew

ithin

the

tank

atan

yin

stant

•Rigid

tank

andW

cv=

0

•Effects

ofK

E&

PE→

0.

•m

2−✟✟✯

0m

1=

∆m

cv=

mi−✟✟✯

0m

e⇒

m2=

mi

•If

tank

isin

sulated

orfilled

rapid

ly,Q

−→

0

•∆Ucv=

✓ ✓✼0

Q−✟✟ ✟✯

0W

cv+∫ei dm

i−∫ee✟

✟ ✟✯0

dm

e→

m2 u

2=

mi h

i=

m2 h

i

⇒u2=

hi

©D

r.M

d.

Zahuru

lH

aq

(BU

ET

)First

Law

ofT

herm

odynam

ics

ME6101

(2021)

22

/27

Tra

nsie

nt

Pro

cesse

s:First

Law

ofT

herm

odynam

ics

CengelEx.

5-1

2⊲

Charg

ing

ofevacu

atedVessel

T120

⇒hi=

h(Stea

m,P

1=

1.0

MPa,T

1=

300oC

)=

3051

kJ/kg

⇒hi=

u2=

u(Stea

m,P

2=

1.0

MPa,T

2=

?)7−→

T2=

456oC⊳

•W

hat

happens

inca

seofCN

Gcy

linderfillin

g?

u2=

hi →

T2=

kTi

©D

r.M

d.

Zahuru

lH

aq

(BU

ET

)First

Law

ofT

herm

odynam

ics

ME6101

(2021)

23

/27

Tra

nsie

nt

Pro

cesse

s:First

Law

ofT

herm

odynam

ics

Mora

nEx.4

-13⊲

Charg

ing

ofa

vessel:Estim

atework

input

required

iffinal

pressure

is4

bar.

���

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�� �������

�� ��������

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�% �������

�% ��������

%

&T1387

•V

=0.2

3m

3

•R

=287

J/kg

-K

•P

1=

Pi=

105

Pa

•P

2=

4×

105

Pa

•PV

=mRT

;V

=m

1 v1=

m2 v

2 ;P

1 v1.4

1=

P2 v

1.4

2

•∆Ucv=

−W

cv+hi ∆

mcv ,

foruniform

flow

,pip

eline

conditio

nis

unch

anged

.

•m

1=

0.3

32,m

2=

0.8

93

kg

•T

2=

163.9

6oC

•W

cv=

hi ∆m

cv−∆Ucv=

hi (m

2−m

1 )−(m

2 u2−m

1 u1 )

=44.7

kJ/kg

©D

r.M

d.

Zahuru

lH

aq

(BU

ET

)First

Law

ofT

herm

odynam

ics

ME6101

(2021)

24

/27

Tra

nsie

nt

Pro

cesse

s:First

Law

ofT

herm

odynam

ics

Discharging

Pro

cess

e835

⇒dm

cv=

✟✟ ✟✯

0dm

i−dm

e=

0−dm

e⇛

dm

cv=

−dm

e

⇒dEcv=

δQ

−δW

cv+(

h+

V22+gz)

i ✟✟ ✟✯

0dm

i−(

h+

V22+gz)

edm

e

⇒dUcv=

−he dm

e=

+hdm

:m

=m

cv

=m

assin

CV

⇒hdm

=dUcv=

d(m

u)=

mdu+udm

⇒(h

−u)dm

=mdu−→

dmm

=du

h−u=

du

Pv

⇒V

=mv→

✟✟✯

0dV

=mdv+vdm

⇒−

dvv=

dmm

⇒du

Pv=

dmm

=−

dvv⇒

du+Pdv

︸︷︷

︸Tds

=0→

ds=

0⇒

s=

const

©D

r.M

d.

Zahuru

lH

aq

(BU

ET

)First

Law

ofT

herm

odynam

ics

ME6101

(2021)

25

/27

Tra

nsie

nt

Pro

cesse

s:First

Law

ofT

herm

odynam

ics

Exam

ple

:⊲

Air

(P1

=10

MPa,

T1

=25o

C)

isallow

edto

escape

from

avessel

of20

m3

to200

KPa.

Estim

atefinal

temperatu

re.

⇒s=

consta

nt⇒

Pvk=

consta

nt⇛

dPP+kdvv=

0

⇒dmm=

−dvv=

1kdPP⇒

m2

m1

=(

P2

P1

)

1/k

=(

T2

T1

)

1/(k−

1)

⇒m

1=

P1V

1

RT

1

=2338

kg

⇒m

2

m1

=(

P2

P1

)

1/k⇒

m2=

143

kg

⇒m

2=

P2V

2

RT

2

⇒T

2=

−175oC⊳

©D

r.M

d.

Zahuru

lH

aq

(BU

ET

)First

Law

ofT

herm

odynam

ics

ME6101

(2021)

26

/27

Tra

nsie

nt

Pro

cesse

s:First

Law

ofT

herm

odynam

ics

Wark

Ex.

5-1

5:⊲

Ifcylin

der

pressure

remain

sco

nstan

t,estim

ate:(a)

mass

entered

thro

ugh

the

valve,(b

)final

temperatu

re.

•m

=PV

RT

•∆m

CV=

m2−m

1=

mi−m

e

•∆ECV=

m2 u

2−m

1 u1=

Q−W

CV+m

i ei−m

e ee

T1079

•air:

cp

=1000,cv

=720

J/kg

K

•m

1=

PV

1 /RT

1=

√,m

2=

PV

2 /RT

2

•m

e=

0→

m2−m

1=

mi

•Q

=0,W

cv=

P(V

2−V

1 )=

√

•ei=

hi ≃

cpTi=

√

•u≃

cvT

,u

1=

√,u

2=

cvT

2

•Energ

yEquatio

n:f(T

2 )only,

→T

2=

√

•m

i=

m2−m

1=

√

©D

r.M

d.

Zahuru

lH

aq

(BU

ET

)First

Law

ofT

herm

odynam

ics

ME6101

(2021)

27

/27