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.
I
I
J δQ = δW
First Law of Thermodynamics
Dr. Md. Zahurul Haq
Professor
Department of Mechanical Engineering
Bangladesh University of Engineering & Technology (BUET)
Dhaka-1000, Bangladesh
Mechanical equivalent of heat, J =
4.1868 kJ/kcal
1.0 in SI unit
zahurul@me.buet.ac.bd
http://teacher.buet.ac.bd/zahurul/
ME 6101: Classical Thermodynamics
T098
Example of a control mass undergoing a cycle.
c Dr. Md. Zahurul Haq
(BUET)
First Law of Thermodynamics
First Law of Thermodynamics
ME 6101 (2013)
1 / 24
Closed (CM) System
H
δW = J δQ
H
H
⇒ δQ = δW
[J = 1.0 in SI unit]
R2
R1
R2
R1
⇒ 1 δQA + 2 δQB = 1 δWA + 2 δWB 1
R2
R1
R2
R1
⇒ 1 δQC + 2 δQB = 1 δWC + 2 δWB 2
R
R
R
R
1 − 2 : 12 δQA − 12 δQC = 12 δWA − 12 δWC
R2
R2
R2
R2
⇒ 1 δQA − 1 δWA = 1 δQC − 1 δWC
Z2
Z2
(δQ − δW )A = (δQ − δW )C = · · ·
H
T038
1
1
R2
As 1 (δQ − δW ) is independent of the path involved 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.
δQ − δW = dE
⇛
Q12 − W12 = ∆E
c Dr. Md. Zahurul Haq
(BUET)
First Law of Thermodynamics
(BUET)
First Law of Thermodynamics
First Law of Thermodynamics
First Law of Thermodynamics for a Change in State
⇒
c Dr. Md. Zahurul Haq
ME 6101 (2013)
3 / 24
ME 6101 (2013)
2 / 24
Closed (CM) System
The physical significance of the property E is that it represents all
the energy of the system in the given state. This energy might be
present in a variety 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
coordinate frame that we select 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 Dr. Md. Zahurul Haq
(BUET)
First Law of Thermodynamics
ME 6101 (2013)
4 / 24
First Law of Thermodynamics
Closed (CM) System
First Law of Thermodynamics
Internal Energy (U): A Thermodynamic Property
Closed (CM) System
E = U + KE + PE + · · ·
⇒ δQ − δW = dU + d(KE ) + d(PE ) + · · ·
⇒
δQ
dt
−
δW
dt
=
dU
dt
+
d(KE )
dt
+
d(PE )
dt
dEcm
dt
+ ··· =
dEcm
dt
_ − W
_
=Q
R2
dU = U2 − U1 = m(u2 − u1 )
R2
⇒ d(KE ) = mVdV =⇒ 1 d(KE ) = 21 m(V22 − V21 )
R2
⇒ d(PE ) = mgdZ =⇒ 1 d(PE ) = mg(Z2 − Z1 ) = mgh
⇒ dU =⇒
T133
Q12 − W12 = (U2 − U1 ) + 21 m(V22 − V21 ) + mg(Z2 − Z1 )
q12 − w12 = (u2 − u1 ) + 12 (V22 − V21 ) + g(Z2 − Z1 ) ≃ (u2 − u1 )
T134
Various forms of microscopic
energies that make up sensible
energy.
c Dr. Md. Zahurul Haq
(BUET)
Internal energy of a system is the
sum of all forms of the microscopic
energies.
First Law of Thermodynamics
First Law of Thermodynamics
ME 6101 (2013)
5 / 24
1
c Dr. Md. Zahurul Haq
Closed (CM) System
(BUET)
First Law of Thermodynamics
First Law of Thermodynamics
ME 6101 (2013)
6 / 24
Closed (CM) System
Enthalpy (H)): A Thermodynamic Property
H ≡ U + PV
⇛
h ≡ u + Pv
⇒ Q12 − W12 = ∆E
⇒ Q12 − W12 = ∆U if KE → 0, PE → 0
R2
⇒ W12 = 1 PdV = P(V2 − V1 )
T129
⇒ Q12 = U2 − U1 + P(V2 − V1 )
⇒ Q12 = (U2 + P2 V2 ) − (U1 + P1 V1 )
⇒ Q12 = H2 − H1
T117
The 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.
T130
c Dr. Md. Zahurul Haq
T131
(BUET)
First Law of Thermodynamics
T132
ME 6101 (2013)
7 / 24
c Dr. Md. Zahurul Haq
(BUET)
First Law of Thermodynamics
ME 6101 (2013)
8 / 24
First Law of Thermodynamics
Closed (CM) System
First Law of Thermodynamics
Closed (CM) System
Joule’s Free Expansion Experiment
⊲ Example: Electric Heating of Gas at Constant Pressure.
⇒ Qnet = −3.7 kJ
⇒ We = −7.2 kJ
⇒ h1 = h (300 kPa, x = 1)
Valve is opened and allowed to
equilibrate.
⇒ h1 = 2724.9 kJ/kg
⇒ h2 = h (300 kPa, T2 )
No change in water temperature.
So, no heat transfer takes place.
⇒ T2 = ?
T118
⇒ Qnet − Wnet = ∆E ≃ U
R2
⇒ Qnet − (We + 1 PdV ) = ∆E ≃ ∆U
T139
⇒ Qnet − We − P(V2 − V1 ) = U2 − U1
⇒ Qnet − We = P(V2 − V1 ) + U2 − U1 = H2 − H1 = m(h2 − h1 )
=⇒ h2 = 2864.9 kJ/kg at 300 kPa & T2 =⇒ T2 = 200o C ⊳
c Dr. Md. Zahurul Haq
(BUET)
First Law of Thermodynamics
First Law of Thermodynamics
ME 6101 (2013)
9 / 24
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 Dr. Md. Zahurul Haq
Closed (CM) System
(BUET)
First Law of Thermodynamics
First Law of Thermodynamics
ME 6101 (2013)
10 / 24
ME 6101 (2013)
12 / 24
Closed (CM) System
Specific Heats
C ≡
1 δQ
m δT
CV ≡
1
m
δQ
δT
δq
δT
δq = du + δw = du + Pdv ⇒
δq
δu
dv
⇒ cV ≡ δT
= δT
+ P δT
=
V
V
δq
du
= dT
=⇒
cV ≡ δT
V
CP ≡
1
m
δQ
δT
Ideal gas models:
P
R=
δu
dv
= δT
+ P δT
δU
δT V + 0
cP (T ) − cV (T ) = R
)
: k (T ) = ccVP (T
(T )
R2
du = cV dT ⇒ u2 − u1 = 1 cV (T )dT
R2
dh = cP dT ⇒ h2 − h1 = 1 cP (T )dT
cV (T ) =
as u = f (T ) for ideal gas
V
Ru
M
δq
δh
dP
= δT
− v δT
δq = du + δw = d(h − Pv ) + Pdv = dh − vdP ⇒ δT
δq
δh
dP
δh
⇒ cP ≡ δT
= δT
− v δT
= δT
−0
P
P
P
δq
dh
as h = f (T ) for ideal gas
= dT
=⇒
cP ≡ δT
R
k (T )−1
T128
P
h = u +Pdv = u +RT ⇒ dh = du +RdT ⇒ cP dT = cV dT +RdT
Specific heat ratio, k ≡ ccVP
=⇒
cP − cV = R
c Dr. Md. Zahurul Haq
(BUET)
cV =
R
k −1
First Law of Thermodynamics
cP =
kR
k −1
ME 6101 (2013)
11 / 24
cT126Dr. Md. Zahurul Haq
(BUET)
T127
First Law of Thermodynamics
First Law of Thermodynamics
Closed (CM) System
First Law of Thermodynamics
Open (CV) System
Conservation of Energy for CV System
T115
First Law of Thermodynamics for closed system
T099
c Dr. Md. Zahurul Haq
(BUET)
First Law of Thermodynamics
First Law of Thermodynamics
ME 6101 (2013)
13 / 24
c Dr. Md. Zahurul Haq
Open (CV) System
(BUET)
First Law of Thermodynamics
First Law of Thermodynamics
ME 6101 (2013)
14 / 24
Open (CV) System
T100
Ecv
dt
_ + m
= Q_ − W
_ i ei − m
_ e ee
V2i
V2e
_
_
+ gzi − m
+ gze
_ i
_ e
ui +
ue +
= Q −W +m
2
2
Flow work, δWflow necessary to push a differential mass, δm into a CV
is the product of force, F = PA and the distance,dx = Vdt under
uniform flow.
_ consists of 3 parts:
W
1
_
Flow work, W
flow , associated with mass crossing the control
surface,
2
_
Shaft work, W
shaft , which can be interchanged between the system
and its surroundings,
3
If the CS is not rigid, displacement work, Wd ≡ PdV occurs.
_
Second and third components of work are lumped together as Wnet .
c Dr. Md. Zahurul Haq
(BUET)
First Law of Thermodynamics
ME 6101 (2013)
T101
15 / 24
⇒ δWflow = Fdx = PAVdt = Pv mdt
_
_
_
⇒ W
flow = Pv m
c Dr. Md. Zahurul Haq
_ = W
_
W
_ i (Pv )i − m
_ e (Pv )e
net + m
(BUET)
First Law of Thermodynamics
ME 6101 (2013)
16 / 24
First Law of Thermodynamics
Open (CV) System
First Law of Thermodynamics
Open (CV) System
⊲ Example: Steam Turbine
First Law of Thermodynamics (FLT) for CV System
dEcv
dt
_ −W
_
=Q
net +
X
X V2
V2
m
_ i
m
_ e
hi + i + gzi −
he + e + gze
2
2
e
i
Closed System: 7→ m
_ i = m
_ e = 0.
dEcv
dt
_ − W
_
=Q
net
Steady-State-Steady Flow (SSSF) System:
P
P
dmcv
_ i =
_ e
:
em
im
dt = 0 =⇒
dEcv
dt
Steady-state Steady-flow:
h
2 2
i
V −V
_ −W
_
_
(h1 − h2 ) + 1 2 2 + g(z1 − z2 )
⇒ 0=Q
net + m
T116
=0
h1 = h (2MPa, 350o C ) = 3137.0 × 103 J/kg
One-inlet, One-exit & Steady-state: 7→ m
_
=m
_ e = m.
_
h
2 2 i
i
V
−V
_ − W
_
_
(h1 − h2 ) + 1 2 2 + g(z1 − z2 )
0=Q
net + m
c Dr. Md. Zahurul Haq
(BUET)
First Law of Thermodynamics
First Law of Thermodynamics
ME 6101 (2013)
h2 = h (0.1MPa, x2 = 1.0) = 2675.5 × 103 J/kg
_
=⇒ W
net = 678.2 kW ⊳
17 / 24
c Dr. Md. Zahurul Haq
Transient Flow
dEcv
dt
0
⇒ ∆mcv
⇒
Rt P
Rt P
_ i ) dt −
_ e ) dt
0( em
0( i m
P R
P R t
t
= mcv (t) − mcv (0) = i 0 m
_ i dt
− e 0m
_ e dt
dt =
∆mcv = mcv (t) − mcv (0) =
X
mi −
i
c Dr. Md. Zahurul Haq
(BUET)
18 / 24
Transient Flow
First Law of Thermodynamics
_
_
= Q − Wcv
X
X V2i
V2e
+ gzi −
+ gze
+
m
_ i
m
_ e
hi +
he +
2
2
e
i
i
dmcv
dt
ME 6101 (2013)
Energy Balance: Transient Flow
X
X
dmcv
m
_ i −
m
_ e
=
dt
e
Rt
First Law of Thermodynamics
First Law of Thermodynamics
Mass Balance: Transient Flow
⇒
(BUET)
X
+
P Rt i 0 h +
V2
2
+ gz
i
m
_ i dt −
⇒ If CV is fixed in space, Ecv = Ucv
R dEcv R
R
⇒
dt = dUcv = d(mu ) = m2 u2 − m1 u1 = ∆Ucv
dt
me
e
ME 6101 (2013)
Rt
Rt
dEcv
_
_
dt = 0 Qdt
− 0W
cv dt
dt
0
P Rt
2
V
m
_ e dt
e 0 h + 2 + gz
e
Rt
19 / 24
⇒ dUcv =
P P 2
2
δQ − δWcv + i h + V2 + gz dmi − e h + V2 + gz dme
i
e
R
R
V2
V2
h + 2 + gz dme
⇒ ∆Ucv = Q − Wcv +
h + 2 + gz dmi −
i
e
R
R
R
⇒ dUcv = d(mu ) = (mdu + udm): alternative form.
c Dr. Md. Zahurul Haq
(BUET)
First Law of Thermodynamics
ME 6101 (2013)
20 / 24
First Law of Thermodynamics
Transient Flow
First Law of Thermodynamics
Transient Flow
Charging of Evacuated Vessel
In the analysis of transient flow system, two models are widely used:
1
Uniform State: the state within CV at any instant is uniform
throughout the CV. However, the state within CV may change
with time. It implies rapid or instantaneous approach to
equilibrium at all times for the mass within CV.
2
Uniform Flow: the state of mass crossing a CS is invariant with
time. However, the mass flow rate across that particular CS may
vary with time. This condition is frequently met when the flow
into transient system is supplied from a very large reservoir of
matter.
c Dr. Md. Zahurul Haq
(BUET)
First Law of Thermodynamics
First Law of Thermodynamics
ME 6101 (2013)
21 / 24
Mass enters only at one section of CS and no efflux of matter.
T119
Assumptions:
Adiabatic, rigid control volume
Uniform state within the tank at any instant
Negligible KE of inflowing gas
Uniform inlet flow
c Dr. Md. Zahurul Haq
Transient Flow
(BUET)
First Law of Thermodynamics
First Law of Thermodynamics
Charging of Evacuated Vessel: CV Analysis
ME 6101 (2013)
22 / 24
Transient Flow
Charging of Evacuated Vessel: CM Analysis
If tank is insulated or filled rapidly, Q −→ 0
Wcv = 0, ∆KE = 0, ∆PE = 0
If tank is insulated or filled rapidly, Qcv −→ 0
Supply from a large reservoir → uniform flow.
P
∆mcv = mi − me ⇒ m2 − m1 = mi − me
Wcv 6= 0, ∆KE = 0, ∆PE = 0
⇒ ∆Ucv = Qcv − Wcv
⇒ m1 = 0, me = 0 ⇒ m2 = mi
⇒ ∆Ucv =
R
R
Qcv − Wcv + (· · · )i dmi − (· · · )e dme = hi mi
T120
⇒ ∆Ucv = m2 u2 − m1 u1 = m2 u2
=⇒ hi = u2
−→ T2 = kTi : for ideal gas.
(BUET)
First Law of Thermodynamics
ME 6101 (2013)
R
⇒ W = PdV = P(V2 − V1 ) = P[Vtank − (Vtank + V1 )] = −mP1 v1
=⇒ hi = u2
⇒ hi = h (Steam, P1 = 1.0 MPa, T1 = 300o C ) = 3051 kJ/kg
c Dr. Md. Zahurul Haq
Pressure in the line near the valve is constant.
T121
⇒ m(u2 − u1 ) = mP1 v1 ⇛ u2 = u1 + P1 v1
⊲ Example: Estimate T2 .
⇒ hi = u2 = u(Steam, P2 = 1.0 MPa, T2 = ?) 7−→ T2 = 456o C ⊳
⇒ ∆Ucv = m(u2 − u1 )
23 / 24
c Dr. Md. Zahurul Haq
: same result.
(BUET)
First Law of Thermodynamics
ME 6101 (2013)
24 / 24