CHAPTER
3
The First Law of
Thermodynamics:
Closed Systems
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Closed system
Energy can cross the boundary of a closed system in two forms: Heat and work
FIGURE 3-1
Specifying the
directions of
heat and work.
3-1
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Heat
Energy transfer by virtue of temperature difference.
If there is no heat transfer during a process, then
the process is called adiabatic.
3-2
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30 kJ heat
Q=30 kJ
M=2 kg
Δt=5 s
Heat transfer per unit mass
Heat transfer rate
(kJ/kg)
(kJ/s or kW)
Q˙=6 kW
q=15 kJ/kg
Sign Convention for Heat
Heat in
Q=+5 kJ
(+) ve if to the system
(-)ve if from the system
Q= - 5kJ
Heat out
3-3
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System boundary
A spring is doing work on the surroundings
Work
An energy interaction which is not
caused by a temperature difference
between a system and its
surroundings.(a rising piston, a
rotating shaft,etc.)
(kJ/kg)
The work done per unit time is called
power and denoted
(kJ/s or kW)
W=5kJ
Sign Convention of Work
W=-5kJ
(+)ve if work done by a system
(-)ve if work done on a system
3-4
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Moving boundary work (P dV work):
The expansion and compression work in a
piston-cylinder device.
FIGURE 3-2
A gas does a differential amount of work dWb
as it forces the piston to move by a differential
amount ds.
3-5
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FIGURE 3-3
The area under
the process
curve on a P-V
diagram
represents the
boundary work.
3-6
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FIGURE 3-4
The net work
done during a
cycle is the
difference
between the
work done by
the
system and the
work done on
the system.
3-7
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First Law of Thermodynamics or the Conservation of energy Principle:
Net energy transfer
to (or from) the system
As heat and work
Q–W
3-9
=
=
Net increase (or decrease)
İn the total energy
of the system
ΔU
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(Remember from chapter 1:
ΔE = ΔU + ΔKE + ΔPE )
Q – W = ΔU + ΔKE + ΔPE
where ΔU = m(
-
ΔKE =1/2 m(
ΔPE =mg(
)
-
-
)
)
Q – W = ΔU
3-10
(kJ)
Most closed systems
encountered in
practice are stationary
i.e. ΔPE =0 ΔKE =0
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Example:
0
System
boundary
Wpw =
18.5kJ
0
Q  W  U  KE  PE
ΔU = m(u2 - u1 )
5 kg of
steam
Wpiston
W=Wpw +Wpiston
Q -(Wpw +Wpiston)= m(u2 - u1 )
Q = 80 kJ
u1  2709.9 kJ/kg
(+80kJ) –( ( -18.5kJ)+ Wpiston=(5kg)(2659.6 -2709.9)kJ/kg
u2  2659.6 kJ/kg
Wpiston =350kJ
3-12
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Example: constant-pressure process, initially saturated water vapor.
Q – Wel =
=
-3.7 kJ – (-7.2kJ)= 0.025 kg (
= 2865.3 kJ/kg
- 2725.3 kJ/kg)
=300 kPa
= 2865.3
3-13
=?
Note: for constant pressure case: Q – Wother= ΔH
=200ºC Table A-6
and W = Wboundary + Wother
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Example: constant volume
T
1
Superheated
steam
2
v
0
Q
0
0
Q  W  U  KE  PE
Q  m (u2  u1 )
Note: for constant volume case:
3-14
Q  Wother  U  KE  PE
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Example: changing volume and pressure
P
Evacuated
system
1
2
Partition
Q
H2O
V
0
0
0
Q  W  U  KE  PE
Q  m (u2  u1 )
note: u1≈ uf@Tsat
u2 = uf +x2ufg
3-15
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Specific Heats
The energy required to raise the temperature of a unit of a substance by one
degree.
Cv : specific heat at constant volume
Cp : specific heat at constant pressure
Helium gas:
V=constant
m=1kg
ΔT=1ºC
Cv=3.13 kJ/kgºC
(2)
(1)
P=const
m=1kg
ΔT=1ºC
Cp=5.2 kJ/kgºC
3.13kJ
5.2kJ
3-16
Cp > Cv
Because at constant
pressure, the energy
required for expansion work
must also be supplied to
system.
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First law for constant volume:
First law for constant pressure:
3-17
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Ideal Gases:
Pv  RT
Joule demonstrated that for ideal
gases
u=u(T)
Cv=Cv(T)
h  u  Pv
Pv  RT
h  u  RT
Since R is constant and u=u(T)
and
3-18
h=h(T)
Cp=Cp(T)
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Three ways of calculating u and h
h  h2  h1
from tables
2
h   c p (T ) dT
1
h  c p ,avg T
Similarly:
3-19
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differentiate
cpdT =cvdT+RdT
÷ dT
On a molar basis
Specific heat ratio
3-20
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Solids and Liquids:
Can be approximated as incompressible:
c p  cv  c
Again, specific heats depend on temperature only.
The change in internal energy between states 1 and 2:
2
u  u2  u1   c(T ) dT
1
u  cavg (T2  T1 )
3-21
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Example:
insulated
Q-W=ΔU or ΔU=0
U  U iron  U water  0
Water
25ºC
mC(T2  T1 )iron  mC(T2  T1 )water  0
m  50kg
mwater 
80 o C
0.5
50 kg(0.45 kJ
T2  25.6 o C
3-22
V
v25 C
0.5 m 3

 500 kg
3
0.001 m kg
Specific heats are determined from
table A-3.


) T2  80o C  500kg(4.18 kJ o )(T2  25o C)  0
kg C
kg C
o