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Thermodynamics
Chapter 5: The First Law ofThermodynamics
Joule’s Experiment
During 1843-1849, Joule conducted several
experiments that subsequently led to the
formulation of the first law of thermodynamics.
Thermodynamics – Chapter 5
James Prescott Joule
Joule's Paddle-Wheel Apparatus,
1849, London Science Museum
Thermodynamics – Chapter 5
� Experiment consists of two process cycle.
� In process 1-2, work W was done on the system
by means of a paddle wheel.
� Heat was transferred from the fluid to the
atmosphere in process 2-1 till the system returns
to the original temperature.
Thermodynamics – Chapter 5
� The amount of work was measured by the
change in potential energy of a weight mg
falling through a height Z.
� The amount of heat transfer Q from the fluid to
the atmosphere was measured.
� By repeating the experiments for different
systems, Joule found that net work input W was
always proportional to the net amount of heat
transfer Q from the system measured in their
conventional units.
δ δJ Q W=∫ ∫� �Thermodynamics – Chapter 5
� Joule found that in the English system of units,
in which historically heat is measured in British
thermal unit (Btu) and work in foot pound
force (ft lbf), the mechanical equivalent of heat
is
J = 778.17 ft lbf/Btu
� In SI units, both heat and work are expressed in
same unit.
Thermodynamics – Chapter 5
Mechanical Equivalent of Heat (J)
Joule experimented on the amount of
mechanical work needed to raise the
temperature of a pound of water by
one degree Fahrenheit and found a
consistent value of 778.17 foot
pound force (4.186 J/cal).
Thermodynamics – Chapter 5
Observations From Joule’s Experiment
� Net work input W is always proportional to the
net amount of heat transfer Q from the system.
� Since the system at the end is restored to its
original state, so, algebraic sum of heat and work
interactions during a thermodynamic cycle is
zero (when same units are used).
� This experiment led to the formulation of the first
law of thermodynamics.
Thermodynamics – Chapter 5
Statement of First Law of thermodynamics
For a system undergoing a thermodynamic cycle
the net heat transfer during the cycle (cyclic
integral of heat added) is equal to the net work
done during the cycle (cyclic integral of work
done).
δ δQ W=∫ ∫� �
Thermodynamics – Chapter 5
� The symbol δδδδ is used to indicate that W and Q
are inexact differentials.
� The constraints of this law: This law applies to a
closed system and only to a thermodynamic
cycle.
� Both heat and work have same units.
� The first law of thermodynamics is also known as
the conservation of energy principle.
Thermodynamics – Chapter 5
Some aspects of the first law
First Law of thermodynamics for a
Change in State of a Control mass
2 1 2 1
1 2 1 2
A B A BQ Q W Wδ δ δ δ+ = +∫ ∫ ∫ ∫
Path 1-A-2-B-1
Path 1-C-2-B-1
( ) ( )2 2
1 1
δ δ δ δA C
Q W Q W− = −∫ ∫
Subtracting and rearranging
2 1 2 1
1 2 1 2
C B C BQ Q W Wδ δ δ δ+ = +∫ ∫ ∫ ∫
P
V2 2 2 2
1 1 1 1
A C A CQ Q W Wδ δ δ δ− = − ⇒∫ ∫ ∫ ∫
Thermodynamics – Chapter 5
� It is seen that for any process between states 1and 2, the quantity (δQ – δW) is always thesame.
� It is independent of the path followed for thischange of state.
� So, it is a point or state function and hence aproperty. This property is called energy (E) ofthe control mass.
� Thus we can write,
------------ (1)dE Q Wδ δ= −
Thermodynamics – Chapter 5
2 1 1 2 1 2E E Q W− = −
Since E is a property, its derivative can be written as
dE. When Eq. (1) is integrated from an initial state 1
to a final state 2,
where E1 and E2 are the initial and final values of
energy of the control mass.
� For a cycle, initial and final state is identical, so
energy increase is zero.
� For isolated system W and Q = 0. Hence E2 = E1
Thermodynamics – Chapter 5
Significance of the Property E
• It represents all the energy of the system
(macroscopic and microscopic forms) in the
given state.
• Macroscopic: Kinetic or Potential energy of the
system as a whole with respect to the chosen
coordinate frame.
• Microscopic: Energy associated with motion and
position of the molecules, and the structure of the
atom is called internal energy (U).
Thermodynamics – Chapter 5
( ) ( ) ----- (2)dE dU d KE d PE= + +
The first law of thermodynamics for a change of state
( ) ( ) --- (3)dE dU d KE d PE Q Wδ δ= + + = −
where2
and( )
( )2
( ) ( )
d mVd KE mVdV
d PE mg dZ d mg Z
= =
= =
Thermodynamics – Chapter 5
• In absence of electrical, magnetic, surface tension
effects
• i.e., for a simple compressible system
E = U + KE + PE
Integrating for a change of state from state 1 to state
2 with constant g,
2 1 1 2 1 2
2 2
2 1
2 1 2 1 1 2 1 22 2
E E Q W
mV mVU U mgZ mgZ Q W
− = −
− + − + − = −
Thermodynamics – Chapter 5
--- (4)dE dU mVdV mgdZ Q Wδ δ= + + = −
Three Observations
� The property E, the energy of the control mass,
was found to exist.
� The net change of the energy of the control mass
is always equal to the net transfer of energy across
the boundary as heat and work.
Thermodynamics – Chapter 5
� This equation can give only changes in internal
energy, kinetic energy, and potential energy.
in out System
Net Energy Transfer by heat and work Change in U, K.E and P.E energies
− = ∆14243 14243E E E
Internal Energy – A Thermodynamic Property
• Total energy contained by a thermodynamic system
• It is an extensive property and can be one of the
independent properties of a pure substance
• The symbol U designates the internal energy of a
given mass of a substance and u as the specific
internal energy.
• In the liquid – vapor saturation region,
(1 ) ( )
liq vap
lq f vap g
f g f g f
f fg
U U U
mu m u m u
u x u xu u x u u
u u xu
= +
= +
= − + = + −
= +
Thermodynamics – Chapter 5
Internal energy
Calculate the specific internal energy of
saturated steam having a pressure of 0.6
MPa and quality of 95%
u=uf + xufg
= 669.9 + 0.95 (1897.5) = 2472/5 kJ/kg
(E 5.4) Determine the missing property (P, T or
x) and v for water at
(a) T=300oC, u = 2780 kJ/kg
From table B.1.1, Psat = 8581kPa, ug = 2563 kJ/kg. So
given u > ug so the state is in the superheated vapor
region at some P < Psat
Searching through table B 1.3 at 300 C (superheated
vapor), u = 2780 kJ/kg at 1600 < P < 1800 kPa.
By linear interpolation, P = 1648 kPa and v = 0.1542 m3/kg
x is undefined (superheated).
Problem analysis and solution technique
1. What is the control mass/volume?
2. What is the initial state (known properties)?
3. What do we know about the final state?
4. What is the process that takes place? Is anything
constant or zero? Any functional relation between
properties?
5. Would a P-v, T-v diagram be useful?
6. What is the thermodynamic model for the behavior of
the substance (steam tables, ideal gas law, etc.)?
7. Use of all of the above information to analyze and solve
the problem
Enthalpy -The Thermodynamic Property
• Consider a control mass
undergoing quasi-equilibrium
constant-pressure process.
• Assume there is no change in
kinetic or potential energy .
• Work done during the process
is that associated with the
boundary movement.
Thermodynamics – Chapter 5
Taking the gas as our control mass and applying the
first law,
1 2 2 1 1 2Q U U W= − +
The work can be calculated from the relation,
2 2
1 2 2 1
1 1
( )W PdV P dV P V V= = = −∫ ∫Therefore,
1 2 2 1 2 2 1 1
2 2 2 1 1 1( ) ( )
Q U U PV PV
U PV U PV
= − + −
= + − +
Thermodynamics – Chapter 5
� Heat transfer during the process is given in terms
of the change in the quantity U + PV between
the initial and final states.
� Because all these quantities are thermodynamic
properties, that is, functions only of the state of
the system, their combination must have these
same characteristics.
� Therefore, it is convenient to define a new
extensive property, the enthalpy,
H = U + PV
Thermodynamics – Chapter 5
� The heat transfer in a constant-pressure, quasi-
equilibrium process is equal to the change in
enthalpy, which includes both the change in U and
the work for this particular process
� Many tables and charts of thermodynamics properties
give values for enthalpy but not for the internal
energy. In that case, u can be determined by
u = h – pv
� Is this valid when analyzing system processes that do
not occur at constant pressure?
� Remember enthalpy is a state function so it can be used to
calculate u irrespective of the process
The enthalpy of a substance in a saturation state and
with a given quality is found in the same way as the
specific volume and internal energy.
(1 )
liq vap
lq f vap g
f g
f fg
H H H
mh m h m h
h x h xh
h h xh
= +
= +
= − +
= +
For substances for which compressed-liquid tables
are not available, the enthalpy is taken as that
saturated liquid at the same temperature.
Thermodynamics – Chapter 5
Engineering Thermodynamics
(5.14) 2 kg of water at 120oC with a quality of 25%
has its temperature raised to 20oC in a
constant volume process. What are the new
quality and specific internal energy ?
State 1, at 120oC
1 1 11 1
3
( )
= 0.001061+ 0.25 (0.89152 - 0.001061)
= 0.22367 m /
f g fx
kg
ν ν ν ν= + −
State 2 has same ν at 140oC
Engineering Thermodynamics
( )2
2
2 2
2
2
2
0.22367 - 0.00108 0.4386
0.50849 - 0.00108
588.72 + 0.4386 1961.3 = 1448.9 kJ/kg
f
fg
f fg
x
u u xu
ν ν
ν
−= = =
= +
= ×
Engineering Thermodynamics
(5.15) 2 kg of water at 200 kPa with a quality of
25% has its temperature raised to 20oC in a
constant pressure process. What is the change
in enthalpy ?
State 1, at 200 kPa
State 2 has same pressure
1 11 1= 504.7 + 0.25 2201.6
= 1055.1 kJ /
f fgh h x h
kg
= + ×
o
220 120.2 + 20 = 140.2 C
satT T= + =
So state 2 is superheated vapor
Engineering Thermodynamics
( )2
140.2 - 120.2 2706.3 + 2768.5 - 2706.3
150 120.2
2748 kJ/kg
h =−
=
2 1 2748 - 1055.1
= 1693 kJ/kg
h h− =
Specific Heat
� It takes different amounts of energy to raise thetemperature of identical masses of differentsubstances by 1 degree.
� This is because different substances havedifferent energy storing capabilities.
Thermodynamics – Chapter 5
� It is thus desirable to have a property that enables
us to compare the energy storage capabilities of
various substances. This property is the specific
heat.
� It is defined as the energy required to raise the
temperature of a unit mass of a substance by one
degree.
Thermodynamics – Chapter 5
� The SI unit of specific heat is kJ/kg.K.
� Denoting specific heat by C, we have by
definition
Mathematical Model
1q Qc
T m T
δ δ
δ δ
= =
Thermodynamics – Chapter 5
Specific Heat at Constant Volume and
Pressure
� Cv is the energy required to raise the temperature
of a unit mass of a substance by one degree as the
volume is maintained constant.
� Cp is the energy required to do the same at
constant pressure is the specific heat at constant
pressure
Thermodynamics – Chapter 5
� Which is greater in value? Cp or Cv?
� The specific heat at constant pressure will alwaysbe greater than that at constant volume becauseat constant pressure the system can changevolume and energy for this expansion must besupplied.
Thermodynamics – Chapter 5
Specific Heat at Constant Volume
From definition of specific heat
1V
V V
q Qc
T m T
δ δ
δ δ
= =
Since for constant vo 0lum
e P dV
dU Q Wδ δ
=
= − (from First Law)
1V
V V V
q u Uc
T T m T
δ
δ
∂ ∂ = = =
∂ ∂
Thermodynamics – Chapter 5
Specific Heat at Constant Pressure
From definition of specific heat
1P
P P
q Qc
T m T
δ δ
δ δ
= =
1 2 2 1 2 2 1 1
2 2 2 1 1 1 2 1
Since for constant pressure process
(
first
( )
)
w)
(
la
Q U U PV PV
U PV U PV H H
= − + −
= + − + = −
Therefore, 1
P
P P P
q h Hc
T T m T
δ
δ
∂ ∂ = = =
∂ ∂
Thermodynamics – Chapter 5
Observations
� Both expressions for specific heats contain
thermodynamic properties and can be written as
cv = f (u, T, ν) and cP = f (h, T, P )
� cv and cp are state functions and hence are
properties.
� cv is a measure of the variation of internal energy
of a substance with temperature.
� cP is a measure of the variation of enthalpy of a
substance with temperature
Thermodynamics – Chapter 5
� Note the equations for cP and cv are property
relations and as such independent of the type of
process. They are valid for any substance
undergoing any process.
Sketch showing two ways in which a given ∆Umay be achieved
Thermodynamics – Chapter 5
dh du c dT≈ ≈
2 1 2 1 2 1( )h h u u c T T− − −� �
Engineering Thermodynamics
Specific Heat for Solids and Liquids
Also, for both of these phases, the specific volumeis very small, such that in many cases
Where c is either the constant-volume or constantpressure specific heat, as the two would be nearlythe same.
Since both of phases are nearly incompressible,
( )dh du d Pv du vdP Pdv= + = + + du vdP≈ +
� In general, u depends on the two independent propertiesspecifying the state
�For a low-density gas, however u depends primarily on T andmuch less on the second term, P or v.
Internal Energy, Enthalpy and Specific
Heat Relations for Ideal Gases
Thermodynamics – Chapter 5
� For an ideal gas, Pv = RT and u = f (T )
0 0 0 or = ⇒ = =
v v v
duc du c dT dU mc dT
dT
Thermodynamics – Chapter 5
� The relation between the internal energy and thetemperature can be given by
∂ =
∂ v
v
uc
T
� Because the internal energy of an ideal gas is nota function of specific volume, for an ideal gas
Where the subscript 0 denotes the specific heat ofan ideal gas
� From definition of enthalpy, we have
� Since R is a constant and u = f (T), it follows that
h = f (T )
0 0 0 or
p
p
p p p
hc
T
dhc dh c dT dH mc dT
dT
∂ =
∂
= ⇒ = =
Thermodynamics – Chapter 5
� The relation between the enthalpy and thetemperature can be given by
h u pv u RT= + = +
Pressure – volume diagram for an ideal gas
Since u and h are f (T ) for an ideal gas, lines of
constant T are also lines of constant u and constant h
No matter what the path, (1-2, 1-2’, 1-2”), change in u
(and h) remains the same.
0 0( ), ( )
v pc f T c f T= =
Thermodynamics – Chapter 5
� For an ideal gases,
Thermodynamics – Chapter 5
� The ideal gas specific heat for a given substance
is often called the zero-pressure specific heat.
� The zero-pressure, constant-volume specific heat
is given by Cvo.
� The zero-pressure, constant-pressure specific
heat is given by Cpo.
( ) ( )0 0
and v p
du c T dT dh c T dT= =
•Heat capacity for some gases as function of temperature.
•These values are determined by the techniques of statistical thermodynamics.
• The principal factor causing c to vary is molecular vibration.
• Complex molecules have multiple vibration modes, so greater T dependency
Thermodynamics – Chapter 5
The change in internal energy or enthalpy for an ideal
gas during a process from state 1 to 2,
To carry out the integrations relation for cP0 as
functions of T is required. There are three possibilities
to examine.
1) Assume constant specific heat, that is, no temperature
dependence (Table A.5).
True for some cases (fig) and a reasonable approximation if
Cp in the relevant T range is used
Thermodynamics – Chapter 5
2 1 0 2 1( )p
h h h C T T∆ = − = −
( )2 1 0ph h h C T dT∆ = − = ∫
2) Use an analytical equation cP0 as a function oftemperature.
� Because the results of specific-heat calculations fromstatistical thermodynamics don’t lend themselves toconvenient mathematical forms, these results have beenapproximated empirically.
� The equations for cP0 as a function of temperature arelisted in Table A.6 (Page 7) for a number of gases.
Thermodynamics – Chapter 5
3) To integrate the results of the calculations ofstatistical thermodynamics from an arbitraryreference temperature to any other temperatureT and to define a function
Thermodynamics – Chapter 5
0
2 1
2 1
0 0
0
2 1 0 0
T
T p
T
T T
p p T T
T T
h C dT
h h C dT C dT h h
=
− = − = −
∫
∫ ∫
and it is seen that the reference temperaturecancels out. This function hT is listed for air inTable A.7 (Page 8).
Specific Heat Relations for Ideal Gases
� It has been noted that for an ideal gas
Thermodynamics – Chapter 5
0 0
0 0
Differentation and substituting for and
= + = +
= +
= +
− =
p v
p v
dh du
h u pv u RT
dh du RdT
C dT C dT RdT
C C R
� The ratio of two specific heats of a gas is called
the specific heat ratio/heat capacity
ratio/adiabatic index and denoted by k or γ where
k = γγγγ = cP0 / cv0
� This ratio also is a function of temperature, but
the variation is mild.
� For monatomic gases, its value is essentially
constant at 1.667.
� Many diatomic gases, including air, have a
specific ratio of about 1.4 at room temperature.
Thermodynamics – Chapter 5
(5.82 /5.83) Use the ideal gas air table A.7 toevaluate the heat capacity Cp at 300 K as a slopeof the curve h(T) by ∆h/∆T. How much larger isit at 1000 K and 1500 K.
Engineering Thermodynamics
320 290
0300 K
320 290
320.58 290.431.005 kJ/kg.K
30
p
h hdh hc
dT T
−∆⇒ = = =
∆ −−
= =
1050 950
01000 K
1050 950
1103.48 989.441.14 kJ/kg.K
100
p
h hdh hc
dT T
−∆⇒ = = =
∆ −−
= =
Engineering Thermodynamics
1550 1450
01500 K
1550 1450
1696.45 1575.41.21 kJ/kg.K
100
p
h hdh hc
dT T
−∆⇒ = = =
∆ −−
= =
Notice an increase of 14%, 21% respectively.
(5.83/5.82) We want to find the change in u forcarbon dioxide between 600 K and 1200 K.
a) Find it from a constant Cvo from table A.5
b) Find it from a Cvo evaluated from equation inA.6 at the average T.
c) Find it from the values of u listed in table A.8
Engineering Thermodynamics
2 1 0 2 1(a) ( )
0.653 kJ/kg.K (1200 600) K
391.8 kJ/kg
vu u u C T T∆ = − ≅ −
= × −
=
Engineering Thermodynamics
1 2
avg
avg
2 3
0 0 1 2 3
2 3
0
0 0
2 1 0 2 1
1200 600(b) 900
2 2
9000.9
1000 1000
(KJ/kg)
0.45 1.67 0.9 1.27 0.9 0.39 0.9
1.2086 kJ/kg.K
1.2086 0.1889 1.0197 kJ/kg.K
( )
1.0197 kJ.kg
p
p
v p
v
T TT
T
C C C C C
C
C C R
u u u C T T
θ
θ θ θ
+ += = =
= = =
= + + +
= + × − × + ×
=
= − = − =
∆ = − = −
= ( ).K 1200 600 K 611.8 kJ/kg− =
The First Law as a Rate Equation
The first law can be expressed as a rate equation i.e.,either the instantaneous or average rate at whichenergy crosses the control surface as heat and workand the rate at which the energy of the control masschanges.
This form of the first law finds extensiveapplications in thermodynamics, fluid mechanics,and heat transfer. ( ) ( )
d KE d PEdUQ W
dt dt dt
dEQ W
dt
+ + = −
= −
& &
& &
Thermodynamics – Chapter 5
)(
)(
.
.
power
HTrate
W
Q
Engineering Thermodynamics
(5.10) A pot of water is boiling on a stove supplying
325 W to the water. What is the rate of mass
(kg/sec) vaporizing, assuming a constant
pressure process ?
=
325 W0.144 g/sec
2257 kJ/kg
fg
fg
fg
Q dE W du PdV dH h dm
Q dmh
dt dt
dm Q
dt h
δ δ
δ
= + + = =
=
= = =
�
Assume that the process takes place at atmospheric
pressure 101 kPa, so T = 100oC
Engineering Thermodynamics
Airplane takeoff from an aircraft carrier is assisted
by a steam-driven piston/cylinder with an average
pressure of 1250 kPa. A 17,500 kg airplane should
be accelerated from zero to a speed of 30 m/sec with
30% of the energy coming from the steam piston.
Find the needed piston displacement volume.
Hint : Take airplane as control mass
No change in internal or potential energy. Find the
energy increase of the control mass
Engineering Thermodynamics
( )2 2
2 1 2
10 0.5 17500 30
2
7875 kJ
E E m V− = − = × ×
=
The work supplied by the piston is 30% of the
energy increase.
2 1
3
0.30( )
0.30 7875 = 2362.5 kJ
2362.5 = 1.89 m
1250
piston avg
avg
W PdV P V E E
WV
P
= = ∆ = −
= ×
∆ = =
∫
Engineering Thermodynamics
(5.35/5.34) A 100-L rigid tank contains nitrogen(N2) at 900 K, 3 MPa. The tank is now cooled to100 K. What are the work and heat transfer forthis process?
2 1
2 1 1 2 1 2
2 1 1 2
( )
C.V. : Nitrogen in tank.
Energy equation :
Process : constant an 0d
m m
m u u Q W
Vv v
mV W
=
− =
= == ⇒
−
=
1 1
31 1
State 1: superheated
Tab
900 K, 3 Mpa
0.090 m /kg, 691.7 kJ/kle B.6.2 : g
T P
v u
= = ⇒
= =
Engineering Thermodynamics
3
31
0.1 m1.111kg
0.090 m /kg
Vm
v= = =
32 2 1 @100K
3
3
2
100 K, 0.090 m /kg
200kPa : 0.1425 m /kg, 71.73 kJ/kg
400kPa
State 2:
superheated
look in T
: 0.06806 m /kg,
able B.6.2 at 1
69.30 kJ/k
00 K
interpolation give
g
341kPas : ,
gT v v v
v u
v u
P u
= = = >
= =
= =
=
⇒
2 70.0 kJ/kg=
Engineering Thermodynamics
( )1 2 2 1 1 2( )
1.111 kg 70.0 691.7 kJ/kg 0
= 690.7 kJ
Q m u u W= − +
= − +
−
Engineering Thermodynamics
(5.135/5.153) A small flexible bag contains 0.1 kgammonia at –10oC and 300 kPa. The bag material issuch that the pressure inside varies linearly withvolume. The bag is left in the sun with an incidentradiation of 75 W, losing energy with an average 25 Wto the ambient ground and air. After a while the bag isheated to 30oC at which time the pressure is 1000 kPa.Find the work and heat transfer in the process and theelapsed time.
Engineering Thermodynamics
Take Ammonia as constant mass.
2 1
2 1 1 2 1 2
Continuity Eq.
Energy equation: ( - )
Process: (linear in V)
m m m
m u u Q W
P A BV
= =
= −
= +
State 1: from table B.2 (page 692/720) compressedliquid P > Psat therefore take saturated liquid atsame temperature.
o 3
1
1
( 10 C) = 0.001534 m /kg,
= 133.96 kJ/kg
f
f
v v
u u
= −
=
Engineering Thermodynamics
State 2: Table B.2.1 (page no. 692/720) at 30oC :
P < Psat so superheated vapor
3
2 2
3
2 2
From superheated table B.2.2 (page no. 696)
= 0.13206 m /kg, = 1347.1 kJ/kg,
= = 0.0132 m
v u
V mv
⇒
Work is done while flexible bag bulge outwards atincreasing pressure, so we get
( ) ( )
1 2 1 2 2 1
1( ) ( )
2
1300 1000 0.1 0.13206 0.001534
2
8.484 J
W P P V V= + −
= + × −
=
Engineering Thermodynamics
Heat transfer is found from the energy equation
1 2 2 1 1 2
.
.
1 2
( )
= 0.1 (1347.1-133.96) + 8.484
= 121.314 + 8.484 = 129.8 kJ
75 25 50 Watts
129800/ 2596 sec = 43.3 min
50
net
net
Q m u u W
Q
t Q Q
= − +
= − =
= = =
Engineering Thermodynamics
(5.28/Q) Find the missing properties.
o 3
2a. H 0, 250 C, = 0.02 m / kg, ? ?T P uν= = =
( ) ( )
oTable B.1.1 at 250 C: 3973 kPa
/ 0.02 - 0.001251 / 0.04887
= 0.38365
1080.37 + 0.38365 1522
= 1664.28 kJ/kg
f g sat
f g
f fg
P
x
u u xu
ν ν ν
ν ν ν
< < ⇒ =
= − =
= + = ×
Engineering Thermodynamics
2c. H O 2 C, 100 kPa ? ?
oT P u ν= − = = =
(5.37/5.39) A cylinder fitted with a frictionlesspiston contains 2 kg of superheated refrigerantR-134a vapor at 350 kPa, 100oC. The cylinder isnow cooled so the R-134a remains at constantpressure until it reaches a quality of 75%.Calculate the heat transfer in the process
Engineering Thermodynamics
2 1
1 2 1 2
C.V.: R 134a
Energy Eq.
Process : P const.
( 2 1)
1W2 PdV P V
P(V2 V1) Pm(v )
2 v1
m m m
m u u Q W
= =
− = −
= ⌠ = ∆
= − = −
−
= =>
Engineering Thermodynamics
State 1: 350 kPa, 100oC Table B.5.2 (pg 711)
h1 = (490.48 + 489.52)/2
= 490 kJ/kg
State 2: 350 kPa, x2=0.75 Table B.5.1 (pg 709)
h2 = 206.75 + 0.75 ×194.57
= 352.7 kJ/kg (P=350.9 kPa, saturation table)
Engineering Thermodynamics
1Q2 = m(u2 - u1) + 1W2
= m(u2 - u1) + Pm(v2 - v1)
= m(h2 - h1)
1Q2 = 2 × (352.7 – 490)= -274.6 kJ
(5.57/5.51) A cylinder having a piston restrained bya linear spring (of spring constant 15 kN/m)contains 0.5 kg of saturated vapor water at 120°C,as shown in Fig. Heat is transferred to the water,causing the piston to rise. If the piston cross-sectional area is 0.05 m2, and the pressure varieslinearly with volume until a final pressure of 500kPa is reached. Find the final temperature in thecylinder and the heat transfer for the process.
Engineering Thermodynamics
C.V. Water in cylinder.
Continuity: m2 = m1 = m ;
Energy : m(u2 - u1) = 1Q2 - 1W2
Engineering Thermodynamics
Process: P2 = P1 + ksm/Ap2 (v2 - v1)
State 2: P2 = 500 kPa
500 kPa = 198.5+15 × 0.5/(0.05)2 (v2 - 0.89186)
⇒ v2 = 0.9924 m3/kg
State 1: 120oC, saturated vapor
Table B.1.1(page 674 )
=> v1 = 0.89186 m3/kg, u1 = 2529.24 kJ/kg
Table B.1.3 => T2 = 803°C; u2 = 3668 kJ/kg
W12 = ⌠ PdV = ((P1 + P2)/2) m(v2 - v1)
= ((198.5 + 500)/2) × 0.5 × (0.9924 - 0.89186)
= 17.56 kJ
Engineering Thermodynamics
1Q2 = m(u2 - u1) + 1W2
= 0.5 × (3668 - 2529.2) + 17.56 = 587 kJ
Engineering Thermodynamics
(5.59/5.60) A 10 m high open cylinder,with Acyl = 0.1m2 contains 20oC waterabove and 2 kg of 20oC water below a198.5 kg thin insulated floating piston.Assume standard g, Po. Now heat isadded to the water below the piston sothat it expands , pushing the piston up,causing the water on top to spill overthe edge. This process continues untilthe piston reaches the top of thecylinder. Find the final state of thewater below the piston (T, P, v) and theheat added during the process.
B
A
Engineering Thermodynamics
( )
2
o
1
3
1 1
(
State 1 is compressed liquid 20 C
From Table B.1.1 (Page no. 674)
0.001002 m / kg 83.94 kJ/kg
0.1 10 998 kg
0.001002mass above the piston
A
A f A f
totaltotal
H O at
T
v v u u
Vm
vm
=
≅ = ≅ =
×= = =
1 ) 1 996 kgB total Am m= − =
( )
2 ( 1 )
1
3
1 1 1
State 1A:
198.5 9.81 + 996 9.81 101 +
0.1 1000 218.2 kPa
0.001002 2 0.002004 m
P H O at B
A o
P
A A A
m g m gP P
A
V v m
+= +
× ×=
×=
= × = × =
Engineering Thermodynamics
( )2
3 322 2
198.5 9.81State 2: 101 +
0.1 1000
120.5 kPa
10.1 10 1 m 0.5 m /kg
2
A o
P
AA cyl A
A
mgP P
A
VV A h v
m
×= + =
×
=
= × = × = = = =
( )
2 2
2 2
2
o 3 3
2
2
From Table B.1.2 120.5 kPa by interpolation
104.8 C, 0.001047 m / kg, 1.43128 m / kg
439.3 kJ / kg, 2070.5 kJ / kg
0.5 0.001047 0.348
1.4312
A A
A A
A
A f fg
f fg
A
P
T v v
u u
x
u
⇒ =
= = =
= =
−= =
2 439.3 0.348 2070.5 1161 kJ/kgA
= + × =
Engineering Thermodynamics
( )( )
2 1 1 2 1 2
1 2 1 2 2 1
Energy equation ( )
1 ( ) ( )
21
= 218.2 120.5 1 0.002 169 kJ2
Am u u Q W
W PdV P P V V
⇒ − = −
= = + −
+ − =
∫
1 2 2 1 1 2 2 1 1 2( ) ( )
= 2(1161 - 83.94) + 169
= 2323.12 kJ
A A A AQ m u u W m u u W= − + = − +
(5.78/5.75) A car with mass 1275 kg drives at60 km/h when the brakes are applied quickly todecrease its speed to 20 km/h. Assume the brakepads are 0.5 kg mass with heat capacity of 1.1kJ/kg.K and the brake discs/drums are 4.0 kgsteel. Further assume both masses are heateduniformly. Find the temperature increase in thebrake assembly.
Car, loses Kinetic energy, gains Internal
energy, No Heat transfer, No wo
C.V:
rk, constant=m
2 1 1 2 1 2
2 2
car 2 1 brake 2 1
Energy Eqn. (5.11): E E 0 0
1m ( ) m ( ) 0
2
Q W
V V u u
− = − = −
− + − =
The brake system mass is two different kinds sosplit it, also use Cv from Table A.3 since we donot have a u table for steel or brake pad material.
steel pad V
2
2 2 2 2
car
2 2
0
C from Table A.3:
m C m C
1 1000m 60 20 ) m s
2 3600
kJ4 0 46 0 5 1 1 T
K
=1275 kg 0 5 3200 0 07716 m s
65 9 C
V
VT T
T
∆ + ∆
= −
× + × ∆
× × ×
⇒ ∆ =
(
( . . . )
. ( . )
.