Download Unit 61: Engineering Thermodynamics

Unit 61: Engineering
Thermodynamics
Lesson 7: First Law of
Thermodynamics
Objective
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• The purpose of this lesson is to consider the
concepts of work and heat.
First Law of
Thermodynamics
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• The first law is often called the conservation of
energy.
• In school physics the conservation of energy
emphasises changes in kinetic and potential
energy and their relationship to work. A more
general form of the conservation of energy
includes the effects of heat transfer and
internal energy – this general form is the first
law of thermodynamics.
First Law of
Thermodynamics applied to a Cycle
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• The first law is not derived or proved it is a statement
based on experimental observations.
• The first law states that for a cycle the net heat transfer
is equal to the net work done for a system undergoing
a cycle i.e.
ΣW = ΣQ or
δW = δQ
• Note: δ is used because there are inexact differentials
• Note the
represents integration around a complete
cycle
First Law of
Thermodynamics
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Consider a weight W attached to a pulley / paddle wheel
Q
W
W
First Law of
Thermodynamics applied to a cycle
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• Let the weight fall through a certain distance
thereby doing work on the system, contained in
the insulated tank shown. Work done is weight
times distance dropped.
• Let the system return to its initial state by
transferring heat to the surroundings Q thereby
reducing the temperature of the system to its
original temperature.
• The first law states that this heat transfer will be
exactly equal to the work which was done by the
falling weight.
First Law of
Thermodynamics applied to a Cycle
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• A spring is stretched a distance of 0.8m and attached
to a paddle wheel. The paddle wheel then rotates
until the spring is un-stretched. Calculate the heat
transfer necessary to return the system to its initial
state (k = 100N/m)
0.8
0.8
W1-2 = fdx = 100xdx = 100(0.82/2) = 32 N.m
0
0
But from first law Q1-2 = W1-2 = 32 N.m
First Law of
Thermodynamics applied to a Process
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• The first law is often applied to a process as
the system changes from one state to another.
Realising that a cycle results when a system
undergoes several processes and returns to
the initial state, we could consider a cycle
composed of two or more processes
represented as shown…
First Law of
Thermodynamics applied to a Process
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P
2
A
1
B
V
First Law of
Thermodynamics applied to a Cycle
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δQA + δQB =
1
1
2
1
2
δWA + δWB
2
1
2
2
2
2
δQA -
δQB =
δWA -
1
1
1
2
2
(δQ – δW)A =
(δQ – δW)B
1
1
2
δWB
1
First Law of
Thermodynamics applied to a Process
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• Thus the change in the quantity Q – W from state 1
to state 2 is the same along path A as along path B.
• Since this change is independent between states a
and 2 we let δQ – δW = dE (note an exact
differential).
• E is an extensive property of the system shown
experimentally to represent the energy of the system
at a particular state. Thus
Q1-2 – W1-2 = E2 – E1
First Law of
Thermodynamics applied to a Process
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• Q1-2 is the heat transferred to the system during the
process from state 1 to state 2.
• W1-2 is the work done by the system on the
surroundings during the process
• E2 and E1 are the values of the property E.
• E represents all the energy – kinetic (KE), potential
(PE) and internal (U) (which includes chemical and
the energy associated with the atom).
Thus Q1-2 – W1-2 = KE2 – KE1 + PE2 – PE1 + U2 – U1 +
Epress
First Law of
Thermodynamics applied to a Process
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• In an open system, the fluid is continuously flowing
in and out of the system and work transfers take
place. Epress is thus the flow or pressure energy =
pressure x volume = PV as the fluid passes through
the system
• Q1-2 – W1-2 = KE2 – KE1 + PE2 – PE1 + (U2 – U1) +
(Epress2 – Epress1)
= (m/2)(V22 –V12)+ mg(z2 – z1) + (U2 – U1)
+ (p2V2 – p1V1)
Enthalpy
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• In the solution of problems involving systems,
certain products or sums or properties occur
with regularity.
• One combination of properties can be
demonstrated by considering the addition of
heat to the constant-pressure situation.
Consider…
Enthalpy
W
W
Gas
W
Gas
Gas
Q
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Heat is added
slowly to the
system (the gas in
the cylinder) which
is maintained at
constant pressure
by assuming a
frictionless seal
between the piston
and the cylinder.
First Law of
Thermodynamics applied to a Process
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• If the KE changes and the potential energy changes
of the system are neglected and all other work
modes are absent, the first law of thermodynamics
requires that…
Q – W = U 2 – U1
• The work done raising the weight for the constant
pressure process is given by…
W = P(V2 – V1)
Thus Q = (U + PV)2 - (U + PV)1
Enthalpy
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• The quantity U + PV is known as enthalpy (H). As this
is a combination of properties, it itself is therefore a
property.
• Specific enthalpy is found by dividing by the mass…
h = u + Pv
• Thus Q1-2 = H2 – H1
• Note: the enthalpy was defined using a constantpressure system with the differences between
enthalpies between two states being the heat
transfer.
Latent Heat
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• The amount of energy that must be transferred in the
form of heat to a substance held at constant pressure in
order that a phase change occurs is called latent heat.
• It is the change in enthalpy of the substance at the
saturated conditions of the two phases
• The heat necessary to melt (or freeze) a unit mass of
substance at constant pressure is the heat of fusion and
is equal to…
hif = hf – hi
• Where hi is the enthalpy of the saturated solid and hf the
enthalpy of the saturated liquid
Latent Heat
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• The heat of vapourisation is the heat required to
completely vapourise a unit mass of saturated liquid
(or condense a unit mass of saturated vapour). It is
equal to…
hfg = hg – hf
• When a solid changes phase directly to a gas,
sublimation occurs. The heat of sublimation is equal
to…
hig = hg – hi
Specific Heats
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• For a simple system only two independent variables
are necessary to establish the state of the system.
Consequently we can consider the specific internal
energy to be a function of temperature and specific
volume i.e.
u = u(T, v)
• Using the chain rule…
du = ∂u/∂T|v dT + ∂u/∂v|TdV
Specific Heats
• Since u, v and T are all properties, the partial
derivative is also a property and is called the
constant-volume specific heat Cv.
Cv = ∂u/∂T|v
…this is because for an ideal gas ∂u/∂v|T = 0
Thus du = CvdT T2
i.e
u2 – u1 = CvdT
T1
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Specific Heats
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• For a known Cv(T) this can be integrated to find the
change in internal energy over any temperature
interval for an ideal gas.
• Likewise, considering specific enthalpy to be
independent on the two variables T and P i.e.
h = h(T,P)we have…
dh = ∂h/∂T|pdT+ ∂h/∂P|T dP
• The constant-pressure specific heat Cp is defined as…
Cp = ∂h/∂T|P
Specific Heats
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• For a known Cv(T) this can be integrated to find the
change in internal energy over any temperature
interval for an ideal gas.
• Likewise, considering specific enthalpy to be
independent on the two variables T and P we have…
dh = ∂h/∂T|pdT+ ∂h/∂P|T dP
• Note for an ideal has ∂h/∂P|T = 0 (see next slide)
• The constant-pressure specific heat Cp is defined as…
Cp = ∂h/∂T|P
Specific Heats
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• From previous h = u + Pv = u + RT.
• Since internal energy u, is only a function of T we see
that h is also only a function of T for an ideal gas. Hence
for an ideal gas…
∂h/∂P|T = 0
Again from previous…
dh = ∂h/∂T|pdT+ ∂h/∂P|T dP
Thus dh = CpdT
Over the temperature rate T1 to T2 this is integrated to
give….
Specific Heats
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T2
h2 – h1 =
CPdT
T1
It is often convenient to specify specific heats on
a per mole basis rather than per-unit-mass
__
__
basis. These molar specific heats are Cv and Cp
thus…
__
__
Cv = MCv
and
Cp = MCp
Where M is the molar mass.
Specific Heats
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The equation for enthalpy can be used to relate
for an ideal gas, the specific heats and the gas
constant i.e.
dh = du + d(Pv)
Introducing the specific heat relationships and
the ideal gas equation gives…
CPdT = CvdT + RdT
Thus Cp = Cv + R
or
R = Cp – Cv
Specific Heat Ratio
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We can define the specific heat ratio as…
γ = Cp/Cv
Thus substituting into…
R = Cp – Cv for Cv gives…
R = C p – Cp/γ = Cp(γ-1)/γ
i.e.
Cp = Rγ/(γ-1)
Similarly Cv = R/(γ-1)
Specific Heats
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• Note 1: Since R is a constant for an ideal gas, the
specific heat ratio will depend only on temperature
• Note 2: For gases, the specific heats slowly increase
with increasing temperature. Since they ado not
vary significantly over fairly large temperature
differences, it often acceptable to treat Cp and Cv as
constants. For such cases…
u2 – u1 = Cv(T2 – T1)
and h2 – h1 = Cp(T2 – T1)
Example 1
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• At entry to a horizontal steady flow system, a
gas has a specific enthalpy of 2000kJ/kg and
possesses 250kJ/kg of kinetic energy. At the
outlet of the system, the specific enthalpy is
1200kJ/kg and there is negligible kinetic
energy. If there I no heat energy transfer
during the process determine the magnitude
and direction of the work done.
Solution
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• Using the specific enthalpy form of the steady
flow energy equation (SFEE)…
Q – W = (h2 – h1) + (1/2)(V22 –V12)+ g(z2 – z1)
We note that g(z2 – z1) = 0 since there is no change in
height between the fluid entry and exit (horizontal).
Also there is negligible fluid kinetic energy at exit thus
(1/2)V22 = 0 and during the process Q = 0. Thus SFEE
gives…
0 – W = (1200 – 2000) + 0 + (0 – 250) = 1050 kJ/kg
As work is positive, work is done by the system.
Example 2
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• Steady flow air enters the inlet of a horizontal
compressor at a velocity of 10m/s, a pressure of 1bar and a temperature of 15oC. It exits the
compressor flowing at 0.6 kg/s with a velocity of
25m/s, a pressure of 7-bar and a temperature of
65oC. If the input power to the compressor is 42kW
and for the air cp = 1.005 kJ/kg.K and R = 0.287
kJ/kg.K determine…
a. The specific heat energy lost to the surroundings
b. The cross sectional area of the compressor exit
duct.
Solution
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Heat (Q)
v1 = 10m/s
h1 = ?
P1 = 1-bar
T1 = 288K
v2 = 25m/s
h2 = ?
P2 = 7-bar
T2 = 338K
System Boundary
Work (W)
Solution
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• Using…
Q – W = (h2 – h1) + (1/2)(V22 –V12)+ g(z2 – z1)
And
h2 – h1 = cp(T2 – T1)
In this case because the potential energy of air = 0 (negligible
difference in height between inlet and outlet of the
compressor) then the specific energy changes imparted to
the air and the heat loss to the surroundings are given by the
relationship…
Q – W = (h2 – h1) + (1/2)(V22 –V12)
i.e. Q – W = cp(T2 – T1) + (1/2)(V22 –V12)
Solution
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• The specific work input done on the air (W) can be
found from the input power to the compressor and
the mass flow rate of the gas from the compressor.
• Thus power in to the air = -42kJ/s (power or work
input per second done on the fluid =.negative) and
the mass flow rate of air at the exit m = 0.6kg/s;
therefore the specific work done on the gas at exit
is…
W = - 42/0.6 = - 70kJ/kg
Solution
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• Thus…
Q – (-70) = 1.005(338 – 288) + [(252 – 102)/2x103]
i.e. Q = -70 + 50.25 + 0.2625 = -19.48 kJ/kg
Thus the heat loss to the surroundings (heat out) =
19.48kJ/kg.
b) PV = mRT
Thus p2VS2 = RT2
So that the specific volume at exit is…
VS2 = (287 x 338)/(7X105) = 0.1386m3/kg
Solution
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But
ρ = 1/Vs = 1/0.1386 = 7.215 kg/m3
Thus,
Area A = m /ρV = 0.6/(7.215x25) = 3.326 x 10-3 m2