MECH 230 –
Thermodynamics 1
Midterm 1 Solutions
CREATED BY JUSTIN BONAL
MECH 230 Midterm 1
Contents
1.0 General Knowledge ................................................................................................................................................. 1
1.1 Unit Analysis .......................................................................................................................................................... 1
1.2 Systems ................................................................................................................................................................... 1
1.3 Pressure ................................................................................................................................................................... 2
2.0 Energies ....................................................................................................................................................................... 3
2.1 Work ......................................................................................................................................................................... 3
2.2 Internal Energy ..................................................................................................................................................... 5
2.3 Heat .......................................................................................................................................................................... 5
2.4 First Law of Thermodynamics ......................................................................................................................... 6
2.5 Piston Cylinder Operating in a Cycle ........................................................................................................... 6
3.0 Plotting ........................................................................................................................................................................ 9
3.1 Phase Diagram ..................................................................................................................................................... 9
3.2 Interpolation....................................................................................................................................................... 11
4.0 Ideal gas ................................................................................................................................................................... 14
4.1 Universal Gas Constant .................................................................................................................................. 14
4.2 Polytropic Work Using Ideal Gas ................................................................................................................ 15
5.0 Specific Heat ........................................................................................................................................................... 16
5.1 Constant Volume Heat Addition ................................................................................................................ 16
5.2 Constant Pressure Heat Addition ............................................................................................................... 16
5.3 Relating cv and cp for an Ideal Gas.............................................................................................................. 17
5.4 Incompressible Assumption ......................................................................................................................... 18
References....................................................................................................................................................................... 22
List of Figures
Figure 1: A useful way to conceptualize measuring pressures[1] ................................................................. 2
Figure 2: Example of a manometer [https://www.researchgate.net/figure/Fig3-9-Simple-U-tubemanometer_fig2_318378486] ..................................................................................................................................... 3
Figure 3: A graphical interpretation of work path dependency .................................................................... 4
Figure 4: A piston cylinder system operating in a cycle ................................................................................... 7
Figure 5: A P-V diagram for the piston cylinder cycle in figure 4 ................................................................. 7
Figure 6: A T-V diagram showing the characteristics of a phase diagram ................................................ 9
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Figure 7: From left to right: Superheated vapor, subcooled liquid, a substance in liquid-vapor
phase (point C) .............................................................................................................................................................. 10
Figure 8: Data for interpolation example ............................................................................................................ 12
List of Tables
Table 1: Useful Derived Units for Thermodynamics ........................................................................................... 1
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MECH 230 Midterm 1
1.0 General Knowledge
Thermodynamics! The class everyone has seen memes about being terrible and horrendous.
Englinks is here to ease the process! If you master the basics, and learn from the ground up, this
class will NOT be as bad as everyone says.
1.1 Unit Analysis
Units are a core part of Thermodynamics, and an easy place to lose marks. They can be tricky, so
make sure to have the derived units found in table 1 on your formula sheet
Table 1: Useful Derived Units for Thermodynamics
Name
Formula
Sign
Composition
Force
m*a
Newton (N)
kg*m/s2
Pressure
F*A
Pascal (Pa)
kg*m3/s2
Energy
F*d
Joule (J)
kg*m2/s2
Density
m/V
rho (𝜌)
kg/m3
Specific Volume
V/m || 1/𝜌
Upsilon (𝜐)
m3/kg
Often, you will need to change from a given unit into fundamental unit (ie atm to Pa). Always
change your values into fundamental units before you perform a calculation. Useful conversions
can be found below.
1 𝑎𝑡𝑚 = 101.326 𝑃𝑎 = 760𝑚𝑚𝐻𝑔|| 1 𝑏𝑎𝑟 = 100,000 𝑃𝑎 = 100𝑘𝑃𝑎 = 1𝑀𝑃𝑎 || 1 𝑚3 = 1000𝐿
1.2 Systems
A system is the area within a defined boundary that will be analyzed. System surroundings are
everything outside of the system boundary. There are a few types of systems:
Open system: A fixed volume in space where energy and mass can be transferred across the
system’s boundary
Closed system: A fixed volume where energy can be transferred in and out, but the total
amount of mass is fixed
Isolated system: There is no interaction with the system surroundings (i.e. no mass or heat
transfer)
There are two types of system properties:
Extensive: How the system is made up (volume, mass, energy, etc.)
Intensive: Is independent of the system’s size (pressure, temperature, etc.), can be thought of as
the system characteristics
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A system’s state is the system as defined by its properties. A steady state system is a system
where none of its properties changes with time. A system in equilibrium is isolated from its
surroundings and has no change in its properties.
Thermodynamics focuses on the changing of a system’s state. When a system’s state changes,
this is called a process. A cycle is a repeated set of processes that start and end at the same
state.
1.3 Pressure
Pressure is defined as force over area and has units of Pascals. There are two ways to measure
pressures.
Absolute pressure (Pabs): Measured relative to 0kPa (a perfect vacuum)
Gauge pressure (Pg): Measured relative to the atmospheric pressure
Atmospheric pressure (Patm ) is the local pressure of the measurement. Figure 1 and equation
[1] are useful to conceptualize these definitions and should be on your formula sheet.
𝑃𝑔 = 𝑃𝑎𝑏𝑠 − 𝑃𝑎𝑡𝑚
[1]
Figure 1: A useful way to conceptualize measuring pressures[1]
There are various ways to measure pressures. One of which is a manometer. An example of a
manometer can be found in Figure 2. In this case, the pressure of the gas is calculated using
equation [2].
Pg = Pabs – Patm = ρliquid ∗ 𝑔 ∗ ℎ
[2]
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Figure 2: Example of a manometer [2]
2.0 Energies
Energy has units of Joules [J]. It is a property that tells you how the system reacts in process. It
is essential to know where the energy in a system is going during a process or cycle to
understand how the system can be used practically.
There are three types of energy that you are going to be working with in Thermodynamics:
Work, Heat, and Internal.
2.1 Work
Although energy is a property of a system, work is not a property of the system. This is
because work depends on the path that it takes from one point in a process to another. So, it
needs to be evaluated at a specific time, not just the start and end.
Work is labeled by W, and the rate at which work is being done, power, is labeled by 𝑊̇ .
Equation [3] shows the most general work equation, and equation [4] the most general power
equation.
2
𝑊1→2 = ∫ 𝐹⃗ ∙ 𝑑𝑥⃗
[3]
1
𝑊̇1→2 = ⃗⃗⃗⃗⃗
𝐹 ∙ 𝑣⃗
[4]
It is important to note the arrows over the F and the x. This means that it is path dependent, and
each differential step matters. So, you cannot calculate work simply by doing 𝑊1→2 = 𝑊2 − 𝑊1 ,
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as this would be an inaccurate calculation. A graphical interpretation of this can be found in
Figure 3. The area under path 2 is greater than that of path 1, therefore more work will be
needed taking path 2, even with the same start and end points.
Figure 3: A graphical interpretation of work path dependency
In general, the work covered in this course will be due to a pressure force. An example of this is
a piston-cylinder assembly. The pressure force will cause a change in volume, so equation [3]
can be simplified to equation [5].
𝟐
𝑾𝟏→𝟐 = ∫ 𝑷𝒈 𝒅𝑽
𝟏
[5]
Where Pg is the pressure of a gas and dV is a differential change in volume. From equation [5] it
can be seen that a constant volume process will have no work.
∆𝑽 = 𝟎 ∴ ∆𝑾 = 𝟎
[6]
Often, the pressure and volume have a polytropic relationship. When you are told that a system
undergoes a polytropic process, that means it’s behaving according to equation [7].
𝑷𝑽𝒏 = 𝒄
[7]
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Where P is pressure, V is volume, c is a constant, and n is the polytropic index. The polytropic
index dictates what path a curve will take on a P-V graph. From this, it is easy to see the
following relation: 𝑃1 𝑉1𝑛 = 𝑃2 𝑉2𝑛 , since c is a constant.
From equation [7], three different ways to calculate work can be derived. Equation [8] shows the
work for when 𝑛 ≠ 1, equation [9] for when n = 1, and equation [10] for when n = 0. When n =
0, this is called a constant pressure process.
𝑾𝟏→𝟐 =
𝑷𝟐 𝑽𝟐 − 𝑷𝟏 𝑽𝟏
𝒇𝒐𝒓 𝒏 ≠ 𝟏
𝟏−𝒏
[8]
𝑽𝟐
) 𝒇𝒐𝒓 𝒏 = 𝟏
𝑽𝟏
[9]
𝑾𝟏→𝟐 = 𝑷𝟏 𝑽𝟏 𝐥𝐧 (
𝑾𝟏→𝟐 = 𝑷∆𝑽 𝒇𝒐𝒓 𝒏 = 𝟎
[10]
Something essential to note about calculating the work in process is the sign convention. Work
done by the system is positive, and work done on the system is negative. When calculating
processes and cycles, an engineer wants often wants to know what they are getting out of it. For
example, you would not say that an engine is giving off negative work when it is pushing you
forward.
2.2 Internal Energy
Internal energy can be thought of the energy that is contained within the fluid in a system, and
is denoted by U. You can think back to the kinetic molecular theory learned in first year
Chemistry.
If a process has no change in temperature, ∆𝑇 = 0, it is called an isothermal process. For an
ideal gas, internal energy is a function of temperature. So, if there is no change in temperature
for an ideal gas, there is no change in internal energy.
∆𝑻 = 𝟎 ∴ ∆𝑼 = 𝟎 ‼ (𝑰𝑫𝑬𝑨𝑳 𝑮𝑨𝑺)‼
[11]
2.3 Heat
Energy transferred by a difference in temperature between the system and surroundings is
called heat, and is denoted by Q. Contrary to work, heat transferred to the system is positive
and heat transferred from the system is negative.
Just like work, heat is not a property. The amount of energy transferred depends on the
characteristics of the process, not just the beginning and the end. Equation [12] shows the
general formula for heat transfer.
2
𝑄1→2 = ∫ 𝛿𝑄
1
[12]
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An adiabatic process is a process where there is no heat exchange from the system to the
surroundings. Therefore ∆𝑸 = 𝟎.
2.4 First Law of Thermodynamics
The first law is something that will follow you everywhere in this course and is essential to
almost every question you will do. Equation [13] states the first law.
𝐸2 − 𝐸1 = 𝑄 − 𝑊
[13]
Where E2 is the energy at the end of a process and E1 is the energy at the beginning. The change
in energy of a system can also be dictated by equation [14].
𝐸2 − 𝐸1 = ∆𝐾𝐸 + ∆𝑈 + ∆𝑃𝐸
[14]
Where KE is kinetic energy and PE is potential energy. Unless otherwise stated in a problem, the
changes in potential and kinetic energy can be neglected. So, combining equations [13] and [14]
yields equation [15].
∆𝑼𝟏→𝟐 = 𝑸𝟏→𝟐 − 𝑾𝟏→𝟐
[15]
Meaning that the change in the internal energy of a process is equal to the heat transfer of a
process minus the work of the process.
Often you will not be given a specific mass in the system. In this case, you will need to work in a
per unit mass basis. This simply means that instead of having units of joules [J] working with a
problem, you will have units of joules per kilogram [J/kg]. The energy symbols will be in
lowercase letters. U -> u, W -> w, Q -> q.
2.5 Piston Cylinder Operating in a Cycle
As previously defined, a cycle is a series of processes that have the same start and end state. An
example of this can be found in figure 4.
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Figure 4: A piston cylinder system operating in a cycle
A standard P-V graph of this cycle can be found in figure 5. It is important to note the direction
that the arrows are pointing in. This directly effects the integrals when calculating energies.
Figure 5: A P-V diagram for the piston cylinder cycle in figure 4
The work from 1 to 2 will be negative, and the work from 3 to 4 will be positive. The work from
2-3 and 4-1 will be zero because these are constant volume processes. The net work (Wnet) of
a cycle is equal to the sum of the total work done in the cycle.
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If Wnet is greater than zero, then the cycle is called a power cycle. If it is less than zero, then it
is called a refrigeration cycle.
QUESTION
Nitrogen gas (N2) in a piston-cylinder assembly undergoes a polytropic process with
pV1.45=c. The gas is initially at 18 bar and has a volume of 0.3 m3 and ends with a
volume of 2.85 m3. Determine the pressure at state 2 in bar and the work of the process
in kJ.
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3.0 Plotting
In all thermodynamics questions, you should draw a plot of the process or cycle that you are
trying to solve. Sometimes it is asked in the questions, sometimes not. It is highly suggested that
even if it is not part of the question’s requirements to get marks that you draw the plot. This will
help you conceptualize the problem, and help you find mistakes if you make any.
3.1 Phase Diagram
There are two main types of plots that you will be working with in Thermodynamics. One is
temperature versus volume [T-V], and another is pressure versus volume [P-V]. Volume in
these two relations can be exchanged for specific volume [𝝊], which is the inverse of density.
You will use specific volume when you are doing calculations on a per mass basis.
The two plots are very similar, with the only difference being the unit on the y-axis. Each plot
contains a saturation (or vapor) dome. Depending where you are relative to the saturation
dome dictates the composition of your substance. If you are to the left (outside) of the dome,
then your substance is in a liquid phase. If you are to the right, then it is in a gaseous phase. If
you are under the dome, then it is in a liquid-vapor phase. If your point is directly on the line
of the saturation dome, then it is in a saturation state. To the left of the peak, this is called the
liquid saturation line, and to the right the vapor saturation line. Figure 6 clearly shows all
these characteristics for a T-V diagram.
Figure 6: A T-V diagram showing the characteristics of a phase diagram
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.
In addition, phase diagrams hold straight lines, which can be seen in Figure 6. For a P-V diagram,
the straight line represents a specific temperature, called an isotherm. For a T-V diagram it
represents a specific pressure, called an isobar. An interesting characteristic of these lines is that
they are completely horizontal under the saturation dome (between the saturation lines). This
means that when a substance is transitioning between a liquid and gaseous state by changing
the temperature, the pressure stays constant. Vice versa, if this phase change is happening due
to a change in pressure, the temperature will stay constant.
The temperature where an isobar crosses the saturation line on a T-V is called the saturation
temperature. Similarly, the pressure where an isotherm passes the saturation line on a P-V
diagram is called the saturation pressure.
This is very useful for understanding what state a substance will be at a specific temperature and
pressure. If the temperature is greater than the saturation temperature, then it is in a
superheated vapor state. If it is less than the saturation temperature, then it is in a subcooled
liquid state. If it is at the saturation temperature, then it is in a liquid-vapor state. These
situations can be seen in Figure 7, from left to right. The same concept can also be applied to a
P-V diagram, except measured relative to the saturation pressure.
Figure 7: From left to right: Superheated vapor, subcooled liquid, a substance in liquid-vapor phase (point C)
When under the saturation dome (liquid-vapor state) calculating quantities like mass can
become difficult. The mixture contains both liquid and vapor, therefore the mixtures mass is a
summation of the two: 𝑀 = 𝑀𝐿 + 𝑀𝑉 , where 𝑀𝑣 is the mass of vapor and 𝑀𝐿 is the mass of
liquid. At any point along the line under the saturation dome, we say that the substance has a
vapor quality, x. The vapor quality is measured using equation [16].
𝑴𝑽
𝑽 +𝑴𝑳
𝒙=𝑴
[16]
The vapor quality is essential for determining other specific properties of the substance while
under the saturation dome. For example, equations [17] and [18] show how to calculate the
specific internal energy, u, and the specific volume, 𝜐, based on the vapor quality of the
substance.
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𝑢 = (1 − 𝑥)𝑢𝐿 + 𝑥 ∗ 𝑢𝑣
[17]
𝜐 = (1 − 𝑥)𝜐𝐿 + 𝑥 ∗ 𝜐𝑣
[18]
Where every variable with a subscript “L” is a property of the liquid, and “v” a property of the
vapor. These calculations are important, and essential to have on your formula sheet.
3.2 Interpolation
Interpolation is a powerful tool for finding data “in between the lines”. The best way to visualize
interpolation is imagining a line, dictated by two points. Each point has a coordinate (x, y). In this
case, these coordinates can be any two specific quantities given in the data tables (specific
volume, internal energy, enthalpy, entropy, pressure, temperature, etc.). These values are given
at regular intervals (1 Mpa, 2 Mpa,3 Mpa, etc.), but what do you do when you have a value at 2.4
Mpa?
Interpolation uses linear analysis to predict a value based on another given value. To solve, set
up a linear equation based on the “coordinates” around the known values. Next, use the
equation to solve for the missing value. The linear equation is given by equation [19], where m is
the slope between these two coordinates, b is an initial value, y is the dependent variable, x is
the independent variable, and x1 is the point we are evaluating from.
𝑦 = 𝑚(𝑥 − 𝑥1 ) + 𝑏
[19]
The equation for slope is given by equation [20]
𝑦2 − 𝑦1
𝑥2 − 𝑥1
[20]
𝑦2 − 𝑦1
] (𝑥 − 𝑥1 ) + 𝑏
𝑥2 − 𝑥1
[21]
𝑚=
Combining these two yields equation [21]
𝑦=[
A quick example will easily show how to solve for a value using interpolation.
Let’s say that you are working with compressed liquid water with a temperature of 50 degrees
Celsius and a pressure of 25 bar. The question asks you to solve for the specific volume of the
liquid under these conditions. First, lets look at Figure 8 for the data needed.
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Figure 8: Data for interpolation example
Look at the section with 25 bar and try to locate 50 degrees Celsius. You quickly realize that only
40 and 80 degrees are on the data table! What are you supposed to do?! Never fear,
interpolation is here.
First, recognize the two coordinates that you are going to be working with. You are trying to
find a value between 40 and 80, therefore your x1 [T1] = 40 and your x2 [T2] = 80. Since you are
looking for specific volume, the corresponding y values are y1 [v1] = 1.0067, and y2 [v2] =
1.0280. Your initial value, b, is always based off the value that you are trying to find, therefore b
[v1] = 1.0067. Equation [22] shows equation 18 all in terms of variables used in this example.
𝜐=
𝜐2 − 𝜐1
(𝑇 − 𝑇1 ) + 𝜐1
𝑇2 − 𝑇1
[22]
Now it’s a simple plug and chug formula, with T = 50, to find 𝜐 under these conditions.
Calculating yields 𝜐 = 1.0120. There are two quick easy ways to make sure that you plugged
everything in right.
First, check to make sure your resulting value is between your two y input values. In this case
y1<1.0120<y2, so check one is good. Next, we expect the value to be closer y1 rather than y2,
because 50 is closer to x1 than it is to x2. For this example, the calculated value is closer to y1, so
both checks pass. Now we can confirm that we did the calculation correctly.
QUIZ 1 2018 QUESTION 1
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4.0 Ideal gas
You most likely have some experience working with the ideal gas law from previous courses. In
thermodynamics, you are going to take the same concepts and apply them on a broader scale.
4.1 Universal Gas Constant
𝐽
𝑘𝐽
The universal gas constant, 𝑅̅ = 8314 𝑘𝑚𝑜𝑙∗𝐾 = 8.314 𝑘𝑚𝑜𝑙 𝐾, is something that is consistent for
all gases. However, in thermodynamics it is often more convenient to use the gas constant, R,
specific to a substance. To find R, simply divide the universal gas constant by the substances
molar mass, 𝜇, as can be seen in equation [23].
𝑅=
𝑅̅
𝜇
[23]
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MECH 230 Midterm 1
𝐽
After dividing by the molar mass, the gas constant has units of 𝑘𝑔∗𝐾 . So, your calculations will be
in terms of mass, rather than mols.
The general ideal gas law is given in equation [24], where P is pressure, V is volume, n is mols,
and T is temperature. Equation [25] shows the ideal gas law in terms of the gas constant, R,
where M is mass. Notice how rather than using mols, mass is used.
𝑃𝑉 = 𝑛𝑅̅ 𝑇
𝑃𝑉 = 𝑀𝑅𝑇
[24]
[25]
Equation [25] can be rearranged to be calculated on a per mass basis by dividing through my M
or V. Equation [26] shows the equation in terms of specific volume (divide by M), and equation
[27] in terms of density (divide by V).
𝑃𝜐 = 𝑅𝑇
[26]
𝑃 = 𝜌𝑅𝑇
[27]
4.2 Polytropic Work Using Ideal Gas
After being told that a gas is ideal, multiple equations become “unlocked” to solve a process.
Remembering back to a polytropic process, the work of the process for n=/ 1 is given by
equation [8]. However, was have just learned that PV=MRT from equation [25]. So, we can make
an easy substitution to yield equation [28].
𝑊1→2 =
𝑀𝑅(𝑇2 − 𝑇1 )
𝑓𝑜𝑟 𝑛 ≠ 1
1−𝑛
[28]
An ideal gas going though a polytropic process with n=1 means it is an isothermal process
(∆𝑻 = 𝟎). Therefore equation [9] can be modified to the following.
𝑉2
𝑊1→2 = 𝑀𝑅𝑇 ln ( ) 𝑓𝑜𝑟 𝑛 = 1
𝑉1
[29]
Equation [26] can be rearranged to yield equation [30]. This relation holds no matter the
temperature, pressure, or specific volume of the substance. Thus, we can use equations [5]
(polytropic) and [30] to easily make relationships from one point to another.
𝑅=
𝑃𝜐
𝑇
[30]
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These relationships simplify to equations [31] and [32], where n is the polytropic index. These
formulas are essential for any ideal polytropic process. However, they can only be used if the
problem says that the substance is an ideal gas. DO NOT USE UNLESS YOU ARE TOLD THAT
THE SUBSTANCE IS AN IDEAL GAS. It is an easy trap to fall into.
𝑛−1
𝑛
𝑇2
𝑃2
=( )
𝑇1
𝑃1
[31]
𝑇2
𝜐2 𝑛−1
=( )
𝑇1
𝜐1
[32]
5.0 Specific Heat
Specific heat, c, is a property that represents the amount of heat required to raise the
temperature of 1 kg of a substance by 1-degree Kelvin, as seen in equation [33]. The units of
𝑘𝐽
specific heat are 𝑘𝑔°𝐾.
𝑐=
∆𝑄
𝑀 ∗ ∆𝑇
[33]
5.1 Constant Volume Heat Addition
For a constant volume, there is no work being done. Therefore, the first law gets reduced to:
∆𝑈 = ∆𝑄
Rearranging equation [33], dividing through by M, and plugging it into the above yields
∆𝒖 = 𝒄𝒗 ∆𝑻
[34]
Noting that 𝑐𝑣 depends on both P and T, equation [34] gets generalized to
2
2
𝑢2 − 𝑢1 = ∫ 𝑑𝑢 = ∫ 𝑐𝑣 (𝑃, 𝑇)𝑑𝑇
1
1
[35]
5.2 Constant Pressure Heat Addition
For a constant pressure heat addition, the change in work is simply ∆𝑊 = 𝑃∆𝑉. So, first law in a
per mass basis can be reduced to
𝑀∆𝑢 = ∆𝑄 − 𝑃(𝑀𝑑𝜐)
[36]
∆𝑞 = 𝑑𝑢 + 𝑃𝑑𝜐
[37]
Which further simplifies to
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Now, we introduce a new property: enthalpy, h
ℎ = 𝑢 + 𝑃𝜐
[38]
At an infinitesimally small scale, and with a change in pressure of zero, the above simplifies to
𝑑ℎ = 𝑑𝑢 + 𝑃𝑑𝜐
[39]
Finally, combining equations [33] and [39] yields
𝒅𝒉 = 𝒄𝒑 𝒅𝑻
[40]
Like 𝑐𝑣 , 𝑐𝑝 is dependent on P and T. So, equation [40] generalizes to
2
2
ℎ2 − ℎ1 = ∫ 𝑑ℎ = ∫ 𝑐𝑝 (𝑃, 𝑇)𝑑𝑇
1
1
[41]
5.3 Relating cv and cp for an Ideal Gas
For an ideal gas, the values of cp and cv are dependent solely on temperature, and not pressure.
The tabulated values for cp for specific substances can be found in table A-20
For an ideal gas, Pv=RT. Therefore, equation [38] reduces to
ℎ = 𝑢 + 𝑅𝑇
[42]
Differentiating equation [42] with respect to T yields
𝒄𝒑 (𝑻) = 𝒄𝒗 (𝑻) + 𝑹
[43]
We know that cv and R are always positive, therefore we know that cp is always greater than cv.
A commonly used term is the specific heat ratio, k. For an ideal gas
𝑘(𝑇) =
𝑐𝑝
𝑐𝑣
[44]
Using equations [43] and [44] yields the following equations, useful for calculating the specific
heats of a substance given its specific heat ratio
𝒄𝒗 (𝑻) =
𝑹
𝑲(𝑻) − 𝟏
[45]
𝒄𝒑 (𝑻) =
𝑲(𝑻)𝑹
𝑲(𝑻) − 𝟏
[46]
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MECH 230 Midterm 1
5.4 Incompressible Assumption
A substance is incompressible if there is very little change in its specific volume with changes in
pressure, such as with liquids and solids. Using this assumption yields that 𝑐𝑣 = 𝑐𝑝 = 𝑐. This
reduces equation [35] to
2
𝑢2 − 𝑢1 = ∫ 𝑐(𝑇)𝑑𝑇
1
[47]
And equation [41] to
2
ℎ2 − ℎ1 = ∫ 𝑐(𝑇)𝑑𝑇 + 𝜐(𝑃2 − 𝑃1 )
1
[48]
QUIZ 1 2016 QUESTION 1a
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MECH 230 Midterm 1
19
MECH 230 Midterm 1
QUIZ 1 2018 QUESTION 2A
20
MECH 230 Midterm 1
QUIZ 1 2018 QUESTION 2B
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MECH 230 Midterm 1
References
[1] All schematics are taken from Prof. Ciccarelli’s notes. Exam questions come from Prof.
Ciccarelli.
[2] https://www.researchgate.net/figure/Fig3-9-Simple-U-tube-manometer_fig2_318378486
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